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eepro - PocketProfessional™ Series EE•Pro®...

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Unformatted text preview: PocketProfessional™ Series EE•Pro® #UQHVYCTG#RRNKECVKQP 1P6+CPF6+2NWU User’s Guide June 1999 © da Vinci Technologies Group, Inc. Rev. 1.0 da Vinci Technologies Group, Inc. 1600 S.W. Western Blvd Suite 250 Corvallis, OR 97333 www.dvtg.com Notice This manual and the examples contained herein are provided “as is” as a supplement to EE•Pro application software available from Texas Instruments for TI-89, and 92 Plus platforms. Da Vinci Technologies Group, Inc. (“da Vinci”) makes no warranty of any kind with regard to this manual or the accompanying software, including, but not limited to, the implied warranties of merchantability and fitness for a particular purpose. Da Vinci shall not be liable for any errors or for incidental or consequential damages in connection with the furnishing, performance, or use of this manual, or the examples herein. © Copyright da Vinci Technologies Group, Inc. 1999. All rights reserved. PocketProfessional and EE•Pro is a registered trademarks of da Vinci Technologies Group, Inc. We welcome your comments on the software and the manual. Forward your comments, preferably by e-mail to da Vinci at [email protected] Acknowledgements The EE•Pro software was developed by Dave Conklin, Casey Walsh, Michael Conway and Megha Shyam with the generous support of TI’s development team. The user’s guide was developed by Michael Conway and Megha Shyam. Many helpful comments from the testers at Texas Instruments and other locations during β testing phase is gratefully acknowledged. Table of Contents 1 Introduction to EE•Pro.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i 1.1 Key Features of EE•Pro . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i 1.2 Download/Purchase Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii 1.3 Manual Ordering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii 1.4 Memory Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii 1.5 Differences between TI-89 and TI-92 plus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii 1.6 Beginning EE•Pro. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii 1.7 Manual Organization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii 1.8 Disclaimer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv 1.9 Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv Part I: Analysis 2 Introduction to Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2.2 Setting up an Analysis Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2.3 Solving Problems in Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.4 Special Function Keys in Analysis Routines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 2.5 Data Fields, Analysis Functions and Sample Problems . . . . . . . . . . . . . . . . . . . . . . . . 5 Example 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Example 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Example 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.6 Session Folders, Variable Names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3 AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Impedance Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Voltage Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Current Divider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Circuit Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 10 11 11 11 12 12 13 14 4 Polyphase Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Wye ↔∆ Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 4.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Balanced Wye Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Balanced ∆ Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 4.3 . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 15 16 16 17 17 18 5 Ladder network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Using Ladder Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Using the Ladder Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 19 22 22 23 6 Filter Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 6.1 Chebyshev Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Example 6.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 6.2 Butterworth Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 Example 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.3 Active Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Example 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 7 Gain and Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Bode Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 8.1 FFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Example 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 8.2 Inverse FFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Example 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 9 Two-Port Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 9.1 Parameter Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Example 9.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 9.2 Circuit Performance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Example 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 9.3 Connected Two-Ports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Example 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 29 29 30 30 31 10 Transformer Calculations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 10.1 Open Circuit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Example 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 10.2 Short Circuit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40 Example 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 10.3 Chain Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 Example 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 11 Transmission Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 11.1 Line Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Example 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 11.2 Line Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Example 11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 11.3 Fault Location Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Example 11.3 . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 11.4 Stub Impedance Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .46 Example 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 12 Computer Engineering. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 12.1 Special Mode Settings, the † Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .48 12.2 Binary Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .50 Example 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 12.3 Register Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Example 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 12.4 Bit Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Example 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 12.5 Binary Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Binary Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Karnaugh Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 54 54 55 55 56 13 Error Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 13.1 Using Error Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Example 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 14 Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 14.1 Using Capital Budgeting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Example 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Part II: Equations 15 Introduction to Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 15.1 Solving a set of Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 15.2 Viewing an Equation or Result in Pretty Print . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 15.3 Viewing a Result in different units. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 15.4 Viewing Multiple Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 15.5 Partial Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 15.6 Copy/Paste. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 15.7 Graphing a Function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 15.8 Storing and recalling variable values in EE•Pro-creation of session folders . . . . . . 5 15.9 solve, nsolve, and csolve and user-defined functions (UDF) . . . . . . . . . . . . . . . . . . 6 15.10 Entering a guess value for the unknown using nsolve. . . . . . . . . . . . . . . . . . . . . . . . 6 15.11 Why can't I compute a solution?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 15.12 Care in choosing a consistent set of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 15.13 Notes for the advanced user in troubleshooting calculations. . . . . . . . . . . . . . . . . . . 7 16 Resistive Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Resistance Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Ohm’s Law and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Temperature Effect on resistance . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 16.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Maximum Power Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 16.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Voltage and Current Source Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 16.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 9 10 10 10 11 11 12 12 13 13 17 Capacitors and Electric Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 17.1 Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Example 17.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 17.2 Long Charged Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 Example 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 17.3 Charged Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Example 17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 17.4 Parallel Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Example 17.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 17.5 Parallel Wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.6 Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.7 Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 17.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 18 18 18 19 19 18 Inductors and Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 18.1 Long Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Example 18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 18.2 Long Strip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Example 18.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 18.3 Parallel Wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Example 18.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 18.4 Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Example 18.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 18.5 Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Example 18.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 18.6 Skin Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Example 18.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 19 Electron Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 19.1 Beam Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Example 19.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 19.2 Thermionic Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Example 19.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 19.3 Photoemission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 Example 19.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 20 Meters and Bridge Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 20.1 A, V, Ω Meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Example 20.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 20.2 Wheatstone Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Example 20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 20.3 Wien Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Example 20.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 20.4 Maxwell Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Example 20.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 20.5 Owen Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Example 20.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 20.6 Symmetrical Resistive Attenuator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Example 20.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 20.7 Unsymmetrical Resistive Attenuator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Example 20.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 21 RL and RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 21.1 RL Natural Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40 Example 21.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 21.2 RC Natural Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Example 21.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 21.3 RL Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Example 21.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 21.4 RC Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 21.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 RL Series-Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 21.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 RC Series-Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 21.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 42 43 43 44 45 22 RLC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 22.1 Series Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Example 22.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 22.2 Parallel Admittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Example 22.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 22.3 RLC Natural Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Example 22.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 22.4 Underdamped Transient Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .49 Example 22.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 22.5 Critically-Damped Transient Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Example 22.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 22.6 Overdamped Transient Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51 Example 22.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 23 AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 23.1 RL Series Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Example 23.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 23.2 RC Series Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Example 23.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 23.3 Impedance↔Admittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Example 23.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 23.4 Two Impedances in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Example 23.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 23.5 Two Impedances in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Example 23.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 24 Polyphase Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 24.1 Balanced ∆ Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Example 24.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 24.2 Balanced Wye Network . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Example 24.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 24.3 Power Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Example 24.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 25 Electrical resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 25.1 Parallel Resonance I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Example 25.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 25.2 Parallel Resonance II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Example 25.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 25.3 Resonance in a Lossy Inductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Example 25.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 25.4 Series Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 Example 25.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 26 OpAmp Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 26.1 Basic Inverter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Example 26.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 26.2 Non-Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Example 26.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 26.3 Current Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Example 26.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 26.4 Transconductance Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Example 26.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 26.5 Level Detector (Inverting) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Example 26.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 26.6 Level Detector (Non-inverting) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Example 26.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 26.7 Differentiator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Example 26.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 26.8 Differential Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .75 Example 26.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 27 Solid State Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 27.1 Semiconductor Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Example 27.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Example 27.1.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 27.2 PN Junctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Example 27.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Example 27.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 27.3 PN Junction Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .84 Example 27.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 27.4 Transistor Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Example 27.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 27.5 Ebers-Moll Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Example 27.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 27.6 Ideal Currents - pnp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 Example 27.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 27.7 Switching Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Example 27.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 27.8 MOS Transistor I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Example 27.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 27.9 MOS Transistor II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Example 27.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 27.10 MOS Inverter (Resistive Load) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Example 27.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 27.11 MOS Inverter (Saturated Load) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Example 27.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 27.12 MOS Inverter (Depletion Load) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Example 27.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 27.13 CMOS Transistor Pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Example 27.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 27.14 Junction FET . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Example 27.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 28 Linear Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.1 BJT (Common Base) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 28.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.2 BJT (Common Emitter) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 102 103 103 104 Example 28.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 28.3 BJT (Common Collector) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Example 28.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 28.4 FET (Common Gate) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Example 28.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 28.5 FET (Common Source) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Example 28.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 28.6 FET (Common Drain) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Example 28.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 28.7 Darlington (CC-CC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Example 28.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 28.8 Darlington (CC-CE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Example 28.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 28.9 Emitter-Coupled Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Example 28.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 28.10 Differential Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .111 Example 28.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 28.11 Source-Coupled JFET Pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Example 28.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 29 Class A, B and C Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 29.1 Class A Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Example 29.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 29.2 Power Transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .115 Example 29.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 29.3 Push-Pull Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Example 29.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 29.4 Class B Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Example 29.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 29.5 Class C Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Example 29.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 30 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1 Ideal Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 30.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 Linear Equivalent Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 30.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 121 121 122 122 123 31 Motors and Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 31.1 Energy Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Example 31.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 31.2 DC Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Example 31.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 31.3 Separately-Excited DC Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Example 31.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 31.4 DC Shunt Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Example 31.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 31.5 DC Series Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Example 31.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 31.6 Separately-Excited DC Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Example 31.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 31.7 DC Shunt Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Example 31.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 31.8 DC Series Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.8 . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.9 Permanent Magnet Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.10 Induction Motor I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.11 Induction Motor II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.12 Single-Phase Induction Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.13 Synchronous Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 134 134 135 135 136 137 138 138 138 139 139 Part III: Reference 32 Part III - Introduction to Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 32.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 32.2 Accessing the Reference Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 32.3 Reference Screens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 32.4 Viewing Reference Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 33 Resistor Color Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 33.1 Using the Resistor Color Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Example 33.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 34 Standard Component Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 Example 34.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 35 Semiconductor Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 36 Boolean Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 36.1 Using Boolean Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Example 36.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 37 Boolean Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 37.1 Using Boolean Algebra Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Example 37.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14 38 Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38.1 Using Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . Example 38.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 38.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 16 17 17 39 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 39.1 Using Constants . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Example 39.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19 40 SI Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20 40.1 Using SI Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 41 Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Appendix Appendix A Frequently Asked Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 A.1 Questions and Answers. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .1 A.2 General Questions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .1 A.3 Analysis Questions . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 A.4 Equation Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 A.5 Reference Questions . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Appendix B Warranty and Technical Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 B.1 da Vinci License Agreement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 B.2 How to contact Customer Support. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Appendix C Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Appendix D TI 89 and TI 92 plus- Display and Keystroke Differences . . . . . . . . . . . . . . . . . . . 9 D.1 Display Property Differences between the TI-89 and TI-92 plus . . . . . . . . . . . . . . . 9 D.2 Keyboard Differences Between TI-89 and TI-92 Plus. . . . . . . . . . . . . . . . . . . . . . . . 9 Appendix E Error Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 E.1 General Error Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 E.2 Analysis Error Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 E.3 Equation Error Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 E.4 Reference Error Messages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 Appendix F System variables and Reserved names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18 Chapter 1 Introduction to EE•Pro Thank you for your purchase of EE•Pro, a member of the PocketProfessional® Pro software series designed by da Vinci Technologies to meet the computational needs of students and professionals in the engineering and scientific fields. Many long hours and late nights have been spent by the designers of this software to compile and organize the subject material in this software. We hope you enjoy EE•Pro and that it serves as a valuable companion in your electrical engineering career. Topics in this chapter include: • • • • Key Features of EE•Pro Download/Purchase Information Manual Ordering Memory Requirements • • • • Differences between the TI 89 and TI 92 plus. Beginning EE•Pro Manual Organization Summary 1.1 Key Features of EE•Pro The manual is organized into three sections representing the main menu headings of EE•Pro. Analysis AC Circuits Polyphase Circuits Ladder Network Filter Design Gain and Frequency Fourier Transforms Two-Port Networks Transformer Calculations Transmission Lines Computer Engineering Error Functions Capital Budgeting Equations Resistive Circuits Capacitors and Electric Fields Inductors and Magnetism Electron Motion Meters and Bridges RL and RC Circuits RLC Circuits AC Circuits Polyphase Circuits Electrical Resonance Operational Amplifier Circuits Solid State Devices Linear Amplifiers Class A, B, & C Amplifiers Transformers Motors and Generators Reference Resistor Color Chart Standard Component Values Semiconductor Data Boolean Expressions Boolean Algebra Transforms Constants SI Prefixes These main topic headings are further divided into sub-topics. A brief description of the main sections of the software is listed below: EE Pro for TI - 89, 92 Plus Introduction to EE•Pro i Analysis-(Chapters 2-14) Analysis is organized into 12 topics and 33 sub-topics. The tools in this section incorporate a wide variety of analysis methods used by electrical engineers. Examples include evaluation of AC circuit performance characteristics, designing signal filters, building and computing ladder network properties, plotting transfer functions, estimating transformer and transmission line characteristics, performing binary arithmetic operations, and estimating pay-back returns for different projects in capital budgeting. Many sections in analysis can perform calculations for numeric as well as symbolic entries. Equation Library (Chapters 15-31) This section contains over 700 equations organized under 16 topics and 105 subtopics. In any sub-topic, the user is able to select a set of equations, enter known values and compute results for unknown variables. The math engine is able to compute multiple or partial solution sets. A built-in unit management feature allows for the entry and expression of values in SI or other established measurement systems. Descriptions of each variable, unit selection, and appropriate diagrams are included in this section of the software. Reference (Chapters 32-41) The Reference section of EE•Pro contains tables of information commonly found in electrical engineering handbooks. Topics include physical and chemical properties of common semiconductor materials, a list of fundamental constants commonly used by electrical engineers, tables of Fourier, Laplace, and ztransforms, and a list of Boolean algebraic expressions. Added features are the ability to perform simple computations, such as estimating standard (or preferred) manufacturer component values for inductors, resistors, and capacitors, in addition to a resistor color chart guide which can compute resistance and tolerance from a resistor’s color band sequence. 1.2 Purchasing, Downloading and Installing EE•Pro The EE•Pro software can only be purchased on-line from the web store at Texas Instruments Inc. at http://www.ti.com/calc. The software can be installed directly from your computer to your calculator using TIGRAPH LINKTM hardware and software (sold separately). Directions for purchasing, downloading and installing EE•Pro software are available from TI’s website. 1.3 Manual Ordering Information Chapters and Appendices of the manual for EE•Pro can be downloaded from TI’s web store and viewed using the free Adobe Acrobat Reader which can be downloaded from http://www.adobe.com (it is recommended that you use the latest version of the Acrobat reader and use the most updated driver for your printer). Printed manuals can be purchased separately from da Vinci Technologies (see address on cover page or visit da Vinci’s website http://www.dvtg.com/ticalcs/docs). 1.4 Memory Requirements The EE•Pro program is installed in the system memory portion of the flash ROM, which is separate from the RAM available to the user. EE•Pro uses RAM to store some of its session information, including values entered and computed by the user. The exact amount of memory required depends on the number of user-stored variables and the number of session folders designated by the user. To view the available memory in your TI calculator, use the ° function. It is recommended that at least 10K of free RAM be available for installation and use of EE•Pro. EE Pro for TI - 89, 92 Plus Introduction to EE•Pro ii 1.5 Differences between TI-89 and TI-92 plus EE•Pro is designed for two models of graphing calculators from Texas Instruments, the TI-92 Plus and the TI-89. For consistency, keystrokes and symbols used in the manual are consistent with the TI-89. Equivalent key strokes for the TI-92 plus are listed in Appendix D. 1.6 Beginning EE•Pro • To begin EE•Pro, start by pressing the key. This accesses a pull down menu. Use the D key to move key on TI-92 the cursor bar to EE Pro Elec. Eng. and press ¸. Alternatively, enter [A] on TI-89, or Plus to get to the home screen of EE•Pro. (Pull down Menu for EE•Pro option is further down the list) Pull down Menu on for (EE•Pro at the end of the list) The EE•Pro home screen is displayed below. The tool bar at the top of the screen lists the titles of the main sections of EE•Pro which can be activated by pressing the function keys. • • • • • Tools: Editing features, information about EE•Pro in A:About „ Analysis-Accesses the Analysis section of the software Equations-Accesses the Equations section of the software. Reference-Accesses the Reference section of the software. Info-Helpful hints on EE•Pro. A selection on any menu can be entered by moving the highlight bar to the item with the arrow key D and pressing ¸ (alternatively, the number or letter of the selection can be typed in). The Analysis, Equation and Reference menus are organized in a directory tree of topics and sub-topics. The user can return to a previous level of EE•Pro by pressing . EE•Pro can be exited at any time by pressing the key. When EE•Pro is restarted the software returns to it’s previous location. 1.7 Manual Organization • • The five sections in the manual, Introduction, Analysis, Equations, Reference, and Appendix, have separate page numbering systems. The manual section, chapter heading and page number appear at the bottom of each page. The first chapter in each of the Analysis, Equations and Reference sections (Chapters 2, 15, and 32) gives an overview of the succeeding chapters and introduces the navigation and computation features common to each of the main sections. For example, Chapter 2 explains the basic layout of the Analysis section, menu navigation principles, and gives general examples of features common to all topics in Analysis. The chapters which follow are dedicated to the specific topics in each section. The titles of these chapters EE Pro for TI - 89, 92 Plus Introduction to EE•Pro iii • correspond to the topic headings in the software menus. The chapters list all the equations used and explains their physical significance. These chapters also contain example problems and screen displays of the computed solutions. The Appendices contain trouble-shooting information, commonly asked questions, a bibliography used to develop the software, and warranty information provided by Texas Instruments. 1.8 Manual Disclaimer • The calculator screen displays in the manual were obtained during the testing stages of the software. Some screen displays may appear slightly different due to final changes made in the software while the manual was being completed. 1.9 Summary The designers of EE•Pro have attempted to maintain the following features: • Easy-to-use, menu-based interface. • Computational efficiency for speed and performance. • Helpful-hints and context-sensitive information provided in the status line. • Advanced EE analysis routines, equations, and reference tables. • Comprehensive manual documentation for examples and quick reference. We hope to continually add and refine the software products in the Pocket Professional series line. If you have any suggestions for future releases or updates, please contact us via the da Vinci website http://www.dvtg.com or write to us at [email protected] Best Regards, da Vinci Technologies Group, Inc. EE Pro for TI - 89, 92 Plus Introduction to EE•Pro iv Chapter 2 Introduction to Analysis The Analysis section of the software is able to perform calculations for a wide range of topics in circuit and electrical network design. A variety of input and output formats are encountered in the different topics of Analysis. Examples for each of the input forms will be discussed in some detail. • • • • • • The unit management feature in Equations is not present in Analysis due the variety of computation methods used in this section. All entered values are assumed to be common SI units (F, A, kg, m, s, Ω) or units chosen by the user (such as len in Transmission Lines). A feature unique to Analysis is the ability to perform symbolic computations for variables (with the exception of Filter Design and Computer Engineering and a few variables in other sections). An entry can consist of a single undefined variable (such as ‘a’ or ‘x’) or an expression of defined variables which can be simplified into a numerical result (such as ‘x+3*y’, where x=-3 and y=2). More information on a particular input can be displayed by highlighting the variable, press and /Type: to show a brief description of a variable and its entry parameters. A variable name cannot be entered which is identical to the variable name (ie.: C for capacitance). If a symbolic calculation using the variable name, leave the entry blank. Variables which accept complex entries (ex: 115+23*i) are followed by an underscore ‘_’ (ex: ZZ1_). 2.1 Introduction Analysis routines have been organized into twelve sections, each containing tools for performing electrical analysis of a variety of circuit types. One can design filters, solve two-port networks problems, calculate transmission line properties, minimize logic networks, perform binary arithmetic at bit and register levels, draw Bode diagrams, and examine capital budget constraints - all with context-sensitive assistance displayed in the status line. 2.2 Setting up an Analysis Problem The Analysis section is located in the home screen of EE•Pro. • To access the home screen of EE•Pro, start by pressing the key. This accesses a pull down menu listing all the topics available. Use the D key to move the cursor bar to EE Pro Elec. Eng. and press ¸. Alternatively, enter [A] on TI-89, or key on TI-92 Plus to get to the home screen of EE•Pro. On TI -89 for (Pull down Menu On EE•Pro option is further down the list) EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis Pull down Menu on TI -89 for (Lower end of the list) 1 • Pressing „ will access the Analysis section of the software and display a pull down menu listing the topics available. There are 12 sections under Analysis. The sections are accessed by using the D key to move the highlight bar to the desired section and pressing ¸. Alternatively, any section can be accessed by entering the number associated with each section. Thus pressing ¨ will display a pop up menu for AC Circuits, while pressing will list topics in Gain and Frequency. Analysis sections are listed and tagged with a number 1, 2, 3, etc. A down arrow (↓) beside a topic at the bottom of a menu indicates there are more choices. Main Screen for EE•Pro Press „ for Analysis Pull down menu for Analysis; down arrow (↓) indicates that there are more items. Press ¨ for topics in AC Circuits. • • Press for topics in Gain and Frequency. Once an Analysis topic is accessed, a pop up menu lists the sub topics available for in the section. For example, when Two-Port Networks is selected, the pop up menu shows three sub topics: 1. Convert Parameters 2. Circuit Performance 3. Connected Two-Ports. Each of these sub topics are tagged with a number 1, 2, 3 as shown in the screen display. These topics are accessed by using the D key to move the highlight bar to the desired choice and pressing ¸. Alternatively, a section can be accessed by typing the number associated with the topic or subtopic. Thus pressing ¨ will display an input screen for Convert Parameters. Pop up menu in Two-Port Networks Input Type EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis z Pop up menu for Input Type Input Screen for data entry The right arrow indicates that there are choices to be made for input type. Pressing or ¸ displays a pop up menu showing the choices for Input Type. To select h parameters for input, press or use the D to move the highlight bar to h and press ¸. 2 Prm 1: z11_ Prm 2: z12_ Prm 3: z21_ Prm 4: z22_ Output Type y4 Parameter z11_; when h parameter is selected this changes to h11_. Parameter z12_; when h parameter is selected this changes to h12_. Parameter z21_; when h parameter is selected this changes to h21_. Parameter z22_; when h parameter is selected this changes to h22_. The right arrow4indicates additional choices for this parameter. Select this using the cursor bar. Pressing or ¸ displays a pop up menu showing the choices for Output Type the screen display shown. To select say z parameters for output, press ¨ or use the D to move the highlight bar to z and press ¸. The input screen for Convert Parameters has several characteristics common to various portions of the EE•Pro software. • The status line contains helpful information prompting the user for action. Choose: Input parameter type Input Type z Prm 1: z11_ Enter: P1 Impedance V1/I1 (I2=0) Prm 2: z12_ Enter: P1 Impedance V1/I2 (I1=0) Prm 3 z21_ Enter: P2 Impedance V2/I1 (I2=0) Prm 4: z22_ Enter: P2 Impedance V2/I2 (I1=0) Output Type y Choose: Output parameter type • The status line contents change if h parameters were chosen for Input Type: Prm 1: h11_ Enter: P1 Impedance V1/I1 (V2=0) Prm 2: h12_ Enter: P1 Parameter V1/V2 (I1=0) Prm 3 h21_ Enter: P2 Parameter I2/I1 (V2=0) Prm 4: h22_ Enter: P2 Admittance I2/V2 (I1=0) 2.3 Solving a Problem in Analysis Continuing the example of Parameter Conversion, h parameters are to be converted to y parameters. • At the input screen choose h for Input Type and y for Output type. Move the highlight bar to h11_ and type in a value of 125.35 and press ¸. • Repeat the above step entering a value of for h12_= .000028, h21_= -200 and h22_= 2.3E-6. The entered data can be real or complex or a variable name that is acceptable to the operating system. • Press „ to solve the problem. • The results of the computation are displayed in the result screen shown below. Inputs entered, ready to solve. EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis Calculated results. 3 Note: If the calculator is turned off automatically or manually while a results screen is being displayed, when EE•Pro is accessed again via the key, the software automatically bypasses the home screen of EE•Pro and returns to the screen result display. 2.4 Special Function Keys in Analysis Routines When Analysis functions are selected, the function keys in the tool bar access or activate features which are specific to the context of the section. They are listed in Table 2-1: Table 2-1 Description of the Function keys Function Key „ Description Labeled "Tools" - contains all the functions available on the TI-89 at the Home screen level. These functions are: 1: Open 2: (save as) 3: New 4: Cut 5: Copy 6: Paste 7: Delete 8: Clear 9: (format) A: About Labeled "Solve" - Is the primary key in various input screens. Pressing this key enables the software to begin solving a selected problem and display any resulting output to the user. Labeled "Graph" - This feature is available in input screens where the solution can be represented in graphical form. A plot can be viewed in the full screen or a split screen mode. This can be performed by pressing the followed by „. Use and to toggle between the data entry screen and graph window. Normally labeled as "View" - This enables the information content highlighted by the cursor to be displayed using the entire screen in Pretty Print format. An example of such a screen is shown in the screen displays shown. This function is also duplicated by pressing the key. If there is no contents to be viewed, then pressing the key has no effect. In some cases is labeled as "Mode", "Split Screen", "Pict", "Cash". • "Mode" is used in Computer Engineering Applications that displays an input screen to select binary parameters such word size, octal, binary, decimal or hexadecimal data. • "Split Screen" is displayed when a graphing solution is being set up. It allows the user to use the right half or the bottom half of the screen to display the graphical representations. • "Pict" is available when the Polyphase Circuits is selected, giving the user a quick glimpse of a picture of the circuit configuration. • "Cash" is used in Capital Budgeting section of the software Labeled "Opts" - This key displays a pop up menu listing the options: 1: View - allows the highlighted item to be viewed using Pretty Print. 2: (type) - Not active 3: Units - Not active 4: (conv) - Not active 5: Icons - presents a dialog box identifying certain Icons used by the software to display content and context of the information. EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis 4 ˆ • • • 6: (know)- Not active 7: Want - Not active “Edit” - Brings in a data entry line for the highlighted parameter. “Choose” in Capital Budgeting enabling the user to select from one of nine projects. “√ Check” requesting the user to press this key to select a highlighted parameter for use in an Analysis computation. Appears only when solving problems in the Ladder Network section and is labeled "Add"; this displays the input screen allowing the user to add new elements to a ladder network. Appears only when solving problems in the Ladder Network Section and is labeled "Del". Pressing will delete an element from the ladder network. 2.5 Data Fields Analysis Functions and Sample Problems The Data Fields available to the user in the Analysis functions falls into four convenient categories. AC Circuits, Polyphase Circuits, Filter Design sections provide the first type of user interface. In these sections, a pop up menu presents the types of analysis available. Once the user has chosen a specific analysis topic, an input form is presented to the user. For example, choosing AC Circuits section followed by Circuit Performance as a topic displays a screen that has all the inputs and output variables. Input screen for circuit. Impedance for Load Type. Press ¸ to display a Pop-up menu for Load Type. Input variables change when Admittance is selected for Load Type. Use the cursor bar to highlight Load Type, press ¸ to display a pop up menu for Impedance or Admittance as a load type. Selecting a different load type will automatically update the variable list in the input screen as shown above. Ladder networks presents a second type of user input interface. A ladder network consists of a load, and a series of ideal circuit elements (16 in variety) that can be added to the load as a rung or the side of a ladder. Circuit ). After selecting the proper element, enter the elements are added to the ladder network via the "Add" key (or value for the element and press „. This produces a description of the ladder as shown in the screen display. Ladder elements can be added at any location by moving the highlight bar to the element just prior to where the new element is needed and pressing . Any element can be removed from the list by pressing . The circuit elements are listed in order of appearance moving from the load (output) to the input. Pop up menu of elements for Ladder Network EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis Edit Screen for Resistor in Ladder Network 5 Updated component list for the Ladder Network The Gain and Frequency section under the Analysis menu offers an example where a problem is set up in one topic area (Transfer Function) and graphed under another topic heading (Bode Diagram). A Transfer Function is set up as in the screen shown below. It is important to note that data for Zeroes and Poles is entered as a list, e.g., numbers entered within curly brackets separated by commas. Once the Transfer Function has been determined, it can be graphed by switching to the Bode Diagram topic by pressing N followed by ©. The software takes full advantage of the graphing engine portion of the operating system of the calculator. Thus when the graphing function is executed using the … key, the tool bar reflects the functions available during a graphing operation. Input Screen for Transfer Function Graphing Parameters for Transfer Function Split Screen Display of Bode Diagram of Gain Function (Hs) The Computer Engineering section, under the Analysis menu, performs calculations involving numbers represented in binary, decimal, octal or hexadecimal formats within the constraints of parameters defined by the user. In any topic of the Computer Engineering section, the function key † opens a dialog box allowing the user to specify parameters such as the base of a number system, its word size, arithmetic using unsigned, 1's or 2's complement methods, setting Carry and Range Flags. Examples of these screens are shown below: Press † to access EE•Pro's Highlight Sign and press B MODE screen. to display a pop-up menu for available options. Capital Budgeting represents a fourth category of an input interface where the user can compare relative financial performance of several projects with relevant data such as interest rate or discount rate (k), IRR, NPV, Payback period. Screen displays shown here illustrate the basic user interface. Press † to display Cash Flow Input Screen for Capital Budgeting for ‘Project 1’ EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis 5 51 Example 2.1 (Numeric Results) Find the electrical Circuit Performance of an AC circuit consisting of a voltage source 110+15*i volts and an impedance of 25-12*i ohms. The load for the example is a capacitive impedance 70-89*i. Pop up menu in AC Circuits 1. 2. 3. 4. 5. 6. Input entry complete Computed output From the home screen of EE•Pro press the „ key labeled Analysis to display the pull down menu listing all the sections available under Analysis. Press ¨ to access AC Circuits section to view a pop up menu of all topics available. Press to enter to the input screen of Circuit Performance. Enter the value of 110+15*i for Vs_, 25-12*i ohms for Zs_ and 70-89*i for load impedance ZL_. Press the Solve key „. The results of the calculations are displayed in the data screen as shown. Example 2.2 (Symbolic Results) Find the parameters of a transmission line given the open circuit impedance is Z0_, the short circuit impedance is Z1_, distance is d1, and frequency of measurement is f1. Pull down menu for Analysis 1. 2. 3. 4. 5. 6. Menu for Transmission Lines Input for Line Parameters From the home screen of EE•Pro press the „ key labeled Analysis to display the pull down menu listing all the sections available under Analysis. Press to access the Transmission Lines section to view a pop up menu of the all available topics. Press to open the input screen for Line parameters. Enter the value of z0 for Zoc_, z1 of Zsc_, d1 for d and f1 for f. Press the Solve key „. The results of the calculations are displayed in the data screen as shown below. EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis 7 Input entered symbolically Calculated Output also symbolic Example 2.3 (Graphical Results) Construct a Bode diagram for a system with pole locations at 1000, 10000, 50000, a zero at 5000, and a proportionality constant of 1000000. From the home screen of EE•Pro, press the „ key to display the pull down menu listing all the sections available under Analysis. 2. Press to access the Gain and Frequency section to view a pop up menu of available topics. 3. Press ¨ to open the input screen for Transfer Functions. 4. Choose Roots for Inputs, enter 1000000 for Constant, {5000} for Zeroes List and {1000, 10000, 50000} for Poles List. 5. Press the Solve key „. 6. The results of the calculations are displayed in the data screen as shown. 7. Press key to revert to the pop up display for Gain and Frequency, and press to access Bode Diagram input screen. 8. Begin entering parameters for graphing the Gain of the Transfer Function. The minimum and maximum values for the horizontal axis show the default settings. Note that log (ω) is the horizontal axis. 9. Move the highlight bar to set ω-Min to 1, and ω-Max to 200000. 10. Move the highlight bar to Auto Scale and press ˆ to select this option. 11. Press to graph the function. 1. Examples of the screen display for this problem are shown here: Pull down menu of Graphing parameters EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis Pop up menu for Gain and Frequency Split Screen Graph 8 Input Screen for Analysis Function Transfer Partial view of Transfer function 2.6 Session Folders, Variable Names EE•Pro automatically stores its variables in the current folder specified by the user in or the HOME screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values for a particular session of EE•Pro, press :/TOOLS, :/NEW and type the name of the new folder (see Chapter 5 of the TI-89 Guidebook for the complete details of creating and managing folders). There are several ways to display or recall a value: • The contents of variables in any folder can be displayed using the °, moving the cursor to the variable name and pressing ˆ to display the contents of a particular variable. • Variables in a current folder can be recalled in the HOME screen by typing the variable name. • Finally, values and units can be copied and recalled using the /Tools 5:COPY and 6:PASTE feature. All inputs and calculated results from Analysis and Equations section are saved as variable names. Previously calculated, or entered values for variables in a folder are replaced when equations are solved using new values for inputs. Overwriting of variable values in graphing When an equation or analysis function is graphed, EE•Pro creates a function for the TI grapher which expresses the dependent variable in terms of the independent variable. This function is stored under the variable name pro(x). When the EE•Pro’s equation grapher is executed, values are inserted into the independent variable for pro(x) and values for the dependent value are calculated. Whatever values which previously existed in either of the dependent and independent variables in the current folder are cleared. To preserve data under variable names which may conflict with EE•Pro’s variables, run EE•Pro in a separate folder. Reserved Variables There is a list of reserved variable names used by the TI operating system which cannot be used as user variable names or entries. These reserved variables are listed in Appendix F. EE Pro for TI - 89, 92 Plus Analysis - Introduction to Analysis 9 Chapter 3 AC Circuits This chapter describes the software in the AC Circuits section and is organized under four topics. These topics form the backbone of AC circuit calculations. ™Impedance Calculations ™Voltage Divider ™Current Divider ™Circuit Performance 3.1 Impedance Calculations The Impedance Calculations topic computes the impedance and admittance of a circuit consisting of a resistor, capacitor and inductor connected in Series or Parallel. The impedance and admittance values are displayed to the user in symbolic, numeric, real or complex form. As stated in Chapter 2, due to the variety of computation methods used in each topic in of Analysis, the unit management feature is not present. All entered values are assumed to be in common SI units (F, A, m, Ω, s, etc.). Symbolic computation is limited to single undefined terms for each entry (such as ‘a’ or ‘x’ where ‘a’ and ‘x’ are undefined ) or an algebraic expression of previously defined terms which can be simplified to a numerical result upon entry (such as 1.5*x-3/y, where x=1+2*i and y=4). Field Descriptions Config: Elements: fr: R: L: C: ZZ_: YY_: (Circuit Configuration) Press ¸ and select Series or Parallel configuration by using D After choosing, press ¸ to display the input screen updated for the new configuration.. (Element Combination) Pressing „ displays the following circuit elements: R, L, C, RL, RC, LC and RLC. The choice of elements determines which of the input fields are available. (Frequency in Hz) Enter a real number or algebraic expression of defined terms. (Resistance in ohms - only appears if R, RL, RC or RLC is chosen in Elements field) Enter a real number or algebraic expression of defined terms. (Inductance in H enry- only appears if L, RL, LC or RLC is chosen in Elements field) Enter a real number or algebraic expression of defined terms. (Capacitance in Farads - only appears if C, LC, RC or RLC is chosen in Elements field) Enter a real number or algebraic expression of defined terms. (Impedance in ohms) Returns a real or complex number. (Admittance in Siemens) Returns a real or complex number. EE Pro for TI-89, 92 Plus Analysis - AC Circuits 10 Example 3.1 Compute the impedance of a series RLC circuit consisting of a 10 ohm resistor, a 1.5 Henry inductor and a 4.7 Farad capacitor at a frequency of 100 Hertz. 1. 2. 3. 4. 5. Input Screen Output Screen Choose Series for Config and RLC for Elements using the procedure described above. Enter 100 for Freq. Enter 10 for R, 1.5 for L, and 4.7 for C. Press „to calculate ZZ_ and YY_. The output screen shows the results of computation. This section demonstrates how to calculate the voltage drop across a load connected to an ideal voltage source. The load consists of impedances or admittances in series. The software computes the current through the load and the voltage across each impedance/admittance. Y1 Z2 3.2 Voltage Divider Z1 Y2 Vs Vs Z3 Y3 Field Descriptions Load Type: (Type of Load) Impedance/Admittance loads in series Press ¸ to display the choices; Impedance (Z) or Admittance (Y). The choice made determines whether the third field is ZZ_ (impedance chain) or YY_ (admittance chain) is displayed . (Source Voltage in V) Enter a real or complex number, variable name, or algebraic expression of defined terms. (Impedances in Series) Enter a list of real or complex numbers, or algebraic expression of defined terms. (Admittances in Series) Enter a list of real or complex numbers, or algebraic expression of defined terms. (Load Current in A) Returns a real or complex number, variable name or algebraic expression. (Element Voltages in V) Returns a list of real or complex numbers, variable names, or algebraic expressions. Vs_: ZZ_: YY_: IL_: V_: Example 3.2 Calculate the voltage drop across a series of loads connected to a voltage source of (110+25*i) volts. The load consists of a 50 ohm resistor, and impedances of (75+22*i), and (125-40*i) ohms. Input Screen EE Pro for TI-89.,92Plus Analysis - AC Circuits Output Screen 11 1. 2. 3. 4. 5. Choose Impedance for Load Type Enter the value 110 + 25*i for Vs_. Enter {50, 75 + 22*i, 125 - 40*i} for ZZ_. Press „ to calculate IL_ and V_. The results of the computation are shown in the screen display above. 3.3 Current Divider This section demonstrates how to calculate individual branch currents in a load defined by a set of impedances or admittances connected in parallel. In addition, the voltage across the load is calculated. ↑ ↑ Z1 Is Is Z3 Z2 Current Divider - Impedances Y1 Y2 Y3 Current Divider - Admittances Field Descriptions Load Type: Is_: ZZ_: YY_: VL_: I_: Press ¸ to select Impedance or Admittance. This sets the third field to be ZZ_ (Impedances) or YY_ (Admittances). (Source Current in A) Enter a real or complex number, variable name or algebraic expression of defined terms. (Impedance in ohms) Enter a list of real or complex numbers, or algebraic expression of defined terms. (Admittances in Siemens) Enter a list of real or complex numbers or algebraic expressions of defined terms. (Load Voltage in V) Returns a real, complex number or algebraic expression. (Currents in A) Returns a list of real or complex numbers or algebraic expressions. (Type of Load) Example 3.3 Calculate the voltage drop across impedances connected in parallel to a current source of (50 + 25*i). The load consists of 50, 75+22*i, 125-40*I Ω. Input Screen 1. 2. 3. 4. 5. Output Screen Partial Pretty Print of I_ Choose Impedance for Load Type Enter the value 50 + 25*i for Is_. Enter the value { 50, 75+22*i, 125-40*i } for ZZ_. Press „ to calculate VL_ and I_. The results of the computation are displayed in the screen shown above along with a Pretty Print display of the expression for I_. EE Pro for TI-89.,92Plus Analysis - AC Circuits 12 3.4 Circuit Performance This section shows how to compute the circuit performance of a simple load connected to a voltage or current source. Performance parameters computed include load voltage and current, complex power delivered, power factor, maximum power available to the load, and the load impedance required to deliver the maximum power. Field Descriptions - Input Screen Load Type: Vs_: Zs_ ZL_: Is_: Ys_: YL_: (Type of Load) Press ¸ to select load impedance (Z) or admittance (Y). This will determine whether the remaining fields Vs_, Zs_, and ZL_ or Is_, Ys_, and YL_ are displayed, respectively. (RMS Source Voltage in V) A real or complex number, variable name, or algebraic expression of defined terms. (Source Impedance in Ω) A real or complex number, variable name or algebraic expression of user-defined terms. A real or complex number, variable name, or algebraic (Load Impedance in Ω) expression of defined terms. (RMS Source Current in A) A real or complex number, variable name, or algebraic expression of defined terms. (Source Admittance in Siemens) Enter a real or complex number, variable name, or algebraic expression of defined terms. (Load Admittance in Siemens) Enter a real or complex number, variable name or algebraic expression of defined terms. Field Descriptions - Output Screen VL_: IL_: (Load Voltage in V) (Load Current in A) Returns a real, complex number or algebraic expression. Returns a real, complex number or algebraic expression. P: (Real Power in W) Returns a real number or algebraic expression. Q: VI_: (Reactive Power in W) (Apparent Power in W) Returns a real number or algebraic expression. Returns a complex number or algebraic expression. θ: (pf Angle in degrees or radians. setting) determined by the (Load Power Factor) (Maximum Power Available in W) (Load Impedance for Maximum Power in Ω - if Impedance is chosen for Load Type at the input screen). (Load Admittance for Maximum Power in Siemens - if Admittance, is chosen for Load Type at the input screen) Returns a real number or algebraic expression. PF: Pmax: ZLopt_: YLopt_: Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real, complex number or algebraic expression. Returns a real, complex number, or algebraic expression. Example 3.4 Calculate the performance parameters of a circuit consisting of a current source (10 - 5*i) with a source admittance of .0025 - .0012*i and a load of .0012 + .0034*i. EE Pro for TI-89.,92Plus Analysis - AC Circuits 13 Input Screen 1. 2. 3. 4. 5. Output: Upper Half Output: Lower Half Choose Admittance for Load Type. Enter the value 10 - 5*i for Is_. Enter the value .0025 - .0012*i for Ys_, and .0012 + .0034*i for a load of YL_. Press „ to calculate the performance parameters. The input and results of computation are shown above. EE Pro for TI-89.,92Plus Analysis - AC Circuits 14 Chapter 4 Polyphase Circuits This chapter describes Wye and ∆ arrangements in Polyphase Circuits. ™Wye ↔ ∆ Conversion ™Balanced Wye Load ™Balanced ∆ Load 4.1 Wye ↔ ∆ Conversion The Wye ↔ ∆ Conversion converts three impedances connected in Wye or ∆ form to its corresponding ∆ or Wye form, i.e., . Wye ↔ ∆ or ∆ ↔ Wye Input Fields Input Type: ZZA_: (∆ Impedance) ZZB_: (∆ Impedance) ZZC_: (∆ Impedance) ZZ1_: (Y Impedance) ZZ2_: (Y Impedance) ZZ3_: (Y Impedance) Z1 Selection choices are ∆→Wye or Wye→∆. This determines whether the next 3 fields (input fields) accept ∆ or Wye Impedances. Real or complex number, variable name, or algebraic Z2 Z3 expression of defined terms. Real or complex number, variable name, or algebraic expression of defined terms. Real or complex number, variable name, or algebraic Fig. 4.1 Wye Network expression of defined terms. Real or complex number, variable name, or algebraic expression of defined terms. Real or complex number, variable name, or algebraic expression of defined terms. Real or complex number, variable name, or algebraic expression of defined terms. Result Fields ZZA_: (∆ Impedance) ZZB_: (∆ Impedance) ZZC_: (∆ Impedance) ZZ1_: (Y Impedance) ZZ2_: (Y Impedance) ZZ3_: (Y Impedance) EE Pro for TI-89, 92 Plus Analysis - Polyphase Circuits Real or complex number, or algebraic expression. Real or complex number, or algebraic expression. Real or complex number, or algebraic expression. Real or complex number, or algebraic expression. Real or complex number or algebraic expression. Real or complex number or algebraic expression. 15 ZC ZB ZA Fig. 4-2 ∆ Network Example 4.1 - Compute the Wye impedance equivalent of a ∆ network with impedances 75+12*i, 75-12*i, and 125 ohms. 1. Select ∆ →Y for Input Type. . 2. Enter the values 75+12*i, 75 -12*i, 125 for ZZA_, ZZB_ and ZZC_. 3. Press „ to calculate ZZ1_, ZZ2_ and ZZ3_. Input Parameters The computation results are: Calculated Output ZZ1_: 34.0909 - 5.45455⋅i ZZ2_: 34.0909 + 5.45455⋅i ZZ3_: 20.9782 4.2 Balanced Wye Load A balanced Wye load refers to three identical impedance loads connected in a Wye configuration. The voltage V12_ represents the line voltage from line 1 to line 2 and is used as the reference voltage throughout this Wye network. The voltages across lines 2 and 3, and across 3 and 1 have the same magnitude as V12_, but are out of phase by 120º and 120º respectively. The software computes the currents I1_, I2_, and I3_ in each leg of the Wye network, the line to neutral voltage in each phase V1N_, V2N_, and V3N_, the power dissipated in each phase P, and the wattmeter readings W12 and W13 connected to the circuit as shown in the Fig 4-3. Input Fields Input Type: V12_: (Reference Voltage in V across lines 1 and 2) ZZ_: (Phase Impedance in Ω) Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. (Voltage in V across lines 2 and 3) (Voltage in V across lines 3 and 1) (Voltage in V across 1N) (Voltage in V across 2N) (Voltage in V across 3N) (Line Current 1 in A) (Line Current 2 in A) (Line Current 3 in A) (Phase Power in W) (Wattmeter reading in W across lines 1 and 2) (Wattmeter reading in W across lines 1 and 3) A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real, complex number, or variable name. A real number or variable name. A real number. A real number. Result Fields V23_: V31_: V1N_: V2N_: V3N_: I1_: I2_: I3_: P: W12: W13: Example 4.2 - A Wye network consists of three impedances of 50 + 25*i with a line voltage of 110 volts across line 1 and 2. Find the line current and power measured in a two-wattmeter measurement system. EE Pro for TI-89, 92 Plus Analysis - Polyphase Circuits 16 1. Select Balanced Wye Load. 2. Enter the value 50 + 25*i for ZZ_. 3. Press „ to calculate performance characteristics of the circuit. 1 I1 Z W12 V12 I3 W13 V31 Z 3 Z V23 I2 N 2 Input Screen Fig. 4.3 Balanced Wye network with 2 wattmeters Output Screen (upper half) Output Screen (lower half) The results of computation are listed below: V23_: V31_: V1N_: V2N_ : V3N_: I1_: I2_: I3_: P: W12: W13: -55 - 95.2628*i -55 + 95.2628*i 55 - 31.7543*i -55 - 31.7543*i 63.5085*i .625966 - .948068*i -1.13403 - .068068*i .508068 + 1.01614*i 64.533 124.744 68.8562 4.3 Balanced ∆ Load A balanced Delta load refers to three identical impedance loads connected in a Delta configuration. The voltage VAB_ represents the line voltage from line A to line B and is used as the reference voltage throughout this Delta network. The voltages across lines B and C, and C and A have the same magnitude as VAB_, but are out of phase by -120º and 120º respectively. The software computes the currents IA_, IB_, and IC_ in each leg of the Delta network, power dissipated in each phase P, wattmeter readings WAB and WAC connected as shown in the Fig 4-4 . Input Fields Input Type: VAB_: ZZ_: (Reference Voltage in V across lines 1 and 2) Enter a real or complex number, variable name, or algebraic expression of defined terms. (Phase Impedance in Ω) Enter a real or complex number, variable name, or algebraic expression of defined terms. Result Fields VBC_: (Voltage in V across lines B and C) EE Pro for TI-89, 92 Plus Analysis - Polyphase Circuits A real or complex number, or algebraic exp. 17 VCA_: IA_: IB_: IC_: P: WAB: WAC: (Voltage in V across lines C and A) (Line Current A in A) (Line Current B in A) (Line Current C in A) (Phase Power in W) (Wattmeter reading in across lines A and B) (Wattmeter reading in W across lines A and C) A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real number or algebraic expression. A real number or algebraic expression. A real number or algebraic expression. Example 4.3 - A Delta network consists of three impedances of 50 - 25*i with a line voltage of 110 volts across line A and B. Find the line current and power measured in a two-wattmeter measurement system. Input Parameters Calculated Output 1. Select Balanced Delta Load. 2. Enter the values 50 - 25*i for ZZ_ and 110_V for VAB_. 3. Press „ to calculate performance characteristics of the circuit. A IA WAC VAB IC VCA Z C IB WAB Z B Z VBC . Fig. 4.4 Wattmeter Measurement in a Delta Circuit The results of computation are listed below: VBC_: VCA_: IA_: IB_: -55 - 95.2628*i -55 + 95.2628*i 3.4021 - .204205*i -1.8779 – 2.8442*i EE Pro for TI-89, 92 Plus Analysis - Polyphase Circuits IC_: P: WAB: WAC: 18 -1.5242 + 3.04841*i 193.6 206.569 374.231 Chapter 5 Ladder Network This chapter describes ladder network analysis - a circuit reduction method by which branches of the circuit are treated as sides (series connection) or rungs (parallel or shunt connection) of a ladder. 5.1 Elements of a Ladder Network In the examples that follow, the left end of the ladder is the input end and the right end of the ladder is the output end, where the load is connected. The elements of the ladder (from 1 to N) are entered from right to left, going from output to input. The input impedance Zin_ is calculated as if you were “looking in” to the left end of the ladder. Field Descriptions - Input Screen Frequency: (Frequency) Enter a real number, or algebraic expression of defined terms. Load_: (Initial Load) Enter a real or complex number, variable name, or algebraic expression of defined terms. 1: (Element 1 - closest to the load or output) ... N: (Element N - furthest from the load or output) Field Descriptions - Element Screen Sixteen different element types are available to build the ladder network. These elements can be inserted in series or parallel configuration. Resistor A resistor can be added as a rung (parallel) or side (series). Choose Series or Parallel for Config, and enter a value for R in ohms. R R Series R Inductor - (ideal inductor) An inductor can be added as a rung (parallel) or side (series)). Choose Series or Parallel for Config, and enter a value for L in henrys. L L Series L EE Pro for TI-89.92 Plus Analysis - Ladder Networks 19 Parallel R Parallel L Capacitor - (ideal capacitor) A capacitor can be added as a rung (parallel) or side (series). Choose Series or Parallel for Config, and enter a value for C in Farads. C C Series C Parallel C L RL An RL series circuit can be added as a rung (parallel) or as an RL parallel circuit as a side (series). Choose Series or Parallel for Config, and enter a value for R in ohms and L in henrys. R R L Parallel RL Series RL LC An LC series circuit can be added as a rung (parallel) or as an LC parallel circuit as a side (series). Choose Series or Parallel for Config, and enter a value for L in henrys and C in farads. L C C L Series LC Parallel LC R RC An RC series circuit can be added as a rung (parallel) or as an RC parallel circuit as a side (series). Choose Series or Parallel for Config, and enter a value for R in ohms and C in farads. C C R Series RC Parallel RC L RLC An RLC series circuit can be added as a rung (parallel) or as an RLC parallel circuit as a side (series). Choose Series or Parallel for Config, and enter a value for R in ohms, L in henrys, and C in farads. R R C L C Series RLC General Impedance An impedance can be added as a rung (parallel) or side (series). Choose Series or Parallel for Config, and enter a value for Z_ in ohms. Parallel RLC Z Z Series Impedance Parallel Impedance n Transformer - (ideal transformer) A transformer can be added only in cascade connection. Specify turns ratio by entering a value for n. α Gyrator- (synthetic inductance filter) A gyrator can be added only in cascade connection. Specify gyrator parameter by entering a value for α . EE Pro for TI-89.92 Plus Analysis - Ladder Networks 20 v Voltage-Controlled I A controlled voltage can be added only in cascade connection. Specify base resistance and transconductance by entering values for rb in ohms and gm in siemens. Current-Controlled I A controlled current can be added only in cascade connection. Specify base resistance and common base current gain by entering values for rb in ohms and β. gm·v rb VCIS I β ·I rb (hie) (hfe) ICIS Transmission Line A transmission line can be added only in cascade connection. Specify characteristic impedance and electrical length by entering values for Z0 in ohms and θ0 in radians. The variation of θ0 with frequency is not taken into account in the ladder network calculation. Caution-be sure to enter a value for the electrical length θ0 which is consistent with the chosen frequency. Open Circuited Stub Can be added only in cascade connection. Specify characteristic impedance and electrical length by entering values for Z0 in ohms and θ0 in radians. Caution-be sure to enter a value for the electrical length θ0 which is consistent with the chosen frequency. Short Circuited Stub Can be added only in cascade connection. Specify characteristic impedance and electrical length by entering values for Z0 in ohms and θ0 in radians. Caution-be sure to enter a value for the electrical length θ0 which is consistent with the chosen frequency. 11 Two-Port Network Can be added only in cascade connection. Choose z, y, h, g, a, or b for Input Parameters, and enter values for ..11, ..12, ..21, and ..22. 12 21 22 Field Descriptions - Output Screen Zin_: I2/V1_: I2/I1_: Pout/Pin: V2/V1_: V2/I1_: (Input Impedance in ohms) Returns a real or complex number, variable name or algebraic expression. (Forward Transfer Admittance Returns a real or complex number, variable name or algebraic in Siemens) expression. (Current Transfer Ratio) Returns a real or complex number, variable name or algebraic expression. (Real Power Gain) Returns a real or complex number, variable name or algebraic expression. (Forward Voltage Transfer Ratio) Returns a real or complex number, variable name or algebraic expression. (Forward Transfer Impedance Returns a real or complex number, variable name or algebraic in ohms) expression. EE Pro for TI-89.92 Plus Analysis - Ladder Networks 21 5.2 Using the Ladder Network General instructions for entering the elements and computing the parameters of a ladder network. 1. The initial screen prompts the user for entry of values for Frequency and Load. 2. Build the ladder by adding elements to it. Press to insert the first element. Choose an element type and press ¸. Enter the appropriate values. Press „ to update the ladder with the new element just added. A second press of the „ key computes the electrical performance of the circuit. 3. New elements can be added or inserted by moving the highlight bar to the location desired and pressing . The new element will appear after the a highlighted element. 4. A circuit element can be deleted from the ladder by moving the highlight bar to the element and pressing .. 5. Press „ to compute the overall ladder network parameters. 6. Previously calculated results are not automatically updated for new element entries; the user must press „ to re-solve for the circuit parameters for a new circuit configuration. Example 5.1 What is the input impedance of the circuit shown below in Fig. 5.1 at 1 MHz and 10 MHz? 50 Ω 10E-6 H 100 pF Element 4 Element 3 Load 50 pF Element 2 50 Ω Element 1 Fig. 5.1 Ladder Network Example Entering Load and Frequency Partial list of Element Choices Typical Edit Screen for an Element List of all the Ladder Components Output Screen at 1 MHz . Output Screen at 10 MHz EE Pro for TI-89.92 Plus Analysis - Ladder Networks 22 1. 2. 3. 4. 5. 6. 7. Enter 1E6 for Frequency. Enter 50 for Load. Press to add the first element and move the highlight bar in the pull down menu to Capacitor and press ¸ to display the input screen for the Capacitor. Select Parallel for Configuration of the capacitor and enter the value 50E-12 for C. Press ¸ to accept the element data and press „ to return a listing of the Ladder Network. to enter the second element in this circuit. Move the highlight bar to 1: Capacitor and press Move the highlight bar to Inductor to display the input screen for the new element. Choose Series for Config and enter the value 10E-6 for L. Press ¸ to accept the value and press „ to update the ladder. Enter the remaining two elements: a 100 pF (100E-12) capacitor in parallel and a 50 ohm resistor in series. Press „ to calculate the results displayed in the output screen as shown above. To delete an element from the network, highlight it and press , the delete key. To calculate the ladder network parameters at a second frequency of 10 MHz: to return to the input screen. 1. Press 2. In the Frequency field, type 10E6. 3. Press „ to calculate the results, as displayed in the output screen shown above. Example 5.2 A transistor amplifier is shown in the figure below. The transistor is characterized by a base resistance of 2500 ohms, a current gain of 100 and is operating at a frequency of 10,000 Hertz. Figure 5.2 and 5.3 show the circuit and its simplified form. VCC 5K 1M 0.638 µ F 318 pF Vs Fig. 5.2 Transistor Amplifier Example This schematic can be reduced to the ladder network that appears in Fig. 5.3 I 0.638 µF 1 MΩ Vs Capacitor Resistor 2500 Ω 100 I Current-controlled 318 pF I Capacitor Fig 5.3. Simplified Transistor Amplifier Circuit Input Screen EE Pro for TI-89.92 Plus Analysis - Ladder Networks Output Screen 23 Load 5K 1. Enter the frequency and load values: Frequency: 10,000 Hz. Load: 5000 ohms. 2. Enter the ladder elements in the following order: Capacitor: Parallel, 318E-12. Current-Controlled I: Enter 2500 ohms for RB and 100 for β. Resistor: Parallel, 1E6 ohms. Capacitor: Series, 0.638E-6 farads. 3. Press „ to compute the results, which are displayed in the output screen above. EE Pro for TI-89.92 Plus Analysis - Ladder Networks 24 Chapter 6 Filter Design This chapter covers a description of the software under the heading Filter Design. Three filter designs are included in this section. Design computations result in the value of component elements comprising the filter. v Chebyshev Filter v Butterworth Filter v Active Filter 6.1 Chebyshev Filters This section of the software computes component values for Chebyshev filters between equal terminations. Inputs are termination resistance, pass band characteristics, attenuation at some out-of-band frequency, and allowable passband ripple as shown in Fig. 6.1. The Chebyshev circuit elements are assumed to be ideal, and are illustrated below. Odd Elements Low Pass High Pass Even Elements Cn Ln Cn Ln Lpn Band Pass Lsn Csn Cpn Lpn Band Elimination Lsn Csn Cpn Fig. 6.1 Chebyshev Filter Elements Field Descriptions - Input Screen Char: R: f0: f1: ∆ dB: Bandwidth: Ripple: (Bandpass Characteristic) Press ¸ to select Low Pass, High Pass, Band Pass, or Band Elimination. (Termination Resistance in ohms) Enter a real number, or algebraic expression of defined terms. (Cutoff Frequency in Hz - for Low Pass and High Pass) Enter a real number. (Center Frequency in Hz - for Band Pass and Band Elimination) Enter a real number. (Attenuation Frequency in Hz) Enter a real number. (Attenuation in dB) Enter a real number. (Bandwidth in Hz - only appears for Band Pass or Band Elimination) Enter a real number. (Pass Band Ripple in dB) Enter a real number. EE Pro for TI-89, 92 Plus Analysis - Filter Design 25 Field Descriptions - Output Screen Element1: number. Element2: …. ElementN: (First element in parallel) Returns a real (Second element in series) Returns a real number. (nth element in series because N is always odd) Returns a real number. Example 6.1 Design a low-pass Chebyshev filter with a cutoff at 500 Hz, a termination resistance of 50 ohm, 3 dB pass band ripple, and a 30 dB attenuation at 600 Hz. Input Screen 1. 2. 3. Output Screen Enter 50 for R, 500 for f0, and 600 for f1. Enter 30 for ∆ dB and 3 for Ripple. Press „ to calculate the results displayed in the output screen above. 6.2 Butterworth Filter This section computes the component values for Butterworth filters between equal terminations. Inputs are termination resistance, pass band characteristics, and attenuation at some out-of-band frequency. The basic form of the filter uses elements shown In Fig. 6.2 below: Element 2 R VS Element 4 Element n Element 3 Element 1 Odd Elements Low Pass R Even Elements Cn Ln High Pass Ln Band Pass Lsn Csn Lpn Band Elimination Lsn Csn Cpn Fig. 6.2 Elements for Butterworth Filter, basic design EE Pro for TI-89, 92 Plus Analysis - Filter Design 26 Field Descriptions - Input Screen Char: (Bandpass Characteristic) R: f0: Press ¸ to select Low Pass, High Pass, Band Pass, or Band Elimination. (Termination Resistance in Ohms) Enter a real number. (Cutoff Frequency in Hz - for Low Pass and High Pass) Enter a real number. (Center Frequency in Hz - for Band Pass and Band Elimination) Enter a real number. (Center Frequency in Hz - for Band Pass and Band Elimination) Enter a real number. f1: ∆ dB: (Attenuation Frequency in Hz) (Attenuation in dB) Bandwidth: (Band width in Hz - for Band Pass or Band Elimination) Enter a real number. Enter a real number. Enter a real number. Field Descriptions - Output Screen Element1: Element2: Element3: Element4: number. ElementN: (First element in parallel) (Second element in series) (Third element in parallel) (Fourth element in series) Returns a real number. Returns a real number. Returns a real number. Returns a real nth element in series (if n is odd) or parallel (if n is even) Returns a real number. Example 6.2 Design a 100 Hz wide Butterworth band pass filter centered at 800 Hz with a 30 dB attenuation at 900 Hz. The termination and source resistance is 50 ohms. Input Screen 1. 2. 3. 4. Output Screen Choose Band Pass for Char. Enter 50 for R and 800 for f0, and 900 for f1. Enter 30 for ∆ dB and 100 for Bandwidth. Press „ to calculate the results, which are displayed in the output screen above. 6.3 Active Filter This topic covers computation of element values for the standard active filter circuits shown below. In each case, five different elements are calculated. R4 Low Pass Filter C5 R1 R3 + C2 EE Pro for TI-89, 92 Plus Analysis - Filter Design 27 C4 High Pass Filter R5 C3 C1 + R2 C4 Band Pass Filter R5 C3 R1 + R2 Fig. 6.3 Active Filter Configurations Field Descriptions - Input Screen Type: f0: A: (Filter Type) (Band Cutoff in Hz) (Midband Gain in dB) Q: (Quality Factor: Q = C: (Capacitor in F) Press ¸ to select Low Pass, High Pass, or Band Pass. Enter a real number or algebraic expression of defined terms. Enter a real number or algebraic expression of defined terms. 1 1 = where α is the peaking factor and ζ is the damping factor) α 2 ⋅ζ Enter a real number or algebraic expression of defined terms. Enter a real number or algebraic expression of defined terms. Field Descriptions - Output Screen Element1: Element2: Element3: Element4: Element5: (First element) (Second element) (Third element) (Fourth element) (Fifth element) Returns a real number. Returns a real number. Returns a real number. Returns a real number. Returns a real number. Example 6.3 Design a High Pass active filter with a cutoff at 10 Hz, a midband gain of 10 dB, a quality factor of 1 and a capacitor of 1 µF. Input Screen 1. 2. 3. 4. Output Screen Choose High Pass for Type. Enter 10 for f0 and 10 for A. Enter 1 for Q and 1E-6 for C. Press „ to calculate the results are displayed on the screen above. EE Pro for TI-89, 92 Plus Analysis - Filter Design 28 Chapter 7 Gain and Frequency This chapter covers the basic principles of circuit analysis using a transfer function model and plots the resulting equations using the classical graphical representation often referred to as a Bode diagram for gain or phase: v Transfer Function v Bode Diagrams 7.1 Transfer Function A transfer function is defined as the ratio of an output to its input signal and is generally modified by a network between the two. In the classic sense, the transfer function is dependent upon the output and input definitions and is represented by a ratio of two polynomials of the complex frequency, s_. The roots of the numerator and denominator polynomials are referred to as zeros and poles, respectively. Transfer functions can be defined by the poles and zeros or by the coefficients of the numerator and denominator polynomials. The results computed include symbolic expressions for the transfer function and its partial fraction expansion. Field Descriptions Inputs: (Type of Input) Constant: Zeros: Poles: Numer list: Denom list: H(s)_: Press ¸ to select Roots or Coefficients. Determines whether the third and fourth fields are Zeros and Poles or Numer and Denom. (Constant Multiplier) Enter a real number. Default is 1. (Numerator Roots - if Roots is chosen for input type) Enter an array or list of real numbers. The number of zeros must be less than the number of poles. (Denominator Roots - if Roots is chosen for input type) Enter an array or list of real numbers. The number of poles must be greater than the number of zeros. (Numerator Coefficients - if Coefficients is chosen for input type) Enter an array or list of real numbers. The number of numerator coefficients must be less than the number of denominator coefficients. (Denominator Coefficients - if Coefficients is chosen for input type) Enter an array or list of real numbers. The number of denominator coefficients must be greater than the number of numerator coefficients. (Transfer Function) Returns a symbolic expression in the following form: EE Pro for TI-89, 92 Plus Analysis - Gain and Frequency 29 FG H FG H PFE_: IJ FG IJ KH K IJ FG IJ KH K s_ s_ 1− ... z1 z2 s_ s_ 1− 1− ... p1 p2 K ⋅ 1− Eq. 7.1.1 (Partial Fraction Expansion) Returns a symbolic expression of the form: FG H K1 K2 K3 + + +... s_ s_ s_ 1− 1− 1− p1 p2 p3 IJ FG KH IJ FG KH IJ K Eq. 7.1.2 Example 7.1 Find the transfer function and its partial fraction expansion for a circuit with a zero located at -10 r/s and three poles located at -100 r/s, -1000 r/s and -5000 r/s. Assume that the multiplier constant is 100000. Input Screen Output screen Partial view of Partial Fraction Expansion form for H(s) 1. 2. 3. Choose Roots for Inputs. Enter 100000 for Constant, {-10} for Zeros, and {-100 -1000 -5000} for Poles. Press „ to calculate H(s)_ and PFE_. 4. To view H(s)_ in Pretty Print format, press and ¸. Alternatively, the key can be pressed to achieve the same result. to return to the Gain and Frequency screen and Now you are ready to go on to the next example. Press select Bode Diagrams. 7.2 Bode Diagrams The behavior of the transfer function, as the frequency of a sinusoidal source varies, is of great interest to engineers. A very effective way to grasp the relationship between transfer function and frequency is to plot the magnitude and the argument of the transfer function on two separate graphs. These plots are often called Bode gain and phase plots. A gain plot shows the magnitude of the transfer function expressed in decibels (dB) as 20*LOG(Magnitude of Transfer Function) as a function of the logarithm of the radian frequency ω on the horizontal scale. The phase plot shows the argument of the transfer function expressed as the phase angle (i.e., ARG (Transfer Function) ) plotted as a function of the logarithm of the radian frequency on the horizontal scale. Field Descriptions Xfer: Indep: Graph Type: (Transfer Function) (Independent Variable (Bode Gain or Bode Phase) Enter a symbolic expression. Enter a global name. Default is ‘s_’. Press ¸ and select Gain or Phase. Determines whether the y axis range fields appear as A-Min and A-Max (for Gain) or θ-Min and θ-Max (for Phase). EE Pro for TI-89, 92 Plus Analysis - Gain and Frequency 30 ω-Max: Autoscale: Label Graph Full Screen A-Min: A-Max: θ-Min: (Maximum Frequency in r/s - X axis) Enter a real number. (Scales the plot: hides A-min and A-max fields) Press ˆ to select. Attaches labels to the graph. Select to graph the results in a full screen display. (Minimum amplitude in dB (Y axis) - if Gain is chosen for Plot Type) Enter a real number. (Maximum amplitude in dB (Y axis) - if Gain is chosen for Plot Type) Enter a real number. (Minimum amplitude in dB (Y axis) - if Phase is chosen for Plot Type) Enter a real number, which will be interpreted as degrees or radians, depending on the current setting in the Custom Settings screen. Example 7.2 Graph the gain and phase plots for the transfer function just computed in the previous example. Graph Dialogue (Upper and Lower displays) Bode Gain Diagram Bode Phase Diagram Graph the Bode Gain vs. Radian frequency Complete the entire example for the Transfer Function section. In the Bode Diagram screen, the Xfer field contains the Transfer Function H(s)_ calculated in the previous example. A choice is available to the user for Graph Type (Gain or Phase). 3. Choose s_ for Indep. (The default is s_ and should rarely be changed, because the Transfer Functions screen always generates transfer functions as functions of lowercase s_). 4. Enter 0.1 for ω-Min as the start of the radian frequency plot. 5. Enter 50000 for ω-Max as the endpoint of the radian frequency plot. 6. Put a check mark in the Autoscale field. Press ˆ if necessary 7. Put a check mark in the Label Plot field. Press ˆ if necessary. 8. Put a check mark on Full Screen graphing mode. If this field is not checked, the graph will default to the right half of the screen. 9. Press to graph the transfer function. 10. Press followed by to toggle between the input screen and the graph window when split-mode is active. 1. 2. Graph the Bode Phase vs. Radian frequency 1. In order to graph the phase of the transfer function, select Phase for Graph Type. Follow graphing sequence as defined in the section above. EE Pro for TI-89, 92 Plus Analysis - Gain and Frequency 31 Chapter 8 Fourier Transforms This section contains software computing discrete “Fast” Fourier transforms and its inverse. v FFT v Inverse FFT 8.1 FFT A physical process can be monitored in two significantly different ways. First, the process can be monitored in time domain in analog or digital form. Second, the data can be collected in the frequency domain in analog or digital form. In a variety of measurement and digital storage devices, data is gathered at regular, discrete time intervals. This data can be converted to its equivalent set in the frequency domain by the use of the so-called FFT algorithm. This algorithm maps a data array of N items to the corresponding array in the frequency domain using the following equation. N −1 Hk = h n ⋅ e-2π j⋅k⋅n N Eq. 8.1.1 n =0 The variable hn is the nth element in the time domain and Hk is the kth element in the frequency domain. The FFT algorithm treats the data block provided as though it is one of a periodic sequence. If the underlying data is not periodic, the resulting FFT-created wave is subject to substantial harmonic distortion. This section does not pad the input array with 0’s when the number of data points is not a power of 2. ∑ Field Descriptions Time: (Time Signal) Freq: (Frequency Spectrum) Enter an array or list of real or complex numbers. Returns spectral coefficients. Example 8.1 Find spectral coefficients for the periodic time signal [1 2 3 4]. 1. Enter [1 2 3 4] for Time. 2. Press „ to calculate and display results in the frequency domain Freq. 3. The screen display of the output and the input are shown below. ←Input Screen Output of Computation→ EE Pro for TI-89, 92 Plus Analysis - FFT 32 8.2 Inverse FFT This section focuses on transforming data from the frequency domain to the time domain. The inverse transform 1 hn = N N −1 ∑H k ⋅ e2π j⋅k⋅n N k =0 Eq. 8.2.1 algorithm uses the relationship displayed in the above equation, where Hk is the kth element in the frequency domain and hn is the nth element in the time domain. Field Descriptions Freq: (Frequency Spectrum) Time: (Time Signal) Enter an array or list of real or complex numbers. Returns time signal. Example 8.2 Find spectral coefficients for the periodic time signal [1 2 3 2 1]. 1. 2. 3. Enter [1 2 3 2 1] for Freq. Press „ to calculate and display results in the time domain Time. The screen display shows the computed results. Input Screen EE Pro for TI-89, 92 Plus Analysis - FFT Output Screen 33 Chapter 9 Two-Port Networks This chapter covers the basic properties of two-port network analysis under three topical headings. ™Parameter Conversion ™Circuit Performance ™Interconnected Two-Ports 9.1 Parameter Conversion input output I2 I1 V1 z, y, h, g, a or b V2 Many electrical or electronic systems are often modeled as two-port networks with four variables, namely input voltage V1, input current I1, output voltage V2, and output current I2. The two-port network is a black-box approach to solving simple to sophisticated problems of electrical and electronic circuits. The schematic representation (Fig. 9.1) shows a two-port network where all four variables are identified. For instance, a two-port network characterized by z parameters is defined by the following pair of equations: Fig. 9.1 Two- Port Network Model V 1 = I 1⋅ Z11 + I 2 ⋅ Z12 V 2 = I 1⋅ Z 21 + I 2 ⋅ Z 22 Eq. 9.1.1 Eq. 9.1.2 The four components of z parameters are defined as follows: Z11 = V1 I1 Z12 = V1 I 2 with I 2 = 0 with I1 = 0 Z 21 = V2 I1 Z 22 = V2 I 2 with I 2 = 0 with I1 = 0 The table below lists the independent variables for two-port circuits, followed by the dependent variables and twoport parameters associated with each set of independent variables. Conversion of parameters from one type to another is covered in this section. Independent Variables I1, I2 V1, V2 I1, V2 I2, V1 V2, I2 Dependent Variables V1, V2 I1, I2 V1, I2 I1, V2 V1, I1 V2, -I2 EE Pro for TI - 89, 92 Plus Analysis - Two-Port Networks V1,-I1 34 Two-Port Parameter Type z y h g b a Field Descriptions - Input Screen The software is configured for the entry of two-port parameters; an input of one type and an output of another type. Input Type: Selecting this item (highlighting and pressing ¸ ) displays a menu of input types available to the user. Selecting z, y, h, g, a, or b automatically updates the names of the remaining parameter to reflect the choice of input type. Thus if z were selected, the inputs would have the labels listed below: Z11_: Z12_: Z21_: Z22_: Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Output Type: Press ¸ to display z, y, h, g, a, or b, use D to move the highlight bar to select the choice and press ¸. Note: The help text (e.g., “Enter P1 Impedance V1/I1 (I2=0)”) shows whether the ratio is an impedance, admittance or dimensionless ratio, which port is being described, and whether it is the current or the voltage that is open. Field Descriptions - Output Screen ..11_: Returns a real or complex number, variable name, or algebraic expression. ..12_: Returns a real or complex number, variable name, or algebraic expression. ..21_: Returns a real or complex number, variable name, or algebraic expression. ..22_: Returns a real or complex number, variable name, or algebraic expression. Menu of available Input types Input screen updates for choice of input type Example 9.1 Convert a resistive two-port network with z11_ = 10, z12_ = 7.5, z21_ = 7.5, z22_ = 9.375 into its equivalent y values. 1. Choose z for Input Type. 2. Enter 10 for z11_, 7.5 for z12_, 7.5 for z21_, and 9.375 for z22_. 3. Choose h for Output Type. 4. Press „ to calculate the results which are displayed in the output screen shown below. Input Screen EE Pro for TI - 89, 92 Plus Analysis - Two-Port Networks Output Screen 35 9.2 Circuit Performance This section computes the circuit performance of a two-port network. Given a voltage source with a finite impedance and a finite load, the software will compute the input and output impedances, the current and voltage gains, the voltage gain with reference to source, the current gain to the source, the power gain, the power available to the load, the maximum power available to the load, and the load impedance for maximum power deliverable to the load. Zs ZL Vs Field Descriptions - Input Screen Parameter Type: Press ¸ to select z, y, h, g, a, or b. ..11_: Enter a real or complex number, variable name, or algebraic expression of defined terms. ..12_: Enter a real or complex number, variable name, or algebraic expression of defined terms. ..21_: Enter a real or complex number, variable name, or algebraic expression of defined terms. ..22_: Enter a real or complex number, variable name, or algebraic expression of defined terms. Vs_: (Source Voltage in V) Enter a real or complex number, variable name, or algebraic expression of defined terms. Zs_: (Source Impedance in Ω) Enter a real or complex number, variable name, or algebraic expression of defined terms. ZL_: (Load Impedance in Ω) Enter a real or complex number, variable name, or algebraic expression of defined terms. Field Descriptions - Output Screen Zin_: (Input Impedance in Ω ) Iout_: (Output Current in A) Vout_: (Thevenin Voltage in V) Zout_: (Thevenin Impedance in Ω) Igain_: (Current Gain) Vgain_: (Voltage Gain) VgainAbs_: (Absolute Voltage Gain) GP: (Power Gain) Pmax: (Maximum Power at ZLopt in W) ZLopt_: (Optimum Load Impedance in Ω) Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real number or algebraic expression. Returns a real or complex number or algebraic expression. Example 9.2 A transistor has the following h-parameters h11_ = 10 ohms, h12_ = 1.2, h21_= -200, h22_= .000035_siemens. The source driving this transistor is a 2 volt source with a source impedance of 25 ohms. The output port is connected to a 50 ohm load. Find the performance characteristics of the circuit. 1. 2. 3. 4. Select h parameters to prepare the input screen to receive input. Enter h11_= 10, h12_= 1.2, h21_= -200, h22_= .000035 Enter Vs_ =2 volts, Zs_ = 25 ohms, and ZL_= 50 ohms. Press „ to solve for the characteristics. 5. The result of the calculations is presented below: EE Pro for TI - 89, 92 Plus Analysis - Two-Port Networks 36 Input Screen Results: Upper Half Zin_ (Input Impedance) Iout_ (Output Current) Vout_ (Thevenin Voltage) Zout_ (Thevenin Impedance) Igain_ (Current Gain) Vgain_ (Voltage Gain) VgainAbs (Absolute Voltage Gain) GP_ (Power Gain) Pmax: (Maximum Power at ZLopt) ZLopt_ (Optimum Load Impedance) Results: Lower Half 11989.03667 ohms -.033236 amps 1.66666 volts .145833 ohms -199.651 .832638 .830906 664.947 4.76188 watts .145833 ohms 9.3 Connected Two-Ports Two-port networks constitute basic building blocks of linear electrical or electronic systems. In the design of large, complex systems, it is easier to synthesize the system by first designing subsections. Large and complex systems can be built using simpler two-port building blocks. Assuming that Brune’s criteria is valid for these networks, the twoport subsections can be interconnected in five ways: 1. Cascade: The output of network 1 is connected directly to the input of network 2. Network 1 Network 2 Cascade Network 1 2. Series-Series: The inputs and outputs of the two networks are both connected in series. Network 2 Series-Series Network 1 3. Parallel-Parallel: The inputs and outputs of the two networks are both connected in parallel. Network 2 Parallel-Parallel Network 1 4. Series-Parallel: The inputs of the two networks are connected in series, while the outputs are connected in parallel. Network 2 Series-Parallel Network 1 5. Parallel-Series: The inputs of the two networks are connected in parallel, while the outputs are connected in series. Network 2 Parallel-Series This section accepts parameters for either network as z, the choice made. EE Pro for TI - 89, 92 Plus Analysis - Two-Port Networks y, h, g, a, or b and yields the output in any form based on 37 Field Descriptions - Input Screen Connection: (Type of Connection) Press ¸ to select Cascade, Series Series, Parallel Parallel, Series Parallel, or Parallel Series. First Input Type: Press ¸ to select parameter type z, y, h, g, a, or b for entry. ..111_: ..112_: ..121_: ..122_: Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Second Input Type: Press ¸ to select parameter type ..211_: ..212_: ..221_: ..222_: Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms. Output Type: Press ¸ to select z, ..11_: ..12_: ..21_: ..22_: z, y, h, g, a, or b for entry. y, h, g, a, or b. Returns a real or complex number, variable name, or algebraic expression of defined terms. Returns a real or complex number, variable name, or algebraic expression of defined terms. Returns a real or complex number, variable name, or algebraic expression of defined terms. Returns a real or complex number, variable name, or algebraic expression of defined terms. Example 9.3 The 2 two-port networks are defined in terms of their z and h parameters. The z parameters are 10, 7.5, 7.5, 9.375 respectively while the h parameters are 25, .001, 100, .0025. If the two-ports are connected in a cascade configuration, compute the y parameters for the resulting equivalent two-port. To solve this problem: 1. 2. 3. 4. 5. 6. 7. Choose Cascade for Connection. Choose z for First Input Type. Enter 10 for z11_, 7.5 for z12_, 7.5 for z21_, and 9.375 for z22_. Choose h for Second Input Type. Enter 25 for h11_, .001 for h12_, 100 for h21_, and .0025 for h22_. Choose y for Output Type. Press „ to complete the computation. The resulting data is shown in the screen displays shown. Input screen showing z parameters for the first two-port. EE Pro for TI - 89, 92 Plus Analysis - Two-Port Networks Input screen showing h parameters for the Second two-port. 38 Output screen showing y parameters of the combined Two-port. Chapter 10 Transformer Calculations This chapter covers the software features used to perform to calculations for electrical transformers. This section is organized under three categories: ™Open Circuit Test ™Short Circuit Test ™Chain parameters 10.1 Open Circuit Test An open circuit test described here is usually performed at rated conditions of the primary or secondary side of a transformer. It is common practice to apply a voltage to the primary side. The measured parameters which include primary and secondary voltages, current, and power determine the core parameters of the transformer. Field Descriptions - Input Screen V1: (Primary RMS Voltage in V) V2: (Secondary RMS Voltage in V) I1: (Primary RMS Current in A) PP1: (Primary Real Power in W) Magnitude only. A real number, variable name, or algebraic expression of defined terms. Magnitude only. A real number, variable name, or algebraic expression of defined terms. Magnitude only. A real number, variable name, or algebraic expression of defined terms. A real number, variable name, or algebraic expression of defined terms. Field Descriptions - Output Screen n: Q1: Gc: Bc: (Primary to secondary turns ratio) (Reactive power in W) (Primary core conductance in S) (Primary core susceptance in S) Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Example 10.1 Perform an open circuit test on the primary side of a transformer using the following data: The input to the primary coils with the secondary side open is 110 volts, and a current of 1 ampere and a power of 45 watts. The secondary open circuit voltage is 440 volts. Find the circuit parameters of the transformer. EE Pro for TI -89, 92 Plus Analysis - Transformer Calculations 39 Input Screen 1. 2. 3. Output Screen Enter the values 110 for V1 and 440 for V2. Enter 1 for I1 and 45 for PP1. Press „ to calculate and display the results, as shown above. 10.2 Short Circuit Test Short circuit tests are often a quick method used to determine the winding impedance of a transformer and are usually reported at rated kVA values. This test consists of placing a short circuit across the secondary windings and applying a small primary voltage to measure the secondary current, and power supplied to the transformer. The calculated circuit parameters (i.e., resistance and reactance of primary and secondary coils) are based on the assumption that the heat dissipation in the primary and secondary windings are equal. Field Descriptions - Input Screen V1: (Primary RMS Voltage in V) I2: (Secondary RMS Current in A) PP1: (Primary Real Power in W) kVA: (kVA rating in kVA) V1R: (Primary Voltage Rating in V) Magnitude only. A real number, variable name, or algebraic expression of defined terms. Magnitude only. A real number, variable name, or algebraic expression of defined terms. A real number, variable name, or algebraic expression of defined terms. A real number, variable name, or algebraic expression of defined terms. A real number, variable name, or algebraic expression of defined terms. Field Descriptions - Output Screen n: QP1: RR1: RR2: XX1: XX2: (Primary to secondary turns ratio) (Primary Reactive Power in W) (Primary Resistance in Ω) (Secondary Resistance in Ω) (Primary Reactance in Ω) (Secondary Reactance in Ω) Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Example 10.2 Short circuit test data is taken on a transformer. A primary input voltage of 5 volts forces 18 amperes of current into the secondary winding under short circuit conditions. The power supplied for the test is 5 watts. The transformer has a kVA rating of 30 and a primary voltage rating of 110 volts. Find the parameters of the of the transformer. Input Screen EE Pro for TI -89, 92 Plus Analysis - Transformer Calculations Output Screen 40 1. 2. 3. Enter the values 5 for V1 and 18 for I2. Enter 5 for PP1, 30 kVA rating and 110 for V1R. Press „ to calculate the results, which are displayed in the output screen. 10.3 Chain Parameters Chain parameters (or the so-called ABCD parameters) are convenient problem-solving tools used in solving transmission and distribution problems. The parameters are expressed essentially as two-port type parameters by the use of the following pair of linear equations: n2·Z2 Z1 Vin = A ⋅ Vout – B ⋅ Iout Bc Iin = C ⋅ Vout – D ⋅ Iout Gc Eq. 10.3.1 Eq. 10.3.2 where Vin and Iin are input voltage and current, and Vout and Iout are output voltage and current. (All quantities in the diagram refer to the primary winding.) This approach is useful as cascaded transformers follow two-port network rules. Field Descriptions - Input Screen ZZ1_: (Primary impedance in Ω) A real or complex number, variable name or algebraic expression of defined terms. ZZ2_: (Secondary impedance in Ω) A real or complex number, variable name, or algebraic expression of defined terms. n: (Primary to secondary turns ratio) Returns a real number, variable name, or algebraic expression. Gc: (Primary core conductance in S) A real number, or algebraic expression of defined terms Gc>0. Bc: (Primary core susceptance in S) A real number, or algebraic expression of defined terms Bc<0. Field Descriptions - Output Screen Param A_: Param B_: Param C_: Param D_: (A Parameter) (B Parameter) (C Parameter) (D Parameter) Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression. Example 10.3 A transformer has a primary and secondary impedance of 250 + 23*i and 50 + 10*i and a turns ratio of 0.2. The conductance and susceptance of the primary coil is .001 and -.005 respectively. Find the A, B, C and D parameters. Input Screen 1. 2. 3. Output Screen Enter the values of 250 + 23*i and 50 + 10*i for ZZ1_ and ZZ2_. Enter .2 for n, .001 and -.005 for Gc and Bc respectively. Press „ to calculate the results, displayed in the output screen above. EE Pro for TI -89, 92 Plus Analysis - Transformer Calculations 41 EE Pro for TI -89, 92 Plus Analysis - Transformer Calculations 42 Chapter 11 Transmission Lines This chapter covers the basic principles of transmission line analysis covered under four categories of topics: ™Line Properties ™Line Parameters ™Fault Location Estimate ™Stub Impedance Matching 11.1 Line Properties This portion of the software computes the characteristic parameters of a transmission line from fundamental properties of the wires forming the transmission line. A variable name can consist of a single undefined variable (such as ‘a’ or ‘x’) or an algebraic expression of defined terms of defined variables which can be simplified into a numerical result (such as ‘x+3*y’, where x=-3 and y=2). All entered values are assumed to be common SI units (F, A, kg, m, s, Ω) or units arbitrarily chosen by the user (such as such as length which can be in m, km, miles or µm). Field Descriptions - Input Screen L: R: G: C: ZL_: d: f: (Series Inductance in H/unit Length) A real number, variable name or algebraic expression of defined terms, L≥0. (Series Resistance in Ω/unit Length) A real number, variable name or algebraic expression of defined terms ,R≥0. (Shunt Conductance in Siemens/unit Length) A real number, variable name or algebraic expression of defined terms, G≥0. (Shunt Capacitance in F/unit Length) A real number, variable name or algebraic expression of defined terms, C≥0. (Load Impedance in Ω) A real or complex number, variable name or algebraic expression of defined terms, ZL_≥0. (Distance to Load - unit length) A real number, variable name or algebraic expression of defined terms, d>0. (Frequency in Hz) A real number, variable name or algebraic expression of defined terms, f>0. Field Descriptions - Output Screen ZZ0_: (Characteristic Impedance in Ω) YY0_: (Characteristic Admittance in Siemens) α: (Neper Constant in 1/unit length) EE Pro for TI 89, 92 Plus Analysis - Transmission Lines Returns a real or complex number, variable name, or algebraic expression. Returns a real or complex number, variable name, or algebraic expression. Returns a real number, algebraic expression. 43 β: (Phase Constant in degrees or radians/unit length) Returns a real number, or algebraic expression. λ: (Wavelength in unit length) vp: (Phase Velocity in unit length/s) Zoc_: (Open Circuit Impedance in Ω) Zsc_: (Short Circuit Impedance in Ω) ρ_ : (Reflection Coefficient) Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real number, or algebraic expression. SWR: (Standing Wave Ratio) Example 11.1 A transmission line has a series inductance of 1 mH/mile, a line resistance of 85.8 ohm/mile, a conductance of .0015 x 10-6 Siemens/mile, and a shunt capacitance of 62 x 10-9 F/mile. For a load impedance of 75 ohms and a frequency of 2000 Hz, compute the line characteristics 3 miles away from the load. Note: Since the unit length is the mile, all entered and calculated values are with respect to miles. Input Screen 1. 2. 3. 4. Output Screen (upper half) Output Screen (lower half) Enter the values 0.001, 85.8, and .0015E-6 for L, R, and G, respectively. Enter the values 62E-9, 75, and 3 for C, ZL_, and d, respectively. Enter 2000 for f. Press „ to calculate the results. The input and output screen displays are shown above. 11.2 Line Parameters This topic computes fundamental parameters of a transmission line from measured data. The algorithm used in this section solves for γ in the equation 11.2.1, where γ = α + jβ. In general γ, Zsc_, and Zoc_ have complex values. In solving for γ, there is a principal value and a set of equivalent values because of the cyclical nature of the equation. Recognizing the fact that physical parameters such as R, L, G, C, and vp are all real and positive numbers, extreme caution should be exercised when entering input data. In particular, d should be less than one wavelength. tan(γ ⋅ d ) = Zsc _ Zoc _ Eq. 11.2.1 Field Descriptions - Input Screen Zoc_: (Open Circuit Impedance in Ω) Zsc_: Enter a real or complex number, variable or algebraic expression of defined terms. Enter a real or complex number, variable or algebraic expression of defined terms. (Short Circuit Impedance in Ω) EE Pro for TI 89, 92 Plus Analysis - Transmission Lines 44 d: (Distance to Load Location/ unit length) f: (Frequency in Hertz) Enter a real number or algebraic expression of defined terms or variable. Enter a real number or variable or algebraic expression of defined terms. Field Descriptions - Output Screen R: L: G: C: ZZ0: (Series Resistance in Ω/unit length) (Series Inductance in H/unit length) (Shunt Conductance in Siemens/unit length) (Shunt Capacitance in F/unit length) (Characteristic Impedance in Ω) YY0: (Characteristic Admittance in Ω) α: β: Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real or complex number or algebraic expression. Returns a real or complex number or an algebraic expression. . (Neper Constant in 1/unit length) Returns a real number or algebraic expression. (Phase Constant in degrees or radians/unit length) vp: Returns a real number or algebraic expression. Returns a real number or algebraic expression. (Phase Velocity in unit length/s) Example 11.2 A transmission line is measured to have an open circuit impedance of 103.6255 - 2.525* i, and an impedance under short circuit conditions of 34.6977 + 1.7896*i, at a distance 1 unit length from the load location. All measurements are conducted at 10 MHz. Compute all the line parameters. Input Screen 1. 2. 3. 4. Output Screen (upper half) Output Screen (lower half) Enter 103.6255 - 2.525*i for Zoc_. Enter 34.6977 + 1.7896*i for Zsc_. Enter 1 and 10E6 for d and f, respectively. Press „ to calculate the results and display them as shown above. 11.3 Fault Location Estimate In this section, the equation set estimates the distance to the location of a fault condition in a transmission line. The calculation is made from a few measurements and is based on the underlying assumption that the transmission line is ideal. Field Descriptions Xin: RR0: β: (Input Reactance in Ω) A real number, variable name, or algebraic expression of defined terms. (Characteristic Resistance in Ω) A real number, variable name, or algebraic expression of defined terms. (Phase Constant - degrees or radians /unit length) EE Pro for TI 89, 92 Plus Analysis - Transmission Lines 45 docmin: dscmin: A real number, variable name, or algebraic expression of defined terms. (Minimum distance to an open circuit fault in unit lengths) Returns a real number, variable name, or algebraic expression. (Minimum distance to short circuit fault in unit lengths) Returns a real number, variable name, or algebraic expression. Example 11.3 A transmission line measures a capacitive reactance of -275 ohms. The characteristic line impedance is 75 ohms, and has a phase constant of 0.025 r/length. Estimate the location of the fault. Input Screen 1. 2. Output Screen Enter the values -275, 75, and 0.025 for Xin, RR0, and β, respectively. Press „ to calculate the results as shown in the screen display above. 11.4 Stub Impedance Matching This topic calculates the parameters for a single-stub impedance-matching device. The location display and the electrical length of an open and short-circuit stub can be computed from the input data. Because the solution is circular in nature, there are two possible stub-locations d1 and d2. Field Descriptions - Input Screen ZL_: (Load Impedance in Ω) RR0: defined (Characteristic Resistance in Ω) A real or complex number, variable name, or algebraic expression of defined terms. A real number , variable name, or algebraic expression of terms. Field Descriptions - Output Screen β*d1: β*d1-sc: β*d1-oc: β*d2: β*d2-sc: β*d2-oc: (Electrical length from a stub at location d1 to the load in degrees or radians) Returns a real or algebraic expression. (Electrical length of a short-circuited shunt stub at distance d1 from load in degrees or radians) Returns a real or algebraic expression. (Electrical length of an open-circuited shunt stub at distance d1 from load in degrees or radians) Returns a real or algebraic expression of defined terms. (Electrical length from a stub at location d2 to the load in degrees or radians) Returns a real or algebraic expression. (Electrical length of a short-circuited shunt stub at distance d2 from load in degrees or radians) Returns a real or algebraic expression. (Electrical length of an open-circuited shunt stub at distance d2 from load in degrees or radians) Returns a real or algebraic expression. EE Pro for TI 89, 92 Plus Analysis - Transmission Lines 46 Example 11.4 A transmission line has a characteristic impedance of 50 ohms and a load of 75 ohms. Estimate the shorting stub location for matching purposes. Input Screen 1. 2. Output Screen Enter the values 50 and 75 for RR0, and ZL_ respectively. Press „ to calculate the results as shown in the screen displays above. EE Pro for TI 89, 92 Plus Analysis - Transmission Lines 47 Chapter 12 Computer Engineering This chapter covers functions of interest in the design of logic systems and circuits. The modules include binary arithmetic, bit operations, comparisons and a form of logic minimization using the Quine-McCluskey algorithm: ™Binary Arithmetic ™Register Operations ™Bit Operations ™Binary Conversions ™Binary Comparisons ™Karnaugh Map Read This! The format of integers in the Computer Engineering section of EE•Pro is (p)nnnn…, where p is the letter prefix b, o, d, or h indicating binary, octal, decimal, and hexadecimal, number systems and n represents the legal digits for the number base. If a number does not begin with a letter prefix in parenthesis, then it is assumed to be in the number base set in the †/Modes dialog. With the exception of Binary Conversion, no numbers with fraction components are allowed to be entered. The TI-89 built-in binary conversion feature supports 32 bit word lengths. EE•Pro has extended this feature to allow the user to modify the word size between 1 and 128 bits. 12.1 Special Mode Settings, the † Key The screen allows the user to set the parameters for solving binary problems. Press the † key to display a user interface dialog box prompting the user to clarify and specify the foundation for digital arithmetic. Digital arithmetic operations are allowed in binary, octal, decimal and hexadecimal number systems. As shown in the screen display below, user input is needed in the following areas: Base: Word size: Sign: Binary, Octal, Decimal and Hexadecimal. Up to 128 bits. Unsigned, 1’s complement and 2’s complement techniques. The software allows three conventions for representing numbers. 1. 2. 3. Unsigned mode - In the unsigned mode, the most significant bit adds magnitude and not the sign to the number. 1's Complement mode - One's complement accommodates an equal number of positive and negative numbers, but has two representations for zero; 0 and -0. 2's complement mode - There is just one representation for zero, but there is always one more negative number than a positive number represented. Leading 0's are used for internal representation, but are not displayed as such. Thus a number entered as 000110011 will be displayed as 110011 EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 48 Carry and Range Conditions - The shifting, rotating, arithmetic and bit manipulation operations can result in Carry and Range Flags being modified. The specific condition for the flags to be set depend upon the operation being performed. Range flag is set if the correct result of the operation cannot be represented in the current word size and the complement mode. When the result is out of range, the lower order bits that fit the word size are displayed. In Table 12-1 below, we have used an example for a binary word size of 5 bits to convey the conventions and its decimal interpretation. Binary 01111 01110 01101 01100 01011 01010 01001 01000 00111 00110 00101 00100 00011 00010 00001 00000 11111 11110 11101 11100 11011 11010 11001 11000 10111 10110 10101 10100 10011 10010 10001 10000 Table 12.1 Decimal Interpretation of a 5 bit Binary 1's Complement 2's Complement Mode Mode 15 15 14 14 13 13 12 12 11 11 10 10 9 9 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 0 0 -0 -1 -1 -2 -2 -3 -3 -4 -4 -5 -5 -6 -6 -7 -7 -8 -8 -9 -9 -10 -10 -11 -11 -12 -12 -13 -13 -14 -14 -15 -15 -16 Unsigned Mode 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 Note: The EE•Pro introduces a new convention for entering the binary integers. Binary integers start with (b), octal numbers start with (o), decimal numbers with (d) and hexadecimal numbers by (h). This convention is introduced to ensure that the machine convention of using 0d would be interpreted properly when in hexadecimal mode. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 49 † activates mode dialog box. Binary Arithmetic Screen Display Once the parameters have been set in the binary mode screen, pressing ¸ accepts the entries, while pressing N cancels the entries made. If an incorrect choice is made in the data entry, a dialog box pops up to alert the user that an error has been detected in the data entered. 12.2 Binary Arithmetic This section demonstrates how to add, subtract, multiply, and divide binary numbers, as well as negate or find the absolute value of a binary integer. Field Descriptions Binary 1: (Input Field) Enter an integer in the number base designated in †/Binary Mode or Binary 2: (Input Field) an integer preceded by the number base in parenthesis (b, d, o, or h). Enter an integer in the number base designated in †/Binary Mode or Operator: (Binary Operation) an integer preceded by the number base in parenthesis (b, d, o, or h). Press ¸ to select. ADD SUB MULT DIV RMD NEG ABS Adds binary or real integer. Affects the range and carry flags. Subtracts the second binary or real integer from the first. Affects both the range And carry flags. Multiply the two binary or real numbers. Affects both the range and the carry flags. Divide the first binary integer by the second and returns the quotient. Affects carry flag only. Divides the first binary integer by the second binary integer and returns the remainder. Does not affect the range or the carry flags. Negate a binary integer. Affects the range flag only. Find the absolute value of a binary integer. Affects the range flag only. Result: (Binary Function Value) Returns an integer result using the number base set in †/Binary Mode. Flags: (State of Carry and Range Flags) Returns Carry if the carry flag is set. Returns Range if the range flag is set. Example 12.2 Multiply two real hexadecimal numbers 25a6 and 128d. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 50 Input Screen 1. 2. 3. 4. Output Screen Press † to select hexadecimal for Base: for data, 128 bits for Wordsize, Unsigned number system and press ¸ to accept the choice. Enter 25a6 for the first Binary 1 number. Enter 128d for the second Binary 2 number. Choose MULT for Operator; press „ to compute the result. The screen displays above show the input screen and the resulting output. Note: When a binary operation results in a quantity which exceeds the word size selected by the user, the range and carry flags are automatically set, and the right-most digits of the result (i.e., the least significant portion of the answer) will be displayed in the word size allocated. For example, if the word size is set to 32 bits (corresponding to an 8 digit hexadecimal number), and two hexadecimal numbers, 87654321 and FEDCBA98 are added, the result is a 9 digit hexadecimal number 18641FDB9 which exceeds the 32 bit (8 digit) word size allocated for the result. 87654321 + FEDCBA98 18641FDB9 The carry flag is set and the least significant eight digits of the actual result are displayed (i.e., 8641FDB9). Retention of the least significant digits mimics the usual operation of microprocessors. 12.3 Register Operations This section allows the user to perform operations on any bit of a binary integer. Such operations include shifting bits to the left or right, rotating bits to the left or right, and shifting or rotating bits through the carry flag. The input screen is updated to reflect the choice of operator. For instance, if we choose SL, SR, RR, RL, ASR no additional input lines are displayed. On the other hand, if SLN, SRN, RLN, RRN are chosen, a new line appears at the input wanting an entry for the “no. of bits” N. When Carry is an integral part of the operation in cases such as RLC, RRC, RLCN and RRCN, then input screen updates to includes Carry and N as needed. Field Descriptions Binary: (Input Field) N: (Number of places for shifting or rotating appears only when SRN, RLN, RLN, RRN, RLCN, or RRCN are selected) (Binary Operation) Press ¸ or B to view options. Use D key to move the highlight bar to the desired operator Operator: Enter an integer in the number base designated in †/Binary Mode or an integer preceded by the number base in parenthesis (b, d, o, or h). Enter a real number. and press ¸ to select. Note: Register operations set the carry flag if a 1 is rotated or shifted off the end; otherwise they carry flag is cleared. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 51 SL SR RL Shifts the input to the left by one bit and shifts a 0 into the right-most position. Shifts the input to the right by one bit and shifts a 0 into the left-most position. Rotates the input to the left by one bit. The left-most bit wraps around to the right-most position. Rotates the input to the right by one bit. The right-most bit wraps around to the left-most position. Rotates bits left by one bit through carry. Loads the left-most bit into the carry and moves the carry bit into the right-most position. Rotates bits right by one bit through carry. Loads the right-most bit into the carry and moves the carry bit into the left-most position. Arithmetic shift right: For 1’s Complement and 2's Complement, it shifts bits to the right, but keeps the left-most bit (and the sign bit) the same. For Unsigned mode, there is no sign bit, a 0 is shifted into the left-most position. Shifts left by N bits. Shifts right by N bits. Rotate left by N bits. Rotate right by N bits. Rotate left by N bits through the carry. Rotate right by N bits through the carry. RR RLC RRC ASR SLN SRN RLN RRN RLCN RRCN Result: (Binary Function Value) Returns an integer result using the number base set in †/Binary Mode. Example 12.3 For a binary word 16 bits long, shift left by 6 bits, the octal number 2275. Input Screen 1. Output Screen Use the † key to access the pop up screen for Binary Mode and select octal for Base: and 16 bits for Word size; press ¸ to accept the settings. 2. 3. 4. Enter 2275 for the Binary number. Enter 6 for N, and choose SLN for the binary operation. Press „ to compute the result. The screen display above shows the input and the resulting output screens. 12.4 Bit Operations This section allows you to perform bit-specific operations. You can set a bit, clear a bit, test a bit or find the total number of bits in a set. Field Descriptions Binary: (Input Field) Bit #: (Bit Position - not active if ΣB is selected) EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 52 Enter an integer in the number base designated in †/Binary Mode or an integer preceded by the number base in parenthesis (b, d, o, or h). Enter a binary integer. Operator: Press ¸ or B to select. (Binary Operation) Note: All bit operations affect the carry flag only. SB CB B? ΣB Result: Sets the bit in the specified position. Clears the bit in the specified position. Tests the bit in the specified position. Returns the number of bits set. (Binary Function Value) Returns an integer result using the number base set in †/Binary Mode. Example 12.4 Find the bit sum of the a hexadecimal number AE34578F which is 32 bits wide. Input Screen Output Screen 1. Use † key to select hexadecimal for Base: for data type and 32 bits for Word size; press ¸ to accept 2. 3. the choices made. Enter AE34578F for the Binary number. Choose ΣB for the Operator. Press „ to compute the result. The screen displays above show the input and the resulting output. 12.5 Binary Conversions This topic demonstrates the process of converting real numbers to binary numbers and vice versa. The software allows for the conversion of real numbers to the IEEE format. In 1985, the Institute of Electrical and Electronic Engineers (IEEE), a professional association, developed standards for Binary Floating Point Arithmetic. This Standard (referred to Standard 754), specifies two basic forms of floating-point formats: single and double precision. The single precision format has 23 bits (23-bit significands), and 32 bits overall. In EE•Pro the computed output is in the single floating-point format. Binary entries in IEEE format are justified to the right with the last binary entry appearing in the 0 bit. 31 sign (1 bit [31]) 30 29 28 27 26 25 24 23 base 2 exponent (8 bit [23-30]) 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 mantissa (23 bit floating point [0-22]) Field Descriptions Binary: (Input Field for B→R16C and IEEE→) Enter an integer in the number base designated in †/Binary Mode or an integer preceded by the number base in parenthesis (b, d, o, or h). EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 53 (Input Field for R→B16C and →IEEE) (Binary Function) Real: Function: Real to Binary Binary to Real Real to IEEE IEEE to Real Result: Enter a real number. Press ¸ to display choices available. Converts a real number to a binary number (binary, octal, decimal or hexadecimal). If a number with a fractional part is entered, the value is rounded to the nearest integer. Affects the range flag only. Converts a binary number (decimal, binary, hexadecimal or octal) to a real number. Converts a real number to the IEEE 32 bit format. The word size is automatically set to 32 bits. Affects the range flag only. Converts a IEEE number to a real number. The word size is automatically set to 32 bits. Affects the carry and range flags. (Binary Function Value) Returns an integer result using the number base set in †/Binary Mode. Example 12.5 Convert a real decimal number 42355 to its IEEE standard. View the number in binary (base two) mode. Input Screen Output Screen Result in Pretty Print B. 1. Use † key to select Binary for Base: and 32 bits for Wordsize. Press ¸ to accept the choices made 2. 3. Enter 42355 for the Real Num. Choose Real to IEEE for the Function: press „ to compute the result. The screen display above shows the input screen and the resulting output. 12.6 Binary Comparisons This feature conducts a comparison of two binary numbers to determine if they are the same value, unequal, greater than, less than, etc. These functions are commonly needed for software developers. Field Descriptions Binary 1: Binary 2: Operator: (Input Field) (Input Field) (Binary Operation) Enter an integer in the number base designated in †/Binary Mode or an integer preceded by the number base in parenthesis (b, d, o, or h). Enter an integer in the number base designated in †/Binary Mode or an integer preceded by the number base in parenthesis (b, d, o, or h). Press ¸ to select. Note: All binary comparison commands affect the carry flag only. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 54 == Compares two binary or real numbers. The result field shows a 1 if they are equal, otherwise 0. Compares two binary or real numbers. The result field shows a 1 if they are NOT equal, otherwise 0. Compares two binary or real numbers. If the number in the first field is less than the number in the second field, then the result field displays a 1, otherwise 0. Compares two binary or real numbers. If the number in the first field is greater than the number in the second field, then the result field displays a 1, otherwise 0. Compares two binary or real numbers. If the number in the first field is less than or equal to the number in the second field, then the result field displays a 1, otherwise 0. Compares two binary or real numbers. If the number in the first field is greater than or equal to the number in the second field, then the result field displays a 1, otherwise 0. ≠ < > ≤ ≥ Result: (Binary Function Value) Returns 1 (True) or 0 (False). Example 12.6 Determine whether the octal value 11215 and the hexadecimal value 128D are unequal. Input Screen 1. 2. 3. Output Screen This test can be performed, regardless of the setting for Base: in the †/Binary mode input screen, by directly entering the number base prefix (b, o, d, or h) in parenthesis before the entered value. The software will immediately convert the number to the base setting (binary, in this case). Enter (o)11215 and (h)128D for Binary 1 and Binary 2, respectively. Choose the ≠ Operator, and press „ to compute the result. The screen display above shows the input and the resulting output. 12.7 Karnaugh Map The program provides a symbolic representation of a minimization method in a “sum of products form.” The variable name is restricted to one character per variable and is case-sensitive. The output is an algebraic expression for the prime implicants. In this representation, a logic variable (e.g., A) represents its true value, while A' is used to represent its logical negation.The algorithm uses a form of minimization developed by W.V. Quine and E.J. McCluskey. An exact minimum is usually not possible to obtain because of the amount of computation involved (the problem is not np complete; Ref. Logic Design principles by E. J. McCluskey, Prentice Hall, 1986, p. 246). Field Descriptions Minterms: (List if Minterms) A list of real positive integers representing the decimal number of the inputs for a true output. Thus if we have 4 inputs, the minterms would be a list such as {2,3,4,6,7,8,15} for which the output is a logical 1. Don’t Care: (List of Don’t Care Terms) A list of real positive integers representing the decimal number of the inputs for which the logical system does not care if the output is true or false. Thus if we have 4 inputs, the Don't Care terms would be a list such as {0,9,10} for which the output can be a 0 or 1. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 55 Vars: (List of Variables) A character string consisting of one letter variable names such as ABCD with no spaces between variable names. Prime Impl: (Prime Implicant Expression) Returns a logical algebraic expression in Sum of Products form. Example 12.7 Minimize a five input function with minterms at 0, 2, 4, 6, 8, 10, 11, 12, 13, 14, 16, 19, 29, 30 where minterms 4 and 6 are Don’t Cares. The input variables are V, W, X, Y and Z. Find the prime implicant expression. Input Screen 1. 2. 3. Output Screen Partial view of the Prime Implicant Expression Enter {0, 2, 8, 10, 11, 12, 13, 14, 16, 19, 29, 30} for Minterms and {4, 6} for Don’t Care. Enter variable names VWXYZ. Press „ to compute the result. The screen displays above show the input screen and the resulting output. A Pretty Print form of the resulting Sum of Products expression is shown. Variables that are associated with a prime (‘) indicates a logical inversion. EE Pro for TI-89, 92 Plus Analysis - Computer Engineering 56 Chapter 13 Error Functions This topic demonstrates the procedure for computing numeric solutions for the Error Function and the Complementary Error Function. ™Error Functions 13.1 Using Error Functions The definitions of the Error Function and Complementary Error Function are: erf ( x ) = 2 π x z e − t dt 2 Eq. 13.1.1 0 2 erfc( x ) = 1 − erf ( x ) = π ∞ z e − t dt 2 Eq. 13.1.2 x Field Descriptions X: Func: (Value) (Error Function type) Enter a real number, global name, or algebraic expression. Press ¸ to select erf or erfc. Result: (Error Function value) Returns a real number or algebraic expression Example 13.1 What is the value of erf(.25)? Input Screen 1. 2. 3. Output Screen Enter .25 for X. Choose ERF in Func. Press „ to calculate Result. EE Pro for TI-89, 92 Plus Analysis - Error Functions 57 Chapter 14 Capital Budgeting This chapter covers the four basic measures of capital budgeting: ™Payback Period ™Net Present Value ™Internal Rate of Return ™Profitability Index 14.1 Using Capital Budgeting This section performs analysis of capital expenditure for a project and compares projects against one another. Four measures of capital budgeting are included in this section: Payback period (Payback), Net Present Value (NPV), Internal Rate of Return (IRR), and Profitability Index (PI). This module provides the capability of entering, storing and editing capital expenditures for nine different projects. The following equations are used in calculations: n NPV = ∑ t =1 n CFt − CFt = 0 (1 + k )t CFt ∑ (1 + IRR) t =1 n PI = t Eq. 14.1.1 − CFt = 0 = 0 Eq. 14.1.2 CFt ∑ (1 + k ) t =1 t CFt = 0 Eq. 14.1.3 CFt: Cash Flow at time t (usually years). Payback: The number of time periods (usually years) it takes a firm to recover its original investment. NPV: The present values of all future cash flows, discounted at the selected rate, minus the cost of the investment. IRR: The discount rate that equates the present value of expected cash flows to the initial cost of the project. PI: The present value of the future cash flows, discounted at the selected rate, over the initial cash outlay. Field Descriptions - Input Screen Project: (Project) Press ¸ to select one of nine unique projects or edit the current name of the project by pressing † for Cash option. k: Payback: NPV: IRR: PI: (Discount Rate per Period in %) (Payback Period) (Net Present Value) (Internal Rate of Return) (Profitability Index) EE Pro for TI - 89, 92 Plus Analysis - Capital Budgeting Enter a real number. Returns a real number. Returns a real number. Returns a real number (%). Returns a real number. 58 Multiple Graphs Activation of this feature enables the overlay of each successive graph (projects) on the same axis. Press ¸ to activate. Full Screen Graph Press ¸ to activate. Field Descriptions - Project Edit Screen Name: t0: t1: .... tn: (Project Name) (Investment at t=0) (Cash flow at t=1) Enter the name of the project. Enter a real number. Enter a positive or negative real number. (Cash flow at t=n) Enter a positive or negative real number. Example 14.1 The following projects have been proposed by ACME Consolidated Inc. What is the Payback period, Net Present Value, Internal Rate of Return, and Profitability Index of each project? Which is the more viable project? Table 14-1 Cash Flow for two projects Name of Project: Investment Outlay: Cost of Capital: Year 0 1 2 3 4 EE Pro for TI - 89, 92 Plus Analysis - Capital Budgeting Plant 1 $75,000 (at t=0) 12% Plant 2 $75,000 (at t=0) 12% Net Cash Flow ($) -75,000 40,000 30,000 20,000 10,000 Net Cash Flow ($) -75,000 10,000 20,000 30,000 40,000 59 Cash Flow Input: plant1 1. Cash Flow Input: plant2 Output Screen With the highlight bar on the Project field, press ¸ to select a project to edit. Select a project that has not been used (this example uses projects 1 and 2). Press ¸ to return to the Capital Budgeting screen. 2. Press † to select Cash option enter the project edit screen and edit the cash flows. 3. Enter “plant1” in the Name field (Note: Cash flow data for this project will be stored in a variable of this name, therefore the entered name must begin with a letter, be no more than 8 characters in length, and contain no embedded spaces). Press ‰ 5 times to add 5 time points and enter the cash flows at each time point from the table on the 4. previous page. When finished, your screen should look like the project edit screen above. Be sure to enter 75,000 as a negative number for t0. Press N to save your changes and return to the Capital Budgeting 5. 6. screen. Enter 12 for k. Press „ to calculate Payback, NPV, IRR, and PI. 7. Move the highlight bar to Multiple Graphs and press ¸ to enable overlaying of successive graphs of 8. each project. Press … to graph the curvilinear relationship between the Net Present Value and the Discount Rate. 9. Press 2 followed by O to enable the graph editing toolbar. The curve indicates where k=0, the Net Present Value is simply cash inflows minus cash outflows. The IRR % is shown at the point where NPV=0. Using the built-in graphing capabilities of the TI 89, you can trace the graph to find the values of these two points. The TI 89 will give you the exact coordinates of any point along the graph. Press Nto return to the Capital Budgeting screen. Repeating steps 1 through 9 for the second project, under the Project field, “plant2” and input the values in Table 14-1. Activating the Multiple Graph feature enables a simultaneous plot of the two projects. This will overlay a second graph on top of a previously plotted function. First plot plant1. After graphing, plot plant2. The first curve to appear is plant1, the second is plant2. The most viable project in terms of discounted cash flows, in this example, is the one with the highest curve. Pressing 3 „ B ¨ ¸ resets the display to full screen. Plot of Project 1 EE Pro for TI - 89, 92 Plus Analysis - Capital Budgeting Overlay of Project 2 60 Chapter 15 Introduction to Equations The Equations section of EE•Pro contains over 700 equations organized into 16 topic and 105 sub-topic menus. • • • • • • The user can select several equation sets from a particular sub-topic, display all the variables used in the set of equations, enter the values for the known variables and solve for the unknown variables. The equations in each sub-topic can be solved individually, collectively or as a sub-set. A unit management feature allows easy entry and display of results. Variables in selected equation sets can be graphed to examine the relationship between each other. Multiple and partial solutions are possible using techniques developed for EE•Pro. More information on a particular input can be displayed by highlighting the variable, press ‡ and ©/Type: to show a brief description of a variable and its entry parameters. 15.1 Solving a Set of Equations • • Equations are accessed from the main level of the EE•Pro by pressing function key … labeled "Equations." This displays a pull-down menu listing all the topics as shown in the screen display below. An arrow to the left of the bottom topic ‘ï‘indicates more items are listed. Pressing 2 D jumps to the • bottom of the menu. Scroll the highlight bar to an item using the arrow key D and press ¸, or type the subject number appearing next to subject heading (Resistive Circuits is selected for this example). • • • A second menu will appear listing more subjects (sub-topics) under the topic heading. Selecting a sub-topic displays a list of equations under the subject heading (Ohm’s Law and Power is selected below). Use the arrow key D to move the highlighter and press ¸ to select an equation or series of equations which are applicable to a specific problem (pressing „ selects all of the equations). • Press „ to display all of the variables in the selected equations. As the cursor bar is moved, a brief • description of each variable will appear in the status line at the bottom of the screen. Enter values for the known parameters, selecting appropriate units for each value using the toolbar menu which appear at the top of the screen. Press „ to compute values for the unknown parameters. • Entered and calculated values are distinguished in the display; ‘é‘ for entered values and ‘‹‘ for computed • results. EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 1 1. Pressing … displays the 2. Press ¨ to display the 3. Press © to display the ‘Equations’ menu. menu in ‘Resistive Circuits’ equations for ‘Ohm’s Law’. 4. Select equations by high- 5. Press „ to display the 6. Press ˆ to compute the lighting and pressing ¸. variables in the selected unknown variables. Note: equations. Enter the known variable values. Use the unit toolbar to select units. Computed results ‘‹‘ are distinguished from entered values ‘é‘. Note: Only values designated as known ‘é‘ will be used in a computation. A result displayed from an earlier calculation will not be used unless the user specifically designates the value by selecting the variable and pressing ¸. Press „ to compute a new result for any input that is changed. 15.2 Viewing an Equation or Result in Pretty Print Sometimes equations and calculated results exceed the display room of the calculator. The TI-89 and TI 92 plus include a built-in equation display feature called Pretty Print which is available in many areas of EE•Pro and can be activated by highlighting a variable or equation and pressing the right arrow key B or pressing the † function key when it is designated as View. The object can be scrolled using the arrow keys AB. Pressing N or ˆ reverts to the previous screen. 1. To view an equation in, Pretty Print, highlight 2. Scroll features, using the arrow keys AB, enable a and press † or B. complete view of a large object. 15.3 Viewing a Result in different units To view a calculated result in units which are different from what is displayed. Highlight the variable, press ‡/Options and y/Conv to display the unit tool bar at the top of the screen. Press the function key to convert the result to the desired units. EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 2 1. Highlight the result to be 2. The unit menu tool bar 3. Press † to convert the. converted (R). Press ‡ to is now displayed at the top of result of ‘R’ from Ω to MΩ. display the Options menu, press y/Conv. the screen. 15.4 Viewing Multiple Solutions The math engine used by EE•Pro is able to manage complex values for variables (where they are permitted) and calculate more than one solution in cases where multiple answers exist for an entered problem. When a multiple solution exists, the user is prompted to select the number of a series of computed answers to be displayed. To view additional solutions, press „ to repeat the calculation and enter another solution number. The user will need to determine which result is most useful to the application. 1. Select an equation by highlighting and pressing ¸. Press „ to display variables. 2. Enter known values for each variable using the tool bar to designate units. Press „ to compute the results. Solution 1: To view another solution, press „ to re calculation and enter the number. of another solution to be viewed. 3. If multiple solution exists, a dialogue box will appear requesting the user to enter the number of a solution to view. Solution 2: Enter a new number for each solve to display a series of. multiple solutions. 15.5 Partial Solutions "One partial useable solution found." or "Multiple partial solutions found." will be displayed in the status line if values for one or more variables in the selected equation set cannot be computed. This situation can occur if there are more unknowns than equations in the selected set, the entered values do not form consistent relationships with EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 3 the selected equations, or if the selected equations do not establish a closed form relationship between all of the entered values and the unknowns. In such a case, only the calculated variables will be displayed. Press „ to select all of the If there are more unknowns ...a partial solution will be equations in Resistive Formulas. than selected equations or relationships between variables are not established from the selected equations... displayed if one or more of the unknown variables are able to be computed from the entered inputs. 15.6 Copy/Paste A computed result and it’s expressed units can be copied and pasted to the HOME screen or any other location of EE•Pro using ƒ:Tools-5:Copy key sequence to copy a value and ƒ:Tools-6:Paste to paste the item in any appropriate context of the TI system. 15.7 Graphing a Function The relationship between two variables in an equation can be graphed on a real number scale if the other variables in the equation are defined. • After solving an equation, or entering values for the non x,y variables in the equation to be plotted, press … • /Graph to display the graph settings. Highlight Eq: and press ¸ to select the equation from the list to graph. • • • • • • Use the same steps as above to select the independent and dependent variables (Indep: and Depnd:) from the equation. Note: all pre-existing values stored in the variables used for Indep: and Depnd: will be cleared when the graphing function is executed. The graphing unit scale for each variable reflect the settings in the Equations section of EE•Pro. Scrolling down the list, specify the graphing ranges for the x and y variables, whether to graph in full or split screen modes, automatically scale the graph to fit the viewing area, and label the graph. Press … to graph the function. Once the graph command has been executed, EE•Pro will open a second window to display the plot. All of the TI graphing features are available and are displayed in the toolbar, including Zoom „, Trace …, Math ‡, etc. The Math feature is extremely useful for determining critical function values such as intercepts, inflections, derivatives, integrals, etc. Peak performance, damped resonance and decay functions are able to be evaluated using this tool. If the split-screen graphing mode is activated, the user can toggle between the EE•Pro graph dialogue display and the TI graph by pressing 2 O. If the full-screen graphing mode is activated, the user can switch between EE•Pro and the graph by pressing O 4:Graph or A:EE•Pro. *Before graphing an equation, be sure to specify values for variables in an equation which are not going to be used as x and y variables. EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 4 1*. Graph an equation by pressing …. Press ¸ 2. Select variables for Independent (x) and Variable units reflect settings in EE•Pro. to choose an equation. Dependent (y) variables. 4. Select graphing options by pressing ¸ 5. Split Screen Mode: Toggle between graph and settings 6: Full Screen Mode: Press O, j and Ñ to return by pressing 2 and O. to EE•Pro. Note: If an error is generated when attempting to graph, be sure that all of the variables in the graphed equation which are not specified as the independent and dependent variables have entered values. In the EE•Pro window, press N to view the equations in the sub-topic, select the equation to be graphed by highlighting and pressing ¸, press „ to display the list of variables in the equation and enter values. Only the dependent (y) and independent (x) variables do not have to contain specified values. Press …to display the graph dialogue and repeat the above steps to graph the function. 15.8 Storing and recalling variable values in EE•Pro-creation of session folders EE•Pro automatically stores its variables in the current folder specified by the user in 3 or the HOME screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values for a particular session of EE•Pro, press ƒ:/TOOLS, ª:/NEW and type the name of the new folder (see Chapter 5 of the TI-89 Guidebook for the complete details of creating and managing folders). There are several ways to display or recall a value: • The contents of variables in any folder can be displayed using the °, moving the cursor to the variable name and pressing ˆ to display the contents of a particular variable. • • Variables in a current folder can be recalled in the HOME screen by typing the variable name. Finally, values and units can be copied and recalled using the ƒ/Tools 5:COPY and 6:PASTE feature. All inputs and calculated results from Analysis and Equations section are saved as variable names. Previously calculated, or entered values for variables in a folder are replaced when equations are solved using new values for inputs. Overwriting of variable values in graphing When an equation or analysis function is graphed, EE•Pro creates a function for the TI grapher which expresses the dependent variable in terms of the independent variable. This function is stored under the variable name pro(x). When the EE•Pro’s equation grapher is executed, values are inserted into the independent variable for pro(x) and values for the dependent value are calculated. Whatever values which previously existed in either of the dependent and independent variables in the current folder are cleared. To preserve data under variable names which may conflict with EE•Pro’s variables, run EE•Pro in a separate folder. EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 5 15.9 solve, nsolve, and csolve and user-defined functions (UDF) When a set of equations is solved in EE•Pro, three different functions in the TI operating system (solve, numeric solve, and complex solve) are used to find the most appropriate solution. In a majority of cases, the entered values are adequate to find numeric solutions using either the solve or csolve functions. However, there are a few instances when functions external to the equation set (user-defined functions) are incorporated into the solving process and nsolve must be used. User defined functions which appear in some of the equation sets of EE •Pro are erfc(x) erf(x), eeGALV(RR2, ….) and ni(TT). In most cases, when all the inputs to a UDF are known, solve or csolve can just pass a computed result to the equation. On the other hand if one is solving for a variable that is an input to the UDF, solve or csolve are unable to isolate the variable in an explicit form, and the operating system resorts to using nsolve. nsolve initiates a series of trial and error iterations for the unknown variable until the solution converges. It should be noted that the solution generated by nsolve is not guaranteed to be unique (i.e. this solving process cannot determine if multiple solutions exist.). Table 15-1 User Defined Functions User-defined Function erf(ts, τp) erfc (x,D,t) eegalv (Rx, RR2, RR3, RR4, Rg, Rs, Vs) ni(TT) Topic Solid State Solid State Meters and Bridges Solid State Sub-topic PN Junction Current Semiconductor Basics Wheatstone Bridge Semiconductor Basics, PN Junctions, PN Junction Current, MOS Transistor I 15.10 Entering a guessed value for the unknown using nsolve To accelerate the nsolve converging process and, if multiple solutions exist, enhance the possibility that nsolve resolves the correct solution, the user can enter a guessed value for the unknown which nsolve will use as an initial value in the first iteration of its solving process. ™Enter guessed a value for the variable in the input dialogue. ™Press ‡/Opts, m/Want. ™Press „/ to compute a solution for the variable. erfc(x,D,t) is a user defined function that appears in the Semiconductor Basics section of Solid State. Only one input to a user defined function can be specified as an unknown. EE•Pro displays a notice if the nsolve routine is used. The user can enter a value for for the unknown and designate it as a guessed value to accelerate the nsolve convergence process. EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 6 15.11 Why can't I compute a solution? If a solution is unable to be computed for an entered problem, you might check the following: 1. 2. 3. Are there at least as many equations selected as there are unknown parameters? Are the entered values or units for the known parameters reasonable for a specific case? Are the selected equations consistent in describing a particular case (for example, the choice of certain equations used in the calculation of diode properties depends on whether the donor density of the doping substance Nd, exceeds the acceptor density, Na in the Semiconductors section of Solid State) 15.12 Care in choosing a consistent set of equations The success of the equation solver in generating a useful solution, or a solution at all, is strongly dependent on the user's insight into the problem and care in choosing equations which describe consistent relationships between the parameters. The following steps are suggested: • Read the description of each set of equations in a topic to determine which subset of equations in a series are compatible and consistent in describing a particular case. • Select the equations from a subset which describe the relationships between all of the known and unknown parameters. • As a rule of thumb, select as many equations from the subset as there are unknowns to avoid redundancy or over-specification. The equations have been researched from a variety of sources and use slightly different approximation techniques. Over-specification (selecting too many equations) may lead to an inability of the equation solver to resolve slight numerical differences in different empirical methods of calculating values for the same variable. 15.13 Notes for the advanced user in troubleshooting calculations When there are no solutions possible, EE•Pro provides important clues via key variables eeinput, eeprob, eeans, and eeanstyp. These variables are defined during the equation setup process by the built-in multiple equation solver. EE•Pro saves a copy of the problem, its inputs, its outputs, and a characterization of the type of solution in the user variables eeprob, eeinput, eeans, and eeanstyp. For the developer who is curious to know exactly how the problem was entered into the multiple equation solver, or about what the multiple equation solver returned, and to examine relevant strings. The contents of these variables may be viewed by using VAR-LINK and examining these variables in the current session. Press ° (2 followed by |), scroll to the variable name in the current folder and press ˆ to view the contents of the variable. The string may be recalled to the author line of the home screen, modified and re-executed, if desired. Table 15.2 Topics and Sub-topics List 1: Resistive Circuits 1: Resistance Formulas 2: Ohm's Law and Power 3: Temperature Effect 4: Max. Power Transfer 5: V, I Source 4: Electron Motion 1: Beam Deflection 2: Thermionic Emission 3: Photoemission 7: RLC Circuits 1: Series Impedance EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 2: Capacitors, E-Fields 1: Point Charge 2: Long Charged Line 3: Charged Disk 4: Parallel Plates 5: Parallel Wires 6: Coaxial Cable 7: Sphere 5: Meters and Bridge Circuits 1: A, V, Ω Meters 2: Wheatstone Bridge 3: Wien Bridge 4: Maxwell Bridge 5: Attenuators - Symmetric R 6: Attenuators - Unsym R 8: AC Circuits 1: RL Series Impedance 7 3: Inductors and Magnetism 1: Long Line 2: Long Strip 3: Parallel Wires 4: Loop 5: Coaxial Cable 6: Skin Effect 6: RL and RC Circuits 1: RL Natural Response 2: RC Natural Response 3: RL Step response 4: RC Step Response 5: RL Series to Parallel 6: RC Series to Parallel 9: Polyphase Circuits 1: Balanced ∆ Network 2: Parallel Admittance 3: RLC Natural Response 4: Under-damped case 5: Critical Damping 6: Over-damped Case 10: Electrical Resonance 1: Parallel Resonance I 2: Parallel Resonance II 3: Lossy Inductor 4: Series Resonance 13: Linear Amplifiers 1: BJT (CB) 2: BJT (CE) 3: BJT (CC) 4: FET (Common Gate) 5: FET (Common Source) 6: FET (Common Drain) 7: Darlington (CC-CC) 8: Darlington (CC-CE) 9: EC Amplifier A: Differential Amplifier B: Source Coupled JFET 16: Motors, Generators 1: Energy Conversion 2: DC Generator 3: Sep. Excited DC Gen. 4: DC Shunt Generator 5: DC Series Generator 6: Sep Excite DC Motor 7: DC Shunt Motor 8: DC Series Motor 9: Perm Magnet Motor A: Induction Motor I B: Induction Motor II C: 1 φ Induction Motor D: Synchronous Machines EE Pro for TI - 89, 92 Plus Equations - Introduction to Equations 2: RC Series Impedance 3: Impedance - Admittance 4: 2 Z's in Series 5: 2 Z's in Parallel 2: Balance Wye Network 3: Power Measurements 11: Op. Amp Circuits 1: Basic Inverter 2: Non-Inverting Amplifier 3: Current Amplifier 4: Transconductance Amplifier 5: Lvl. Detector Invert 6: Lvl. Detector Non-Invert 7: Differentiator 8: Diff. Amplifier 12: Solid State Devices 1: Semiconductor Basics 2: PN Junctions 3: PN Junction Currents 4: Transistor Currents 5: Ebers-Moll Equations 6: Ideal Currents - pnp 7: Switching Transients 8: MOS Transistor I 9: MOS Transistor II A: MOS Inverter R Load B: MOS Inverter Sat Load C: MOS Inverter Depl. Ld D: CMOS Transistor Pair E: Junction FET 15: Transformers 1. Ideal Transformer 2: Linear Equiv. Circuit 14: Class A, B, C Amps 1: Class A Amplifier 2: Power Transistor 3: Push-Pull Principle 4: Class B Amplifier 5: Class C Amplifier 8 Chapter 16 Resistive Circuits This software section performs routine calculations of resistive circuits. The software is organized in a number of topics listed below. ™Resistance and Conductance ™Ohm’s Law and Power ™Temperature Effects ™Maximum Power Theorem ™V and I Source Equivalence Variables A complete list of all the variables used, a brief description and applicable base unit is given below. Variable A G I Il Is len P Pmax R Rl Rlm RR1 RR2 Rs T1 T2 V Vl Vs α ρ σ EE Pro for TI - 89, 92 Plus Equations - Resistive Circuits Description Area Conductance Current Load current Current source Length Power Maximum power in load Resistance Load resistance Match load resistance Resistance, T1 Resistance, T2 Source resistance Temperature 1 Temperature 2 Voltage Load voltage Source voltage Temperature coefficient Resistivity Conductivity Unit m2 S A A A m W W Ω Ω Ω Ω Ω Ω K K V V V 1/°K Ω*m S/m 9 16.1 Resistance Formulas Four equations in this topic represent the basic relationship between resistance and conductance. The first equation links the resistance R of a bar with a length len and a uniform crosssectional area A with a resistivity ρ. The second equation defines the conductance G of the same bar in terms of conductivity σ, len and A. The third and fourth equations show the reciprocity of conductance G resistance R, resistivity ρ and conductivity σ. R= ρ ⋅ len A Eq. 16.1.1 σ⋅A len 1 G= R 1 σ= ρ G= Eq. 16.1.2 Eq. 16.1.3 Eq. 16.1.4 Example 16.1 - A copper wire 1500_m long has a resistivity of 6.5_ohm*cm and a cross sectional area of .45_cm2. Compute the its resistance and conductance. Solution - Upon examining the problem, two choices are noted. Equations 16.1.1, 16.1.2 and 16.1.4 or 16.1.1 and 16.1.3 can be used to solve the problem. The second choice was made here. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. The computed results are shown in the screen display shown here. Entered Values Computed results -PQYP8CTKCDNGUNGPAOρAQJO EO#AEO %QORWVGF4GUWNVU4'AQJO)'AUKGOGPU 16.2 Ohm’s Law and Power The fundamental relationships between voltage, current and power are presented in this section. The first equation is the classic Ohm's Law, computes the voltage V in terms of the current I, and the resistance R. The next four equations describe the relationship between power dissipation P, voltage V, current I, resistance R and conductance G in a variety of alternate forms. The final equation represents the reciprocity between resistance R and conductance G. EE Pro for TI - 89, 92 Plus Equations - Resistive Circuits 10 V = I ⋅R Eq. 16.2.1 P =V ⋅I Eq. 16.2.2 P = I2 ⋅R Eq. 16.2.3 V2 R Eq. 16.2.4 P = V 2 ⋅G Eq. 16.2.5 1 G Eq. 16.2.6 P= R= Example 16.2 - A 4.7_kohm load carries a current of 275_ma. Calculate the voltage across the load, power dissipated and load conductance. Entered Values Computed results Solution - Upon examining the problem, several choices are noted. Either Equations 16.2.1, 16.2.2 and 16.2.6, or 16.2.2, 16.2.3 and 16.2.5 or 16.2.2, 16.2.3 and 16.2.6 or 16.2.1, 16.2.2 and 16.2.5 or all the equations. Choose the last option, press „ to open the input screen, enter all the known variables and press „ to solve. -PQYP8CTKCDNGU+AOC4AMΩ  %QORWVGF4GUWNVU8A82A9)AUKGOGPU 16.3 Temperature Effect This equation models the effect of temperature on resistance. Electrical resistance changes from RR1 to RR2 when the temperature change from T1 to T2 is modulated by the temperature coefficient of resistance α. b gh RR2 = RR1⋅ 1 + α ⋅ T 2 − T1 c Eq. 16.3.1 Example 16.3 - A 145_Ω resistor at 75_°F reads 152.4_Ω at 125_ºC. Find the temperature coefficient of resistance. Solution - Since there is only one equation in this topic, there is no need to make a choice of equation. Press „ to display the input screen. Enter the variable values and press „ to solve for the unknown variable. EE Pro for TI - 89, 92 Plus Equations - Resistive Circuits 11 User entered Variables Computed results -PQYP8CTKCDNGU44AΩ44AΩ6Aº(6Aº%  %QORWVGF4GUWNVUαAu- 16.4 Maximum DC Power Transfer The equations under this topic are organized to compute load voltage Vl, load current Il, power dissipation in the load P, maximum power available in the load Pmax, and load impedance Rlm needed for maximum power deliverable to the load. The first equation finds the load voltage Vl of circuit with a voltage source Vs, source resistance Rs, and a load resistance Rl. The next equation defines the load current Il in terms of Vs, Rs and Rl. The power dissipation DC Voltage source with load in the load is defined by the equation relating P with Il and Vl. The next equation links Pmax to Vs and Rs. The last equation represents load resistance needed for a maximum power. Vs ⋅ Rl Rs + Rl Vs Il = Rs + Rl P = Il ⋅Vl Vs 2 P max = 4 ⋅ Rs Rlm = Rs Vl = Eq. 16.4.1 Eq. 16.4.2 Eq. 16.4.3 Eq. 16.4.4 Eq. 16.4.5 Example 16.4 - A 12_V car battery has a resistive load of .52_ohm. The battery has a source impedance of 0.078_ohm. Find the maximum power deliverable from this battery, and the power delivered to this resistive load. Entered Values Computed results Solution - The second, third and fourth equations are needed to compute the solution for this problem. Select these by highlighting and pressing the ¸ key. Press „ to display the input screen, enter the known variables and press „ to solve the unknowns. EE Pro for TI - 89, 92 Plus Equations - Resistive Circuits 12 -PQYP8CTKCDNGU8UA84UAQJO4NAQJO  %QORWVGF4GUWNVU2OCZA92A9 16.5 V and I Source Equivalence The two equations in this topic show the equivalence between a voltage source and a current source. A voltage source Vs with an internal series resistance of Rs is equivalent in all its functionality to a current source Is with a source resistance Rs connected across it. Vs Rs Vs = Is ⋅ Rs Is = Eq. 16.5.1 Voltage and current sources and its equivalence Eq. 16.5.2 Example 16.5 - Find the short circuit current equivalent for a 5_V source with a 12.5_ohm source resistance. Entered Values Computed results Solution - Either form of the equation can be used to solve the equation. Press „ to display the user interface, enter the values of all known inputs, and press „ to solve for Is. -PQYP8CTKCDNGU8UA84UAΩ %QORWVGF4GUWNVU+UA# EE Pro for TI - 89, 92 Plus Equations - Resistive Circuits 13 Chapter 17 Capacitors, Electric Fields This section covers seven topics to compute electric field properties and capacitance of various types of structures. When the section is accessed, the software displays the topics in a pop up menu shown above. ™Point Charge ™Long Charged Line ™Charged Disk ™Parallel Plates ™Parallel Wires ™Coaxial Cable ™Sphere Variables A complete list of all the variables used in this section is given below. Variable A C cl d E Er Ez F Q r ra rb V Vz W z εr ρl ρs Description Area Capacitance Capacitance per unit length Separation Electric field Radial electric field Electric field along z axis Force on plate Charge Radial distance Inner radius, wire radius Outer radius Potential Potential along z axis Energy stored z axis distance from disk Relative permittivity Line charge Charge density EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields Unit m2 F F/m m V/m V/m V/m N C m m m V V J m unitless C/m C/m2 14 17.1 Point Charge The two equations in this topic calculate the radial electric field Er and the potential V at a point located a distance r away from a point change Q. The first equation shows the inverse square relationship between Er and r, while the second equation shows the inverse relationship between the potential V and distance r. The equations have been generalized to include εr, the relative permittivity of the medium. Q 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r 2 Q V= 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r Er = Eq. 17.1.1 Eq. 17.1.2 Example 17.1 - A point charge of 14.5E-14_coulomb is located 2.4_m away from an instrument measuring electric field and absolute potential. The permittivity of air is 1.08. Compute the electric field and potential. Entered Values Computed results Solution - Both equations are needed to solve this problem. Press „ to display the input screen, enter all the known variables, and press „ to solve for the unknown values. Note that ε0, the permittivity of free space does not appear as one of the variables that needs to be entered. It is entered automatically by the software. However, εr, the relative permittivity must be entered as a known value. -PQYP8CTKCDNGU3'AEQWNQODTAOεr = 1.08 %QORWVGF4GUWNVU'TA8O8A8 17.2 Long Charged Line An infinite line with a linear charge density, ρl, (coulombs per unit length) exerts a radial electric field, Er, a distance r away from the line. The equation has been generalized to include εr, the relative permittivity of the medium. Er = ρl 2 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r Eq. 17.2.1 Example 17.2 - An aluminum wire suspended in air carries a charge density of 2.75E-15_coulombs/m. Find the electric field 50_cm away. Assume the relative permittivity of air to be 1.04. EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 15 Entered Values Computed results Solution - Press „ to display the input screen, enter all the known variables, and press „ to solve the selected equation set. The screen display above shows the computed results. -PQYP8CTKCDNGU ρN'AEQWNQODUOTAEOεT %QORWVGF4GUWNVU 'TA8O 17.3 Charged Disk These two equations describe the electric field and potential along the vertical axis through the center of a uniformly charged disk. The first equation defines the electric field along the z-axis of the disk with a radius ra and charge density of ρs, a distance z from the plane of the disk. The second equation computes the electrostatic potential Vz at an arbitrary point along the z-axis. Ez = z ρs ⋅ 1− 2 2 ⋅ ε 0 ⋅ εr ra + z 2 FG H IJ K Eq. 17.3.1 Vz = ρs ⋅ 2 ⋅ ε 0 ⋅ εr j Eq. 17.3.2 e ra 2 + z 2 − z Example 17.3 - A charged disc 5.5_cm in radius produces an electric field of .2_V/cm 50_cm away from the surface of the disc. Assuming that relative permittivity of air is 1.04, what is the charge density on the surface of the disc? Entered Values Computed Results Solution - Select the first equation by pressing ¸ key, press „ to display the input screen for this equation, enter all the known variables, and press „. The computed results are shown in the screen display above. -PQYP8CTKCDNGUTCAEOεT'\A8EO\AEO %QORWVGF4GUWNVUρU'AEQWNQODO@ EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 16 17.4 Parallel Plates The five equations listed in this topic describe the electrical and mechanical forces in a parallel plate capacitor. Two plates are separated by a distance d which is small compared to the lateral dimensions so fringing field effects can be ignored. The first equation computes the electric field E at the plate for a potential difference V between the plates separated by a small distance d. The second equation calculates capacitance C with a dielectric given the relative permittivity εr and area A. The third equation shows the charge Q on each parallel plate. The last two equations compute the mechanical properties associated with this parallel plate capacitor such as the Force F on the plates and energy W stored in the capacitor. V d ε 0 ⋅ εr ⋅ A C= d Q = C ⋅V E= Eq. 17.4.1 Eq. 17.4.2 Eq. 17.4.3 1 V ⋅C F=− ⋅ 2d 12 W = ⋅V ⋅ C 2 2 Eq. 17.4.4 Eq. 17.4.5 Example 17.4 - A silicon dioxide insulator forms the insulator for the gate of a MOS transistor. Calculate the charge, electric field and mechanical force on the plates of a 5_V MOS capacitor with an area of 1250_µ2 and a thickness of .15_µ. Use a value of 3.9 for permittivity of SiO2. Entered Values Computed results Solution - All of the equations are needed to compute the solution to this problem. Press „ to display the input screen, enter all the known variables and press „ to solve. The computed results are shown above. -PQYP8CTKCDNGU 8A8εTF.15_µ. #_µ2 %QORWVGF4GUWNVU %'A(''A8O (A09'A, 17.5 Parallel Wires The equation listed under this topic represents the calculation of capacitance per unit length, cl, of a pair of transmission lines of radius ra and center to center spacing d in a dielectric medium with a relative permittivity of εr. EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 17 cl = π ⋅ ε 0 ⋅ εr d cosh −1 2 ⋅ ra FG H Eq. 17.5.1 IJ K Example 17.5 - Compute the capacitance per unit length of a set of power lines 1_cm radius, and 1.5_m apart. The dielectric medium separating the wires is air with a relative permittivity of 1.04. Entered Values Calculated Output Solution - Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. -PQYP8CTKCDNGUεTTCAEOFAO %QORWVGF4GUWNVUEN'A(O 17.6 Coaxial Cable These three equations describe capacitive and electric field properties of a coaxial cable. The first two equations compute the voltage between the two conductors of the cable carrying a charge of ρl per unit length, and an insulator with a relative permittivity εr. The inner conductor has a radius ra while the outer conductor has a radius rb. The last equation computes the cable capacitance cl per unit length based on mechanical properties of the cable and insulator. V= ρl rb ⋅ ln 2 ⋅ π ⋅ ε 0 ⋅ εr ra Er = FG IJ HK Eq. 17.6.1 V Eq. 17.6.2 F rb I r ⋅ lnG J H ra K 2 ⋅ π ⋅ ε 0 ⋅ εr cl = F rb I lnG J H ra K Eq. 17.6.3 Example 17.6 - A coaxial cable with an inner cable radius of .3_cm, and an outer conductor with an inside radius of .5_cm has a mica filled insulator with a permittivity of 2.1. If the inner conductor carries a linear charge of 3.67E-15_coulombs/m, find the electric field at the outer edge of the inner conductor and potential between the two conductors. Compute the capacitance per m of the cable. EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 18 Solution - After examining the problem, all the three equations need to be selected to solve the problem. Press „ to display the input variable screen. Enter all the known variables, and press „ to solve the selected equation set. The computed results are shown in the screen display below. Input Values for the example Computed results -PQYP8CTKCDNGUTCAEOTDAEOεTTAEOρN'AEQWNQODUO %QORWVGF4GUWNVUEN'A(O'TA8OCPF8A8 17.7 Sphere The first equation in this topic computes the potential between two concentric spheres of radius ra and rb, with a charge Q, and separated by a medium with a relative permittivity of εr. The second equation computes the electric field outside a sphere at a distance r from the center of the sphere. The last equation computes the capacitance between the spheres. FG H 11 Q ⋅ − 4 ⋅ π ⋅ ε 0 ⋅ εr ra rb Q Er = 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r 2 V= C= IJ K Eq. 17.7.1 Eq. 17.7.2 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ ra ⋅ rb rb − ra Eq. 17.7.3 Example 17.7 - Two concentric spheres 2_cm and 2.5_cm radius, are separated with a dielectric with a relative permittivity of 1.25. The inner sphere has a charge of 1.45E-14_coulombs. Find the potential difference between the two spherical plates of the capacitor as well as the capacitance. Entered Values  Computed results Solution - Upon examining the problem, equations 17.7.1 and 17.7.3 are needed to compute a solution. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 19 screen. Enter all the known variables, and press „ to solve the selected equation set. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGUTCAEOTDAEOεTCPF3'AEQWNQODU %QORWVGF4GUWNVU8A8%'A( EE Pro for TI - 89, 92 Plus Equations - Capacitors & Electric Fields 20 Chapter 18 Inductors and Magnetism Topics in this section focus on electrical and magnetic properties of physical elements. ™Long Line ™Long Strip ™Parallel Wires ™Loop ™Coaxial Cable ™Skin Effect Variables A complete list of all the variables used in the various topics of this section are listed below along with the default units used for those variables. Variable θ µr a B bl Bx By D d f Fw I I1 I2 Is L L12 Ls r ra rb Reff rr0 T12 x y z Description Angle Relative permeability Loop radius or side of a rectangular loop Magnetic field Width of rectangular loop Magnetic field, x axis Magnetic field, y axis Center-center wire spacing Strip width Frequency Force between wires/unit length Current Current in line 1 Current in line 2 Current in strip Inductance per unit length Mutual inductance Loop self-inductance Radial distance Radius of inner conductor Radius of outer conductor Effective resistance Wire radius Torque x axis distance y axis distance Distance to loop z axis EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism 21 Unit radian unitless m T m T T m m Hz N/m A A A A/m H/m H H m m m Ω m N*m m m m δ ρ Skin depth Resistivity m Ω*m 18.1 Long Line The magnetic field B from a current I in an infinite wire in an infinitely long line is computed at a distance r from the line. B= µ0 ⋅ I 2⋅π⋅r Eq. 18.1.1 Example 18.1 - An overhead transmission line carries a current of 1200_A, 10_m away from the surface of the earth. Find the magnetic field at the surface of the earth. Input Screen Calculated Results Solution - Since there is only equation, press „ to display the input screen. Enter all the known variables, and press „ to solve the equation. The computed results are shown in the screen display above. -PQYP8CTKCDNGU+A#TAO %QORWVGF4GUWNVU$A6 18.2 Long Strip A thin conducting ribbon strip of width d is infinitely long and carrying a current Is amperes per meter. The x and y component of the magnetic field Bx and By are dependent upon the location described by (x, y) coordinates. F − µ 0 ⋅ Is G ⋅ tan Bx = 2 ⋅π G GH F x + d I F x − d II GG 2 JJ − tan GG 2 JJ JJ GH y JK GH y JK JK F y + FG x + d IJ I H 2 K JJ µ 0 ⋅ Is G ⋅ lnG By = 4 ⋅π GG y + FG x − d IJ JJ H H 2K K −1 −1 Eq. 18.2.1 2 2 2 2 EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism 22 Eq. 18.2.2 Example 18.2 - A strip transmission line 2_cm wide carries a current of 16025_A/m. Find the magnetic field values 1_m away and 2_m from the surface of the strip. Input Screen Calculated Results Solution - Both equations need to be used to compute the solution. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen display above. -PQYP8CTKCDNGU+UA#OFAEOZAO[AO %QORWVGF4GUWNVU$ZA6$[A6 18.3 Parallel Wires Two thin parallel wires of infinite length carrying currents I1 and I2 separated by a distance D exert a force F newtons/meter between them. The second equation computes the magnetic field Bx between the two parallel wires at a distance x from the line carrying current I1. The final equation in this set computes the inductance L from the two wires of diameter a, with a spacing of D. Fw = Bx = L= µ 0 ⋅ I1⋅ I 2 2 ⋅π ⋅ D µ0 I1 I2 ⋅ − 2 ⋅π x D − x FG H IJ K µ0 µ0 D + ⋅ cosh −1 4 ⋅π π 2⋅a FG IJ HK Eq. 18.3.1 Eq. 18.3.2 Eq. 18.3.3 Example 18.3 - A pair of aluminum wires 1.5_cm in diameter are separated by 1_m and carry currents of 1200_A and 1600_A in opposite directions. Find the force of attraction, the magnetic field generated midway between the wires and the inductance per unit length resulting from their proximity. Input Screen EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism Calculated Results 23 Solution - Upon examining the problem, all the three equations are needed. Press „ to display the input screen. Enter all the known variables, and press „ to solve the equation set. The computed results are shown in the screen display above. -PQYP8CTKCDNGU+A#+A#ZAEO&AOCAEO %QORWVGF4GUWNVU(A0O.A*O$ZA6 18.4 Loop The first two equations consider the magnetic properties of wire with a radius rr0, bent into a circular loop of radius a, carrying a current, I. The equation for the magnetic field B is computed at any point along the z-axis through the center of the loop at a distance z from the plane of the loop. The second equation computes the self-inductance, Ls, of the loop. Circular Loop The final two equations calculate the torque, T12 and the inductance, L12, of a rectangular loop carrying a current, I2, in the proximity of an infinitely long wire carrying a current, I1. The rectangular loop has a width, bl, parallel to its axis of rotation, and a length, a, perpendicular to the axis of rotation. The loop axis of rotation and the infinitely long wire intersect at a 90 degree angle. The loop’s angle of tilt, θ, is relative to the plane containing the loop plane and the infinite wire. In the side of the loop closest to the infinite wire, the current flows in the opposite direction Rectangular Loop when θ = 0. The distance d between wire and the closest edge of the of the loop is measured along the loop axis. B= µ0 ⋅ I ⋅ a 2 2⋅ e a +z 2 2 j Eq. 18.4.1 3 FG FG 8 ⋅ a IJ − 2IJ H H rr 0 K K µ 0 ⋅ a ⋅ cosbθ g F bl + d I ⋅ lnG L12 = H d JK 2 ⋅π Ls = µ 0 ⋅ a ⋅ ln T12 = bg µ 0 ⋅ a ⋅ sin θ bl + d ⋅ I 1⋅ I 2 ⋅ ln 2 ⋅π d FG H Eq. 18.4.2 Eq. 18.4.3 IJ K Eq. 18.4.4 Example 18.4 - Calculate the torque and inductance for a rectangular loop of width 7 m and length 5 m, carrying a current of 50 A, separated by a distance of 2 m from a wire of infinite length carrying a current of 30 A. The loop angle of incidence is 5 degrees relative to the parallel plane intersecting the infinite wire. EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism 24 Input Screen Calculated Results SolutionUpon examining the problem, the last two equations are needed. Select these using the ¸| key and press „to display the input screen. Enter the known variables and press „ to solve the selected equation set. The screen display of the input and calculated results are shown below. -PQYP8CTKCDNGU CAODNAOFAO+A#+A#θAFGI %QORWVGF4GUWNVU .AJGPT[6A0 O 18.5 Coaxial Cable A coaxial transmission line with an outer radius of inn the inner conductor of ra and inner radius of the outer conductor of rb is characterized by the inductance L per unit length. L= µ0 µ0 rb + ⋅ ln 8⋅π 2 ⋅π ra FG IJ HK Eq. 18.5.1 Example 18.5 - A coaxial cable has an inner conductor radius of 2_mm and the outer conductor radius of .15_in. Find its inductance per meter. Input Screen Calculated Results Solution - Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGUTCAOOTDAKP  %QORWVGF4GUWNVU.A*O 18.6 Skin Effect These two equations represent the effect of high frequency on the properties of a conductor. The first equation relates the skin depth, δ, with the frequency f and the resistivity ρ, while the second equation EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism 25 computes the effect of higher frequencies on resistance, Reff, in ohms. Since the skin effect is a direct consequence of internal magnetic fields in the conductor, the relative permeability, µr, influences these properties. δ= 1 π ⋅ f ⋅ µ 0 ⋅ µr ρ Eq. 18.6.1 Re ff = π ⋅ f ⋅ µ 0 ⋅ µr ⋅ ρ Eq. 18.6.2 Example 18.6 - Find the effect on depth of signal penetration for a 100 MHz signal in copper with a resistivity of 6.5E-6 _ohm*cm. The relative permeability of copper is 1.02. Input Screen Calculated Results Solution - Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGUρ'AQJO EOHA/*\µT %QORWVGF4GUWNVU4GHHAQJOUUSWCTGδAµ EE PRO for TI - 89, 92 Plus Equations - Inductance & Magnetism 26 Chapter 19 Electron Motion This section covers equations describing the trajectories of electrons under the influence of electric and magnetic fields. These equations are divided into three topics. ™Electron Beam Deflection ™Thermionic Emission ™Photoemission Variables The table below lists all the variables used in this chapter. Variable A0 B d f f0 I L Ls r S T v Va Vd y yd z φ Description Richardson’s constant Magnetic field Deflection tube diameter, plate spacing Frequency Critical frequency Thermionic current Deflecting plate length Beam length to destination Radius of circular path Surface area Temperature Vertical velocity Accelerating voltage Deflecting voltage Vertical deflection Beam deflection on screen Distance along beam axis Work function Unit A/(m2*K2) T m Hz Hz A m m m m2 K m/s V V m m m V 19.1 Electron Beam Deflection An electron beam that is subjected to an accelerating voltage Va achieves a velocity v as defined by the first equation. The second equation calculates the radius of curvature r as these electrons in the beam move with a velocity v passing through a magnetic field B. The third equation calculates the y-deflection yd at distance Ls from the center of deflection plates separated by a distance d and length L making the approximation that the Ls>>L. The final equation calculates the vertical displacement y inside the deflection region with distance z from entry into the plate region and subject to a deflecting voltage Vd. EE Pro for TI-89, 92 Plus Equations - Electron Motion 27 v = 2⋅ r= Eq. 19.1.1 me ⋅ v q⋅B Eq. 19.1.2 L ⋅ Ls ⋅Vd 2 ⋅ d ⋅Va Eq. 19.1.3 q ⋅Vd ⋅ z2 2 2 ⋅ me ⋅ d ⋅ v Eq. 19.1.4 yd = y= q ⋅Va me Example 19.1- An electron beam in a CRT is subjected an accelerating voltage of 1250_V. The screen target is 40_cm away from the center of the deflection section. The plate separation is 0.75_cm and the horizontal path length through the deflection region is .35 cm. The deflection region is controlled by a 100_V voltage. A magnetic field of 0.456_T puts the electrons in the beam in a circular orbit. What is the vertical deflection distance of the beam when it reaches the CRT screen? Display of Input Values Calculated Results Solution - The first three equations are needed to solve this problem. Select these by highlighting and pressing ¸. Press „ to display the input screen, enter values of all known variables and press „ to solve the equation. The computed results are shown in the screen display above. -PQYP8CTKCDNGU8CA8$A68FA8.UAEO.AEO FAEO  %QORWVGF4GUWNVUX'AOU[FAOTAO EE Pro for TI-89, 92 Plus Equations - Electron Motion 28 19.2 Thermionic Emission When certain electronic materials are heated to a high temperature T, the free electrons gain enough thermal energy, forcing a finite fraction to escape the work function barrier φ and contribute to the external current I. The current also depends directly on the surface area S and the so-called Richardson’s constant A0. I = A0 ⋅ S ⋅ T 2 ⋅ e − q ⋅φ k ⋅T Eq. 19.2.1 Example 19.2 - A cathode consists of a cesium coated tungsten with a surface area of 2.45_cm2. It is heated to 1200_°K in a power vacuum tube. If the Richardson's constant is 120_A/( m2*K2), and the work function is 1.22_V, find the current available from such the cathode. Display of Input Values Calculated Results Solution - Since there is only equation, press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGU#A# O@ -@ 5AEO@6A°-φA8 %QORWVGF4GUWNVU+A# 19.3 Photoemission The two equations in this topic represent the behavior of electrons when excited by photon energy. A light beam with a frequency f generates an RMS velocity v for electrons that have to overcome a work function φ. The second equation shows the threshold frequency for the light beam to extract electrons from the surface of a solid. 1 h ⋅ f = q ⋅ φ + ⋅ me ⋅ v 2 2 q ⋅φ f0= h Eq. 19.3.1 Eq. 19.3.2 Example 19.3 - A red light beam with a frequency of 1.4E14_Hz, is influencing an electron beam to overcome a barrier of 0.5_V. What is the electron velocity, and find the threshold frequency of the light. EE•Pro display of Multiple Solution Sets Display of Input Values EE Pro for TI-89, 92 Plus Equations - Electron Motion 29 Display of Solution 1 Display of Solution 2 Solution - Both equations are needed to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. EE•Pro displays a notice that multiple solution sets exist for the entered data, in which case the user needs to select a solution which is meaningful to the application. When viewing one of a multiple solution set, the number of a solution should be entered (1, 2, 3...etc.) followed by pressing ¸ twice. The process of solving and choosing a solution set should be repeated each time the user wishes to view a different solution. In this case, a positive electron velocity has physical significance. -PQYP8CTKCDNGUφA8H'A*\ %QORWVGF4GUWNVUH'A*\XAOU EE Pro for TI-89, 92 Plus Equations - Electron Motion 30 Chapter 20 Meters and Bridge Circuits This section covers a variety of topics on meters, commonly used bridge and attenuator circuits. These equations are organized under seven titles. ™Amp, Volt, and Ohmmeter ™Owen Bridge ™Wheatstone Bridge ™Symmetrical Resistive Attenuator ™Wien Bridge ™Unsymmetrical Resistive Attenuator ™Maxwell Bridge Variables The following is a list of all the variables used in this section. Variable a b c CC3 CC4 Cs Cx DB f Ig Imax Isen Lx Q Radj Rg Rj Rk Rl Rm Rr RR1 RR2 RR3 RR4 Rs Rse Description Resistance multiplier Resistance Multiplier Resistance Multiplier Capacitance, arm 3 Capacitance, arm 4 Series capacitor Unknown capacitor Attenuator loss Frequency Galvanometer current Maximum current Current sensitivity Unknown inductance Quality Factor Adjustable resistor Galvanometer resistance Resistance in L pad Resistance in L pad Resistance from left Meter resistance Resistance from right Resistance, arm 1 Resistance, arm 2 Resistance, arm 3 Resistance, arm 4 Series resistance Series resistance EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits Unit unitless unitless unitless F F F F unitless Hz A A A unitless unitless Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω 31 Rsh Rx Vm Vmax Vs Vsen ω Shunt resistance Unknown resistance Voltage across meter Maximum voltage Source voltage Voltage sensitivity Radian frequency A Ω V V V V r/s 20.1 Amp, Volt, and Ohmmeter The three equations in this section describe the use of resistors in extending the range of ammeters, voltmeters and ohmmeters. A shunt resistor Rsh increases the range of an ammeter with a current sensitivity Isen and a maximum range Imax. A series resistance Rse can extend the range of a voltmeter. The third equation extends the range of a series ohmmeter with an internal resistance Rm and internal voltage Vs, with an adjustable resistor Radj. In a practical setup, Radj is usually set at its midpoint to compensate for variations in the component values, resulting in a systematic error in the measured result. Rsh = Rs = Rm ⋅ Isen Im ax − Isen V max − Vsen Isen = Isen Vs Eq. 20.1.1 Eq. 20.1.2 Eq. 20.1.3 Radj Rs + Rm + 2 Example - 20.1 What resistance can be added to a voltmeter with a current sensitivity of 10 µA, and a voltage sensitivity of 5 V to read 120 V? Input Variables Computed Results Solution - The second equation needs to be selected to solve this problem. Enter the known values for Isen, Vmax, and Vsen and press „ to solve the equation. -PQYP8CTKCDNGU+UGPAµ#8OCZA88UGPA8 %QORWVGF4GUWNVU4U/Ω EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 32 20.2 Wheatstone Bridge A Wheatstone bridge with four resistor elements Rx, RR2, RR3 and RR4 is the foundation of modern measuring systems. When the bridge is balanced, there is no current in the galvanometer circuit. The first equation defines the requirement for a balanced bridge. The voltage across the bridge Vm and the galvanometer current Ig are calculated in as follows. A special function GALV calculates the voltage across the bridge, and is a complex function of Vs, Rx, RR2, RR3, RR4, Rg and Rs. Eq. 20.2.1 Rx R 3 = R2 R4 b Vm = eegalv Rx , RR 2, RR 3, RR 4, Rg , Rs,Vs Ig = g Vm Rg Eq. 20.2.2 Eq. 20.2.3 Example 20.2- A Wheatstone bridge circuit has a resistor RR2 of 100Ω on the unknown side of the bridge and two 1000 Ω resistors connected on the known side of the bridge. A resistor of 99 Ω was connected to the bridge in the location where the unknown resistor would normally be present. The bridge is supplied by a 10 V source with a resistance of 2.5 Ω. The galvonemetric resistance is 1 MΩ. Find the voltage across the meter and the galvanometric current. Entered Values (Lower portion of display) Calculated Results (Upper portion of display) Solution - The second and third equations are needed to solve the problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen display above. -PQYP8CTKCDNGU 44AΩ, RR3= 1000._Ω, RR4=1000._Ω, Rs=2.5_Ω, Rg=1._MΩ 8UA84ZA Ω %QORWVGF4GUWNVU 8OA8+IA# EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 33 20.3 Wien Bridge A Wien bridge circuit is designed to measure an unknown capacitance Cx using a bridge arrangement shown here. The circuit has two methods of varying parameters of the bridge to achieve null. In the first approach, Cx can be computed in terms of RR3, RR1, Rs and Rx while in the second equation, the source frequency can be used as a key controlling parameter. The bridge can also be used to measure the frequency f with the third equation after setting Cx=Cs, Rx=Rs, and RR3=2*RR1. The final equation relates f to radian frequency ω. FG H IJ K Cx RR 3 Rs = − Eq. 20.3.1 Cs RR1 Rx Cs ⋅ Cx = f= 1 ω ⋅ Rs ⋅ Rx Eq. 20.3.2 2 1 Eq. 20.3.3 2 ⋅ Cs ⋅ Rs π ω = 2 ⋅π ⋅ f Eq. 20.3.4 Example -20.3 A set of measurements obtained using a Wien bridge is based on the following input. All measurements are carried out at 1000_Hz. The known resistors RR1 and RR3 are 100 Ω each, the series resistance is 200 Ω, and Cs is 1.2_µF. Find the values of the unknown RC circuit components and the radian frequency. Entered Values Calculated Results Solution - Use the first, second and fourth equations to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve for the unknowns. The computed results are shown in the screen display above. -PQYP8CTKCDNGU%UAµ(HA*\44AΩ44AΩ4UA Ω  %QORWVGF4GUWNVU%ZAµ(4ZAΩωATU EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 34 20.4 Maxwell Bridge A Maxwell bridge is designed to measure the inductance Lx and its series resistance Rx in a bridge circuit. The input stimulus to the bridge circuit is an AC source with a variable frequency and an AC meter detecting a null. The first two equations measure the unknown inductance Lx and its resistance Rx by varying the capacitance Cs, and its parallel resistance Rs and bridge arm resistances RR1 and RR2. The third and fourth equations measure the quality factor Q by using either measured values of Lx and Rx or Cs and Rs. The final equation links the radian frequency ω to the frequency f. Lx = RR2 ⋅ RR3 ⋅ Cs RR 2 ⋅ RR 3 Rs Lx Q=ω⋅ Rx Eq. 20.4.1 Eq. 20.4.2 Rx = FG IJ HK Eq. 20.4.3 Q = ω ⋅ Cs ⋅ Rs Eq. 20.4.4 ω = 2 ⋅π ⋅ f Eq. 20.4.5 Example 20.4- Find the inductance and resistance of an inductive element using the Maxwell bridge. The bridge are resistors are 1000_Ω each with a .22 µF capacitor and 470 Ω parallel resistance. Compute Lx and Rx. Entered Values Calculated Results Solution - The first two equations are needed to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to compute the solution. The calculated results are shown in the screen display above. -PQYP8CTKCDNGU%UAµ44AΩ44AΩ %QORWVGF4GUWNVU.ZAJGPT[4ZAΩ 20.5 Owen Bridge The Owen bridge circuit is an alternative AC bridge circuit used to measure an inductance and its series resistance. The input stimulus to the bridge circuit is typically an AC source with variable frequency and an AC meter detecting a null. The inductance Lx is measured in terms of the capacitor CC3, and the branch resistance RR1 and RR4. The series resistance Rx is measured by a ratio of the capacitors CC3 and CC4 and tempered by RR2, the variable resistance in the Lx arm of the bridge. EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 35 Lx = CC3⋅ RR1⋅ RR4 Rx = Eq. 20.5.1 CC3⋅ RR1 − RR2 CC4 Eq. 20.5.2 Example 20.5- A lossy inductor is plugged into an Owen bridge to measure its properties. The resistance branch has 1000_Ω resistors and a capacitor of 2.25µF on the non-resistor leg and 1.25_µF capacitor on the resistor leg of the bridge. A series resistance of 125_Ω connects the CC4 leg to balance the inductive element. Entered Values Calculated Results Solution - Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGU %%Aµ(%%Aµ(44AΩ44AΩ %QORWVGF4GUWNVU .ZAJGPT[4ZAΩ 20.6 Symmetrical Resistive Attenuator Tee Pad Pi Pad Bridged Pad Balanced Pad Balanced resistive networks are commonly used as attenuators in transmission line circuits. The three equations below form the basis of the design for resistive attenuators in a Tee pad, a Pi pad, a bridged Tee pad or a balanced pad configuration. These design equations compute the value of the multiplier in the circuit for a given an attenuation loss dB, in decibels. The values of a, b or c define the attenuator network. The figure above show the four key various configurations for the attenuator pads. For example, if the terminating resistance is 100 ohms, and a 20 db attenuation is desired, computing a and b will be adequate to design a Tee pad or a Pi pad. The resistance of each leg is computed by the terminating resistance by the multiplier factor. EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 36 FG10 H a= 10 b= DB 20 DB 20 2 ⋅ 10 10 FG H DB 10 IJ K −1 Eq. 20.6.1 +1 DB 20 Eq. 20.6.2 −1 DB IJ K c = 10 20 − 1 Eq. 20.6.3 Example 20.6- Design a symmetrical and Bridges Tee attenuator for a 50 Ω load and a 6 DB loss. Entered Values Calculated Results Solution - All three equations are needed. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown above. -PQYP8CTKCDNGU&$ %QORWVGF4GUWNVUCDE Having computed the values of the multipliers a and b, a Tee pad can be constructed using the values Ro*a (or 16.6395_ohms) for the two sides of the Tee, and a calue Ro*b (or 66.931_ohms) for the vertical leg of the network. 20.7 Unsymmetrical Resistive Attenuator This section contains equations for unsymmetrical resistive attenuator design. These equations compute the resistance values for a minimum loss L pad of an unsymmetrical network with an impedance Rl to the left of the L-network and an impedance Rr to the right of the network. The first two equations calculate Rj and Rk, the resistor values of the L network. The third equation determines the minimum loss of signal strength DB. Rj = Rl − Rk = EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits Rk ⋅ Rr Rk + Rr Rl ⋅ Rr 2 Rl − Rr 37 Eq. 20.7.1 Eq. 20.7.2 DB = 20 ⋅ log10 ⋅ F GH Rl − Rr Rl + Rr Rr I JK Eq. 20.7.3 Example 20.7- A network needs to be patched by an unsymmetrical attenuator. The network to the right of the attenuator presents a resistive load of 125 Ω, while the network to the left of the attenuator possesses an impedance of 100 Ω. What is the expected loss in dB? Entered Values Calculated Results Solution - The last equation is needed to compute the signal attenuation. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen display above. -PQYP8CTKCDNGU4NAΩ4TAΩ %QORWVGF4GUWNV&$ Rj and Rk can be computed from the first two equations above. EE Pro for TI-89, 92 Plus Equations - Meters and Bridge Circuits 38 Chapter 21 RL and RC Circuits This chapter covers the natural and transient properties of simple RL and RC circuits. The section is organized under six topics. ™RL Natural Response ™RC Natural Response ™RL Step Response ™RC Step Response ™RL Series to Parallel ™RC Series to Parallel Variables The table below lists all the variables used in this chapter, along with a brief description and appropriate units. Variable C Cs Cp f iC iL I0 L Lp Ls Qp Qs R Rp Rs τ t vC vL V0 Vs ω W EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits Description Capacitor Series capacitance Parallel capacitance Frequency Capacitor current Inductor current Initial inductor current Inductance Parallel inductance Series inductance Q, parallel circuit Q, series circuit Resistance Parallel resistance Series resistance Time constant Time Capacitor voltage Inductor voltage Initial capacitor voltage Voltage stimulus Radian frequency Energy dissipated Unit F F F Hz A A A H H H unitless unitless Ω Ω Ω s s V V V V r/s J 39 21.1 RL Natural Response These four equations define all the key properties for the natural response of an RL circuit with no energy sources. The first equation shows the characteristic time constant τ in terms of the resistance R and the inductance L. The second equation computes the decay of the voltage vL across the inductor with an initial current I0. The third equation displays the decay of the inductor current iL. The final equation expresses the energy dissipation characteristic W of the circuit for the specified conditions. τ= L R Eq. 21.1.1 vL = I 0 ⋅ R ⋅ e iL = I 0 ⋅ e W= − − t τ Eq. 21.1.2 t τ Eq. 21.1.3 F GH 2⋅t − 1 ⋅ L ⋅ I 02 ⋅ 1 − e τ 2 I JK Eq. 21.1.4 Example 21.1 An RL circuit consists of a 400_mH inductor and a 125_Ω resistor. With an initial current of 100_mA, find the inductor current and voltage across the inductor 1_ms and 10_ms after the switch has been closed. Entered Values Calculated Results (1 ms) Solution - Upon examining the problem, the first three equations are needed to solve the problem. Select these equations using the highlight bar and pressing ¸, press „ to display the input screen, enter all the known variables and press „ to solve. Perform the computations for a time of 1_ms, write down all the results, enter 10 ms for t and press „ to recalculate the results for the new time entry. -PQYP8CTKCDNGU +AO#.AO*4AΩVAOUAOU  %QORWVGF4GUWNVU K.A#X.A8CHVGTAOU K.AO#X.A8CHVGTAOU 21.2 RC Natural Response These four equations define the natural response of an RC circuit with no energy sources. The first equation specifies the characteristic time constant τ in terms of the resistance R and the EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits 40 capacitance C. The second equation computes the decay of the voltage vC across the capacitor with an initial voltage of V0. The third equation shows the decay of the capacitor current iC. The final equation computes the energy dissipation W. τ = R ⋅C Eq. 21.2.1 t − τ vC = V 0 ⋅ e V 0 − τt ⋅e iC = R 2⋅t − 1 W = ⋅ C ⋅V 0 2 ⋅ 1 − e τ 2 FG H Eq. 21.2.2 Eq. 21.2.3 IJ K Eq. 21.2.4 Example 21.2- An RC circuit consists of a 1.2_µfarad capacitor and a 47_Ω resistor. The capacitor has been charged to 18_V. A switch disconnects the energy source. Find the voltage across the capacitor 100_µs later. How much energy is left in the capacitor? Entered Values Calculated Results Solution - Upon examining the problem, all of the equations are needed to solve the problem. Press „ to display the input screen. Enter all the known variables and press „ to solve the set of equations. -PQYP8CTKCDNGU%Aµ(4AΩ8A8VAµU %QORWVGF4GUWNVUX%A8K%A#9A,τAU 21.3 RL Step Response These equations describe the response of an inductive circuit to a voltage step stimulus. The first equation calculates the time constant τ in terms of the inductance L and the resistance R. The last two equations compute the inductor voltage vL and current iL in terms of the step stimulus Vs, initial condition I0, time t, and time constant τ. τ= L R Eq. 21.3.1 b g vL = ⋅ Vs − I 0 ⋅ R ⋅ e iL = FG H IJ K − t τ Vs Vs + I0− ⋅e R R Eq. 21.3.2 t − τ Eq. 21.3.3 Example 21.3 - An inductor circuit consisting of 25_mH inductance and 22.5_Ω resistance. Prior to applying a 100_V stimulus, the inductor carries a current of 100_mA. Find the current in and the voltage across the inductor after .01_s. EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits 41 Entered Values Calculated Results Solution - Upon examining the problem, all three equations are need to be solve the problem. Press „ to display the input screen, enter all the known variables and press „ to compute the solution. -PQYP8CTKCDNGU.AO*4AΩ+AO#VAU8UA8 %QORWVGF4GUWNVUK.A#X.A8 21.4 RC Step Response These three equations describe the step response properties of an RC circuit. The first equation defines the characteristic time constant, τ, in terms of the resistance R and the capacitance C. The last two equations compute the capacitor voltage vC and current iC in terms of the step stimulus voltage Vs, the initial capacitor voltage V0, time t, and time constant τ. τ = R ⋅C Eq. 21.4.1 b g vC = Vs + V 0 − Vs ⋅ e iC = bVs − V 0g ⋅ e − t τ Eq. 21.4.2 t − τ Eq. 21.4.3 R Example 21.4- A 10_V step function is applied to an RC circuit with a 7.5_Ω resistor and a 67_nfarad capacitor. The capacitor was charged to an initial potential of –10_V. What is the voltage across the.1_µs after the step function has been applied? Entered Values Calculated Results Solution - All three equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve. -PQYP8CTKCDNGU8A8%APHCTCF4AΩV.1_µs, Vs = 10_V %QORWVGF4GUWNVUX%A8K%A# EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits 42 21.5 RL Series to Parallel The equations in this topic show the equivalence relationship between a series RL circuit (Rs and Ls) and its parallel equivalent circuit, Rp, Lp. The first equation translates the frequency, f, to its radian frequency, ω. The second equation specifies the quality factor Qs for Rs and Ls. The third and fourth equations define the values of Rp and Lp in terms of Rs, Ls and ω. The fifth equation computes Qp in terms of Rp, Lp and ω. The sixth and seventh equations calculate Rs and Ls in terms of Rp, Lp and ω. The eighth and ninth equations compute Rp and Lp in terms of Rs, Ls and Qs. The last two of equations calculate Rs and Ls in terms of Rp, Lp and Qp. ω = 2 ⋅π ⋅ f Eq. 21.5.1 ω ⋅ Ls Rs Rs 2 + ω 2 ⋅ Ls 2 Rp = Rs 2 Rs + ω 2 ⋅ Ls 2 Lp = ω 2 ⋅ Ls Rp Qp = ω ⋅ Lp ω 2 ⋅ Lp 2 ⋅ Rp Rs = Rp 2 + ω 2 ⋅ Lp 2 Qs = Eq. 21.5.2 Eq. 21.5.3 Eq. 21.5.4 Eq. 21.5.5 Eq. 21.5.6 Rp 2 ⋅ Lp Ls = Rp 2 + ω 2 ⋅ Lp 2 Eq. 21.5.7 Rp = Rs ⋅ 1 + Qs 2 Eq. 21.5.8 ch F 1 IJ Lp = Ls ⋅ G 1 + H Qs K Eq. 21.5.9 2 Rp 1 + Qp 2 Qp 2 ⋅ Lp Ls = 1 + Qp 2 Rs = Eq. 21.5.10 Eq. 21.5.11 Example 21.5 - A 24_mH inductor has a quality factor of 5 at 10000_Hz. Find its series resistance and the parallel equivalent circuit parameters. EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits 43 Entered Values Calculated Results Solution - Upon examining the problem, the first six equations need to be solved as a set. Select these equations, and press „ to display the input screen enter all the known variables and press „ to solve. The computed results are shown in the screen display shown here. -PQYP8CTKCDNGU.UAO*3UHA*\ %QORWVGF4GUWNVU.RA*4UAΩ4RAΩωTU 21.6 RC Series to Parallel The equations in this topic show the equivalence between a series RC circuit (Rs and Cs) and its parallel equivalent circuit with values Rp, and Cp. The first equation converts frequency, f, to its radian equivalent radian frequency, ω. The second equation computes the quality factor Qs in terms of ω, Rs and Cs. The next two equations compute the parallel equivalent values as a function of Rs, Cs and ω. The fifth equation defines Qp in terms of Rp, Cp and ω. The sixth and seventh equations compute Rs and Cs in terms of Rp, Cp and ω. The last four equations describe the relationships between Rs, Cs, Rp, Cp, Qs and Qp in a symmetrical and complementary form. ω = 2 ⋅π ⋅ f 1 Qs = ω ⋅ Rs ⋅ Cs FG H Eq. 21.6.1 Eq. 21.6.2 1 ω ⋅ Rs2 ⋅ Cs2 Cs Cp = 2 1 + ω ⋅ Cs2 ⋅ Rs2 Rp = Rs ⋅ 1 + 2 IJ K Eq. 21.6.3 Eq. 21.6.4 Qp = ω ⋅ Rp ⋅ Cp Eq. 21.6.5 Rs = Rp 1 + ω ⋅ Rp 2 ⋅ Cp 2 Eq. 21.6.6 Cs = 1 + ω 2 ⋅ Rp 2 ⋅ Cp 2 ω 2 ⋅ Rp 2 ⋅ Cp Eq. 21.6.7 2 Rp = Rs ⋅ 1 + Qs 2 c EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits h Eq. 21.6.8 44 Cs 1 1+ 2 Qs Rp Rs = 1 + Qp 2 Cp = Cs = Eq. 21.6.9 Eq. 21.6.10 c Cp ⋅ 1 + Qp 2 Qp h Eq. 21.6.11 2 Example 21.6 - A parallel RC Circuit consists of a 47_µfarad and 150000_Ω at 120000_Hz. Find its series equivalent. Entered Values Calculated Results Solution - Upon examining the problem, use equations 21.6.1, 21.6.3, 21.6.4, 21.6.6, 21.6.7 are needed to solve the problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve. -PQYP8CTKCDNGU%R_P(, 4R = 150000_Ω, HA*\ %QORWVGF4GUWNVU4U'_Ω%U'A( EE PRO for TI -89, 92 Plus Equations - RL and RC Circuits 45 Chapter 22 RLC Circuits The essential equations for computing impedance and admittance, natural response, and transient behavior of RLC circuits are organized into six topics. ™Series Impedance ™Parallel Admittance ™RLC Natural Response ™Underdamped Transient ™Critically-Damped Transient ™Overdamped Transient Variables All the variables used here are listed along with a brief description and units. Variable α A1 A2 B B1 B2 BC BL C D1 D2 f G I0 L θ R s1 s2 s1i s1r s2i s2i t v V0 ω ωd EE Pro for TI-89, 92 Plus Equations – RLC Circuits Description Neper’s frequency Constant Constant Susceptance Constant Constant Capacitive susceptance Inductive susceptance Capacitance Constant Constant Frequency Conductance Initial inductor current Inductance Phase angle Resistance Characteristic frequency Characteristic frequency Characteristic frequency Characteristic frequency Characteristic frequency Characteristic frequency Time Capacitor voltage Initial capacitor voltage Radian Frequency Damped radian frequency 46 (imaginary) (real) (imaginary) (real) Unit rad/s V V S V V S S F V/s V Hz S A H rad Ω rad/s rad/s rad/s rad/s rad/s rad/s s V V rad/s rad/s ω0 X XXC XL Ym Zm Classical radian frequency Reactance Capacitive reactance Inductive reactance Admittance – magnitude Impedance – magnitude rad/s Ω Ω Ω S S 22.1 Series Impedance The series impedance of an RLC circuit is calculated using the first two equations. The magnitude |Zm| and phase angle θ of the impedance is calculated from the resistance the R and reactance X in the first two equations. The reactance X in the third equation is calculated in terms of the inductive reactance XL. In the fifth equation, the capacitive reactance XXC is calculated from the frequency ω and capacitance C. In the final equation, ω is expressed in terms of the oscillation frequency, f. c Zm h 2 = R2 + X 2 θ = tan −1 Eq. 22.1.1 FG X IJ H RK Eq. 22.1.2 X = XL + XXC Eq. 22.1.3 XL = ω ⋅ L −1 XXC = ω ⋅C ω = 2 ⋅π ⋅ f Eq. 22.1.4 Eq. 22.1.5 Eq. 22.1.6 Example 22.1 - A circuit consists of a 50 ohm resistor in series with a 20_mhenry inductor and 47_µfarad capacitor. At a frequency of 1000_Hz, calculate the impedance, phase angle of impedance, . Entered Values Calculated Results Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen ,enter all of the known variables and press „ to solve for the unknowns. -PQYP8CTKCDNGU.AOJGPT[4AQJO%AµHCTCFHA*\ %QORWVGF4GUWNVUωATCFU:.AΩ::%AΩ :AΩ<OAΩθATCF EE Pro for TI-89, 92 Plus Equations – RLC Circuits 47 22.2 Parallel Admittance The admittance of a parallel RLC circuit consists of a magnitude, Ym, and phase angle θ. Both can be calculated in terms of the conductance G and susceptance B. The conductance G is expressed in terms of resistance R, while the susceptance B is expressed in terms of the inductive and capacitive components BL and BC. c Ym h 2 = G2 + B2 θ = tan −1 Eq. 22.2.1 FG G IJ H BK Eq. 22.2.2 1 R B = BL + BC −1 BL = ω⋅L G= Eq. 22.2.3 Eq. 22.2.4 Eq. 22.2.5 BC = ω ⋅ C Eq. 22.2.6 ω = 2 ⋅π ⋅ f Eq. 22.2.7 Example 22.2 - A parallel RLC Circuit consists of a 10,000 ohm resistor, 67 µhenry and .01 µfarads. Find the circuit admittance parameters at a frequency of 10 MHz. Input and calculated results (upper half) Input and calculated results (lower half) Solution - All of the equations need to be used to solve this problem. Press „ to display the input screen and enter the values of all known variables. Press „ to compute the unknown parameters. -PQYP8CTKCDNGUHA/*\4AQJO.AµJGPT[%AµHCTCF %QORWVGF4GUWNVU;OAUKGOGPU)AUKGOGPU$.AUKGOGPU $%AUKGOGPUθATCFωATCFU 22.3 RLC Natural Response These equations compute the complex frequencies of an RLC circuit. In general, every RLC circuit has two complex frequencies s1 and s2 with real and imaginary parts s1r, s1i, s2r, and s2i. These frequencies are complex conjugates of each other which are computed from the resonant frequency ω0, and Neper’s frequency α, defined by the final two equations. EE Pro for TI-89, 92 Plus Equations – RLC Circuits 48 s1r = real −α + α 2 − ω 02 e s1i = image −α + s2r = reale −α − s2i = image −α − j α −ω0 j α −ω0 j α −ω0 j Eq. 22.3.1 2 2 Eq. 22.3.2 2 2 Eq. 22.3.3 2 2 Eq. 22.3.4 1 L⋅C 1 α= 2⋅ R⋅C ω0 = Eq. 22.3.5 Eq. 22.3.6 Example 22.3 – A series RLC circuit of Example 22.1 is used to compute the circuit parameters. Input and calculated results (upper half) Input and calculated results (lower half) Solution - All of the equations are needed to solve the parameters from these given set of variables. Press „ to access the input screen and enter all known variables, press „ to solve for the unknowns. -PQYP8CTKCDNGU.AOJGPT[4AQJO%AµHCTCF %QORWVGF4GUWNVUUTATCFUUKATCFU ωATCFUUTATCFUUKATCFU 22.4 Underdamped Case The equations in this section represent the transient response of an underdamped RLC circuit. The classical radian frequency ω0 is calculated from the inductance, L and the capacitance, C in Equation 22.4.1. The Neper’s frequency α is shown by the second equation. The condition for an underdamped system is that ω0 >α. The damped resonant frequency ωd is expressed in equation 22.4.3 in terms of ω0 and α. The voltage across the capacitor v, is defined in terms of two constants B1 and B2. B1 is equivalent to the initial capacitor voltage V0, and B2 is related to the initial inductor current I0, C, ωd and α resistance R. The voltage v has an oscillation frequency of ωd. 1 L⋅C 1 α= 2⋅ R⋅C ω0 = EE Pro for TI-89, 92 Plus Equations – RLC Circuits Eq. 22.4.1 Eq. 22.4.2 49 ωd = ω 02 − α 2 Eq. 22.4.3 b g b v = B1⋅ e −α ⋅t ⋅ cos ωd ⋅ t + B2 ⋅ e −α ⋅t ⋅ sin ωd ⋅ t B1 = V 0 B2 = − g Eq. 22.4.4 Eq. 22.4.5 α ⋅ V 0 − 2 ⋅ I 0⋅ R ωd b g Eq. 22.4.6 Example 22.4 - A parallel RLC circuit is designed with a 1000 Ω resistor, a 40 mH inductor and a 2 µF capacitor. The initial current in the inductor is 10 mA and the initial charge in the capacitor is 2.5 V. Calculate the resonant frequency and the voltage across the capacitor 1 µs after the input stimulus has been applied. Calculated Results (Upper display) Calculated Results (Lower display) Solution - All of the equations need to be selected to solve this problem. Press „ and enter the known variables followed by a second press of „ to solve for the unknowns. -PQYP8CTKCDNGU%Aµ(+AO#.AO*4AΩVAµU %QORWVGF4GUWNVUαATU$A8$A8XA8 ωFATUωATU 22.5 Critical-Damping The equations in this section represent the RLC transient response of a critically-damped circuit. The condition for a critically-damped system is ω0 = α. The first two equations define the Neper’s frequency α and the classical resonant frequency ω0. The transient response to a step function stimulus of a voltage across the capacitor v, is expressed in terms of constants D1 and D2, α, and time t. The constants D1 and D2 are defined in terms of the capacitor voltage V0 and the initial inductor current I0. 1 2⋅ R⋅C 1 ω0 = L⋅C α= Eq. 22.5.1 Eq. 22.5.2 v = D1⋅ e −α ⋅t ⋅ t + D2 ⋅ e −α ⋅t EE Pro for TI-89, 92 Plus Equations – RLC Circuits 50 Eq. 22.5.3 D1 = I0 + α *V 0 C Eq. 22.5.4 D2 = V 0 Eq. 22.5.5 Example 22.5 - A critically damped RLC circuit consists of a 100 Ω resistor in series with a 40 mH inductor and a 1 µF capacitor. The initial inductor current is 1 mA and the initial capacitor charge is 10 V. Find the voltage across the capacitor after 10 µs. Calculated Results (Upper display) Calculated Results (Lower display) Solution - All of the equations need to be selected to solve this problem. Press „ and enter the known variables followed by a second press of „ to solve for the unknowns. -PQYP8CTKCDNGU%Aµ(+AO#.AO*4AΩVAµU 8A8 %QORWVGF4GUWNVU&A8U&A8XA8ωATU αATU 22.6 Overdamped Case These equations show the transient performance of an overdamped RLC circuit. The condition for an overdamped system is that α > ω0. The first two equations define the characteristic frequencies s1 and s2 in terms of the Neper’s frequency α and the classical resonant frequency ω0. The transient response to a step function stimulus of the voltage is found by the constants A1, A2, s1 and s2, and time t. The constants A1 and A2 relate to the initial capacitor voltage V0, the initial inductor current I0, s1 and s2: s1 = −α + α 2 − ω 2 Eq. 22.6.1 s2 = −α − α 2 − ω 02 Eq. 22.6.2 1 L⋅C 1 α= 2⋅ R⋅C ω0 = Eq. 22.6.3 Eq. 22.6.4 v = A1⋅ e s1⋅t + A2 ⋅ e s 2⋅t EE Pro for TI-89, 92 Plus Equations – RLC Circuits 51 Eq. 22.6.5 FG H A1 = FG H FG H Eq. 22.6.6 IJ IJ KK 1 V0 ⋅ + I0 CR s2 − s1 − V 0 ⋅ s1 + A2 = IJ K 1 V0 ⋅ + I0 CR s2 − s1 V 0 ⋅ s2 + Eq. 22.6.7 Example 22.6 - An overdamped RLC circuit consists of a 10 Ω resistor in series with a 40 mH inductor and a 1 µF capacitor. If the initial inductor current is 0 mA and the capacitor is charged to a potential of 5 V, find the voltage across the capacitor after 1 ms. Calculated Results (Upper display) Calculated Results (Lower display) Solution - All of the equations need to be selected to solve this problem. Press „ and enter the known variables followed by a second press of „ to solve for the unknowns. The solver takes about five minutes to solve this problem. -PQYP8CTKCDNGU%Aµ(+AO#.AO*4AΩVAOU 8A8 %QORWVGF4GUWNVUUATUUATUXA8ωATU EE Pro for TI-89, 92 Plus Equations – RLC Circuits 52 Chapter 23 AC Circuits This chapter covers equations describing the properties of AC circuits. ♦ ♦ ♦ ♦ ♦ RL Series Impedance RC Series Impedance Impedance ↔ Admittance Two Impedances in Series Two Impedances in Parallel Variables All the variables here are listed with a brief description and appropriate units. Variable C f I Im L θ θ1 θ2 R RR1 RR2 t V VC VL Vm VR ω X XX1 XX2 Y_ Z1m Z2m Z_ Zm EE Pro for TI-89, 92 Plus Equations - AC Circuits Description Capacitance Frequency Instantaneous current Current amplitude Inductance Impedance phase angle Phase angle 1 Phase angle 2 Resistance Resistance 1 Resistance 2 Time Total voltage Voltage across capacitor Voltage across inductor Maximum voltage Voltage across resistor Radian frequency Reactance Reactance 1 Reactance 2 Admittance Impedance 1 magnitude Impedance 2 magnitude Complex impedance Impedance magnitude Unit F Hz A A H rad rad r Ω Ω Ω s V V V V V r/s Ω Ω Ω S Ω Ω Ω Ω 53 23.1 RL Series Impedance The equations in this section describe the relationships of an RL series circuit. The first equation shows the sinusoidal behavior of current, I, defined by the amplitude, Im, radian frequency ω, and time t. The second equation defines the magnitude of impedance Zm in terms of the resistance R, the inductance L and ω. The voltages VR and VL across the resistor and inductor are defined by the next two equations. The fifth equation calculates the total voltage drop V of the circuit as the sum of the individual voltage drops of the resistor and inductor. The next two equations compute the amplitude Vm and phase θ of the voltage across the circuit. The final equation relates the radian frequency and frequency f. Note: I, VR, and VL are functions of time. bg I = Im⋅ sin ω ⋅ t bg Eq. 23.1.1 abs Zm = R 2 + ω 2 ⋅ L2 2 Eq. 23.1.2 b g bg b g bg VR = Zm ⋅ Im⋅ sin ω ⋅ t ⋅ cos θ VL = Zm ⋅ Im⋅ cos ω ⋅ t ⋅ sin θ V = VR + VL Vm = Zm ⋅ Im ω⋅L θ = tan −1 R ω = 2 ⋅π ⋅ f FG H Eq. 23.1.3 Eq. 23.1.4 Eq. 23.1.5 Eq. 23.1.6 IJ K Eq. 23.1.7 Eq. 23.1.8 Example 23.1 - An RL circuit consists of a 50 Ω resistor and a 0.025 henry inductor. At a frequency of 400 Hz, the current amplitude is 24 mA. Find the impedance of the circuit and the voltage drops across the resistor and inductor after 100 µs. Upper Display Lower Display Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUHA*\+OAO#.AJGPT[4AΩVAµU %QORWVGF4GUWNVU+A#θATCF8A88.A8 8OA884A8ωATU<OAΩ EE Pro for TI-89, 92 Plus Equations - AC Circuits 54 23.2 RC Series Impedance The equations in this section describe the key relationships involved in an RC series circuit. The first equation shows the sinusoidal behavior of current, I, defined by the amplitude Im, the radian frequency ω, and time t. The magnitude Zm of the impedance of the circuit is calculated in terms of the resistance R, capacitance C, and ω in the second equation. The voltage VR across R and the voltage VC across C are given in the next two equations. The total RC voltage drop V is expressed in the fifth equation. The amplitude Vm and phase θ of the voltage across the circuit is expressed in equations 6 & 7. The last equation is relationship between the radian frequency ω and actual frequency f. Note: I, VR, and VC are functions of time. bg I = Im⋅ sin ω ⋅ t Eq. 23.2.1 1 ω ⋅ C2 VR = Zm ⋅ Im⋅ sin ω ⋅ t ⋅ cos θ bg abs Zm = R 2 + 2 Eq. 23.2.2 2 b g bg VC = Zm ⋅ Im⋅ cosbω ⋅ t g ⋅ sinbθ g Eq. 23.2.3 Eq. 23.2.4 V = VR + VC Vm = Zm ⋅ Im Eq. 23.2.5 Eq. 23.2.6 θ = tan −1 Eq. 23.2.7 FG −1 IJ H ω ⋅C ⋅ RK ω = 2 ⋅π ⋅ f Eq. 23.2.8 Example 23.2 - An RC circuit consists of a 100 Ω resistor in series with a 47 µF capacitor. At a frequency of 1500 Hz, the current peaks at an amplitude of 72 mA. Find all the parameters of the RC circuit and the voltage drop after 150µs. Computed Results Upper Display Computed results Lower Display Solution - Use all of the equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU%Aµ(HA*\+OAO#4AΩVAµU %QORWVGF4GUWNVU+A#θATCF8A88%A8 8OA884A8ωATU<OAΩ EE Pro for TI-89, 92 Plus Equations - AC Circuits 55 23.3 Impedance ↔ Admittance The equation is designed to convert impedances to admittances with unit management built-in. As shown in the figure to the right, an impedance Z consists of a real and reactive components (R and X) to describe it. The admittance Y consists of real and reactive components (G and B) to describe it. The variables Z and Y have an _ attached to them to add emphasize that they are complex in general and have units attached. Y_ = 1 Z_ Eq. 23.3.1 Example 23.3 - Find the admittance of an impedance consisting of a resistive part of 125 Ω and a reactance part of 475_ Ω. Entered Value Calculated Result Solution -Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU<  K AΩ %QORWVGF4GUWNVU;   K A5KGOGPU 23.4 Two Impedances in Series These equations combine two impedances Z1 and Z2 in series with real and imaginary parts RR1 and XX1, RR2 and XX2, respectively. The impedances Z1 and Z2 are expressed by their magnitudes Z1m and Z2m, and phase angles θ1 and θ2 respectively. The combined result of the two impedances in series is an impedance with a magnitude Zm and a phase angle θ. bg abs Zm = R 2 + X 2 2 θ = tan −1 Eq. 23.4.1 FG X IJ H RK Eq. 23.4.2 R = RR1 + RR 2 Eq. 23.4.3 X = XX 1 + XX 2 Eq. 23.4.4 bg absb Z 2mg abs Z1m = RR12 + XX 12 2 EE Pro for TI-89, 92 Plus Equations - AC Circuits 2 = RR 2 2 + XX 2 2 56 Eq. 23.4.5 Eq. 23.4.6 FG XX 1IJ H RR1 K F XX 2 IJ θ 2 = tan G H RR2 K θ 1 = tan −1 Eq. 23.4.7 −1 Eq. 23.4.8 Example 23.4 - Two impedances consisting of resistances of 100 Ω and 75 Ω and reactive components 75 and 145, respectively are connected in series. Find the magnitude and phase angle of the combination. Input Variables Computed results Solution - Select the first four equations to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU44AΩ44AΩ::AΩ::AΩ %QORWVGF4GUWNVU<OAΩθATCF4A Ω:A Ω 23.5 Two Impedances in Parallel Two impedances Z1 and Z2 when connected in parallel result in an equivalent parallel impedance Zp, represented by a real part R and a reactive part X. The impedances Z1 and Z2 have real and reactive parts RR1, XX1 and RR2, XX2 respectively. The first two equations show a numeric expression for magnitude Zm and phase θ. Simpler versions of R and X are shown in the third and fourth equations. The last four equations calculate the magnitude and phase of Z1 and Z2 as Z1m and θ1, and Z2m and θ2, respectively. b RR1⋅ RR2 − XX 1⋅ XX 2g + b RR1⋅ XX 2 + RR2 ⋅ XX 1g absb Zmg = b RR1 + RR2g + b XX 1 + XX 2g F XX 1⋅ RR2 + RR1⋅ X 2 IJ − tan FG XX 1 + XX 2 IJ θ = tan G H RR1⋅ RR2 − XX 1⋅ X 2 K H RR1 + RR2 K R = Zm ⋅ cosbθ g X = Zm ⋅ sinbθ g absb Z1mg = RR1 + XX 1 absb Z 2mg = RR 2 + XX 2 2 2 2 −1 −1 2 2 EE Pro for TI-89, 92 Plus Equations - AC Circuits 2 2 2 2 2 57 2 Eq. 23.5.1 Eq. 23.5.2 Eq. 23.5.3 Eq. 23.5.4 Eq. 23.5.5 Eq. 23.5.6 FG XX 1IJ H RR1 K F XX 2 IJ θ 2 = tan G H RR2 K θ 1 = tan −1 Eq. 23.5.7 −1 Eq. 23.5.8 Example 23.5 - For two impedances in parallel possessing values identical to the previous example, calculate the magnitude and phase of the combination. Entered Values Calculated Results Solution - Use the first and second equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU44AΩ44AΩ::AΩ::AΩ %QORWVGF4GUWNVU<OAΩθATCF EE Pro for TI-89, 92 Plus Equations - AC Circuits 58 Chapter 24 Polyphase Circuits This chapter covers equations used in Polyphase circuits. The equations have been divided into three sections. ™Power Measurements ™Balanced ∆ Network ™Balanced Wye Network Variables All the variables are listed below with a brief description and units. Variable IL Ip P PT θ VL Vp W1 W2 Description Line current Phase current Power per phase Total power Impedance angle Line voltage Phase voltage Wattmeter 1 Wattmeter 2 Unit A A W W rad V V W W 24.1 Balanced ∆ Network These equations describe the essential features of a Balanced ∆ Network. The line voltage VL is defined in terms of phase voltage Vp. The second equation expresses the line current IL using the phase current Ip. The third equation computes the power in each phase P from Vp, Ip and the phase delay θ between voltage and current. The last two equations represent the total power PT delivered to the system in terms of IL, VL Vp and Ip. VL = Vp Eq. 24.1.1 IL = 3 ⋅ Ip Eq. 24.1.2 bg P = Vp ⋅ Ip ⋅ cos θ bg 3 ⋅VL ⋅ IL ⋅ cosbθ g Eq. 24.1.3 PT = 3 ⋅Vp ⋅ Ip ⋅ cos θ Eq. 24.1.4 PT = Eq. 24.1.5 Example 24.1 - Given a line current of 25 A, a phase voltage of 110 V, and a phase angle of 0.125 rad, find the phase current, power, total power and line voltage. EE Pro for TI – 89, 92 Plus Equations – Polyphase circuits 59 Input variables Computed results Solution - Upon examining the problem, all equations are needed. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen display above. -PQYP8CTKCDNGU+.A#θATCF8RA8 %QORWVGF4GUWNVU+RA#2A926A98.A8 24.2 Balanced Wye Network These equations describe the relationship for a Balanced Wye Network. The first equation computes the line voltage VL from the phase voltage Vp. The second equation calculates the line current IL from the phase current Ip. The power/phase P, is defined in terms of Vp, Ip and phase delay θ. The last two equations are used to estimate the total power PT delivered to the system from the parameters VL, IL or Vp and Ip. VL = 3 ⋅Vp Eq. 24.2.1 IL = Ip Eq. 24.2.2 bg P = Vp ⋅ Ip ⋅ cos θ bg 3 ⋅VL ⋅ IL ⋅ cosbθ g Eq. 24.2.3 PT = 3 ⋅Vp ⋅ Ip ⋅ cos θ Eq. 24.2.4 PT = Eq. 24.2.5 Example 24.2 - Using the known parameters in the previous example for the Balanced ∆ Network, find the phase current, power, total power and line voltage. Input variables Computed results Solution - All of the equations are needed to compute the solution. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen display above. EE Pro for TI – 89, 92 Plus Equations – Polyphase circuits 60 -PQYP8CTKCDNGU+.A#θATCF8RA8 %QORWVGF4GUWNVU+RA#2A926A98.A8 24.3 Power Measurements These three equations for a two-watt meter connection are used to measure the total power of a balanced network. The first two equations determine the watt meter readings W1 and W2 are expressed from the line current IL, line voltage VL, and phase delay θ between the voltage and the current. The final equation represents the total power, PT, delivered to the three-phase load. FG π IJ H 6K F πI W 2 = VL ⋅ IL ⋅ cosGθ − J H 6K PT = 3 ⋅VL ⋅ IL ⋅ cosbθ g W1 = VL ⋅ IL ⋅ cos θ + 24.3.1 Eq. Eq. 24.3.2 Eq. 24.3.3 Example 24.3 - Given a line voltage of 110 V and a line current of 25 A and a phase angle of 0.1 rad, find the wattmeter readings in a 2 wattmeter meter system. Input variables Computed results Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen display above. -PQYP8CTKCDNGU+.A#θATCF8.A8 %QORWVGF4GUWNVU26A99A99A9 EE Pro for TI – 89, 92 Plus Equations – Polyphase circuits 61 Chapter 25 Electrical Resonance The equations in this section describe the electrical properties of resonance in circuits composed of ideal circuit elements. The section is organized under four topics: ™Parallel Resonance I ™Parallel Resonance II ™Resonance in Lossy Inductor ™Series Resonance Variables The following variables are listed with a brief description and their appropriate units. Variable α β C Im L θ Q R Rg Vm ω ω0 ω1 ω2 ωd ωm Yres Z Zres Description Damping coefficient Bandwidth Capacitance Current Inductance Phase angle Quality factor Resistance Generator resistance Maximum voltage Radian frequency Resonant frequency Lower cutoff frequency Upper cutoff frequency Damped resonant frequency Frequency for maximum amplitude Admittance at resonance Impedance Impedance at resonance Unit r/s r/s F A H rad unitless Ω Ω V r/s r/s r/s r/s r/s r/s S Ω Ω 25.1 Parallel Resonance I These ten equations describe resonance properties in parallel resonance circuits. The first equation expresses Vm, the voltage across the circuit, in terms of Im, the magnitude of the supplied current and the equivalent impedance of the parallel circuit consisting of an inductor L, a capacitor C and a resistor R at the radian frequency ω. The second equation computes the phase angle, θ, between Vm and Im. The third equation defines the resonant frequency ω0 from the EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance 62 reactive parameters L and C. The terms ω1 and ω2 represent the lower and upper cutoff frequencies beyond ω0, where the impedance is half the impedance at resonance. The bandwidth, β, is the difference between ω1 and ω2. The last three equations calculate the quality factor Q in terms of R, C, L and ω0. Im Vm = F1 F 1 II GGH R + GHω ⋅ C − bω ⋅ LgJK JJK 1I I FF θ = tan G G ω ⋅ C − H H ω ⋅ L JK ⋅ RJK Eq. 25.1.1 2 2 −1 ω0 = 1 L⋅C ω1 = −1 + 2⋅ R⋅C ω2 = 1 + 2⋅ R⋅C Eq. 25.1.2 Eq. 25.1.3 1 b2 ⋅ R ⋅ C g 2 1 b2 ⋅ R ⋅ C g 2 + 1 L⋅C Eq. 25.1.4 + 1 L⋅C Eq. 25.1.5 β = ω 2 − ω1 Eq. 25.1.6 ω0 β Eq. 25.1.7 Q= Q = R⋅ C L Eq. 25.1.8 Q = ω0⋅ R ⋅C Eq. 25.1.9 Example 25.1 - Calculate the resonance parameters of a parallel resonant circuit containing a 10,000 Ω resistor, a 2.4 µF capacitor and a 3.9 mH inductor. The amplitude of the current is 10 mA at a radian frequency of 10,000 rad/s. Input Values Calculated Results (Upper and Lower Screens) Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. The computed results are shown in the screen displays above. EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance 63 -PQYP8CTKCDNGU%Aµ(+OAO#4AΩ.O*ωATU %QORWVGF4GUWNVUβATUθATCF3 8OA8ωATUωATUωATU 25.2 Parallel Resonance II These equations represent an alternative method of expressing the properties of a resonant circuit in terms of the quality factor Q. The first equation links Q with the resonant frequency ω0 and the bandwidth β. The lower and upper cutoff frequencies, ω1 and ω2 are defined in terms of ω0 and Q. The fourth and fifth equations compute α, the damping coefficient, either from R and C, or ω0 and Q. The final two equations express the damped resonant frequency ωd in terms of α and ω0 or ω0 and Q. Q= ω0 β Eq. 25.2.1 F −1 + 1 + 1I GH 2 ⋅ Q b2 ⋅ Qg JK F 1 + 1 + 1I ω2 = ω0⋅G H 2 ⋅ Q b2 ⋅ Qg JK ω1 = ω 0 ⋅ 2 Eq. 25.2.2 2 Eq. 25.2.3 α= 1 2⋅ R⋅C Eq. 25.2.4 α= ω0 2⋅Q Eq. 25.2.5 ωd = ω 02 − α 2 ωd = ω 0 ⋅ 1 − Eq. 25.2.6 1 4 ⋅ Q2 Eq. 25.2.7 Example 25.2 - A parallel resonant circuit has a 1000 Ω resistor and a 2 µF capacitor. The Quality Factor for this circuit is 24.8069. Find the band-width, damped and resonant frequencies. Input Values EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance Calculated Results (Upper and Lower Screens) 64 Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU%Aµ(34AΩ %QORWVGF4GUWNVUαβATUωATUωATU ωATUωFATU 25.3 Resonance in Lossy Inductor These equations characterize the properties of a lossy inductor in parallel with an ideal capacitor and a current source with an impedance of Rg. At the resonant frequency ω0, the admittance of the parallel circuit is purely conductive. Yres and Zres represent the impedance and admittance of the circuit at resonance. ωm represents the frequency that occurs when the amplitude of the voltage across the resonant circuit is at maximum. FG IJ HK 2 1 R − L⋅C L L + Rg ⋅ R ⋅ C Yres = L ⋅ Rg 1 Zres = Yres ω0 = FG 1 IJ ⋅ FG1 + 2 ⋅ R IJ + FG R IJ H L ⋅ C K H Rg K H L K 2 ωm = Eq. 25.3.1 Eq. 25.3.2 Eq. 25.3.3 2 FG IJ HK 2 R ⋅ − L⋅C L 2 Eq. 25.3.4 Example 25.3 - A power source with an impedance Rg of 5 Ω is driving a parallel combination of a lossy 40 µH inductor with a 2 Ω loss resistance, and a capacitor of 2.7 µF. Find the frequency of resonance and the frequency for maximum amplitude. Entered Values Calculated Results Solution - Upon examining the problem, all equations are needed to solve for a solution. Press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU%Aµ(4AΩ.Aµ*4IAΩ EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance 65 %QORWVGF4GUWNVUωATUωOATU;TGUAUKGOGPU <TGUAΩ 25.4 Series Resonance These equations characterize the properties of a series resonance circuit. The first equation connects the resonant frequency ω0 and the reactive elements, the inductance L, and the capacitance, C. The second and third equations compute the magnitude Z and phase θ of the impedance, respectively.. The equations for ω1 and ω2 represent the upper and lower cutoff frequencies beyond resonance, where the impedance is half the impedance at resonance. The expressions for β represents the bandwidth of the resonant circuit. The last equation determines the quality factor Q in terms of R, C, L, and ω0. ω0 = 1 L⋅C Eq. 25.4.1 1I F Z = R + Gω ⋅ L − H ω ⋅ C JK Fω ⋅ L − 1 I G ω ⋅ C JJ θ = tan G GH R JK −R F R IJ + 1 +G ω1 = H 2 ⋅ LK L ⋅C 2⋅ L R F R IJ + 1 ω2 = +G H 2 ⋅ LK L ⋅C 2⋅ L 2 2 −1 Eq. 25.4.2 Eq. 25.4.3 2 Eq. 25.4.4 2 β = ω 2 − ω1 Eq. 25.4.6 R L ω0⋅ L Q= R β= Q= Eq. 25.4.5 Eq. 25.4.7 Eq. 25.4.8 1 L ⋅ RC Eq. 25.4.9 Example 25.4 - Find the characteristic parameters of a series-resonant circuit with R = 25Ω, L = 69 µH, C = 0.01 µF and a radian frequency of 125,000 rad/s. EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance 66 Entered Values Calculated Results (Upper and Lower Screens) Solution - Upon examining the problem, all equations are needed to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU%Aµ(4AΩ.Aµ*4AΩωATU %QORWVGF4GUWNVUωATUωATUωATU <AΩ EE Pro for TI – 89, 92 Plus Equations – Electrical Resonance 67 Chapter 26 OpAmp Circuits Seven commonly-used OpAmp circuits are presented in this section. An OpAmp is a direct-coupled high-gain amplifier that can be configured with the use of feedback to circuit elements to achieve overall performance characteristics. There are two inputs labeled ‘+’ and ‘-’. The manner in which input signals are connected to these terminals defines the inverting or non-inverting properties of the circuit. ™Basic Inverter ™Non-Inverting Amplifier ™Current Amplifier ™Transconductance Amplifier ™Level Detector (Inverting) ™Level Detector (Non-Inverting) ™Differentiator ™Differential Amplifier Variables All the variables used here are listed with a brief description and proper units. Variable Acc Aco Ad Agc Aic Av CC1 Cf CMRR Cp fcp fd f0 fop IIf RR1 RR2 RR3 RR4 Rf Rin Rl Ro Rout Rp EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits Description Common Mode current gain Common Mode gain from real OpAmp Differential mode gain Transconductance Current gain Voltage gain Input capacitor Feedback capacitor CM rejection ratio Bypass capacitor 3dB bandwidth, circuit Characteristic frequency Passband, geometric center 3dB bandwidth, OpAmp Maximum current through Rf Input resistor Current stabilizor Feedback resistor Resistor Feedback resistor Input resistance Load resistance Output resistance, OpAmp Output resistance Bias current resistor 68 Unit unitless unitless unitless S unitless unitless F F unitless F Hz Hz Hz Hz A Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Rs tr ∆VH VL Vomax VR Vrate VU Vz1 Vz2 Ω s V V V V V/s V V V Voltage divide resistor 10-90% rise time Hysteresis Detection threshold, low Maximum circuit output Reference voltage Maximum voltage rate Detection threshold, high Zener breakdown 1 Zener breakdown 2 26.1 Basic Inverter These equations define the properties of a basic inverter. The first equation relates the voltage gain Av to the feedback resistance Rf and input resistance RR1. The optimum value of Rp is defined by the second equation to minimize output-voltage offset due to input bias current. The first pole frequency fcp is defined by the third equation. Small signal rise time tr (10 to 90%) is defined by the fourth equation. Av = − Rf RR1 Eq. 26.1.1 Rp = RR1⋅ Rf RR1 + Rf Eq. 26.1.2 fcp = fop ⋅ ( − Av ) ⋅ tr = FG RR1IJ H Rf K Eq. 26.1.3 .35 ⋅ Rf fop ⋅ ( − Av ) ⋅ RR1 Eq. 26.1.4 Example 26.1 - Find the gain of an inverter and its optimum value for bias resistance given an input resistance of 1 kΩ and a feedback resistance of 20 kΩ. Entered Values Calculated Results Solution - Use the first and second equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU44AMΩ, Rf=20._kΩ %QORWVGF4GUWNVU#X4RA Ω EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 69 26.2 Non-Inverting Amplifier These equations define the properties of a non-inverting amplifier. The first equation expresses the voltage gain Av in terms of the feedback resistor Rf and resistor RR1. The second equation gives the value of Rp needed in the input circuit to minimize offset current effects. Av = 1 + Rf RR1 Eq. 26.2.1 Rp = RR1⋅ Rf RR1 + Rf Eq. 26.2.2 Example 26.2 - Find the DC gain of a non-inverting amplifier with a feedback resistance of 1 MΩ and a resistance to the load of 18 kΩ. Find the gain and the optimum value for a bias resistor. Entered Values Calculated Results Solution - Use the first and second equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU4HA/Ω44AMΩ %QORWVGF4GUWNVU#X4RAΩ 26.3 Current Amplifier This section describes the properties of a current amplifier. The first equation shows the relationship between the current gain Aic with feedback resistance Rf, load resistance Rl, output resistance of OpAmp Ro, voltage divide resistor Rs, and voltage gain Av. The remaining equations define the input resistance Rin and output resistance Rout of the system. Aic = b Rs + Rf g ⋅ Av Rl + Ro + Rs ⋅ b1 + Av g Eq. 26.3.1 Rin = Rf 1 + Av Eq. 26.3.2 EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 70 b Rout = Rs ⋅ 1 + Av g Eq. 26.3.3 Example 26.3 - A current amplifier with a 200 kΩ feedback resistance has a voltage gain of 42. If the source resistance is 1 kΩ, the load resistance is 10 kΩ and the output resistance of the OpAmp is 100 Ω, find the current gain, input and output resistances. Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU#X4HAMΩ4NAMΩ4QAΩ4UAMΩ %QORWVGF4GUWNVU#KE4KPAΩ4QWVAΩ 26.4 Transconductance Amplifier The two equations in this section specify a closed loop transconductance Agc and output resistance Rout in terms of the resistance Rs and voltage gain Av. 1 Rs Rout = Rs ⋅ 1 + Av Agc = b Eq. 26.4.1 g Eq. 26.4.2 Example 26.4 - Find the transconductance and output resistance for a transconductance amplifier with a voltage gain of 48 and an external resistance of 125 Ω Entered Values Calculated Results Solution - Use both equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen display above. -PQYP8CTKCDNGU4UAΩ#X EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 71 %QORWVGF4GUWNVU#IEAUKGOGPU4QWVAΩ 26.5 Level Detector (Inverting) The first equation in this section computes the value of the resistor RR1 attached to an OpAmp inverting input. The second equation calculates the hysteresis (or memory) ∆VH of the level detector circuit. The third and fourth equations define the upper and lower trip voltages VU and VL for an ideal inverting level detector, assuming a reference voltage VR and breakdown voltages Vz1 and Vz2, and in terms of Rp and Rf. RR1 = Rp ⋅ Rf Rp + Rf ∆VH = bVz1 + Vz2g ⋅ Rp Eq. 26.5.2 Rp + Rf VR ⋅ Rf + Rp ⋅Vz1 Rf + Rp Eq. 26.5.3 VR ⋅ Rf − Rp ⋅Vz 2 Rf + Rp Eq. 26.5.4 VU = VL = Eq. 26.5.1 Example 26.5. - An inverting level detector possesses two zener diodes to set the trip level. The setting levels are 4 V and 3 V, respectively, for the first and second diodes. The reference voltage is 5 V, the OpAmp is supported by a 10 kΩbias resistor and a 1 MΩ feedback resistor. Find the hysteresis, the upper and lower detection thresholds, and the input resistance. Upper Display Lower Display Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU4HA/Ω4RAMΩ84A88\A88\A8 %QORWVGF4GUWNVU∆8*A844AMΩ8.A887A8 EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 72 26.6 Level Detector (Non-Inverting) This section computes the value of the resistor RR1 attached to an OpAmp non-inverting input. The next equation calculates the hysteresis (or memory) ∆VH of the level detector circuit. The next two equations define the upper and lower trip voltages VU and VL for an ideal inverting level detector, a reference voltage VR and breakdown voltages Vz1 and Vz2, in terms of Rp and Rf. RR1 = Rp ⋅ Rf Rp + Rf ∆VH = bVz1 + Vz2g ⋅ Rp Eq. 26.6.1 Eq. 26.6.2 Rp + Rf b b g g VR ⋅ Rf + Rp + Rp ⋅Vz 2 Rf VR ⋅ Rp + Rf − Rp ⋅Vz1 VL = Rf VU = Eq. 26.6.3 Eq. 26.6.4 Example 26.6 - For a non-inverting level detector with the same specifications as the inverting level detector in the previous example, compute the hysteresis, the upper and lower detection thresholds, and the input resistance. Entered Values Calculated Results Solution - Use the first three equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU4HA/Ω4RAMΩ84A88\A88\A8 %QORWVGF4GUWNVU∆8*A844AΩ87A8 26.7 Differentiator These equations define all the components required for a differentiator. The first equation defines the feedback resistor Rf in terms of the maximum output voltage Vomax and current IIf. Typically, IIf is of the order of 0.1 - 0.5 mA. The second equation computes the value for the resistor Rp used to cancel the effects of OpAmp input bias current. CC1 is the input capacitor required for the differentiator, and RR1 is the resistor utilized for stability. The characteristic frequency of the differentiator fd is expressed by the fifth equation. The last two equations compute the bypass capacitor Cp and the feedback capacitor Cf. EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 73 Rf = Vo max IIf Eq. 26.7.1 Rp = Rf Eq. 26.7.2 Vo max Rf ⋅Vrate 1 RR1 = 2 ⋅ π ⋅ fd ⋅ CC1 1 fd = 2 ⋅ π ⋅ Rf ⋅ CC1 CC1 = Eq. 26.7.3 Eq. 26.7.4 Eq. 26 7.5 Cp = 10 2 ⋅ π ⋅ f 0 ⋅ Rp Eq. 26.7.6 Cf = 1 4 ⋅ π ⋅ f 0 ⋅ Rf Eq. 26.7.7 Example 26.7 - A differentiator circuit designed with an OpAmp has a slew rate of 1.5 V/µs. If the maximum output voltage is 5 V, and the feedback resistor is 39 kΩ, what input capacitor and resistor are needed for the amplifier with a characteristic frequency of 50 kHz? Entered Values Calculated Results Solution - Use the third and fourth equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUHFAM*\8TCVGA8µU8QOCZA84HAMΩ %QORWVGF4GUWNVU%%'A(44AMΩ 26.8 Differential Amplifier These four equations describe the primary relationships used in designing of a differential amplifier. The first equation computes the differential gain Ad in terms of the input and feeback resistors RR1 and RR3. The second equation shows the common-mode gain Aco in terms of RR3, RR1, and the common-mode rejection ratio CMRR. The third equation expands the definition of Ad from the first equation to accomodate a practical OpAmp with a finite voltage gain Av. The final equation shows the common-mode gain due to resistor mismatching Acc. EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 74 Ad = RR3 RR1 Aco = Ad = Acc = Eq. 26.8.1 RR 32 RR3 ⋅ RR1 + RR3 ⋅ CMRR b g Av ⋅ RR3 Eq. 26.8.3 RR12 ⋅ Av 2 + RR32 RR4 ⋅ RR1 − RR2 ⋅ RR3 RR1⋅ RR2 + RR4 b Eq. 26.8.2 g Eq. 26.8.4 Example 26.8 - Find the differential mode gain and the current gain for a differential amplifier with bridge resistors RR1, RR2, RR3 and RR4 of 10 kΩ, 3.9 kΩ, 10.2 kΩ and 4.1 kΩ, respectively. Assume a voltage gain of 90. Entered Values Calculated Results Solution - Use the third and fourth equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU#X44AMΩ44AMΩ44AMΩ44AMΩ %QORWVGF4GUWNVU#EE#F EE Pro for TI - 89, 92 Plus Equations - OpAmp Circuits 75 Chapter 27 Solid State Devices This section covers a variety of topics in solid state electronics. Read this! Note: The equations in this section are grouped under topics which describe general properties of semiconductors or devices. Equations for a variety of specific cases and are listed together under a sub-topic heading and are not necessarily a set of consistent equations which can be solved together. Choosing equations in a subtopic w/o regard as to whether the equations represent actual relationships could generate erroneous results or no solution at all. Read the description of each equation set to determine which equations in a sub-topic form a consistent subset before attempting to compute a solution. ™Semiconductor Basics ™PN Junctions ™PN Junction Currents ™Transistor Currents ™Ebers-Moll Equations ™Ideal Currents - pnp ™Switching Transients ™MOS Transistor I ™MOS Transistor II ™MOS Inverter (Resistive) ™MOS Inverter (Saturated) ™MOS Inverter (Depletion) ™CMOS Transistor Pair ™Junction FET Variables A complete list of all the variables used in this section along with a brief description and appropriate units is given below. Variable α aLGJ A A1 A2 αf Aj αr β b βf βr Cj EE PRO for TI-89, 92 Plus Equations - Solid State Devices Description CB current gain Linearly graded junction parameter Area EB junction area CB junction area Forward α Junction area Reverse α CE current gain Channel width Forward β Reverse β Junction capacitance 77 Unit unitless 1/m4 m2 m2 m2 unitless m2 unitless unitless m unitless unitless F CL Cox D DB DC DE Dn Dp εox εs Ec EF Ei Ev ffmax γ gd gm gmL Go I I0 IB IC ICB0 ICE0 ICsat ID IDmod ID0 IDsat IE IIf Ir Ir0 IRG IRG0 Is λ Is kD kL kn kn1 kN kP KR L LC LD LE LL EE PRO for TI-89, 92 Plus Equations - Solid State Devices Load capacitance Oxide capacitance per unit area Diffusion coefficient Base diffusion coefficient Collector diffusion coefficient Emitter diffusion coefficient n diffusion coefficient p diffusion coefficient Oxide permittivity Silicon Permittivity Conduction band Fermi level Intrinsic Fermi level Valence band Maximum frequency Body coefficient Drain conductance Transconductance Transconductance, load device Conductance Junction current Saturation current Base current Collector current CB leakage, E open CE leakage, B open Collector I at saturation edge Drain current Channel modulation drain current Drain current at zero bias Drain saturation current Emitter current Forward current Reverse current E-M reverse current component G-R current Zero bias G-R current Saturation current Modulation parameter Saturation current MOS constant, driver MOS constant, load MOS constant MOS process constant MOS constant, n channel MOS constant, p channel Ratio Transistor length Diffusion length, collector Drive transistor length Diffusion length, emitter Load transistor length 78 F F/m2 m2/s m2/s m2/s m2/s m2/s m2/s unitless unitless J J J J Hz V.5 S S S S A A A A A A A A A A A A A A A A A A 1/V A A/V2 A/V2 A/V2 A/V2 A/V2 A/V2 unitless m m m m m LLn lNN Lp lP µn µp mn mp N Na nnC Nd nE ni N0 npo p pB φF φGC pno Qtot Qb Qb0 Qox Qsat ρn ρp Rl τB τD τL τo τp τt t TT tch tdis tox tr ts tsd1 tsd2 Ttr V1 Va Vbi VBE VCB VCC EE PRO for TI-89, 92 Plus Equations - Solid State Devices Diffusion length, n n-channel length Diffusion length, p p-channel length n (electron) mobility p (positive charge) mobility n effective mass p effective mass Doping concentration Acceptor density n density, collector Donor density n density, emitter Intrinsic density Surface concentration n density in p material p density p density, base Fermi potential Work function potential p density in n material Total surface impurities Bulk charge at bias Bulk charge at 0 bias Oxide charge density Base Q, transition edge n resistivity p resistivity Load resistance lifetime in base Time constant Time constant Lifetime Minority carrier lifetime Base transit time Time Temperature Charging time Discharge time Gate oxide thickness Collector current rise time Charge storage time Storage delay, turn off Storage delay, turn off Transit time Input voltage Applied voltage Built-in voltage BE bias voltage CB bias voltage Collector supply voltage m m m m m2/(V*s) m2/(V*s) unitless unitless 1/m3 1/m3 1/m3 1/m3 1/m3 1/m3 1/m3 1/m3 1/m3 1/m3 V V 1/m3 unitless C/m2 C/m2 C/m2 C Ω*m Ω*m Ω s s s s s s s K s s m s s s s s V V V V V V 79 VCEs VDD VDS VDsat VEB VG VGS VIH Vin VIL VL VM VOH VOL Vo Vp VSB VT VT0 VTD VTL VTL0 VTN VTP W WB WD WL WN WP x xd xn xp Z CE saturation voltage Drain supply voltage Drain voltage Drain saturation voltage EB bias voltage Gate voltage Gate to source voltage Input high Input voltage Input low voltage Load voltage Midpoint voltage Output high Output low Output voltage Pinchoff voltage Substrate bias Threshold voltage Threshold voltage at 0 bias Depletion transistor threshold Load transistor threshold Load transistor threshold n channel threshold p channel threshold MOS transistor width Base width Drive transistor width Load transistor width n-channel width p-channel width Depth from surface Depletion layer width Depletion width, n side Depletion width, p side JFET width V V V V V V V V V V V V V V V V V V V V V V V V m m m m m m m m m m m 27.1 Semiconductor Basics The nine equations listed under this sub-topic describe the basic properties used semiconductor technology. The first four equations are a subset which describes the basic properties of free carriers in semiconductors such as resistivity, mobility and diffusion properties. Since the main semiconductor material of commercial use is silicon, a special function ni(TT) was developed to calculate the intrinsic carrier density as a function of temperature TT. The first two equations define the resistivities ρn and ρp of n and p-type semiconductors in terms of the electron and hole mobilities, µn and µp, and doping densities, Nd and Na. The next two equations are often called the Einstein equations connecting the electron and hole diffusion coefficients Dn and Dp, to their mobilities µn and µp and the temperature TT. 1 q ⋅ µn ⋅ Nd 1 ρp = q ⋅ µp ⋅ Na ρn = EE PRO for TI-89, 92 Plus Equations - Solid State Devices Eq. 27.1.1 Eq. 27.1.2 80 Dn = k ⋅ TT ⋅ µn q Eq. 27.1.3 Dp = k ⋅ TT ⋅ µp q Eq. 27.1.4 The Fermi level EF is a measure of the chemical potential in silicon and is used to estimate the doping density. For instance in a n-type semiconductor, equation 27.1.5 is used to determine the location of the Fermi level, while in a ptype material equation 27.1.6 is used to establish the Fermi level. In some cases, when both donor and acceptor levels are specified, one has to choose either equation 27.1.5 or 27.1.6 whether Nd>Na. At normal temperatures, if Nd>Na, the material is defined as n-type and 27.1.6 would be used provided Nd in the equation is replaced by Nd-Na. The equation 27.1.5 would be used if Nd<Na. The intrinsic Fermi level is always defined by the equation 21.1.7. The Fermi level equations define the relative location of EF, with respect to the intrinsic Fermi Level Ei, in terms of temperature TT, Na or Nd, and the intrinsic carrier density function ni(TT). Ei is calculated in the seventh equation in terms of the conduction and valence band levels Ec and Ev, temperature TT, the effective mass of electrons mn and holes mp. F Na I GH nibTT gJK F Nd I EF = Ei + k ⋅ TT ⋅ lnG H nibTT gJK Ei = EF + k ⋅ TT ⋅ ln Ei = Eq. 27.1.5 Eq. 27.1.6 b Ec + Evg + 3 ⋅ k ⋅ TT ⋅ lnFG mp IJ H mn K 2 4 Eq. 27.1.7 The final two equations are diffusion properties of dopants in silicon based on two distinctly different conditions. The equation 27.1.8 covers the diffusion from an infinite source while the equation 27.1.9 covers diffusion from a finite source. The diffusion of impurities in a semiconductor subject to an infinite source with a surface concentration N0 at a depth x below the surface after a time t, given the diffusion coefficient D is shown in the next equation. The final equation details the diffusion from a finite impurity source Q over a surface area A with a classic Gaussian distribution. FG H x N = erfc N0 2⋅ D⋅t IJ K Eq. 27.1.8 2 x − Qtot ⋅ e 4⋅ D⋅t N= A⋅ π ⋅ D⋅t Eq. 27.1.9 Example 27.1.1 - Find the intrinsic and actual Fermi levels for silicon at 300_°K if the conduction band is 1.12 eV above the valence band. The donor density is 8 x 10-17 cm-3. The effective masses for electrons and holes are 0.5 and 0.85. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 81 Entered Values Computed Results Solution - Since the dopant is a donor, use equations 27.1.6 and 27.1.7 to find compute a solution. Select these equations and press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. To convert the results of Ef and Ei to units of electron volts, highlight each value press ‡/Opts and y/Conv to display the unit menu in the tool bar. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU'EAG8'XA,66A-OPOR  0FZ'AEO@ %QORWVGF4GUWNVU'(AG8CPF'KAG8 Example 27.1.2 - Find the diffusion penetration depth after one hour for phosphorus atoms with a diffusion coefficient of 1.8 x 10-14 cm2/s. The carrier density at the desired depth is 8 x 1017 cm-3 while the surface density is 4 x 1019 cm-3. Entered Values Notice of nsolve routine Calculated Results Solution - Equation 21.1.8 is needed to compute the solution for this problem. Select it by highlighting and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The nsolve routine is used since x is an input for the user defined function erfc (see Chapter 15: Introduction to Equations for more information about nsolve and user-defined functions). The computed results are shown in the screen displays above. -PQYP8CTKCDNGU&'AEO@U0'AEO@0'AEO@ VAU %QORWVGF4GUWNVUZ'Aµ 27.2 PN Junctions These equations describe the properties of PN junctions. They can be classified in two four distinct categories. The first equation calculates the built-in voltage Vbi for a step junction in terms of temperature TT, the doping densities Nd and Na, and the intrinsic density ni(TT). EE PRO for TI-89, 92 Plus Equations - Solid State Devices 82 Vbi = F I GH b g JK k ⋅ TT Nd ⋅ Na ⋅ ln 2 q ni TT Eq. 27.2.1 Equations 27.2.2 - 27.2.4 compute the depletion layer widths xn and xp in the p and the n regions of the junction in terms of the dielectric constant εs, doping densities Nd and Na, built-in voltage Vbi and the applied voltage Va; xd is the total depletion region width for a given applied voltage. 2 ⋅ εs ⋅ ε 0 ⋅ Vbi − Va ⋅ Na q ⋅ Nd ⋅ Na + Nd xn = b Eq. 27.2.2 g Nd ⋅ xn Na Eq. 27.2.3 xd = xn + xp Eq. 27.2.4 xp = Equation 27.2.5 calculates the capacitance Cj of a PN junction in terms of εs, junction area Aj and xd. Cj = εs ⋅ ε 0 ⋅ Aj xd Eq. 27.2.5 The last two equations (27.2.6 and 27.2.7) calculate the built-in voltage Vbi and depletion layer width xd for a linearly-graded junction with a gradient parameter aLGJ. Vbi = FG H 2 ⋅ k ⋅ TT aLGJ ⋅ xd ⋅ ln 2 ⋅ ni (TT ) q F 12 ⋅ εs ⋅ ε 0 ⋅ Vbi − Va IJ xd = G K H q ⋅ aLGJ IJ K Eq. 27.2.6 1 3 Eq. 27.2.7 Example 27.2.1 - A PN step junction is characterized by an acceptor doping density of 6 x 1016 cm-3 and a donor doping density of 9 x 1017 cm-3. The junction area is 100 µm2 at room temperature. For an applied voltage of -5 V, find the built-in potential and junction capacitance. Use a value of 11.8 for the relative permittivity of silicon. Results: Upper Half Results: Lower Half Solution - Use the first five equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU#LAµ@εU0C'AEO@0F'AEO@ EE PRO for TI-89, 92 Plus Equations - Solid State Devices 83 66A-8CA8 %QORWVGF4GUWNVU%L'A(8DKA8ZF'AO ZP'AOZR'AO Example 27.2.2 - A linearly graded junction has an area of 100 µ2, a built-in voltage of 0.8578 V, and an applied voltage of -5.V. The relative permittivity of silicon is 11.8. Under room temperature conditions, what is the junction capacitance, depletion layer width, and the linear-graded junction parameter? Entered Values Calculated Results Solution - Use equations 27.2.5-7 to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU8CA88DKA8#LAµ@εU66A°- %QORWVGF4GUWNVUC.),'AO@%L'A(ZFAµ 27.3 PN Junction Currents These equations characterize the relationships for computing currents in PN junctions. They can be classified into four categories. The first three equations define the junction currents. First, the junction current I is expressed in terms of the junction area Aj, diffusion coefficients Dn and Dp, diffusion lengths LLn and Lp, equilibrium densities of minority carriers npo and pno, applied bias Va, and temperature TT. The second equation is a simplified form the first equation where the current I0 is defined as the multiplier of the exponential term. In this form, it is often called Shockley equation. The third equation calculates this saturation current I0 in terms of the junction area Aj, diffusion coefficients Dn and Dp, diffusion lengths LLn and Lp, equilibrium densities of minority carriers npo and pno. It is used to simplify the first equation. FG Dn ⋅ npo + Dp ⋅ pnoIJ ⋅ FG e Lp H LLn KH F − 1IJ I = I 0⋅Ge H K F Dn ⋅ npo + Dp ⋅ pnoIJ I 0 = q ⋅ Aj ⋅ G Lp H LLn K I = q ⋅ Aj ⋅ q ⋅Va k ⋅TT EE PRO for TI-89, 92 Plus Equations - Solid State Devices 84 q ⋅Va k ⋅TT IJ K −1 Eq. 27.3.1 Eq. 27.3.2 Eq. 27.3.3 The so-called Generation-Recombination current IRG0 at 0 bias is calculated by the fourth equation in terms of Aj, average recombination time τo, intrinsic density ni, depletion width xd. The fifth equation shows that applying an external voltage Va, the generation recombination current IRG increases exponentially. IRG 0 = bg −q ⋅ Aj ⋅ ni TT ⋅ xd 2 ⋅ τo bg Eq. 27.3.4 IJ K FG H q ⋅Va q ⋅ Aj ⋅ ni TT ⋅ xd 2⋅k ⋅TT ⋅e −1 IRG = 2 ⋅ τo Eq. 27.3.5 The small signal conducatnce of the junction is defined as Go, and is computed in terms of temperature TT and currents I and I0. The last two equations compute the charge storage time ts when the diode current is switched externally from IIf to Ir. It is seen from these two equations that ts depends strongly upon the minority carrier lifetime τ p. q ⋅ I + I0 k ⋅ TT IIf ts = τp ⋅ ln 1 + Ir b Go = FG H 1 = erf g Eq. 27.3.6 IJ K Eq. 27.3.7 F ts I GH τp JK Eq. 27.3.8 Ir 1+ IIf Example 27.3.1 - A PN Junction is characterized as having a junction area of 100 µm2, an applied voltage of 0.5 V, and diffusion coefficients for electrons and holes of 35 cm2/s and 10 cm2/s, respectively. The diffusion lengths for electrons and holes are 25 µm and 15 µm. The minority carrier densities are 5 x 106 cm-3 (electrons) and 25 cm-3 (holes). Find the junction current and the saturation current for room temperature conditions. Entered Values Calculated Results Solution - Use the equations 27.3.1 and 27.3.2 or 27.3.1 and 27.3.3 to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU#LAµ@&PAEO@U&RAEO@U.NPAµ.RAµ PRQ'AEO@RPQAEO@66A°-8CA8 %QORWVGF4GUWNVU+A#+'A# Example 27.3.2 -Find the generation-recombination current at room temperature for a pn junction biased at 0.85 V, a junction area of 10 µm2, a depletion layer width of 0.5 µm, a carrier life time of 1.5 x 10-9 s. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 85 Entered Values Calculated Results Solution - Use the fifth equation to solve this problem. Select this by highlighting the equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU#LA µ^τAPU66A-8CZFAµ %QORWVGF4GUWNVU+4)A# 27.4 Transistor Currents The seven equations describe top-level relationships between the emitter, base and collector currents IE, IB and IC, respectively. The first equation defines the common base current gain α as the ratio of IC to IE. The second equation defines the common emitter current gain β in terms of α. IC IE α β= 1− α α= Eq. 27.4.1 Eq. 27.4.2 The third equation represents Kirchoff’s current law for the bipolar junction transistor. IE = IB + IC Eq. 27.4.3 The next three equations represent alternate forms of the collector current in terms of α, IE, IB, β and the leakage currents ICE0, and ICB0. IC = α ⋅ IE + ICB 0 α ICB 0 ⋅ IB + IC = 1− α 1− α Eq. 27.4.4 Eq. 27.4.5 IC = β ⋅ IB + ICE 0 Eq. 27.4.6 The final equation links ICE0 and ICB0 in terms of β. bg ICE 0 = ICB 0 ⋅ β + 1 EE PRO for TI-89, 92 Plus Equations - Solid State Devices Eq. 27.4.7 86 Example 27.4 - A junction transistor has the following parameters: α is 0.98, the base current is 1.2 µA while ICBO is 1.8 pA. Find the β, emitter and collector currents. Entered Values Calculated Results Solution - A few different choices are available, however the results might differ slightly due to the combination of equations used. The second, third and fourth equations can be used to solve this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUα+$Aµ#+%$AR# %QORWVGF4GUWNVUβ+%A#+'A# 27.5 Ebers-Moll Equations These ten equations show a collection of relevant relationships developed by J. J. Ebers and J. L. Moll in the mid-1950s recognizing the reciprocal behavior of bipolar function transistors. The first three equations connect the emitter, collector and base currents IE, IC and IB in terms of forward and reverse current gain αf and αr and the forward and reverse currents IIf and Ir. IE = IIf − αr ⋅ Ir Eq. 27.5.1 IC = αf ⋅ IIf − Ir Eq. 27.5.2 b g b g IB = 1 − αf ⋅ IIf + 1 − αr ⋅ Ir Eq. 27.5.3 The corresponding common emitter current gains in the forward and reverse directions are given by βf and βr, in terms of αf and αr in the fourth and fifth equations. αf 1 − αf αr βr = 1 − αr βf = Eq. 27.5.4 Eq. 27.5.5 The reciprocity relationships between αf and αr, IIf and Ir, and the saturation current Is are defined by the next two equations. The recognition of this reciprocity relationship has been the basis of computing switching characteristics of a transistor. αf ⋅ IIf = Is EE PRO for TI-89, 92 Plus Equations - Solid State Devices Eq. 27.5.6 87 αr ⋅ Ir = Is Eq. 27.5.7 The last three equations define ICE0 and ICB0 in terms of αf , αr, βf and Ir0. b g ICB 0 = 1 − αr ⋅ αf ⋅ Ir 0 b Eq. 27.5.8 g ICE 0 = ICB 0 ⋅ βf + 1 ICE 0 = b Ir 0 ⋅ 1 − αf ⋅ αr 1 − αf Eq. 27.5.9 g Eq. 27.5.10 Example 27.5.1 - A junction transistor has a forward and reverse α of 0.98 and 0.10 respectively. The collector current is 10.8 mA while the forward current is 12.5 mA. respectively. Compute the base, saturation and reverse currents, in addition to the forward and the reverse β. Display (Upper-half) Display (Lower-half) Solution - The second through sixth equations are needed to solve this problem. Select these using the highlight bar and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUαHαT+%AO#++HAO# %QORWVGF4GUWNVUβHβT+$µ#+TA#+UA# 27.6 Ideal Currents - pnp The four equations in this set form the basis of transistor action resulting in emitter, base and collector currents in a pnp transistor. The first three equations show the emitter, collector and base currents IE, IC, and IB in terms of emitter base area A1, diffusion coefficients DE, DB, and DC, the minority carrier densities nE, pB, and nC, emitter and collector diffusion lengths LE and LC, base width WB, emitter-base and collector base voltages VEB and VCB, base collection junction A2 and temperature TT. The last equation shows the relationship between α, DB, pB, WB, DE, nE and LE. The corresponding equations for an npn transistor can be derived from this equation set by proper use of sign conventions. FG DE ⋅ nE + DB ⋅ pB IJ ⋅ FG e − 1IJ − q ⋅ A2 ⋅ DB ⋅ pBFG e H LE WB K H K WB H I q ⋅ A1⋅ DB ⋅ pB F F DC ⋅ nnC + DB ⋅ pB IJ ⋅ FG e ⋅Ge − 1J − q ⋅ A2 ⋅ G IC = H LC WB K H WB K H q ⋅VEB k ⋅TT IE = q ⋅ A1⋅ q ⋅VCB k ⋅TT q ⋅VEB k ⋅TT EE PRO for TI-89, 92 Plus Equations - Solid State Devices 88 IJ K −1 q ⋅VCB k ⋅TT Eq. 27.6.1 IJ K −1 Eq. 27.6.2 IB = FG H IJ K FG H IJ K q ⋅VBE q ⋅VCB q ⋅ A2 ⋅ DC q ⋅ A1⋅ DE ⋅ nE ⋅ e k ⋅TT − 1 + ⋅ nnC e k ⋅TT − 1 LC LE DB ⋅ pB WB α= DB ⋅ pB DE ⋅ nE + WB LE Eq. 27.6.3 Eq. 27.6.4 Example 27.6 - Find the emitter current gain α for a transistor with the following properties: base width of 0.75 µm, base diffusion coefficient of 35 cm2/s, emitter diffusion coefficient of 12 cm2/s, and emitter diffusion length of 0.35 µm. The emitter electron density is 30,000 cm-3 and the base density is 500,000 cm-3. Entered Values Calculated Results Solution - Use the last equation to compute the solution for this problem. Select this by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU&$AEO@U&'EO@U.'AµP'AEO@ R$AEO@9$Aµ %QORWVGF4GUWNVUα 27.7 Switching Transients These six equations compute the key relationships in determining switching response times of bipolar transistors. The work of Ebers and Moll was supplemented by Gummel to model transistor behavior in several different ways. The concept of charge in the base of the transistor controlling switching times became an important contribution to switching theory. The first equation expresses the base charge at the edge of saturation Qsat in terms of the collector saturation current ICsat and the base transit time τt. The collector saturation is determined (approximately) by the second equation in terms of supply voltage VCC and load resistance Rl. Qsat = ICsat ⋅ τt Vcc ICsat = Rl Eq. 27.7.1 Eq. 27.7.2 The third equation calculates the turn-on transient time tr in terms of base recombination time τB, ICsat, base current IB and base transit time τt. The fourth equation computes the storage delay tsd1 when the bipolar transistor is switched from the saturation region to cutoff by changing the base current from IB to 0. The penultimate equation shows the storage delay tsd2 when the base current is switched from IB to -IB. The final equation computes the so called saturation voltage VCEs, the voltage drop between the collector and the emitter under full saturation, in terms of the collector and base currents IC and IB and the forward and reverse α’s αf and αr. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 89 F I GG 1 JJ tr = τB ⋅ ln ⋅ GH 1 − ICsatτBτt JK IB ⋅ Eq. 27.7.3 FG IB ⋅τB IJ H ICsat ⋅τt K F I G JJ 2 ⋅ IB ⋅ τB tsd 2 = τB ⋅ lnG IB ⋅ GH ICsat ⋅ τt ⋅ FGH1 + ICsatτBτt IJK JK ⋅ I F JJ GG IC ⋅ b1 − αr k ⋅ TT G GG1 + F IB IC g I JJJ ⋅ ln VCEs = q GG αr ⋅ G1 − IB ⋅ b1 − αf gJ JJ JJ J αf GH GGH KK tsd 1 = τB ⋅ ln Eq. 27.7.4 Eq. 27.7.5 Eq. 27.7.6 Example 27.7 - Find the saturation voltage for a switching transistor at room temperature when a base current of 5.1 mA is used to control a collector current of 20 mA. The forward and reverse α's are 0.99 and 0.1 respectively. Entered Values Calculated Results Solution - Use the last equation to solve this problem. Select the equation by highlighting and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGUαHαT+$AO#+%AO#66A%QORWVGF4GUWNVU8%'UA8 EE PRO for TI-89, 92 Plus Equations - Solid State Devices 90 27.8 MOS Transistor I The seven equations in this section form the basic equations of charge, capacitance and threshold voltage for a MOS transistor. The first equation shows the Fermi potential φF defined in terms of temperature TT, the intrinsic carrier density ni, and the hole density p. φF = FG b gIJ HK ni TT k ⋅ TT ⋅ ln q p Eq. 27.8.1 The second equation shows the depletion layer xd at the surface of a p-type semiconductor in terms of the relative dielectric constant εs, Fermi potential φF, and doping density Na. xd = b 2 ⋅ εs ⋅ ε 0 ⋅ 2 ⋅ φF q ⋅ Na g Eq. 27.8.2 The third equation computes the charge density Qb0 accumulated at the surface of the semiconductor due to band bending at a substrate bias of 0_V. The fourth equation shows how surface charge density Qb is influenced by the substrate bias VSB. Qb0 = − 2 ⋅ q ⋅ Na ⋅ εs ⋅ ε 0 ⋅ 2 ⋅ φF Eq. 27.8.3 Qb = − 2 ⋅ q ⋅ Na ⋅ εs ⋅ ε 0 ⋅ −2 ⋅ φF + VSB Eq. 27.8.4 A thin oxide layer with a thickness tox on the surface of the semiconductor results in a capacitance Cox per unit area in the fifth equation. Cox = εox ⋅ ε 0 tox Eq. 27.8.5 The sixth equation defines the body coefficient γ in terms of Cox, Na, and εs. The final equation computes the threshold voltage for a MOS system with a work function potential of φGC and residual oxide charge density Qox. γ= 1 ⋅ 2 ⋅ q ⋅ Na ⋅ εs ⋅ ε 0 Cox VT 0 = φGC − 2 ⋅ φF − Eq. 27.8.6 Qb0 Qox − Cox Cox Eq. 27.8.7 Example 27.8 - A p-type silicon with a doping level of 5 x 1015 cm-3 has an oxide thickness of 0.01 µm and oxide charge density of 1.8 x 10-10_C/cm^2. A -5 V bias is applied to the substrate which has a Fermi potential of 0.35 V. Assume the relative permittivity of silicon and silicon dioxide is 11.8 and 3.9, respectively, and the work function is 0.2 V. Display (Upper-half) EE PRO for TI-89, 92 Plus Equations - Solid State Devices Display (Lower-half) 91 Solution - Use the second through last equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUεQZεU0C'AEO@φ(A8φ)%8 3QZ'A%EO@VQZAµ85$A8 %QORWVGF4GUWNVU%QZA(O@γA√83DA%O@ 3DA%O@86A8ZFAµ 27.9 MOS Transistor II These equations describe the performance characteristics of a MOS transistor. The first two equations give two alternate forms for the process constant kn1 in terms of electron mobility µn, oxide capacitance per unit area Cox, relative oxide permittivity εox, and oxide thickness tox. The third equation links the process constant kn1 to the device constant kn, device length L, and width W. kn1 = µn ⋅ Cox µn ⋅ εox ⋅ ε 0 kn1 = tox W kn = kn1⋅ L Eq. 27.9.1 Eq. 27.9.2 Eq. 27.9.3 The fourth equation defines IDmod the drain current, when the transistor is operating under saturation, in terms of kn, gate voltage VGS, threshold voltage VT, modulation parameter λ, and drain voltage VDS. The basic physics behind the increase in drain current comes from the channel widths being non-uniform under the gate because f a finite potential difference between the source and the drain terminals. IDmod = kn 2 ⋅ VGS − VT ⋅ 1 + λ ⋅VDS 2 b gb g Eq. 27.9.4 The fifth equation computes the drain current ID under linear or saturated conditions in terms of kn, VGS, VT, and VDS. R kn ⋅ c2 ⋅ bVGS − VT g ⋅VDS − VDS h,VGS − VT ≤ VDS U | Eq. 27.9.5 |2 ID = S V | | kn ⋅ bVGS − VT g , else | |2 W T 2 2 The expression for the threshold voltage VT is defined in terms of zero substrate bias threshold voltage VTO, body coefficient γ, substrate bias VSB, and Fermi potential φF. VT = VT 0 + γ ⋅ e −2 ⋅ φF + VSB − 2 ⋅ φF j Eq. 27.9.6 The last four equations calculate performance parameters transconductance gm, transit time through the channel Ttr, maximum frequency of operation ffmax, and drain conductance gd. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 92 The last four equations calculate performance parameters transconductance gm, transit time through the channel Ttr, maximum frequency of operation ffmax, and drain conductance gd. b g b g gm = kn ⋅ VGS − VT 42 ⋅L 3 Ttr = µn ⋅ VGS − VT Eq. 27.9.7 Eq. 27.9.8 gm 2 ⋅ π ⋅ Cox ⋅ W ⋅ L gd = kn ⋅ VGS − VT ff max = b Eq. 27.9.9 g Eq. 27.9.10 Example 27.9 - An nMOS transistor has a 6 µ width and 1.25µ gate length. The electron mobility is 500 cm2/V/s. The gate oxide thickness is 0.01µm. The oxide permittivity is 3.9. The zero bias threshold voltage is 0.75 V. The bias factor is 1.1 V1/2. The drain and gate voltages are 5 V, and the substrate bias voltage is -5 V. Assuming that λ is 0.05 V-1 and φF is 0.35 V, find all the relevant performance parameters. Upper Display Middle Display Lower Display Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUεQZγA√8λA8.µµPAEO@8Uφ(A8 VQZAµ8&5A88)5A885$A886A89Aµ %QORWVGF4GUWNVU%QZA(O@HHOCZ'A*\IFAUKGOGPU IOAUKGOGPU+&A#+&OQFA#MPA#8@ MPA#8@6VT'AU86A8 27.10 MOS Inverter (Resistive Load) This section lists the design equations for a MOS inverter with a resistive load. The first equation specifies the device constant kD for the driver transistor in terms of its gate capacitance Cox, mobility µn, width WD, and channel length LD. kD = µn ⋅ Cox ⋅ WD LD Eq. 27.10.1 The second equation specifies the output high voltage VOH when the input to the driver VDD is below the threshold voltage VT. The third equation determines VOL, the output low voltage, Vo when the input is driven high. This equation is a quadratic in VOL, and the solution is meaningful for positive values of VOL. The next equation computes VIH in the linear region of the drain current equation. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 93 VOH = VDD Eq. 27.10.2 FG 1 + VDD − VT IJ ⋅VOL + 2VDD = 0 H kD ⋅ Rl K kD ⋅ Rl kD bVDD − Vog ⋅ c2 ⋅ bVIH − VT g ⋅Vo − Vo h = VOL2 − 2 ⋅ 2 2 Rl Eq. 27.10.3 Eq. 27.10.4 The final equation computes the midpoint voltage VM for which the driver transistor is in saturation. kD ⋅ VM − VT 2 b − g = bVDDRl VM g 2 Eq. 27.10.5 Example 27.10.1 - Find the driver device constant, output and mid-point voltages for a MOS inverter driving a 100_kΩ resistive load. Driver properties include a 3 µm wide gate, a length of 0.8 µm, Cox of 345313 pF/cm2. The electron mobility is 500 cm2/V/s, VIH=2.8 V, VT = 1_V and VDD = 5_V. Solution 1: Upper Display Lower Display Solution 2: Upper Display EE PRO for TI-89, 92 Plus Equations - Solid State Devices Lower Display 94 Solution 3: Upper Display Lower Display Solution 4: Upper Display Lower Display Solution 5: Upper Display Lower Display Solution 6: Upper Display Lower Display Solution 7: Upper Display EE PRO for TI-89, 92 Plus Equations - Solid State Devices Lower Display 95 Solution 8: Upper Display Lower Display Solution - Use all of the equations to solve for the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. Eight complete solutions are generated, in this case, requiring the user to select an answer most relavent to the situation (ie: physical significance, realistic values for components, currents, etc). Mathematically setting up and solving the problem takes some time. Anticipate about 20-30 seconds for EE¹Pro to display a dialogue box prompting the user to select a solution set to be displayed. Select the number of a solution to be viewed (ie: 1, 2, 3...etc.) and press ¸ twice. To view another solution, press „ to resolve the problem and select the number of another available solution. The computed results for all four solutions are shown in the screen displays above. In this example VOH, VOL, VIL, Vo and VM have to be positive and between 0 and VDD. -PQYP8CTKCDNGU%QZAR(EO@.&Aµµ0AEO@8U4AMΩ8&&A8 8+*A886A89&Aµ %QORWVGF4GUWNVU 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 5QNWVKQPM&A#8@8/A881*A881.A88QA8 27.11 MOS Inverter (Saturated Load) The features of a MOS inverter with a saturated enhancement transistor load are described in this section. The first two equations define the device constants for the load transistor (kL, WL, LL) and the driver transistor (kD, WD, LD) in terms of the process parameters, namely mobility µn and gate capacitance per unit area Cox. The third equation defines the geometry ratio KR of the load and drive transistors. kL = µn ⋅ Cox ⋅ WL LL Eq. 27.11.1 kD = µn ⋅ Cox ⋅ WD LD Eq. 27.11.2 KR = kD kL Eq. 27.11.3 EE PRO for TI-89, 92 Plus Equations - Solid State Devices 96 The output high voltage VOH is calculated in terms of the drain supply voltage VDD, the threshold voltage at zero bias VT0, the fermi potential φF and the body coefficient γ in the fourth equation. The fifth equation defines the input voltage Vin in terms of the ratio KR between the load kL and drive kD MOS constants, VDD, the threshold of the load and drive transistors VTL and VTD. The sixth equation defines the threshold voltage of the load transistor, VTL. VOH = VDD − VT 0 + γ ⋅ e cb e bVOH + 2 ⋅ φF g − g hb 2 ⋅ φF KR ⋅ 2 ⋅ Vin − VTD ⋅Vo − Vo 2 = VDD − Vo − VTL VTL = VT 0 + γ ⋅ d Vo + 2 ⋅ φF − 2 ⋅ φF i g 2 jj Eq. 27.11.4 Eq. 27.11.5 Eq. 27.11.6 The equation that follows computes the input high voltage VIH in terms of VDD, VTL, KR and VT0. The eighth equation computes the output voltage Vo in terms of VDD, VTL, VT0 and KR. The last five equations show the performance parameters of the inverter circuit. VIH = Vo = b 2 ⋅ VDD − VTL 3 ⋅ KR + 1 g + VT 0 dVDD − VTL + VT 0 + VT 0 ⋅ Eq. 27.11.7 KR i 1 + KR Eq. 27.11.8 The equation for gmL defines the transconductance of the load circuit while the equation for τL defines the characteristic time to charge the load capacitance CL. b gmL = kL ⋅ VDD − VTL τL = g Eq. 27.11.9 CL gmL Eq. 27.11.10 The charging time tch is the time required for the output rise to move from Vo to V1. The final two equations focus on the characteristic time τD and discharge time tdis for the circuit. tch = τL ⋅ τD = FG V 1 − 1IJ H Vo K Eq. 27.11.11 CL kD ⋅ V 1 − VT 0 b g F 2 ⋅VTD + lnF 2 ⋅ bV 1 − VTDg − 1I I tdis = τD ⋅ G JK JK H V 1 − VTD GH Vo Eq. 27.11.12 Eq. 27.11.13 Example 27.11 - A MOS Inverter with a saturated MOS transistor as its load. The driver has a length of 1_µ and a width of 6µ while the load has a length of 3_µ and a width of 6_µ. The Fermi level for the substrate material is 0.35_V, a zero-bias threshold of 1.00 V. Assume a drain supply voltage of 5_V, and EE PRO for TI-89, 92 Plus Equations - Solid State Devices 97 an output voltage of 3.0_ V find the output high voltage, the input high voltage, and the threshold of the load device. Assume a input voltage of 2.5_V. Upper Display Middle Display Lower Display Solution - Use Equations 27.11.1-27.11.4 and 27.11.6-17.11.7 to get a complete solution to the problem on hand. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. It takes time to solve this problem. -PQYP8CTKCDNGU%QZAR(EO@µPAEO@ 8 U ..Aµ.&Aµ9.Aµ 9&AµγA√8φ(A88&&A88QA886A8 %QORWVGF4GUWNVUM.A#8@M&A#8@-4 8+*A881*A886.A8 27.12 MOS Inverter (Depletion Load) This section lists the design equations for a MOS inverter with a depletion load. The first two equations compute device constants kL and kD for the load and the driver transistors in terms of their geometries WD, WL, LD, and LL. µn ⋅ Cox ⋅ WL LL µn ⋅ Cox ⋅ WD kD = LD kL = Eq. 27.12.1 Eq. 27.12.2 At the output low and output high, VOL and VOH, the driver is in the linear region while the load device is saturated. The next equation finds the threshold voltage VTL for the load device in terms of its zero bias threshold VTL0, Fermi potential φF, and body coefficient γ. kD kL ⋅ 2 VOH − VT 0 ⋅VOL − VOL2 = ⋅VTL2 2 2 Eq. 27.12.3 VTL = VTL0 + γ ⋅ Eq. 27.12.4 cb EE PRO for TI-89, 92 Plus Equations - Solid State Devices g d Vo + 2 ⋅ φF − h 2 ⋅ φF 98 i The charging time tch for the CL is defined next. The current in the depletion load I0 is given by the last equation. tch = CL ⋅VL I0 Eq. 27.12.5 I 0 = kL ⋅VTL2 Eq. 27.12.6 Example 27.12 - A MOS inverter with a depletion mode transistor as the load has a driver transistor 5_µ wide and 1_µ long while the load is a depletion mode device with a 0 bias threshold of -4_V, 3_µ long and 3_µ wide. Given an electron mobility of 500_cm^2/(V*s) and a depletion threshold of -4_V; for the load device, compute VOH and VTL when the output voltage is 2.5_V. Assume VOL to be .4_V and .5 for γ. Solution 1: Upper Display Solution 1: Lower Display Solution 2: Upper Display Solution 2: Lower Display Solution - The problem can be solved when equations 27.12.1 - 27.12.4 are selected. Highlight each of these and press ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. Two complete solutions are computed for this case and are shown in the screen displays above. -PQYP8CTKCDNGU%QZAR(EO@γA8@.&Aµ..Aµ µPAEO@ 8 U φ(A881*A88QA886A8 86.A89&Aµ9.Aµ %QORWVGF4GUWNVU 5QNWVKQPM&A#8@M.A#8@81.A886.A8 5QNWVKQPM&A#8@M.A#8@81.A886.A8 27.13 CMOS Transistor Pair These five equations describe the properties of a CMOS inverter. The first two equations compute device parameters for n and p channel devices. The second pair of equations compute input voltages VIH and VIL. The last equation computes Vin when the n-channel driver is in saturation and the p-channel device is in the linear region. EE PRO for TI-89, 92 Plus Equations - Solid State Devices 99 kP = µp ⋅ Cox ⋅ WP lP Eq. 27.13.1 kN = µn ⋅ Cox ⋅ WN lNN Eq. 27.13.2 kP ⋅ VDD − VTP VIH = 2 ⋅Vo + VTN + kN kP 1+ kN c h FG 2 ⋅Vo − VDD − VTP + kN ⋅VTN IJ H K kP VIL = kN 1+ kP kN ⋅ Vin − VTN 2 b g 2 = {VDD > VTP } Eq. 27.13.3 {VDD ≤ VTP } Eq. 27.13.4 kP ⋅ VDD − Vin − VTP 2 c h 2 Eq. 27.13.5 Example 27.13 - Find the transistor constants for an N and P MOS transistor pair given: N transistor: WN=4 µm, lNN=2. µm, µn=1250 cm2/V/s, Cox=34530 pF/cm2, VTN=1 V P transistor: VTP= -1V, Wp=10 µm, µp=200 cm2/V/s, lP=2µm Upper Display Lower Display Solution - The solution can be calculated by selecting the first four equations. Select these equations by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen display above. -PQYP8CTKCDNGU90AµN00AµµPAEO@8U%QZAR(EO@ 860A8862A8N2Aµ8&&A8µRAEO@8U92Aµ %QORWVGF4GUWNVUM0A#8@M2A#8@8+.A88QA8 27.14 Junction FET These five equations describe the characteristics of a symmetrical junction field effect transistor. The first equation stipulates the drain current ID in terms of the electron mobility µn, doping density Nd, channel width b, channel length L, channel depth Z, supply voltage VDD, pinch-off voltage Vp, gate voltage VG, and built-in voltage Vbi. In the next equation, the channel height b is related to the dielectric EE PRO for TI-89, 92 Plus Equations - Solid State Devices 100 constant εs, Nd, Vbi, and the drain saturation voltage VDsat. The last two equations display the relationship for drain voltage and drain current upon saturation. F GGH F IJ − FGVbi −VGIJ IJIJ b g GHFGH K H Vbi −Vp K KJK FFVDD+Vbi −VGI −FVbi −VGI II 2⋅ q ⋅ Z ⋅ µn⋅ Nd ⋅b F 2 IDsat = ⋅ GVDsat − ⋅ bVbi −Vpg⋅ GG GH 3 L HH Vbi −Vp JK GH Vbi −Vp JK JKJJK 2 R U 2ε 0 ⋅ εs whenlVG < Vpqand SVDsat < (Vbi − Vp) V b= ⋅ 1Vbi + VDsat − VG 6 3 T W q ⋅ Nd 2 U R VDsat = VG − Vp whenlVG > Vpqand SVDsat > (Vbi − Vp) V 3 W T F VG IJ IDsat = ID0 ⋅ G 1 − H Vp K 2 ⋅ q ⋅ Z ⋅ µn ⋅ Nd ⋅ b 2 VDD +Vbi − VG ⋅ VDD − ⋅ Vbi −Vp ⋅ ID = 3 L Vbi −Vp 1.5 15 . 1.5 Eq. 27.14.1 15 . Eq. 27.14.2 Eq. 27.14.3 Eq. 27.14.4 2 Eq. 27.14.5 Example 27.14 - Find the saturation current when the drain current at zero bias is 12.5 µA, the gate voltage is 5 V, and the Pinchoff voltage is 12 V. Entered Values Calculated Results Solution - Use the third equation to solve this problem. Select the equation by highlighting and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGUDAµεU0F'AEO@8DKA88)A8 %QORWVGF4GUWNVU8&UCVA8 EE PRO for TI-89, 92 Plus Equations - Solid State Devices 101 Chapter 28 Linear Amplifiers This section covers linear circuit models (i.e., small signal models) used in making first order calculations using bipolar or junction transistors in amplifier circuits. These circuit models are referred to by many different names such as small signal circuit model, AC circuit model, linear circuit model. In addition, popular device configurations such as the Darlington pair, emitter-coupled pair, differential amplifier, and a source-coupled pair topics have been included. ™BJT (Common Base) ™BJT (Common Emitter) ™BJT (Common Collector) ™FET (Common Gate) ™FET (Common Source) ™FET (Common Drain) ™Darlington (CC-CC) ™Darlington (CC-CE) ™Emitter-Coupled Amplifier ™Differential Amplifier ™Source-Coupled JFET Pair Variables The table lists all variables used in this section along with a description and appropriate units. Variable α0 Ac Ad Ai Aov Av β0 CMRR gm µ rb rrc rd re RBA RCA RDA REA RG Ric Rid EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers Description Current gain, CE Common mode gain Differential mode gain Current gain, CB Overall voltage gain Voltage gain, CC/CD Current gain, CB Common mode reject ratio Transconductance Amplification factor Base resistance Collector resistance Drain resistance Emitter resistance External base resistance External collector resistance External drain resistance External emitter resistance External gate resistance Common mode input resistance Differential input resistance 102 Unit unitless unitless unitless unitless unitless unitless unitless unitless S unitless Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Ω Rin Rl Ro Rs Ω Ω Ω Ω Input resistance Load resistance Output resistance Source resistance 28.1 BJT (Common Base) These six equations represent properties of a transistor amplifier connected in the common base configuration at mid frequencies. The first equation relates the common base current gain α0 with the common emitter current gain β0. The second equation computes the input impedance Rin at the input terminals of the amplifier from the emitter and base resistances, re and rb. The third equation equates the output resistance Ro to the collector resistance rrc. The fourth equation represents the current gain Ai. The fifth equation calculates the voltage gain Av from re, rb, α0, β0, and the load resistance, Rl. The last equation computes the overall voltage gain for the amplifier system Aov from Rin, rrc, re, α0, β0, and the source impedance Rs. α0 1− α0 rb Rin = re + β0 Ro = rrc β0 = Eq. 28.1.1 Eq. 28.1.2 Eq. 28.1.3 Ai = α 0 Av = Eq. 28.1.4 α 0 ⋅ Rl rb re + β0 α 0 ⋅ rrc ⋅ Aov = Eq. 28.1.5 FG Rin IJ H Rin + Rs K Eq. 28.1.6 rb re + β0 Example 28.1 - A common base configuration of a linear amplifier has an emitter resistance of 35Ω, collector and base resistances of 1 MΩ and 1.2 kΩ resistances, respectively. The load resistor is 10 kΩ. If the source resistance is 50 Ω and α0 is 0.93, find β0 and the gains for this amplifier. Upper Screen Display Lower Screen Display Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 103 -PQYP8CTKCDNGUαTDAMΩTTEA/ΩTGAΩ4NAMΩ4UAΩ %QORWVGF4GUWNVU#K#QX#Xβ4KPAΩ 4Q'AΩ 28.2 BJT (Common Emitter) This section contains the equations for an amplifier, at mid frequencies, connected in the common emitter configuration. The first equation displays the current gain α0 in relation to the common current gain β0. The second equation computes the input Rin, in terms of base resistance rb, emitter resistance re, and the current gain β0. The output resistance Ro is defined in terms of collector resistance, rrc in the third equation. The next equation defines current gain Ai in terms of β0. The fifth equation computes the voltage gain Av from β0, the load, source and input resistances Rl, Rs, and Rin. The last equation calculates the overall voltage gain Aov from the source impedance Rs, Rl, Rin and β0. α0 1− α0 Rin = rb + β 0 ⋅ re β0 = Eq. 28.2.1 Eq. 28.2.2 Ro = rrc Eq. 28.2.3 Ai = − β 0 − β 0 ⋅ Rl Av = β 0 ⋅ re + rb − β 0 ⋅ Rl Aov = Rs + Rin Eq. 28.2.4 Eq. 28.2.5 Eq. 28.2.6 Example 28.2 - Using the same inputs as in the previous problem, with the exceptions that the load is a 1 KΩ resistor and the output resistance is 1 MΩ, find the gain parameters. Upper Screen Display Lower Screen Display Solution - All of the equations are needed to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUαTDAMΩTTEA/ΩTGAΩ4NAMΩ4Q'AΩ 4UAΩ %QORWVGF4GUWNVU#K#QX#Xβ4KPAΩ EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 104 28.3 BJT (Common Collector) These six equations describe the properties of a transistor amplifier connected in a common collector configuration at mid frequencies. The first equation couples the common emitter current gain α0 with the common base current gain β0. The second equation computes the input impedance Rin in terms of base resistance rb, emitter resistance re, β0, and load resistance Rl. Ro represents the output resistance in terms of the source resistance, common base current gain β0 and the load resistance Rl. The current gain Ai is shown in the fourth equation in terms of the collector resistance rrc, α0, re, and Rl. The final two equations cover the voltage gain Av and overall voltage gain Aov for the amplifier system. Aov includes the effect of source impedance Rs. β0 = α0 1− α0 Eq. 28.3.1 b g Rin = rb + β 0 ⋅ re + β 0 + 1 ⋅ Rl Ro = re + Eq. 28.3.2 b Rs + rbg Eq. 28.3.3 β0 Ai = rrc rrc ⋅ 1 − α 0 + Rl + re Eq. 28.3.4 Av = α 0 ⋅ Rl re + Rl Eq. 28.3.5 b Aov = g bβ 0 + 1g ⋅ Rl Rs + Rin + bβ 0 + 1g ⋅ Rl Eq. 28.3.6 Example 28.3 - An amplifier in a common collector configuration has a gain α0 of 0.99. The emitter, base and collector resistances are 25 Ω, 1000 kΩ, and 100,000 MΩrespectively. If the source resistance is 25 Ωfind all the mid-band characteristics. Upper Screen Display Lower Screen Display Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUαTDAMΩTTEA/ΩTGAΩ4NAΩ4UAΩ %QORWVGF4GUWNVU#K#QX#Xβ4KP'AΩ4QAΩ EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 105 28.4 FET (Common Gate) The equations in this section focus on an FET amplifier in the common gate configuration. The amplification factor µ is described in the first equation in terms of the transconductance gm and the drain resistance rd. In the second equation the input resistance Rin is described as a function of load resistance Rl, rd and µ. The voltage gain Av is defined by the third equation in terms of Rl, rd and µ. The final equation computes the output resistance Ro in terms of rd, µ, and the external gate resistance RG. µ = gm ⋅ rd Rin = Av = Eq. 28.4.1 b Rl + rd g Eq. 28.4.2 µ +1 bµ + 1g ⋅ Rl Eq. 28.4.3 rd + Rl bg Ro = rd + µ + 1 ⋅ RG Eq. 28.4.4 Example 28.4 - An FET amplifier connected in a common gate mode has a load of 10 kΩ. The external gate resistance is 1 MΩ, and the drain resistance is 125 kΩ. The transconductance is 1.6 x 10-3 siemens. Find the midband parameters. Entered Values Calculated Results Solution - All of the equations need to be used to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUIOAUKGOGPUTFAMΩ4)A/Ω4NAMΩ %QORWVGF4GUWNVU#Xµ4KPAΩ4Q'AΩ 28.5 FET (Common Source) These four equations represent the key properties of an FET amplifier in the mid frequency range. The first equation defines the amplification factor µ in terms of transconductance gm and drain resistance rd. The second equation computes input resistance Rin as a function of load resistance Rl, rd and µ. The voltage gain Av is defined in the third equation. The final equation computes the output resistance, Ro. µ = gm ⋅ rd EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers Eq. 28.5.1 106 b Rl + rd g µ +1 F rd ⋅ Rl IJ Av = − gm ⋅ G H rd + Rl K Rin = Eq. 28.5.2 Eq. 28.5.3 Ro = rd Eq. 28.5.4 Example 28.5 - Find the voltage gain of an FET configured as a common-source based amplifier. The transconcductance is 2.5 x 10-3 siemens, a drain resistance of 18 kΩ and a load resistance of 100 kΩ. Find all the parameters for this amplifier circuit. Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen display above. -PQYP8CTKCDNGUIOAUKGOGPUTFAMΩ4NAMΩ %QORWVGF4GUWNVU#Xµ4KPAΩ4QAΩ 28.6 FET (Common Drain) The first equation defines the amplification factor µ in terms of transconductance gm and drain resistance rd. The second equation computes input resistance Rin as a function of load resistance Rl, rd and µ. Voltage gain Av is defined in the third equation. The final equation computes the output resistance, Ro. µ = gm ⋅ rd Rl + rd Rin = µ +1 µ ⋅ Rl Av = µ + 1 ⋅ Rl + rd rd Ro = µ +1 b Eq. 28.6.1 g Eq. 28.6.2 Eq. 28.6.3 bg Eq. 28.6.4 Example 28.6 - Compute the voltage gain for a common-drain FET amplifier as configured in the previous example. The transconductance is 5 x 10-3 siemens, the drain resistance is 25 kΩ, and the load resistance is 100 kΩ. EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 107 Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen display above. -PQYP8CTKCDNGUIOAUKGOGPUTFAMΩ4NAMΩ %QORWVGF4GUWNVU#Xµ4KPAΩ4QAΩ 28.7 Darlington (CC-CC) The first two equations yield the input and output resistances Rin and Ro, computed in terms of emitter resistance re, load resistance Rl, current gain β0, base resistance rb, and source resistance Rs. The final equation computes overall current gain Ai for the transistor pair in terms of β0, re, Rl and the external base resistance RBA. b gh c cβ 0 ⋅ bre + rbg + Rsh Ro = re + Rin = β 0 ⋅ re + β 0 ⋅ re + Rl Eq. 28.7.1 Eq. 28.7.2 β 02 Ai = β 02 ⋅ RBA RBA + β 0 ⋅ Rl + re b Eq. 28.7.3 g Example 28.7 - Transistors in a Darlington pair having a β0 value of 100 are connected to a load of 10 kΩ. The emitter, base and source resistances are 25 Ω, 1500 kΩ and 1kΩ, respectively. The external base resistance is 27 kΩ. Upper Screen Display Lower Screen Display Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 108 -PQYP8CTKCDNGUβTDAMΩTGAΩ4$#AMΩ4NAMΩ4UAMΩ %QORWVGF4GUWNVU#K4KP'AΩ4QAMΩ 28.8 Darlington (CC-CE) The Darlington configuration connected as a common collector-common emitter configuration is described in this section. The first two equations define the input resistance Rin and output resistance Ro, in terms of base resistance rb, emitter resistance re, collector resistance rrc, and current gain β0. The final equation calculates the voltage gain Av, in terms of the emitter and load resistances, the source impedance, and the current gain β0. Rin = rb + β 0 ⋅ re rrc Ro = β0 − Rl Av = Rs re + 2 β0 Eq. 28.8.1 Eq. 28.8.2 Eq. 28.8.3 Example 28.8 - An amplifier circuit has a base, emitter, and load resistance of 1.5 kΩ, 25 Ω, and 10 kΩ, respectively. The configuration has a value of β0 equal to 100. The source and collector resistances are 1 kΩ and 100 kΩ. Find the voltage gain, input and output resistances. Upper Screen Display Lower Screen Display Solution - Use all of the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUβTDAMΩTTEAMΩTGAΩ4NAMΩ4UAMΩ %QORWVGF4GUWNVU#X4KPAΩ4QAMΩ EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 109 28.9 Emitter-Coupled Amplifier Two classes of emitter-coupled amplifiers are covered in this section. The first equation shows the general relationship between β0 and α0; the current gains under common base and common emitter configurations. The next three equations show the input resistance Rin, output resistance Ro, and voltage gain Av for a common collector-common base method of connection. The last three equations correspond to cascade configuration of the transistors, which is a combination of common emitter-common base configuration resulting in a current gain Ai with corresponding input resistance Rin and output resistance Ro. β0 = α0 1− α0 Av = Rl ⋅ Eq. 28.9.1 FG β 0 IJ H 2 ⋅ β 0 ⋅ re + Rl K Eq. 28.9.2 Ai = −α 0 ⋅ β 0 Eq. 28.9.3 Rin = β 0 ⋅ re + rb Eq. 28.9.4 Ro = rrc Eq. 28.9.5 Example 28.9 - An emitter coupled pair amplifier is constructed from transistors with α0=0.98. The emitter, base and collector resistances are 25 Ω, 2000 Ω, and 56 kΩ, respectively. If the load resistance is 10 kΩ, find the mid-band performance factors. Upper Screen Display Lower Screen Display Solution - Use all of the equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUαTDAΩTTEAMΩTGAΩ4NAMΩ %QORWVGF4GUWNVU#K#Xβ4KPAΩ4QAΩ EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 110 28.10 Differential Amplifier The gain Ad in the differential mode of operation is given by the first equation. The common mode gain, Ac is defined in terms of the external collector and emitter resistances RCA and REA and the emitter resistance re. The last two equations show input resistance for differential and common mode inputs Rid & Ric. 1 Ad = − ⋅ gm ⋅ RCA 2 −α 0 ⋅ RCA Ac = 2 ⋅ REA + re Rid = 2 ⋅ rb + β 0 ⋅ re b Eq. 28.10.1 Eq. 28.10.2 g Eq. 28.10.3 Ric = β 0 ⋅ REA Eq. 28.10.4 Example 28.10 - A differential amplifier pair has a transconductance of 0.005 siemens, α0=0.98, β0=49. The external collector and external emitter resistances are 18 kΩ and 10 kΩ respectively. If the emitter resistance is 25 Ω and the base resistance is 2 kΩ, find the common mode, differential resistance and gains. Upper Screen Display Lower Screen Display Solution - Use all of the equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUαβIOAUKGOGPUTDAMΩTGAMΩ4%#AMΩ 4'#AMΩ %QORWVGF4GUWNVU#E#F4KEAΩ4KFAΩ 28.11 Source-Coupled JFET Pair The first two equations describe the differential Ad and common mode Ac gains for a source-coupled JFET pair in terms of the external drain, drain and source resistances RDA, rd and Rs. The third equation shows the amplification factor µ, in terms of the transconductance gm and the drain resistance. The final equation calculates the common mode rejection ratio CMRR. 1 − ⋅ gm ⋅ rd ⋅ RDA Ad = 2 rd + RDA − µ ⋅ RDA Ac = µ + 1 ⋅ 2 ⋅ Rs + rd + RDA b g Eq. 28.10.1 Eq. 28.10.2 bg EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 111 µ = gm ⋅ rd Eq. 28.10.3 CMRR = gm ⋅ Rs Eq. 28.10.4 Example 28.11 - Find the gain parameters of a source-coupled JFET pair amplifier if the external drain resistance is 25 kΩ, and the source resistance is 100 Ω. The drain resistance is 12 kΩ and the transconductance is 6.8 x 10-3 siemens. Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. The computed results are shown in the screen displays above. -PQYP8CTKCDNGUIOAUKGOGPUTFAMΩ4&#AMΩ4UAΩ %QORWVGF4GUWNVU#E#F%/44µ EE PRO for TI - 89, 92 Plus Equations - Linear Amplifiers 112 Chapter 29 Class A, B and C Amplifiers This chapter covers the section called Class A, B and C Amplifiers. These amplifier circuits forms the basis of a class of power amplifiers used in a variety of applications in the industry. Read this! Note: The equations in this section are grouped under topics which describe general properties of Class A, B, and C amplifiers. Equations for a variety of specific cases and are listed together under a sub-topic heading and are not necessarily a set of consistent equations which can be solved together. Choosing equations in a subtopic w/o regard as to whether the equations represent actual relationships could generate erroneous results or no solution at all. Read the description of each equation set to determine which equations in a sub-topic form a consistent subset before attempting to compute a solution. ™Class A Amplifier ™Power Transistor ™Push-Pull Principle ™Class B Amplifier ™Class C Amplifier Variables The variables used in this section are listed along with a brief description and units. Variable gm hFE hOE I IB IC ∆IC ICBO ICQ Idc Imax K m ς n N1 N2 Pd Pdc Description Transconductance CE current gain CE output conductance Current Base current Collector Current Current swing from operating pt. Collector current EB open Current at operating point DC current Maximum current Constant Constant Efficiency Turns ratio # turns in primary # turns in secondary Power dissipated DC power input to amp EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 113 Unit S unitless S A A A A A A A A unitless 1/K unitless unitless unitless unitless W W Po PP Q θJA R Rl RR0 RR2 RB Rrc Rxt S TA TJ ∆Tj V0 V1 VBE VCC ∆VCE VCEmx VCEmn Vm VPP XXC XC1 XC2 XL Power output Compliance Quality factor Thermal resistance Equivalent resistance Load resistance Internal circuit loss Load resistance External base resistance Coupled load resistance External emitter resistance Instability factor Ambient temperature Junction temperature Change in temperature Voltage across tank circuit Voltage across tuned circuit Base emitter voltage Collector supply voltage Voltage swing from operating pt. Maximum transistor rating Minimum transistor rating Maximum amplitude Peak-peak volts, secondary Tuned circuit parameter π equivalent circuit parameter π equivalent circuit parameter π series reactance W V unitless W/K Ω Ω Ω Ω Ω Ω Ω unitless K K K V V V V V V V V V Ω Ω Ω Ω 29.1 Class A Amplifier The eight equations in this section form the basis for analyzing a Class A amplifier with an ideal transformer coupled to a resistive load Rl. The first equation specifies the equivalent load resistance R from the load resistance Rl in the secondary winding of the transformer with a turns ratio n. The second equation defines the AC current swing ∆IC in terms of the voltage swing ∆VCE and R. The third equation computes the maximum collector current Imax in terms of current at the operating point ICQ and ∆IC. These three equations are internally consistent and can be used as a set. R = n 2 ⋅ Rl ∆VCE ∆IC = R Im ax = ICQ + ∆IC Eq. 29.1.1 Eq. 29.1.2 Eq. 29.1.3 The DC power available Pdc is shown in the fourth equation. The DC power measurement is based on the supply voltage and quiscent operating current. Pdc = VCC ⋅ ICQ EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers Eq. 29.1.4 114 The compliance PP is defined as the full voltage swing across the emitter and collector and is expressed in terms of the minimum and maximum transformer ratings VCEmx and VCEmn. VPP represents the peak to peak voltage in the secondary transformer. The final two equations compute the output power Po and the conversion efficiency ζ. PP = VCEmx − VCEmn Eq. 29.1.5 VPP = n ⋅ PP Eq. 29.1.6 ∆IC 2 ⋅ R 8 Po ζ= Pdc Po = Eq. 29.1.7 Eq. 29.1.8 Example 29.1 - A Class A power amplifier is coupled to a 50 Ω load through the output of a transformer with a turn ratio of 2. The quiescent operating current is 60 mA, and the incremental collector current is 50 mA. The collector-to-admitter voltage swings from 6 V to 12 V. The supply collector voltage is 15 V. Find the power delivered and the efficiency of power conversion. The maximum current is 110 mA. Upper Display Lower Display Solution - Use all of the equations to solve this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU∆+%O#+%3AO#+OCZAO#P4NΩ8%%A8 8%'OZA88%'OPA8 %QORWVGF4GUWNVU∆8%'A8ζ2FEA92QA922A84AΩ 4NAΩ822A8 29.2 Power Transistor Power amplifiers generate heat needing rapid transfer to ambient surroundings. The six equations in this section focus on thermal problems in terms of the junction temperature TJ, transistor currents IB and IC, and the instability factor S. The first equation defines the junction temperature TJ as linearly related to the power dissipation Pd and thermal resistance θJA and TA, the ambient temperature. TJ = TA + θJA ⋅ Pd Eq. 29.2.1 The next two equations focus on the collector current IC and base current IB in terms of the current gain hFE, leakage current ICBO, external emitter resistance Rxt and external base resistance RB. b g IC = hFE ⋅ IB + 1 + hFE ⋅ ICBO EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 115 Eq. 29.2.2 IB = b − IC ⋅ Rxt − VBE Rxt + RB g Eq. 29.2.3 The fourth equation expresses a more exact form for the collector current IC in terms of hFE, Rxt, RB, ICBO, and VBE. In using this equation, care must be taken to ensure that Eq.29.2.3 and Eq. 29.2.4 are not selected at the same time. Such a choice will lead to the inability of the solver engine to perform the computation accurately. IC = b g hFE ⋅ Rxt + RB − hFE ⋅VBE + ⋅ ICBO hFE ⋅ Rxt ⋅ RB hFE ⋅ Rxt + RB Eq. 29.2.4 The instability factor S is given by the fifth equation. Stability is a performance measure for the health of the amplifier. The final equation computes IC in terms of hFE, ICBO, a parameter m, S, and the change in junction temperature ∆Tj. FG1 + RB IJ ⋅ hFE H Rxt K S= Eq. 29.2.5 RB hFE + Rxt b IC = −hFE ⋅ IB + S ⋅ ICBO ⋅ 1 + m ⋅ ∆Tj g Eq. 29.2.6 Example 29.2 - A power transistor has a common emitter current gain of 125. A 750 Ω base resistance is coupled to an external emitter resistance of 10 kΩ. The ambient temperature is 75 °F and the thermal resistance of the unit is 10 °C/W. The power that needs to be dissipated is 12.5W. The base emitter voltage is 1.25V while ICBO is 1_ma. Find the junction temperature, collector current and the instability factor. Input variables Computed results Solution - We note from the equation set that IC is computed in three different ways. To make the calculations consistent given the data, we use Equations 1, 2, 4 and 5 to solve for this problem. Select these equations by highlighting the equation with the cursor bar and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen display above. -PQYP8CTKCDNGUJ('+%$AO#2FA9θ,#AQ%94$AΩ 4ZVAMΩ6#AQ(8$'A8 %QORWVGF4GUWNVU+$Aµ#+%A#56,A- 29.3 Push-Pull Principle These equations introduce the push-pull principle. Two transistors have their collector outputs connected to the center-tapped primary winding of a EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 116 transformer. The secondary winding is connected to a load RR2. The first equation computes an equivalent resistance R based on the maximum current supplied to the load Imax and the collector supply voltage VCC. The power output Po is computed by the second equation in terms of VCC and R. The final equation computes the power Po in terms of the load resistance RR2 and the transformer windings N1 and N2. Care must be exercised in selecting the equations. If you select to solve all the equations, ensure that appropriate inputs are selected. VCC Im ax VCC 2 Po = 2⋅ R 2 N2 ⋅VCC 2 2 ⋅ N1 Po = 2 ⋅ RR2 R= FG H Eq. 29.3.1 Eq. 29.3.2 IJ K Eq. 29.3.3 Example 29.3 - Find the output power for a push-pull circuit with a collector voltage of 15 V, a load resistance of 50 Ω. The push-pull transformer secondary winding amplifies voltage by a factor of 2.5. Input variables Computed results Solution - Use the third equation to compute the solution for this problem. Select the equation using the highlight bar and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGU0044AΩ8%%A8 %QORWVGF4GUWNVU2QA9 29.4 Class B Amplifier Power transistors that are connected in a push-pull mode and biased to cutoff, operate under the Class B condition where alternate half-cycles of input are of forward polarity for each transistor. The nine equations in this section define the characteristic properties of this class of amplifiers. The first equation represents the power output Po at any signal level in terms of the constant K, supply voltage VCC, and an equivalent resistance R. The second equation defines the DC current Idc as the average value of a sinusoidal half-wave adjusted by K. K 2 ⋅VCC 2 2⋅ R 2 ⋅ K ⋅ Im ax Idc = π Po = Eq. 29.4.1 Eq. 29.4.2 The next two equations focus on the DC power Pdc in terms of VCC, K, R, and the maximum current Imax. The power calculations are possible in two ways as shown by these equations. EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 117 2 ⋅ K ⋅ Im ax ⋅VCC π 2 ⋅ K ⋅VCC 2 Pdc = π ⋅R Pdc = Eq. 29.4.3 Eq. 29.4.4 The efficiency of power conversion ζ is given by the fifth and sixth equations. The power dissipated by the circuit Pd is computed in the seventh equation. The eighth equation calculates the voltage V1 across a tuned RLC circuit in terms of the transconductance gm, load resistance Rl, and output conductance hOE. The final equation calculates the average collector current IC for a half-sine wave from gm, hOE, Rl and the amplitude of the voltage Vm. Po Pdc π ⋅K ζ= 4 2 ⋅VCC 2 K 2 ⋅π ⋅ K− Pd = 4 π ⋅R ζ= FG H Eq. 29.4.5 Eq. 29.4.6 IJ K F GG GH 1 gm ⋅ Rl ⋅Vm V1 = hOE ⋅ Rl 2⋅ 2 1+ 2 F GG GH 1 gm ⋅Vm IC = ⋅ hOE ⋅ Rl π 1+ 2 Eq. 29.4.7 I JJ JK Eq. 29.4.8 I JJ JK Eq. 29.4.9 Example 29.4 - A Class B amplifier provides 5 W to an effective load of 50 Ω. The collector voltage is 25 V. If the peak current is 500 mA, find the average DC current and the efficiency of power conversion. Input variables Computed results Solution - Use the first, second, fourth and fifth equations to compute the solution for this problem. Select these by highlighting each equation and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU+OCZAO#2QA94AΩ8%%A8 %QORWVGF4GUWNVU+FEA#-ζ2FEA9 EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 118 29.5 Class C Amplifier These six equations outline the properties of a Class C amplifier. The first equation defines the efficiency of conversion ζ in terms of the current I, the coupled-in load Rrc, and the equivalent internal circuit loss resistance RR0. The next equation computes the tuned circuit parameters which have a capacitive reactance of XXC, which is given in terms of the load voltage V0, quality factor Q, and power Po. XL is expressed in terms of XXC,Q, load resistance Rl, and resistance RR2 in the fourth and sixth equations. The remaining two equations calculate the load harmonic suppression resistance values in the output circuit XC1 and XC2. Remember the equations to compute XL have two distinct forms. If this equation is part of your selection, be advised to ensure that the proper inputs are specified. ζ= I 2 ⋅ Rrc I 2 ⋅ Rrc + RR 0 b Eq. 29.5.1 g V 02 Q ⋅ Po XXC ⋅ Q 2 XL = Q2 + 1 Rl XC1 = − Q XXC = 1 ⋅ Rl + Rl + RR2 Q − RR 2 XC 2 = Q XL = d Eq. 29.5.2 Eq. 29.5.3 Eq. 29.5.4 i Eq. 29.5.5 Eq. 29.5.6 Example 29.5 - A Class C amplifier is supplying a tuned circuit, with a quality factor of 5. If the output voltage is 15 V and the power delivered is 75_W, find the capacitive reactance of the circuit needed in the tank circuit. Input variables Computed results Solution - Use the second equation to compute the solution for this problem. Select this by highlighting the equation with the cursor bar and pressing the ¸ key. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed result is shown in the screen displays above. -PQYP8CTKCDNGU2QA938A8 %QORWVGF4GUWNVU::%AΩ EE PRO for TI - 89, 92 Plus Equations - Class A, B, C Amplifiers 119 Chapter 30 Transformers The contents in this chapter is divided into two topics. ™Ideal Transformer ™Linear Equivalent Circuit Variables All variables used in this section are listed here with a brief description and units. Variable I1 I2 N1 N2 RR1 RR2 Rin Rl V1 V2 XX1 XX2 Xin Xl Zin ZL Description Primary current Secondary current # primary turns # secondary turns Primary resistance Secondary resistance Equiv. primary resistance Resistive part of load Primary voltage Secondary voltage Primary reactance Secondary reactance Equivalent primary reactance Reactive part of load Primary impedance Secondary load Unit A A unitless unitless Ω Ω Ω Ω V V Ω Ω Ω Ω Ω Ω 30.1 Ideal Transformer Four equations describe the properties of an ideal transformer. The first equation relates the primary and secondary voltages V1 and V2 in terms of the primary and secondary turns N1 and N2. The second equation shows the corresponding relationship between the primary and secondary currents I1 and I2. In the same fashion, the third equation relates primary and secondary power. The final equation calculates the effect of a load impedance ZL experienced at the primary winding terminal with a primary impedance Zin EE PRO for TI - 89, 92 Plus Equations - Transformers 121 V 1 N1 = V2 N2 Eq. 30.1.1 I 1⋅ N 1 = I 2 ⋅ N 2 Eq. 30.1.2 V 1⋅ I 1 = V 2 ⋅ I 2 Eq. 30.1.3 F N 1 IJ Zin = G H N 2K Eq. 30.1.4 2 ⋅ ZL Example 30.1 - An ideal transformer has 10 primary turns and 36 secondary turns. The primary side draws 500 mA when subjected to a 110 V input. If the load impedance is 175 Ω, find the input impedance at the primary side of the transformer in addition to the voltage and current on the secondary end. Entered Values Calculated Results Solution - Use all of the equations to solve this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen display above. -PQYP8CTKCDNGU +AO#008A8<.AΩ %QORWVGF4GUWNVU +A#8A8<KPAΩ 30.2 Linear Equivalent Circuit The first two equations define the primary voltage and current V1 and I1 in terms of V2 and I2. The last two equations expand the equivalent resistance Rin and reactance Xin at the primary terminals in terms of the primary winding resistance RR1, secondary winding resistance RR2, load resistance Rl, reactances XX1 and XX2 and load reactance Xl. V1 = FG N 1 IJ ⋅V 2 H N 2K Eq. 30.2.1 I1 = I2⋅ N2 N1 Eq. 30.2.2 F N 1 IJ ⋅ b RR2 + Rlg Rin = RR1 + G H N 2K 2 EE PRO for TI - 89, 92 Plus Equations - Transformers 122 Eq. 30.2.3 F N 1 IJ ⋅ b XX 2 + Xlg Xin = XX 1 + G H N 2K 2 Eq. 30.2.4 Example 30.2 - The transformer in the above problem has a primary and secondary resistance of 18 Ω and 5 Ω, respectively. The corresponding coils have a reactance of 6 and 2.5 Ω. The secondary side is loaded with an impedance of 12.5 kΩ. Find the voltage and current on the secondary side in addition to the equivalent impedance on the primary side. Upper Screen Display Lower Screen Display Solution - All of the equations are used to solve this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU +AO#0044AΩ44AΩ4NAMΩ 8A8::AΩ::AΩ:NAΩ %QORWVGF4GUWNVU +A#4KPAΩ8A8:KPAΩ EE PRO for TI - 89, 92 Plus Equations - Transformers 123 Chapter 31 Motors and Generators This section has thirteen topics covering various aspects of motors and generators. The topics are organized under these headings. Read this! Note: The equations in this section are grouped under topics which describe general properties of semiconductors or devices. Equations for a variety of specific cases and are listed together under a sub-topic heading and are not necessarily a set of consistent equations which can be solved together. Choosing equations in a subtopic w/o regard as to whether the equations represent actual relationships could generate erroneous results or no solution at all. Read the description of each equation set to determine which equations in a sub-topic form a consistent subset before attempting to compute a solution. ™Energy Conversion ™DC Generator ™Separately-Excited DC Generator ™DC Shunt Generator ™DC Series Generator ™Separately-Excited DC Motor ™DC Shunt Motor ™DC Series Motor ™Permanent Magnet Motor ™Induction Motor I ™Induction Motor II ™Single-Phase Induction Motor ™Synchronous Machines Variables All the variables used in this section are listed with a brief description and units. Variable A ap B Ea Ef Ema Es Eta F H Ia IIf IL Ir Description Area # parallel paths Magnetic induction Average emf induced in armature Field voltage Phase voltage Induced voltage Average emf induced per turn Magnetic pressure Magnetic field intensity Armature current Field current Load current Rotor current per phase EE Pro for TI-89, 92 Plus Equations - Motors and Generators 124 Unit m2 unitless T V V V V V Pa A/m A A A A Isb Isf K Kf KM L θ N Ns ρ φ p P Pa Pma Pme Pr RR1 Ra Rd Re Rel Rf Rl Rr Rs Rst s sf sm T Tb Tf Tgmax TL Tloss Tmmax TTmax Ts Va Vf Vfs Vt ωm ωme ωr ωs Wf XL Backward stator current Forward stator current Machine constant Field coefficient Induction motor constant Length of each turn Phase delay Total # armature coils # stator coils Resistivity Flux # poles Power Mechanical power Power in rotor per phase Mechanical power Rotor power per phase Rotor resistance per phase Armature resistance Adjustable resistance Ext. shunt resistance Magnetic reluctance Field coil resistance Load resistance Equivalent rotor resistance Series field resistance Stator resistance Slip Slip for forward flux Maximum slip Internal torque Backward torque Forward torque Breakdown torque Load torque Torque loss Maximum positive torque Pullout torque Shaft torque Applied voltage Field voltage Field voltage Terminal voltage Mechanical radian frequency Electrical radian frequency Electrical rotor speed Electrical stator speed Magnetic energy Inductive reactance EE Pro for TI-89, 92 Plus Equations - Motors and Generators A A unitless A/Wb unitless m r unitless unitless Ω/m Wb unitless W W W W W Ω Ω Ω Ω A/Wb Ω Ω Ω Ω Ω unitless unitless unitless N*m N*m N*m N*m N*m N*m N*m N*m N*m V V V V r/s r/s r/s r/s J Ω 125 31.1 Energy Conversion The four equations in this section describe the fundamental relationship amongst electrical, magnetic and mechanical aspects of a system. For example, the first two equations show two ways of computing energy density Wf stored in a magnetic field. The first equation uses the field intensity H and flux density B in a magnetic region with length L and area A. The second an electric analogy to the magnetic circuit as it uses the magnetic reluctance Rel and flux φ to compute Wf. 1 ⋅ H ⋅ B⋅ L⋅ A 2 1 Wf = ⋅ Re l ⋅ φ 2 2 Wf = Eq. 31.1.1 Eq. 31.1.2 The third equation defines the mechanical pressure F due to the flux density B. F= B2 2 ⋅ µ0 Eq. 31.1.3 The last equation shows the r.m.s. value of the emf Es induced by Ns turns moving with an angular velocity ωs sweeping a magnetic flux of φ. Es = Ns ⋅ ωs ⋅ φ 2 Eq. 31.1.4 Example 31.1 - A conductor having a length of 15 cm and a cross sectional area of 0.5 cm2 is subjected to a magnetic induction of 1.8 T and a field intensity of 2.8 A/m. The magnetic reluctance is 0.46 A/Wb. The conductor has 32 turns and is moving at a rotational speed of 62 rad/s. Find the magnetic flux, the magnetic energy, the induced electric field and the mechanical pressure on the coil. 1st Solution: Upper Half 1st Solution: Lower Half 2nd Solution: Upper Half 2nd Solution: Lower Half Solution - All of the equations are needed to solve this problem. Press „ to display the input screen, enter all the known variables and press „ to compute the solution. Since the flux is a squared term in the second equation, there are two equal and opposite results calculated for φ and Es. -PQYP8CTKCDNGU %QORWVGF4GUWNVU EE Pro for TI-89, 92 Plus Equations - Motors and Generators #AEO@$A6*A#O.AEO4GNA#9D 0UωUATU 'UA8 QTA8 ('A2C φA9D QTA9D 9HA, 126 31.2 DC Generator The first equation describes the relation between electrical radian frequency ωme, the mechanical radian frequency ωm, and the number of poles in the generator p. The next equation expresses the emf generated per turn Eta with the relative motion of the coil with respect to the magnetic field φ. p ⋅ ωm 2 p Eta = ⋅ ωm ⋅ φ π ωme = Eq. 31.2.1 Eq. 31.2.2 The next two equations illustrate two ways to express the induced armature emf Ea as a function of number of armature coils N, the number of parallel paths ap, number of poles p, the mechanical radian frequency ωm, a machine constant K, and flux φ. The machine constant K, is seen to be dependent purely on the characteristics of the machine. Np ⋅ ⋅ ωm ⋅ φ ap π Ea = K ⋅ ωm ⋅ φ N⋅p K= ap ⋅ π Ea = Eq. 31.2.3 Eq. 31.2.4 Eq. 31.2.5 The sixth equation shows the conversion of mechanical energy available as torque T and mechanical angular velocity ωm to its electrical counterpart – namely, the emf and current in the armature Ea, and Ia and the voltage and current in the field windings Ef and If. The next equation for torque connects T with K, φ, and the current Ia. T ⋅ ωm = Ea ⋅ Ia + Ef ⋅ IIf T = K ⋅ φ ⋅ Ia Eq. 31.2.6 Eq. 31.2.7 The armature resistance is given by the equation for Ra in terms of N, ap, coil length L, area A and its resistivity ρ. Ra = ρ⋅ N ⋅ L ap 2 ⋅ A Eq. 31.2.8 Vf represents the voltage across the field winding carrying a current IIf and a resistance Rf. The terminal voltage Vt represents the induced voltage minus the IR drop in the armature.. Vf = Rf ⋅ IIf Vt = K ⋅ ωm ⋅ φ − Ra ⋅ Ia Eq. 31.2.9 Eq. 31.2.10 The final equation represents the shaft torque Ts needed to generate the induced emf, assuming a given value for equivalent loss of torque Tloss Ts = K ⋅ φ ⋅ Ia + Tloss Eq. 31.2.11 Example 31.2 - A six-pole DC generator rotates at a mechanical speed of 31 rad/s. The armature sweeps across a flux of 0.65 Wb. There are eight parallel paths and 64 coils in the armature. The armature current is 12 A. The field is supplied by a 25 V source delivering a current of 0.69 A. Find the torque and the voltages generated in the armature. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 127 Display: Upper Half Display: Lower Half Solution - Choose the first six equations. Select these by highlighting each equation and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU CR'HA8+CA#++HAO#0φA9DR ωOATU %QORWVGF4GUWNVU 'CA8'VCA8-6A0O ωOGATU 31.3 Separately-Excited DC Generator The equations in this section describe the properties of a separately excited DC generator. The first equation computes the field current IIf in terms of field voltage Vfs, external shunt resistance re, and field coil resistance Rf. The next equation evaluates armature induced voltage Ea as a function of machine constant K, mechanical radian frequency ωm, and flux φ. Vfs Re+ Rf Ea = K ⋅ ωm ⋅ φ IIf = Eq. 31.3.1 Eq. 31.3.2 The third and fourth equations are alternate forms of expressing terminal voltage Vt in terms of load current IL, load resistance Rl, armature resistance Ra. Vt = IL ⋅ Rl Eq. 31.3.3 Vt = Ea − Ra ⋅ IL Eq. 31.3.4 Finally the armature current IL in terms of K, φ, ωm, Ra and Rl. IL = K ⋅ φ ⋅ ωm Ra + Rl Eq. 31.3.5 Example 31.3 - A DC generator with a machine constant of 3.8 is driving a load of 46 kΩ and rotates at a speed of 31 rad/s. The magnetic flux is 1.6 Wb. The field is driven by a 24 V source. The field coil resistance is 10 Ω. The armature resistance is 13 Ω in series with an external resistance of 55 Ω. Find the field current, armature induced voltage and the terminal voltage. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 128 Display: Upper Half Display: Lower Half Solution - Use all the equations to compute the solution for this problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU -φA9D4CAΩ4GAΩ4HAΩ4NAMΩ 8HUA8ωOATU %QORWVGF4GUWNVU 'CA8++HAO#+.A#8VA8 31.4 DC Shunt Generator The first equation in this section expresses the induced armature voltage Ea in terms of the machine constant K, the mechanical angular frequency ωm, and flux φ. Ea = K ⋅ ωm ⋅ φ Eq. 31.4.1 The second equation defines terminal voltage Vt in terms of the field current IIf, external resistance Re, and field coil resistance Rf. The third equation computes Vt in terms of load current IL and load resistance Rl. The fourth equation expresses Vt as the induced emf Ea minus armature IR drop. Vt = ( Re + Rf ) ⋅ IIf Eq. 31.4.2 Vt = IL ⋅ Rl Eq. 31.4.3 Vt = Ea − Ra ⋅ Ia Eq. 31.4.4 The armature current Ia is the sum of the load current IL and field current IIf. Ia = IL + IIf Eq. 31.4.5 The final equation is an alternate form of expression for Ea. Ea = Ra ⋅ Ia + (Re + Rf ) ⋅ IIf Eq. 31.4.6 Example 31.4 - Find the machine constant of a shunt generator running at 31 rad/s and producing 125 V with a 1.8 Wb flux. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 129 Entered Values Computed Results Solution - Use the first equation to solve this problem. Select this by pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGU 'CA8φA9D, ωOATU %QORWVGF4GUWNV - 31.5 DC Series Generator The two equations in this section describe the properties of a series DC generator. The first equation specifies the field current and the armature current to be the same. The second equation computes the terminal voltage Vt in terms of the induced emf Ea, load current IL, armature resistance Ra, and series field windings Rs. Ia = IIf Eq. 31.5.1 Vt = Ea − ( Ra + Rs) ⋅ IL Eq. 31.5.2 Example 31.5 - Find the terminal voltage of a series generator with an armature resistance of 0.068 Ω and a series resistance of 0.40 Ω. The generator delivers a 15 A load current from a generated voltage of 17 V. Entered Values Computed Results Solution - Use the second equation to solve this problem. Select this with the highlight bar and press ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGU 'CA8+.A#4CAΩ4UAΩ %QORWVGF4GUWNVU 8VA8 EE Pro for TI-89, 92 Plus Equations - Motors and Generators 130 31.6 Separately-Excited DC Motor These equations form the working foundation for a separately excited motor. The first equation calculates the field voltage Vf in terms of the field current IIf and field coil resistance Rf. Vf = Rf ⋅ IIf Eq. 31.6.1 The second equation computes the terminal voltage Vt in terms of the machine constant K, magnetic flux φ, mechanical radian frequency ωm, armature current Ia, and armature resistance Ra. Vt = K ⋅ φ ⋅ ωm + Ra ⋅ Ia Eq. 31.6.2 The load torque TL, in the third equation is defined in terms of K, φ, Ia and Tloss. TL = K ⋅ φ ⋅ Ia − Tloss Eq. 31.6.3 Ea, the back emf induced in the rotor, is calculated by the next equation. Torque T links with K, φ, and Ia. Ea = K ⋅ ωm ⋅ φ Eq. 31.6.4 T = K ⋅ Ia ⋅ φ Eq. 31.6.5 The reciprocal power relationship between ωm and φ by the inverse quadratic relationship. The next set of equations show the relationship between T, the torque lost due to friction Tloss and the torque load TL. The last equation in this set shows relationship of power with torque T and angular velocity ωm. ωm = Vt Ra ⋅ T − 2 K ⋅φ K ⋅φ Eq. 31.6.6 bg T = Tloss + TL Eq. 31.6.7 P = T ⋅ ωm Eq. 31.6.8 Example 31.6 - Find the terminal voltage, field current and machine constant for a motor with an armature current 0.5 A and resistance of 100 Ω rotating at an angular velocity of 31 r/s. The back emf is 29 V. The field is driven by a 15 V source driving a 50 Ωload. The flux available in the armature is 2.4 Wb. Display: Upper Half Display: Lower Half Solution - Solve the first, second, fourth and fifth equations. Select these by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. The computed results are shown in the screen display above. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 131 -PQYP8CTKCDNGU 'CA8+CA#φA9D4CAΩ4HAΩ8HA8 ωOATU %QORWVGF4GUWNVU ++HAO#-6A0O8VA8 31.7 DC Shunt Motor These seven equations describe the principal characteristics of a DC shunt motor. The first equation expresses the terminal voltage Vt in terms of the field current IIf and field resistance Rf along with the external field resistance Re. The second equation defines the terminal voltage Vt in terms of the back emf (expressed in terms of the machine constant K, flux swept φ, and angular velocity ωm) and the IR drop in the armature circuit. b g Vt = Re + Rf ⋅ IIf Eq. 31.7.1 Vt = K ⋅ φ ⋅ ωm + Ra ⋅ Ia Eq. 31.7.2 The third equation refers to the torque available at the load TL due to the current Ia in the armature minus the loss of torque Tloss due to friction and other reasons. TL = K ⋅ φ ⋅ Ia − Tloss Eq. 31.7.3 The fourth equation gives the definitive relationship between the back emf Ea, K, φ and ωm. Ea = K ⋅ ωm ⋅ φ Eq. 31.7.4 The next equation displays the reciprocal quadratic relationship between ωm, Vt, K, φ, armature resistance Ra, adjustable resistance Rd and T. ωm = b g Ra + Rd ⋅ T Vt − 2 K ⋅φ K ⋅φ Eq. 31.7.5 bg The last two equations compute torque T in terms of Tloss, load torque TL, flux φ, Ia, and K. T = Tloss + TL Eq. 31.7.6 T = K ⋅ φ ⋅ Ia Eq. 31.7.7 Example 31.7 - Find the back emf for a motor with a machine constant of 2.1, rotating at 62 rad/s in a flux of 2.4 Wb. Entered Values EE Pro for TI-89, 92 Plus Equations - Motors and Generators Calculated Results 132 Solution - Use the fourth equation to solve this problem. Select the equation with the cursor bar and press ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed result is shown in the screen display above. -PQYP8CTKCDNGU -φA9DωOATU %QORWVGF4GUWNV 'CA8 31.8 DC Series Motor These eight equations describe the performance characteristics of a series DC motor. The first equation links the terminal voltage Vt to the back emf (Ea defined by the third equation) and the IR drop through the armature due to armature resistance Ra, adjustable resistance Rd, and series resistance Rs. The second equation calculates the load torque TL with the machine constant K, flux φ, load current IL, and the torque loss Tloss. b g Vt = K ⋅ φ ⋅ ωm + Ra + Rs + Rd ⋅ IL Eq. 31.8.1 TL = K ⋅ φ ⋅ IL − Tloss Eq. 31.8.2 The third equation defines the back emf in the armature Ea in terms of K, φ, and mechanical frequency ωm. The fourth equation shows torque generated at the rotor due the magnetic flux φ and current IL. Ea = K ⋅ ωm ⋅ φ Eq. 31.8.3 T = K ⋅ φ ⋅ IL Eq. 31.8.4 The next equation shows a reciprocal quadratic link between ωm, Vt, K, φ, Ra, Rs, Rd, and torque T. ωm = b g Ra + Rs + Rd ⋅ T Vt − 2 K ⋅φ K ⋅φ bg Eq. 31.8.5 The sixth equation computes the torque generated T as the sum of load torque TL and lost torque Tloss. The last two equations show the connection between K, φ, a field constant Kf, load current IL, and torque T. T = Tloss + TL Eq. 31.8.6 K ⋅ φ = Kf ⋅ IL Eq. 31.8.7 T = Kf ⋅ IL2 Eq. 31.8.8 Example 31.8 - A series motor, with a machine constant of 2.4, rotating at 62 rad/s, is supplied with a terminal voltage of 110 V and produces a torque of 3 Nm. The armature resistance is 10 Ω, the series resistance is 5 Ω, and the adjustable resistance is 0.001 Ω. Find the average voltage induced in the armature, the flux, and the load current. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 133 Solution 1: Upper Display Solution 1: Lower Display Solution 2: Upper Display Solution 2: Lower Display Solution - The first, third and fifth equations are needed to compute a solution. Select these by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. There are two possible solutions for this example. Type the number of the solution set to be viewed and press ¸ twice. To view another solution set, press „ to and select another number. The computed results are shown in the screen displays above. -PQYP8CTKCDNGU -4CΩ4FAΩ4UAΩ6A0O8VA8 ωOATU %QORWVGF4GUWNVU 'CA8 A8 +.A# A#  φA9D A9D 31.9 Permanent Magnet Motor These five equations characterize the basic features of a permanent magnet motor. The first equation shows the back emf Ea in terms of machine constant K, flux φ, and radian velocity ωm. The second equation shows the connection between generated torque T, K, φ and armature current Ia. The terminal voltage Vt is the sum of back emf Ea and the inductive-resistance drop in the armature. The fourth equation shows conservation of various torques T, TL and Tloss. The final equation displays the quadratic relationship of ωm in terms of K, Vt, φ, T and Ra. Ea = K ⋅ φ ⋅ ωm Eq. 31.9.1 T = K ⋅ φ ⋅ Ia Eq. 31.9.2 Vt = Ea + Ra ⋅ Ia Eq. 31.9.3 T = Tloss + TL Eq. 31.9.4 EE Pro for TI-89, 92 Plus Equations - Motors and Generators 134 ωm = Vt Ra ⋅ T − 2 K ⋅φ K ⋅φ Eq. 31.9.5 bg Example 31.9 – Find the machine constant for a permanent motor rotating at 62.5 rad/s in a magnetic flux field of 1.26 Wb. Assume a 110 V back emf. Entered Values Calculated Results Solution - The first equation is needed to compute the solution. Select it by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. -PQYP8CTKCDNGU 'CA8φA9DωOATU %QORWVGF4GUWNV - 31.10 Induction Motor I These eleven equations define the relationships amongst key variables used in evaluating the performance of an induction motor. The first equation expresses the relationship between the radian frequency induced in the rotor ωr, the angular speed of the rotating magnetic field of the stator ωs, the number of poles p, and the mechanical angular speed ωm. ωr = ωs − p ⋅ ωm 2 Eq. 31.10.1 The second, third and fourth equations describe the slip s using ωr and ωs, ωm, p, the induced rotor power per phase Pr, and the power transferred to the rotor per phase Pma. s = 1− p ωm ⋅ 2 ωs Eq. 31.10.2 Pr =s Pma Eq. 31.10.3 ωr = s ⋅ ωs Eq. 31.10.4 Pma is defined in the fifth equation in terms of the rotor current Ir and the rotor phase voltage Ema. Pma = 3 ⋅ Ir ⋅ Ema Eq. 31.10.5 The sixth and seventh equations account for the mechanical power Pme in terms of p, ωm, ωs, Pma, and torque T. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 135 Pme = 3 ⋅ p ωm ⋅ ⋅ Pma 2 ωs Eq. 31.10.6 Pme = T ⋅ ωm Eq. 31.10.7 The eighth equation expresses torque in terms of p, Pma, and ωs. T = 3⋅ p Pma ⋅ 2 ωs Eq. 31.10.8 The last three equations show an equivalent circuit representation of induction motor action and links the power Pa with rotor resistance Rr, rotor current Ir, slip s, rotor resistance per phase RR1 and the machine constant KM. Pma = Rr ⋅ Ir 2 + 1− s ⋅ Rr ⋅ Ir 2 s Eq. 31.10.9 Pa = 1− s ⋅ Rr ⋅ Ir 2 s Eq. 31.10.10 Rr = RR1 KM 2 Eq. 31.10.11 Example 31.10 – Find the mechanical power for an induction motor with a slip of 0.95, a rotor current of 75 A , and a resistance of 1.8 Ω. Entered Values Calculated Results Solution - Choose the next to last equation to compute the solution. Select by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. -PQYP8CTKCDNGU +TA#4TAΩU %QORWVGF4GUWNV 2CA9 31.11 Induction Motor II These equations are used to perform equivalent circuit analysis of an induction motor. The first equation shows the power in the rotor per phase Pma, defined in terms of the rotor current Ir, rotor resistance Rr, and slip s. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 136 Pma = Rr 2 ⋅ Ir s Eq. 31.11.1 The second equation shows the expression for torque T in terms of poles p, Pma and radian frequency of the induced voltage in the stator ωs. The third equation is an alternate representation of torque in terms of the applied voltage Va, stator resistance Rst, Rr, inductive reactance XL, and ωs. T= 3 Pma ⋅ p⋅ 2 ωs T= 3 p Rr ⋅⋅⋅ 2 ωs s Eq. 31.11.2 Va 2 FG Rst + Rr IJ H sK Eq. 31.11.3 2 + XL 2 The equation for Tmmax represents the maximum positive torque available at the rotor, given the parameters of the induction motor stator resistance Rst, XL, Va, p, and ωs. 3p Va 2 ⋅⋅ 4 ωs Rst 2 + XL2 + Rst Tm max = Eq. 31.11.4 The maximum slip sm in the fifth equation represents the condition when dT/ds=0. sm = Rr Eq. 31.11.5 Rs + XL2 2 The sixth equation defines the so-called breakdown torque Tgmax of the motor. The final equation relates Rr with machine constant KM and the rotor resistance per phase RR1. 3p Va 2 Tg max = − ⋅ ⋅ 4 ωs Rs2 + XL2 − Rst Rr = RR1 KM 2 Eq. 31.11.6 Eq. 31.11.7 Example 31.11 – An applied voltage of 125 V is applied to an eight pole motor rotating at 245 rad/s . The stator resistance and reactance is 8 and 12 Ω respectively. Find the maximum torque. Entered Values Calculated Results Solution - Use the fourth equation to compute the solution. Select by moving the cursor bar, highlighting, and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 137 -PQYP8CTKCDNGU R4UVAΩ8CA8ωUATU:.AΩ %QORWVGF4GUWNV 6OOCZA0O 31.12 Single-Phase Induction Motor These three equations describe the properties of a single-phase induction motor. The first equation defines the slip for forward flux sf with respect to the forward rotating flux φ, the radian frequency of induced current in the stator ωs, the number of poles p, and the angular mechanical speed of the rotor ωm. The final two equations represent the forward and backward torques Tf and Tb for the system with respect to sf, the number of poles p, the electrical stator speed ωs, the equivalent rotor resistance Rr and the currents Isf and Isb. The forward torque is given by the power dissipated in the fictitious rotor resistor. p ωm ⋅ 2 ωs p 1 Isf 2 ⋅ Rr Tf = ⋅ ⋅ 2 ωs 2 ⋅ sf sf = 1 − Tb = − p 1 Isb 2 ⋅ Rr ⋅⋅ 2 ωs 2 ⋅ 2 − sf b Eq. 31.12.1 Eq. 31.12.2 Eq. 31.12.3 g Example 31.12 – Find the forward slip for an eight pole induction motor with a stator frequency of 245 rad/s, and a mechanical radian frequency of 62.5 rad/s. Entered Values Calculated Results Solution - The first equation is needed to compute the solution. Select by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. -PQYP8CTKCDNGU RωOATUωU ATU %QORWVGF4GUWNVU UH 31.13 Synchronous Machines These five equations focus on the basic properties of a synchronous machine. The first equation relates the radian mechanical and electrical speeds ωm and ωs with the number of poles p. The second equation shows maximum torque TTmax (sometimes referred to as pull out torque) in terms of current IIf, applied voltage Va, p, and ωs. Pma represents the power produced in the load with a phase delay of θ. The last two equations show torque relationships with mechanical power Pme, ωm, Pma, and ωs. EE Pro for TI-89, 92 Plus Equations - Motors and Generators 138 ωm = 2 ⋅ ωs p TT max = 3 ⋅ Eq. 31.13.1 p IIf ⋅Va ⋅ 2 ωs Eq. 31.13.2 bg Pma = Va ⋅ Ia ⋅ cos θ T= Eq. 31.13.3 Pme ωm T = 3⋅ Eq. 31.13.4 p Pma 2 ωs Eq. 31.13.5 Example 31.13 – Find the stator radian frequency and the maximum torque for a synchronous machine with a mechanical rotational velocity of 31 rad/s. The motor has eight poles, a field current of 1.8 A, and experiences an applied voltage of 130 V. Entered Values Calculated Results Solution - The first and second equations are needed to compute the solution. Select these using the cursor bar and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equations. -PQYP8CTKCDNGU ++HA#R8CA8ωOATU %QORWVGF4GUWNVU 66OCZA0OωUATU EE Pro for TI-89, 92 Plus Equations - Motors and Generators 139 Chapter 32 Introduction to Reference This chapter guides the user through the Reference Part of EE•Pro. The information in the Reference section of the software is organized in a similar fashion as Analysis and Equations, except the information is generally noninteractive. 32.1 Introduction The Reference Part is organized in nine sections that include topics the following topics: Resistor Color Chart, Standard (or preferred) Component Values, Semiconductor Properties, Boolean Expressions, Boolean Algebra Properties, Fourier, Laplace and z transforms properties, commonly used Fundamental Constants, SI prefixes, and the Greek Alphabet. Pull down menu on TI - 89 and TI - 92 Plus Unlike Analysis and Equations, the screen formats for the topics in the Reference section can differ significantly, depending on the information presented. Some Reference tables, such as the Resistor Color Chart and Standard Component Values, are dynamically interactive and perform calculations. Many sections include pictures for more clarification. 32.2 Finding Reference The Reference Part is accessed by starting EE•Pro. 1. Start EE•Pro: • TI 89 and TI 92 Plus: Press O key to display the pull down menu. Use D key to move the high-light bar to EE•Pro and press ¸. Alternatively, type in [A] when the pull down menu appears. EE Pro for TI - 89, 92 Plus Reference - Introduction to Reference Section 1 • Pressing † accesses the menu for the Reference section listing the topics. EE•Pro is structured with a hierarchy of screens for choosing a specific topic. Select a topic of Reference by moving the highlight bar to the desired section using the D key and pressing 2. ¸. Alternatively, type in the number corresponding to the section desired. For example, press ª to access Semiconductor Data, or press { to access the Transforms section. Semiconductor Data Menu Transforms Menu 32.3 Reference Screens The semiconductor data section has been chosen to illustrate how to navigate within a topic of the Reference section. • When accessing the Semiconductor Data section, a dialog box appears listing the available topics. Use D key to move the highlight bar to 3-5 and 2-6 Compounds and press ¸. This displays the electronic, and physical properties of the compound gallium phosphide, GaP. Sections in Semiconductor Data Properties of GaP Materials available in 3-5, 2-6 Compounds section Properties of CdS • • Note that GaP has an arrow ! to its right indicating that there are other materials whose properties are also listed. Move the highlight bar to GaP, press ¸ to view the other materials. The list of other materials includes GaSb, InAs, InP, InSb, CdS, CdSe, CdTe, ZnS, ZnSe, ZnTe. To display the properties of CdS, use the D key to move the high light bar to CdS and press ¸. Alternatively, type in { when the pull down menu appears. The data displayed automatically updates to list the properties of CdS as shown above. EE Pro for TI - 89, 92 Plus Reference - Introduction to Reference Section 2 • Other properties available in the section include Donor and Acceptor levels in Silicon shown in the screen displays below. Silicon Donor Level Silicon Acceptor Levels The last topic in Semiconductor Properties includes the colors observed for Silicon dioxide and Silicon Nitride thickness. These colors are arranged in order of thickness (µm). Some colors appear multiple times due to multiple diffraction orders. SiO2/Si3N4 Color Choices Properties of SiO2/Si3N4 (thickness in µm) 32.4 Using Reference Tables The Transform section is used as an example of viewing reference tables. Transforms allows the user to inspect Fourier, Laplace or z-transforms. Each of these topics are divided into three sub-topics, Definitions, Properties and Transform Pairs. For example, navigate from Transforms → Fourier Transforms → Transform Properties. A screen display below shows the lists the equations displaying the fundamental properties of Fourier transform pairs. Note that the name of the selected transform equation appears in the status line. For example, if the highlight bar is moved to the sixth equation, the status line displays "Rectangular Pulse" as the description of the property. Inverse View (press †) Normal View To view the equation in Pretty Print format, press B. The contents on the right side of the colon (:) are displayed in Pretty Print, while the contents to the left of the colon are displayed in regular type above the status line. To reverse this display (display the inverse of the property), press N to exit Pretty Print mode, press † to display the inverse form of the transform property, and B to view the inverse transform in PrettyPrint. EE Pro for TI - 89, 92 Plus Reference - Introduction to Reference Section 3 Regular View of Laplace Transform EE Pro for TI - 89, 92 Plus Reference - Introduction to Reference Section Inverse View of Laplace Transform 4 Chapter 33 Resistor Color Chart This section of EE•Pro allows the user to enter the color sequence of a physical resistor and compute its value and tolerance. Most physical resistors come with a band of colors to help identify its value. There are 3 variations of color bands used in practice: 3, 4 or 5 band of colors. The table below identifies the hierarchy used in practice. A picture of the color chart is included in the software and is displayed when the function key † is pressed. Table 33-1 Description of Colors in resistors Band positions represent: Band 1 Band 2 Band 3 Band 4 3-Band digit digit multiplier N/A 4-Band digit digit multiplier tolerance 5-Band digit digit digit multiplier 4-Band digit digit multiplier tolerance N/A 5-Band digit digit digit multiplier tolerance MULTIPLIER ÷E2 ÷E1 xE0 xE1 xE2 xE3 xE4 xE5 TOLERANCE 10% 5% 1% 2% 0.5% Band positions represent: Band 1 Band 2 Band 3 Band 4 Band 5 3-Band digit digit multiplier N/A N/A Colors represent: SILVER GOLD BLACK BROWN RED ORANGE YELLOW GREEN EE Pro for TI - 89, 92 Plus Reference - Resistor Color Chart DIGIT 0 1 2 3 4 5 5 BLUE VIOLET GREY WHITE 6 7 8 9 xE6 xE7 xE8 xE9 0.25% 0.1% 0.05% - Field Description Input Fields Num. of Bands: (Number of Bands) Band 1: (see table above) Band 2: (see table above) Band 3: (see table above) Band 4: (see table above) Band 5: (see table above) Pressing ¸ displays choice of 3, 4, or 5 bands. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Output Fields Value: (Resistor Value) Tolerance: (Resistor Tolerance) Returns a Resistor Value. Returns a percent. 33.1 Using the Resistor Color Chart Select this topic from the Reference section and press ¸. 1. 2. Select the number of bands on the resistor (display automatically updates for the entry). In each Band field, select a color using B and pressing ¸. 3. The results are displayed in the Value and Tolerance lines of the display. Example 33.1- Find the value and tolerance of a resistor with band colors yellow, black, red and gold. Using the steps outlined above. 1. 2. 3. 4. 5. 6. Enter 4 for Num. of Bands. For the 1st Digit Band, select Yellow color. For the 2nd Digit Band, select Black color. For the Multiplier Band, select Red color. For Tolerance Band, select Gold. The results show 4000_Ω in the Value field, and ± 5% in the Tolerance field as shown to the right. EE Pro for TI - 89, 92 Plus Reference - Resistor Color Chart 6 Chapter 34 Standard Component Values In this section, the software computes the inverse of the color chart computation described in the previous chapter (i.e.: given a value and a tolerance for a resistor, a color sequence is generated). As a side benefit the calculation algorithm also allows the user to estimate suitable “off-the-shelf” standard components for needed resistors, inductors and capacitors. These are also referred to as "Preferred Values" of components available from manufacturers. Field Descriptions Input Field Value: (Desired Value or Design Spec.) Tolerance: (Tolerance of Component) Enter a real number. Press ¸ to display selection. The values range from tolerance ±20% down to ±.0.05%. Press ¸ to display component choices: Resistor, Inductor, or Capacitor. Component: (Type of Component) Output Field Value: (Closest Standard Value to the Desired Value) Bands: (Resistor Color Bands - if the Component is a Resistor ) Returns a "Preferred Value". Returns the color bands in a resistor. Example 34.1 - A design calculation yields 2.6 microfarads for a capacitor. Find the closest preferred value with a tolerance of 1%. Use the following directions: 1. Press the 3 setting and set Display Digits to FLOAT 8, press ¸. 2. 3. In the Value field, enter 2.6E-6. In the Tolerance field, press ¸ to display choices; use D key 4. to move the highlight bar ±1 % and press ¸. In the Component field, press ¸ to display; use D key to 5. move the highlight bar to "Capacitor" and press ¸ to select. The Standard Component Value is displayed as 2.61E-6_F. EE Pro for TI - 89, 92 Plus Reference - Standard Component Values 7 Chapter 35 Semiconductor Data Physical, chemical, electrical, electronic and mechanical properties of common semiconductors are presented in this section. The information is organized under five (5) topics listed in detail in Table 35-1. All properties are listed at 300 °K, unless otherwise specifically stated. Details of how to access the information is included in the table. Table 35-1 Semiconductor Data Section Label Data Fields Semiconductors Input Field: Semiconductor: Atoms:(Atoms) At Wt:(Atomic Weight) Br Fld: (Breakdown Field) xtal:(Crystal Structure) ρ: (Density) εr: (Relative permittivity) Nc: (Density of states, CB) Nv: (Density of states, VB) mle: (Longitudinal e- mass) mte: (Transverse e-mass) mlh:(Light hole mass) mhh:(Heavy hole mass) φ: (Electron affinity) EG:(Band gap) ni: (Intrinsic Density) a: (Lattice constant) αth:(Thermal expansion coefficient) MP:(Melting Point) τ: (Carrier lifetime) µn: (Electron mobility) µp: (Hole Mobility) Raman E: (Raman Photon Energy) Sp Ht: (Specific heat) Th. Cond: (Thermal conductivity) Diff Cons: (Diffusion Constant) Vapor Pr: (Vapor Pressure 1600°C) Vapor Pr: (Vapor Pressure 930°C) Work Fn: (Work Function) EE Pro for TI-89, 92 Plus Reference - Semiconductor data 8 Description Press ¸ to display options (Si or GaAs), use D to move the highlighter to Si or GaAs and press ¸ to select 1/cm3 g/mol V/cm structure name g/cm3 unitless 1/cm3 1/cm3 unitless # # # V eV 1/cm3 nm 1/°K °C s 2 cm /(V*s) cm2/(V*s) eV J/g W/(cm*°K) 2 cm /s torr torr V III-V, II-VI Compounds GaP ! Press ¸ to display pull down menu listing the III-V and II-VI semi-conducting compounds. They are: 1:GaP 2:GaSb 3:InAs 4:InP 5:InSb 6:CdS 7:CdSe 8:CdTeœ 9:ZnS A:ZnSe B:ZnTe Input Field: Compound: To select a compound, use the D key to move to highlight bar to the item and press ¸ (or enter the item number). Si Donor Levels EE Pro for TI-89, 92 Plus Reference - Semiconductor data EG: (Energy Gap) µn: (Electron mobility) µp: (Hole mobility) mn: (Electron Effective mass) mp: (Hole Effective mass) a: (Lattice constant) MP: (Melting point) εr: (Dielectric constant) ρ: (Density) Output Field: Silicon Donor levels are displayed relative to the conduction band. The donor list includes the following: Li: Sb: P: As: Bi: Te: Ti: C: Se: Se: Cr: Ta: Ta: Cs: Ba: S: Mn: Mn(VB): 9 eV 2 m /(V*s) m2/(V*s) unitless unitless nm °C unitless 3 g/cm All values are displayed in electron volts (eV) relative to the conduction band. In some cases there is a (VB) designation next to the element. In these cases, the location of the donor level is referenced relative to the valence band. For example, the donor level for Te is displayed as 0.14_eV indicating that the donor level is 0.14 eV below the conduction band. On the other hand Gold has (VB) appended. Thus the value 0.29_eV displayed reflects that the donor level of Gold (i.e., Au) is 0.29 eV above the valence band. Some elements have multiple energy levels and therefore appear more than once. Acceptor Levels EE Pro for TI-89, 92 Plus Reference - Semiconductor data Ag(VB): Pt(VB): Si(VB): Si(VB): Na(VB): Au(VB): V: Mo: Mo(VB): Mo(VB): Hg(VB): Hg(VB): Sr: Sr(VB): Ge: Ge(VB): K: K(VB): Sn: W: W: W: W(VB): W(VB): Pb: O: O: Fe: Fe: Fe(VB): Output Field: Silicon Acceptor levels are displayed relative to the valence band. The acceptor list includes the following: Mg(CB): Mg(CB): Cs: Ba: S: Mn: Ag(CB): Cd(CB): Cd(CB): Cd: Pt(CB): Pt: Si: B: Al: Ga: In: Tl: 10 All values in are in electron volts (eV) relative to the valence band. For some elements, there is a (CB) appended to the element name. In these cases, the location of the acceptor level is referenced relative to the conduction band. For example, the acceptor level for Mg is displayed as 0.11_eV. (CB) follows the element name Mg indicating that the acceptor level is 0.11_eV below the conduction band. On the other hand, Platinum (Pt) has a value of 0.36 eV, with no additional information. This is an indicator that Pt produces an acceptor level 0.36 eV above the valence band. Some elements have multiple levels SiO2/Si3N4 Colors Pd: Be: Be: Zn(CB): Zn: Au(CB): Co(CB): Co: Co: V: Ni(CB): Hg(CB): Hg(CB): Cu: Cu: Cu: Sn: Pb: O(CB): O: Input Field: Compound: and appear more than once. Silicon ! Press ¸ to display the pull down menu listing the color of silicon with SiO2 or Si3N4 deposited on the surface. Once the specific color choice has been made, the display fields show the information for the following parameters: SiO2: Value in µm Si3N4: Value in µm Order: Value as a number 1:Silicon 2:Brown 3:Golden brown 4:Red 5:Deep Blue 6:Blue 7:Pale Blue 8:Very Pale Blueœ 9:Silicon A: Light Yellow ZnSe B: Yellow C: Orange-Red D: Red E: Dark red F: Blue G: Blue Green H: Light Green I: Orange Yellow J: Red To select a specific compound, enter the item number (or use D key) and press ¸ EE Pro for TI-89, 92 Plus Reference - Semiconductor data 11 Chapter 36 Boolean Expressions This section covers the Boolean expressions reference table, which includes 16 commonly-used Boolean expressions. This section also contains a diagram of the most commonly-used logic components. 36.1 Using Boolean Expressions This section displays a list of rules for Boolean algebra for two variables x and y: "+" is used to indicate the OR function. "." is used to indicate AND function. " ' " represents a logical inversion. The logic element symbols for AND, OR, NOR, BUFFER, NAND, INVERTER, XOR and XNOR, INVERTER are shown in this section and can be viewed by pressing the function key †. The Boolean algebra rules are sixteen in number and are listed as functions F0-F15. They are also identified by their name as shown in the Table 36-1. Example 36.1 - Find the properties of an Exclusive OR (XOR). 1. Use the arrow key D to move the highlight bar down the list. Continue 2. scrolling until the status line reads "Exclusive OR (XOR)". To view the expression in Pretty Print, press ‡ to display the pull down menu and press either the ¸ key or ¨. This will display the 3. Boolean expression for XOR in Pretty Print form. Press N to revert to the previous level. Table 36-1 Boolean Expressions Function Name F0 F1 F2 F3 F4 F5 EE Pro for TI-89, 92 Plus Reference - Boolean Expressions Boolean expression 0 x.y x.y' x x'.y y 12 Description of Function Null AND Inhibition Transfer Inhibition Transfer F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 EE Pro for TI-89, 92 Plus Reference - Boolean Expressions (x.y')+(x'.y) x+y (x+y)' (x.y)+(x'.y') y' x+y' x' x'+y (x.y)' 1 13 Exclusive OR (XOR) OR NOT OR (NOR) Equivalence (XNOR) Complement NOT Implication Complement (NOT) Implication NOT AND (NAND) Identity Chapter 38 Transforms This section accesses a series of tables containing transforms of common interest to electrical engineers. The transforms are listed as three topics - Fourier, Laplace and z-Transforms. Each topic contains information categorized under three sub-topics - Definitions, Properties and Transform pairs. All formulae can be viewed in Pretty Print equation-display format. These sub-topics are not interactive, i.e., one cannot specify an arbitrary expression and expect to compute a transformed result. 38.1 Using Transforms When the Transform section is selected, a dialog box is appears. To choose a topic, move the highlight bar using D or Ckeys and press ¸ (alternatively, press the number associated with the topic). 1. Select Fourier Transforms, Laplace Transforms, or z-Transforms. 2. Select Definitions, Properties or Transform Pairs 3. Use the D or C keys to move to the transform line 4. desired. The forward transforms are displayed by default. Pressing † toggles between the forward and inverse formats. 5. Press B to view the transform property in Pretty Print. • Information is presented on either side of the colon as shown below. The term on the left side of the colon, F(ω) represents the function in the frequency domain, while the right side of the colon represents the exact definition for F(ω) in terms of the time domain function f(t), integrated over all time modulated by e -i⋅ω⋅t. Forward and Inverse Formats • The information can be displaced in the inverse (as opposed to forward) form, meaning that the information on either side of the colon changes positions when the Inverse key † is pressed. A ‘é‘ symbol appears in the † tool bar to indicate the inverse form of the transform function is being displayed. +∞ b g z c f bt g ⋅ e hdt Fω: − i ⋅ω ⋅t −∞ EE Pro for TI - 89, 92 Plus Reference - Transforms 16 Forward Fourier Transform f (t ): 1 2π +∞ zc b g F w ⋅ e − i⋅ω ⋅t dw h Inverse Fourier Transform −∞ Status Line Message • The status line gives a description of the equation highlighted. The descriptions use standard mathematical terminology such as Modulation, Convolution, Frequency Integration, etc. Example 38.1 -What is the definition of the Inverse Fourier transform? 1. In the main Transforms screen, move the highlight bar to Fourier Transforms and press ¸. 2. Press ¸ a second time to access Definitions. 3. Move the highlight bar to the second definition and press B to display the equation in Pretty Print format. 4. Press A or B to scroll. Press any key to return to the previous screen. Fourier Definitions Pretty Print of Fourier (Time domain Transform) Example 38.2 View the Laplace transform of the time function f(t)=t. 1. In the initial Transforms screen, move the highlight bar to Laplace Transforms and press ¸. 2. Move the highlight bar to Transform Pairs and press ¸. 3. Scroll down to t: 1/s^2 and press B to view the equation in Pretty Print format. Laplace Transform Pairs EE Pro for TI - 89, 92 Plus Reference - Transforms Pretty Print of ‘t’ 17 Chapter 39 Constants This Constants Reference Table section lists the values and units for 43 commonly-used universal constants. These constants are embedded in equations in the Equations section of EE•Pro and are automatically inserted during computations. 39.1 Using Constants The Constants section in Reference is designed to give a quick glance for commonly used constants. It lists values of accuracy available by the standards of measurement established by appropriate international agencies. This section does not include any information of the uncertainty in measurement. Table 39-1. Constants Reference Table Const. π e γ φ α c ε0 F G Const. µq φ0 µB µe µN µp µµ a0 R∞ Description mass µ / mass emagnetic flux quantum Bohr magneton e- magnetic moment nuclear magneton p+ magnetic moment µ magnetic moment Bohr radius Rydberg constant c1 1st radiation constant h hb k Description circle ratio Napier constant Euler constant golden ratio fine structure speed of light permittivity Faraday constant Newton’s Gravitational constant acceleration due gravity Planck constant Dirac constant Boltzmann constant c2 b σ µ0 q em me mn mp mµ pe re permeability e- charge e- charge / mass e- rest mass n rest mass p+ rest mass µ rest mass mass p+ / mass eclassical e- radius χc χn χp SP ST Vm NA R 2nd radiation constant Wien displacement Stefan-Boltzmann constant e- Compton wavelength n Compton wavelength p+ Compton wavelength standard pressure standard temperature molar volume at STP Avogadro constant molar gas constant g EE Pro for TI - 89, 92 Plus Reference - Constants 18 These constants were arranged in the following order: universal mathematical constants lead the list followed by universal physical constants, atomic and quantum mechanical constants, radiation constants, standard temperature and pressure, universal gas constant and molar constants. To view a constant, use the arrow key D key to move the highlight bar to the value and press the View key †. The status line at the bottom of the screen gives a verbal description of the constant. Example 39.1- Look up the value of π. 1. 2. Pi is the first value to appear in the constant sections. Make sure it is selected by the highlight bar using the arrow keys. Access the View function by pressing key †. 3. Press any key to return to the constants screen. The number of significant digits displayed in Pretty Print can be changed in the 3 setting. EE Pro for TI - 89, 92 Plus Reference - Constants 19 Chapter 40 SI Prefixes The SI Prefixes section displays the prefixes used by the Systeme International [d’Unit[eacute]s] (SI). 40.1 Using SI Prefixes The prefixes are listed in the order shown in Table 40-1. The D key is used to move the highlight bar to select a SI prefix multiplier. The name of the prefix is displayed in the status line. The prefix and multiplier can be viewed by pressing the † key. Table 40-1 SI Prefix Table Prefix Y: (Yotta) Z: (Zetta) E: (Exa) P: (Peta) T: (Tera) G: (Giga) M: (Mega) k: (Kilo) h: (Hecto) da: (Deka) EE Pro for TI -89, 92 Plus Reference - SI Prefixes Multiplier 1E24 1E21 1E18 1E15 1E12 1E9 1E6 1E3 1E2 1E1 Prefix d: (deci) c: (Centi) m: (Milli) µ: (Micro) n: (Nano) p: (Pico) f: (Femto) a: (Atto) z: (Zepto) y: (Yocto) 20 Multiplier 1E-1 1E-2 1E-3 1E-6 1E-9 1E-12 1E-15 1E-18 1E-21 1E-24 Chapter 41 Greek Alphabet This section displays the Greek Alphabet and their names. There are several Greek letters supported by the TI - 89. To enter the Greek letters, the sequential keystrokes are listed in the TI-89 manual. They are repeated here for convenience of the user. Alternatively, 2 «(or ¿) followed by ¨ will access an internal menu listing several Greek characters. Key stroke Sequence Table 40-1 Greek Key stroke Sequence Letter ¥cjÁ α ¥cjc β ¥cjb δ ¥cje ε ¥cjÍ φ ¥cjm γ ¥cjy λ ¥cjz µ ¥cj§ π ¥cj© ρ ¥cjª σ ¥cjÜ τ ¥cj¶ ω ¥cjÙ ξ ¥cjÚ ψ ¥cjÛ ζ Greek Letter EE Pro for TI -89, 92 Plus Reference - Greek Alphabet ¥cjÎb ∆ ¥cjÎm Γ ¥ c j Χ Π ¥ c j Ϊ Σ ¥ c j ζ Ω 21 Appendix A Frequently Asked Questions A complete list of commonly asked questions about the EE•Pro are listed here. Review this list for your questions prior to calling for Technical support. You might save yourself a phone call! The material is covered under four general headings. A.1 Questions and Answers ™General Questions ™Analysis Questions ™Equations Questions ™Reference Questions A.2 General Questions The following is a list of questions about the general features of EE•Pro: Q. Where can I find additional information about a variable? A. A brief description of a highlighted variable appears in the status line at the bottom of the screen. More information, including its allowable entry parameters (i.e.: whether complex, symbolic or negative values can be entered, etc.) can be accessed by pressing ‡/Opts and 2:Type. Q. What does the underscore “_” next to a variable mean? A. This designates a variable which allows entry or expression of complex values. Q . I am in the middle of a computation and nothing seems to be occurring. How can I halt this process? A. Some computations can take a long time, particular if many equations and unknowns are being solved or a complex analysis function has been entered. Notice if the message in the status line at the bottom-right of the screen reads BUSY. This indicates that the TI math engine is attempting to solve the problem. Pressing the ´ key usually halts a computation and allows the user to regain control of the software. If, for some reason, the calculator locks up and does not allow user intervention, a “cold start” will have to be performed. This can be done by holding down the three keys; 2, A, and B, and pressing ´. WARNING: This will delete folders containing any defined variables or stored programs. Use it as a last resort. A "cold start" will not delete EE•Pro from your calculator. Q What do three dots (...) mean at the end of an item on the screen? A. The three dots (an ellipsis) indicate the item is too wide to fit on the available screen area. To view an item in its entirety, select it by moving the highlight bar and press † (or B, in some cases) to view the item in Pretty Print. Press A or B to scroll the item back and forth across the screen to view the entire object. Q. How can I recall, or view values of a previously computed problem? EE Pro for TI-89, 92 Plus Appendix A - Frequently Asked Questions 1 A. EE•Pro automatically stores its variables in the current folder specified by the user in 3 or the HOME screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values for a particular session of EE•Pro, press ƒ:/TOOLS, ª:/NEW and type the name of the new folder (see Chapter 5 of the TI-89 Guidebook for the complete details of creating and managing folders). There are several ways to display or recall a value: • The contents of variables in any folder can be displayed using the °, moving the cursor to the variable name and pressing ˆ to display the contents of a particular variable. • • Variables in a current folder can be recalled in the HOME screen by typing the variable name. Finally, values and units can be copied and recalled using the ƒ/Tools 5:COPY and 6:PASTE feature. All inputs and calculated results from Analysis and Equations section are saved as variable names. Previously calculated, or entered values for variables in a folder are replaced when equations are solved using new values for inputs. Q. Why is it that some of the values of variables saved earlier are cleared when I graph an equation or analysis function which uses the variable name(s)? A. When an equation or analysis function is graphed, EE•Pro creates a function for the TI grapher which expresses the dependent variable in terms of the independent variable. This function is stored under the variable name pro(x). When the EE•Pro’s equation grapher is executed, values are inserted into the independent variable for pro(x) and values for the dependent value are calculated. Whatever values which previously existed in either of the dependent and independent variables in the current folder are cleared. To preserve data under variable names which may conflict with EE•Pro’s variables, run EE•Pro in a separate folder using the guidelines above. Q. An item which is supposed to be displayed in a menu doesn’t appear. A. Some menus have more than eight items. If an arrow ï appears next to the digit 8, use the arrow key D to scroll the menu and view the remaining topics or press 2 D jump to the bottom of the menu. Q. Is there a help section in the software? A. There is a short series (slides) of general hints which can be accessed from the main screen of EE•Pro under ‡/Info. A different message appears each time ‡ is pressed. We’ve attempted to keep most of the explanation of certain topics to the manual in an effort to keep the software compact. Consult the chapter corresponding to the appropriate section of the software. If your are still in need of clarification, contact Texas Instruments (contact information in the Warranty and Technical Support section of the manual) A compiled list of the received questions and answers will be posted periodically on the da Vinci website. http://www.dvtg.com/faq/eepro A.3 Analysis Questions These are some commonly asked questions about the Analysis section of EE•Pro. Q. The screen display of computed results does not look identical to the example in the manual. A. The 3 setting, which controls the number of floating point digits displayed in a value and whether an answer appears in exact or approximate form, may have been set differently on the calculator used to make the screen displays for the example problem. Press the 3 key to view or change the mode settings. The first page will display the number of floating point digits, whether the display is in NORMAL, SCIENTIFIC, OR ENGINEERING exponential formats. Pressing „will display whether computed answers are displayed in APPROXIMATE or EXACT formats. The default mode setting for EE•Pro is Float 6, Radian and Approximate calculations. EE Pro for TI-89, 92 Plus Appendix A - Frequently Asked Questions 2 Q. The calculated angle or radian frequency result isn’t correct. A. Check to be sure the 3 settings list degrees or radians for angle units and make sure it matches the units of your entered value or desired value. Secondly, if the result is greater than 2π or less than -2π (result ≥ 360o), the TI solve(...) function may be generating a non principal solution. A principal solution is defined as a value between 0 and 2π (or between 0 and 360o). A non principal solution can be converted to a principal solution by adding or subtracting integer multiples of 2π (or 360o) until the remainder is within the range of 0 and 2π (or 0 and 360o). The remainder is the principal solution. Example: Imagine solving the equation sin( x ) = 0.5 . Non principle solutions include: 30°, 390°, -330°, 750°, etc., but the principal solution is 30°. Q. The solution for an analysis function is expressed in symbolic terms or variables, whereas a numerical value was expected. Why? A. If a variable name is entered as an input, EE•Pro will calculate a solution in symbolic form. In cases where a parameter entry is left blank, the variable’s name will be used in a symbolic computation. A name cannot be entered which is identical to the variable name (i.e.: C for capacitance) instead, leave the entry blank. Q. Units are not displayed in Analysis computations. Why? A. The Analysis section incorporates methods for calculation which are unique to each topic in analysis (as opposed to Equations which uses similar methods of calculation in all of the sections). Unit management was omitted for simplicity. All entries and computed results in Analysis are assumed to be in common SI units (F, m, A, Ω, etc.) which are stated in the status line for each parameter. In some cases, variables will be expressed in units arbitrarily chosen by the user (example: The variable len in Transmission Lines can be entered in km, m, miles however the answers will be expressed in units of len). If a value is entered that is inconsistent with the expected data type, an error dialog will appear which lists the entry name, the description, and the expected data type(s). See Appendix E for details on error messages. Q. What is the multiple graph feature in Capital Budgeting and how do I get it to plot several projects simultaneously? A. Activating the multiple graph feature allows successive graphs to be overlayed on the same. To do this, the graphing execution must be repeated each time a new project is plotted. 1. Make sure the cash flows for all the named projects have been entered. 2. Enable the Multiple Graphs feature by highlighting and pressing the ¸ key. 3. Select the name of a project you wish to graph and press …. 4. To overlay a second project on the first, select a different project name and press … to graph. 5. Repeat step 4 each time a new project is to be graphed on top of previously plotted functions. A.4 Equations Questions These are some common questions about the Equations section of EE•Pro. Q. There are already values stored in some of my variables. How do I clear or use those values? A. These values remain from previous solving operations. It is okay to ignore the values. As long as they aren’t selected ‘é‘ they will be overwritten by new solutions. If you want to reset the values, clear one or all of the variables. A value can be re-used in a computation by highlighting the displayed value and pressing ¸ twice. Q. Why do the values of the entered or calculated results change when the Units feature is deactivated in the ‡ options menu? EE Pro for TI-89, 92 Plus Appendix A - Frequently Asked Questions 3 A. When the Units feature is on, values can be entered and saved in any unit. When units are off, values can be entered in any unit, but the values will automatically be displayed on the screen in the default SI units. This is necessary so that when a series of equations are solved, all the values are consistent. Q. When solving a set of equations “Too many unknowns to finish solving.” is displayed. Why? A. Sometimes the solver doesn’t have enough to solve for all the remaining, unknown variables. In some cases, a Partial Solution set will be displayed. If the unknown value(s) is not calculated, more known values (or selected equations) will need to be selected to compute the solution. Q. If I view the value of a variable in Pretty Print, I notice that the units contain an extra character (such as ‘∆‘). A. In a few cases EE•Pro and the TI-89 and TI-92 operating systems use slightly different conventions for displaying units. The unit system in EE•Pro is designed to conform to the convention established by SI, however, in order to CUT and PASTE a value and units from EE•Pro to another area of the TI operating system, EE•Pro must insert extra characters in the units to match TI’s syntax. This causes extra characters to appear or symbols to appear differently in Pretty Print. Q. There are already values stored in some of my variables. How do I clear those values? A. The values can be accessed via VAR-LINK menu. To delete variables in VAR-LINK, use the file management tool provided (use ƒ key to access file management tools), check † the variables you want deleted and delete the variables. Q. The solution to my problem is clearly wrong! An angle might be negative or unreasonably large. Why? A. This is most likely to happen when angles are involved in the equation(s) you are solving. The TI 89 may have found a non-principal solution to your equation, or may have displayed the angle in radians (see the answer to the second question in the Analysis section). If a non-principal solution is found, it may then be used to solve other equations, leading to strange results. Example: Imagine solving the equation x + y = 90° . If x is 30°, then y should be 60°. But if a non-principal solution for x was found, such as 750°, then the value of y will be -660°, which although technically correct, is also not a principal solution. A.5 Reference For more information, see Chapter 33, “Reference: Navigation Guide” and your TI 89 User’s Guide. Reference : Standard Component Values • ±23 The message "Out of range" will occur if an entered component value is not in the range of 10 EE Pro for TI-89, 92 Plus Appendix A - Frequently Asked Questions 4 . Appendix B Warranty, Technical Support B.1 da Vinci License Agreement By downloading/installing the software and/or documentation you agree to abide by the following provisions. 1. License: da Vinci Technologies Group, Inc. (“da Vinci”) grants you a license to use and copy the software program(s) and documentation from the linked web/CD page (“Licensed Materials”). 2. Restrictions: You may not reverse-assemble or reverse-compile the software program portion of the Licensed Materials that are provided in object code format. You may not sell, rent or lease copies that you make. 3. Support: Limited technical support is provided through Texas Instruments Incorporated. Details are listed below. 4. Copyright: The Licensed Materials are copyrighted. Do not delete the copyright notice, trademarks or protective notice from any copy you make. 5. Warranty: da Vinci does not warrant that the program or Licensed Materials will be free from errors or will meet your specific requirements. The Licensed Materials are made available “AS IS” to you are any subsequent user. Although no warranty is given for the License Material, the media (if any) will be replaced if found to be defective during the first three (3) months of use. For specific instructions, contact the TI webstore via www.ti.com. THIS PARAGRAPH EXPRESSES da Vinci’s MAXIMUM LIABILITY AND YOUR SOLE AND EXCLUSIVE REMEDY. 6. Limitations: da Vinci makes no warranty or condition, either express or implied, including but not limited to any implied warranties of merchantability and fitness for a particular purpose, regarding the Licensed Materials. 7. Restricted Rights: The Licensed Materials are provided with Restricted Rights. Use, duplication or disclosure by the United States Government is subject to restrictions as set forth in subparagraph [c](1)(ii) of the Rights in Technical Data and Computer Software clause at DFARS 252.227-7013 or in subparagraph [c](1) and (2) of the Commercial Computer Software – Restricted Rights at 48 CFR 52.227-19, as applicable. In no event shall da Vinci or its suppliers, licensers, or sublicensees be liable for any indirect, incidental or consequential damages, lost of profits, loss of use or data, or interruption of business, whether the alleged damages are labeled in tort, contract or indemnity. EE Pro for TI - 89, 92 Plus Appendix B: Warranty and Technical Support 5 Some states do not allow the exclusion or limitation of incidental or consequential damages, so the above limitation may not apply. B.2 How to Contact Customer Support Customers in the U.S., Canada, Puerto Rico, and the Virgin Islands For questions that are specific to the purchase, download and installation of EE•Pro, or questions regarding the operation of your TI calculator, contact Texas Instruments Customer Support: phone: 1.800.TI.CARES (1-800-842-2737) e-mail: [email protected] For questions specific to the use and features of EE•Pro, contact da Vinci Technologies Group, Inc. phone: 1-541-757-8416 Ext. 109, 9 AM-3 PM, P.S.T. (Pacific Standard Time), Monday email: thru Friday (except holidays). [email protected] Customers outside the U.S., Canada, Puerto Rico, and the Virgin Islands For questions that are specific to the purchase, download and installation of EE•Pro, or questions regarding the operation of your TI calculator, contact Texas Instruments Customer Support: e-mail: [email protected] Internet: www.ti.com For questions specific to the use and features of EE•Pro, contact da Vinci Technologies Group, Inc. e-mail: [email protected] Internet: www.dvtg.com EE Pro for TI - 89, 92 Plus Appendix B: Warranty and Technical Support 6 Appendix C Bibliography In developing EE•Pro a number of resources were used. The primary sources we used are listed below. In addition, a large list of publications, too many to list here, were used as additional references. 1. Campsey, B. J., and Eugene F. Brigham, Introduction to Financial Management, Dryden Press, 1985, pp. 392-400 2. Van Horne, James C., Financial management and Policy, 8th Edition, Prentice-Hall, 1989, pp. 126-132 3. Edminister, Joseph A., Electric Circuits, 2nd Edition, Schaum’s Outline Series in Engineering, McGraw-Hill Book Company, 1983 4. Nilsson, James W., Electric Circuits, 2nd Edition, Addison-Wesley Publishing Company, Reading MA 1987 5. Plonsey, Robert and Collins, Robert E., Principles and Applications of Electromagnetic Fields, McGraw-Hill Book Company, Inc. 1961. 6. Ryder, John D., Electronic Fundamentals and Applications, 5th Edition, Prentice-Hall Inc., Englewood Cliffs, NJ, 1980 7. Smith, Ralph J., Electronic Circuits and Devices, 2nd Edition, John Wiley and Sons, New York, NY 1980 8. Boylestad, Robert L., Introduction to Circuit Analysis, 3rd Edition, Charles E Merrill Publishing Company, Columbus, OH 1977 9. Jordan, Edward C., Reference Data for Radio Engineers: Radio Electronics, Computers and Communications, 7th Edition, Howard and Sams, Indianapolis IN 1985. 10. Coughlin, Robert F and Driscoll, Frederick F.., Operational Amplifiers and Linear Integrated Circuits, Prentice-Hall, Englewood Cliffs, NJ 1991. 11. Johnson, David E., and Jayakumar, V., Operational Amplifier Circuits - Design and Applications, Prentice-Hall, Englewood Cliffs, 1982 12. Stout, David F., Handbook of Operational Amplifier Design, McGraw-Hill Book Company, New York, 1976 13. Ghausi, Mohammed S., Electronic Devices and Circuits: Discrete and Integrated, Holt, Rienhard and Winston, New York, NY 1985 14. Slemon G. R., and Straughen, A., Electric Machines, Addison-Wesley, Reading, MA 1980 15. Mano, M.M., Digital Logic and Computer Design, Prentice-Hall Inc., Englewood Cliffs, NJ 1979 16. Cohen and Taylor, Rev. Modern Physics, Vol. 59, No. 4, October 1987, pp. 1139-1145 17. Sze, Simon, Physics of Semiconductors, John Wiley and Son, New Jersey, 1981 18. Mueller, Richard S., and Kamins, Theodore I., Device Electronics for Integrated Circuits, 2nd Edition, John Wiley and Son, New jersey, 1986 19. Hodges, David A., Jackson, Horace A., Analysis of Digital Integrated Circuits, McGraw-Hill, New York, 1988 EE Pro for TI - 89, 92 Plus Appendix C: Bibliography 7 20. Elmasry, Mohamed I., Editor, Digital MOS Integrated Circuits, IEEE Press, New York, 1981 21. Kuo, Benjamin C., Automatic Control Systems, Prentice Hall, New jersey, 1991 22. Stevenson Jr., William D., Elements of Power Systems Analysis, McGraw-Hill International, New York, 1982 23. Wildi, Theodore, Electrical Power Technology, John Wiley and Son, New Jersey, 1981 24. Yarborough, Raymond B., Electrical Engineering Reference Manual, Professional Publications, Belmont, CA 1990 25. Schroeder, Dieter K., Advanced MOS Devices, Addison Wesley, Reading MA 1987 26. Neudeck, Gerold W., The PN Junction Diode, Addison Wesley, Reading MA 1983 27. Pierret, Robert F., Semiconductor Fundamentals, Addison Wesley, Reading MA 1983 28. Pierret, Robert F., Field Effect Devices, Addison Wesley, Reading MA 1983 29. Neudeck, Gerold W., The Bipolar Junction Transistor, Addison Wesley, Reading MA 1983 30. Irwin, David J., Engineering Circuit Analysis, McGraw-Hill, New York, 1987 EE Pro for TI - 89, 92 Plus Appendix C: Bibliography 8 Appendix D: TI-89 & TI-92 PlusKeystroke and Display Differences D.1 Display Property Differences between the TI-89 and TI-92 Plus The complete display specifications for both the TI-89 and TI-92 Plus calculators are displayed below. Table D-1 TI-89 and TI-92 Plus display specifications. Property Display size Pixel Aspect ratio Full Screen TI-89 TI-92 Plus 240 x 128 1.88 40 characters/line 13 Lines 236 x 51 pixels 39 characters, 6 lines Vertical Split Screen (1/3rd) 160 x 100 1.60 26 characters/line 10 lines 156 x 39 pixels 25 characters 4 lines 77 x 80 pixels 12 characters 10 lines Not supported Vertical Split screen (2/3rd) Not supported Horizontal Split Screen (1/3rd) Not supported Horizontal Split Screen (2/3rd) Not supported Key legends 16 pixel rows Horizontal Split Screen Vertical Split Screen 117 x 104 pixels 19 characters, 13 lines 236 x 33 pixels 39 characters, 4 lines 236 x 69 pixels 39 characters, 8 lines 77 x 104 pixels 12 characters, 13 lines 157 x 104 pixels 26 characters, 13 lines 20 pixel row D.2 Keyboard Differences Between TI-89 and TI-92 Plus The keystrokes in the manual for EE•Pro are written for the TI-89. The equivalent keystrokes for the TI-92 Plus are listed in the following tables. EE Pro for TI - 89, 92 Plus Appendix D: TI-89 &TI 92 Plus: Display and Keystrokes Differences 9 Table D-2 Keyboard Differences, Representation in Manual Function Specific Key Function Keys Trig Functions TI-92 Plus key strokes F1 F2 F3 F4 F5 F6 F7 F8 ƒ „ … † ‡ 2ƒ 2„ 2… ƒ „ … † ‡ ˆ ‰ Š ƒ „ … † ‡ ˆ ‰ Š Sin 2Ú W W Cos 2Û X X Tan 2Ü Y Y Sin-1 Cos-1 ¥Ú 2W ¥Û 2X [SIN-1] [COS-1] Tan-1 ¥Ü 2Y [TAN-1] A jÁ Ñ A B jc Ò B C jd Ó C D jb D E je E F jÍ G jm G H jn H I jo I J jp J K Alphabet keys TI-89 Key strokes j^ jy K L M Ô Representation in the manual F L M jz N j{ Õ N O j| Ö O P j§ P Q j¨ Q R j© R S jª Ü T U j« EE Pro for TI - 89, 92 Plus 10 Appendix D: TI-89 &TI 92 Plus: Display and Keystrokes Differences × S T U V jµ W j¶ Ù Ú Û j· X Y Z Space Function V W Ù X Y Z  TI-92 Plus key strokes LN 2Ù x x ¥Ù 2x s π 2Z 2Z T ¥Z · 2½ Ï Ï ¥½ · 2I 2J · ) * Y= ¥ƒ ¥W # Window Graph ¥„ ¥… ¥E ¥R $ % Cut ¥2 5 Copy Paste Delete Special Characters TI-89 Key strokes ex Log, EXP Specific Key Ø ¥¤ ¥N θ Negation i ∞ Graphing Functions Editing Functions Representation in the manual ¥0 2N ¥0 6 7 8 20 2§ § 0 20 K / 2§ § 0 £ § 0 ( c c ) { } [ d 2c 2d 2b 2e d 2 2 2 2 Addition « « « Subtraction Multiplication Division | p e | p e | p e Quit Insert Recall Store Backspace Parenthesis, Brackets Math Operations EE Pro for TI - 89, 92 Plus 11 Appendix D: TI-89 &TI 92 Plus: Display and Keystrokes Differences c c d b e d [ \ g h Raise to power Enter Exponent for power of 10 Equal Integrate Differentiate Function Specific Key Math Operations cont. Less than Greater than Absolute value Angle Square root Approximate Tables TblSet Table Modifier Keys 2nd Diamond Shift Alphabet Alphabet lock Special Areas Math Mem Var-Link Units Char Ans Entry Z ^ Z 2¨ Z ^ Á 2m 2n Á 2m 2n Á < = TI-89 Key strokes TI-92 Plus key strokes 2µ 2¶ Í 2^ 2p ¥¸ 2µ 2¶ 2K 2p ¥¸ Representation in the manual Â Ã Í ’ ‘ ¥† ¥‡ ¥T ¥Y & ' 2 ¥ Î j 2j 2 ¥ ¤ 2 ¥ Î j ™ 2z 2{ 2| 2ª 2« 2· 2¸ 2z 2{ 2| 2« 2· 2¸ I ¯ ° 9 ¿ ± ² Number keys Single Quote 2Á Double Quote Back slash Underscore Colon Semicolon Special Characters 2¨ 2© ¥3 2y 2o 2Ï 2M É Ì  Ë Ê ¨ © ª y z { m n ¨ © ª y z { m n 1 2 3 4 5 6 7 8 One Two Three Four Five Six Seven Eight EE Pro for TI - 89, 92 Plus 12 Appendix D: TI-89 &TI 92 Plus: Display and Keystrokes Differences È 2L 2Á Nine Zero Comma Decimal point Main Functions Specific Key Main Functions cont. Mode Catalog Clear Custom Enter ON OFF ESCAPE Application o µ b ¶ ¥Q Home Function o µ b ¶ 9 0 , . " TI-92 Plus key strokes ´ 2´ N O 3 2© M 2ª ¸ ´ 2´ N O Top C C C Right Left Bottom Cursor Movement TI-89 Key strokes B A D B A D B A D EE Pro for TI - 89, 92 Plus 13 Appendix D: TI-89 &TI 92 Plus: Display and Keystrokes Differences Representation in the manual 3 ½ M ¾ ¸ ´ ® N O Appendix E Error Messages ™General Error Messages ™Analysis Error Messages ™Equations Error Messages ™Reference Error Messages E.1 General Error Messages 1. NOTE: Make sure the settings in the 3 screen do not have the following configuration. Angle: DEGREE Complex Format: POLAR EE•Pro works best in the default mode settings of your calculator (ie. Complex Format: REAL, or Angle: RADIAN). If one of a set of error messages appears which includes “An error has occurred while converting....”, Data Error”, “Domain Error”, and/or “Internal Error”, check to see if the above settings in the 3 screen exists. If it does, change the or reset your calculator to the default mode settings (2 ¯ ƒ ª ¸). 2. “Syntax Error” -- occurs if the entered information does not meet the syntax requirements of the expected entry. Check to make sure extra parenthesis are removed and the entered value meets the legal rules for number entry. 3. "Insufficient Table Space" or "Insufficient Memory" can occur when the system is low on available memory resources. Consult your TI-89 manual on methods of viewing memory status and procedures for deleting variables and folders to make more memory available. 4. The message "Unable to save EE•Pro data" will be displayed if EE•Pro is unable to save information of its last location in the program before exiting due to low memory availability. Consult your TI-89 manual under the index heading: Memory-manage. 5. "The variable prodata1 was not created by EE•Pro..." EE•Pro uses a variable called “prodata1” to recall its last location in the program when it is re-accessed. If this variable list is changed to a format which is non-recognizable to EE•Pro, it displays this message before overwriting. 6. "Data length exceeds buffer size. The variable name will be displayed instead. The variable's value may be viewed with VAR-LINK using ˆ or recalled to the author line of the HOME screen." 7. "An error has occurred while converting this variable's data for display. (The name of the variable is in the title of this dialog box.) There may be something stored in the variable that EE•Pro can't make sense of. You may be able to correct the problem by deleting the variable." EE Pro for TI-89, 92 Plus Appendix E - Error Messages 14 8. “storage error...” This message is set to occur if the user attempts to enter a value into a variable which is locked or archived, or a memory error has occured. Check the current status of the variable by pressing ° and scrolling to the variable name, or check the memory parameters by pressing ¯. 9. “Invalid variable reference. Conflict with system variable or reserved name.” This can occur if a variable name is entered which is reserved by the TI operating system. A list of reserved variable names is included in Appendix F. E.2 Analysis Error Messages AC Circuits Error Messages 1. “No impedances/admittances defined” This message is displayed if an entry for impedance ZZ_or YY_admittance is not a real, complex number or a defined variable in the Voltage Divider and Current Divider sections of AC Circuits. Ladder Network Error Messages 1. When displaying a list of ladder network elements, an element may be named "Unknown Element" if it contains erroneous element-type information. This may occur if the variable "ladnet", which contains information of the components and arrangement of the ladder network, has been altered or corrupted. View "ladnet" in °, and delete it if necessary. If deleted, the user will 2. 3. 4. need to re-enter the ladder network information. A dialog displaying "LadNet invalid. Delete and try again." may occur if a ladder network contains one or more elements with erroneous data. See above. A dialog displaying "No ladder elements exist." will occur if an attempt is made to solve a ladder network that contains no defined elements. Add elements to the network by pressing ‰. “Invalid variable reference. Conflict with system variable or reserved name.” This error is displayed if a variable name is entered as an input which is identical to a variable name used by EE•Pro. This message is also displayed if an entered variable name is a reserved system variable. A complete list of variables reserved by the TI operating system is listed in Appendix F. Filter Design Error Messages 5. A dialog displaying "Invalid freq. ratios." will occur if an attempt is made to design a passive filter using frequency data that does not apply to the particular filter topology. Check your entered values, you may need to try a different filter design. Gain and Frequency: Transfer Function/Bode Diagrams 6. The computed transfer function H(s) may contain “[Entry error]” rather than an actual transfer function. This will occur when the number of zero (or numerator) terms greater than or equal to the number of pole (or denominator) terms. 7. An empty plot or the following dialogue may occur if an attempt is made to graph the transfer function before one has been defined: "Undefined variable. Usually this error is caused by graphing the Bode diagram before the transfer function Hs has been defined. F2:Analysis 5:Gain and Frequency/1:Transfer Function calculates Hs." Computer Engineering: 7. An "Invalid number format” message will be displayed if an entered number does not follow the legal rules of the number base, or if a decimal term is entered. The format of numbers in the Computer Engineering section of EE•Pro is (p)nnnn…, where p is the letter prefix b, o, d, or h indicating binary, octal, decimal, and hexadecimal, number systems and n represents the legal digits EE Pro for TI-89, 92 Plus Appendix E - Error Messages 15 for the number base. If a number does not begin with a letter prefix in parenthesis, then it is assumed to be in the number base set in the †/Mode dialog. 8. “Entry truncated” will appear if an entered value exceeds the binary word size allocated in the †/Mode dialog. For example, the decimal number 100 is expressed as a 7 bit binary number:1100100. If a word size of 6 bits is designated in the †/Mode dialog, the latter six digits of the number (100100) will be entered into the available word space. Computer Engineering : Binary Arithmetic 9. The message "Undefined Result" will be displayed if an attempt is made to divide by zero. Computer Engineering: Karnaugh Map 10. “Two few unique letters in Vars.” This occurs when n, the number of unique, single-letter variables in Vars does not meet the criteria that 2n must be greater than the largest number appearing in the Minterms or Don’t Care series. Thus if the largest number is 15, there must be no less than four unique, single-letter variables since 24>15. Capital Budgeting 11. When changing the name of a project, the error "Duplicate project variable name." will occur if the entered name is already in use by an existing project. 12. The "Too few cash flows" message will be displayed if an attempt is made to solve a project which contains too few or no cash flow entries. 13. The message "Data Error. Reinitializing project." will occur if the project data has been altered or is inconsistent with current state. EE•Pro will restart. E.3 Equation Messages If a value is entered that is inconsistent with the expected data type, an error dialog will appear which lists the entry name, the description, and the expected data type(s), and the expected units. If an error occurs during a computation that involves temperature, "temp conversion err" or "deg/watt conversion err" will be displayed. When solving equation sets, several messages can be displayed. These messages include: • “One or more equations has no unknowns.....” This message occurs if one or more of the selected equations in a solution set has all of its variables defined by the user. This can be remedied by pressing N, deselecting the equation(s) where all of the variables are defined and resolving the solution set by pressing „ twice. To determine which equation has all of its variables defined, press N to view the equations, select an equation in question by highlighting the equation and pressing ¸, and pressing „ to view the list of variables. A ‘é‘ next to a variable indicates a value has been specified for that variable by the user. If all of the variables in an equation are marked with a ‘é‘, no unknown variables exist for that equation. This equation should not be included in the solution set. Press N to view the list of equations. Select the equations to be solved, excluding the equation with no unknowns, and press „ twice to resolve the set of equations. EE Pro for TI-89, 92 Plus Appendix E - Error Messages 16 • • • • • • • • • • "Unable to find a solution in the time allowed. Examine variables eeinput and eeprob to see the exact statement of the problem. EE\xB7Pro sets Exact/Approx mode to AUTO during solve.” “No equations have been selected. Please select either a single equation to solve by itself, or several equations to solve simultaneously.” "Too many unknowns to finish solving"-generally occurs if the number of equations is less than the number of unknowns "It may take a long time to find a complete solution, if one can be found at all. You may abort the calculation at any time by pressing the ON key." -this occurs if there are many unknowns or multiple solutions. “No input values provided....” occurs if none of the variables have values designated when solving an equation set. "The nsolve command will be used. The existing value for the unknown, if any, will be used as an initial guess." The nsolve function is used when a single unknown exists in the equation and the unknown variable is an input in a user defined function (an example is the error function erf in Semiconductor Basics of Solid State or the eegalv in Wheatstone Bridge in Meters and Bridges). The nsolve function will not generate multiple solutions and the solution which nsolve converges upon may not be unique. It may be possible to find a solution starting from a different initial guess. To specify an initial guess, enter a value for the unknown and then use F5:Opts/7:Want to designate it as the variable to solve for. More information on the differences between the solve, nsolve and csolve functions is listed in the TI-89 manual. "One complete useable solution found." All of the unknown variables were able to be solved in the selected equations. "One partial useable solution found." Only some of the variables in the selected equations were able to be solved. “Multiple complete useable solns found." One or more variables in the selected equations have two possible values "Multiple partial useable solns found." One or more variables in the selected equations have two possible values, however not all of the unknown variables were able to be solved. The following messages can appear when attempting to graph equation set functions: • • • • "Independent and dependent variables are the same." "Unable to define Pro(x)"-cannot resolve the dependent and independent variables. "Undefined variable" too many dependent variables or dependent variable unable to be defined in terms of the independent variable. "Error while graphing." E.4 Reference Error Messages For more information, see Chapter 33, “Reference: Navigation Guide” and your TI 89 User’s Guide. Reference : Standard Component Values • ±23 The message "Out of range" will occur if an entered component value is not in the range of 10 EE Pro for TI-89, 92 Plus Appendix E - Error Messages 17 . Appendix F: System Variables and reserved names The TI-89 and TI-92 Plus has a number of variable names that are reserved for the Operating System. The table below lists all the reserved names that are not allowed for use as variables or algebraic names. Graph y1(x)-y99(x)* xt(t)-xt99(t)* ui1-ui99* tc xfact xmax ymax ∆x zsc1 ncontour tmin tplot Estep nmax y1’(t)-y99’(t)* yt(1)-y99(t)* xc rc yfact xsc1 ysc1 ∆y eye0 θmin tmax ncurves fldpic plotStrt Yi1-yi99* z1(x,y)-z99(x,y) yc θc zfact xrid ygrid zmin eyeφ θmax tstep difto1 fldres plotStep r1(θ)-r99(θ)* u1(n)-u99(n)* zc nc xmin ymin xres zmax eyeϕ θstep t0 dtime nmin sysMath Graph Zoom Zxmax Zymax zθmin ztmax ztstepde zzsc1 znmin Zxscl Zyscl zθmax ztstep ztplotde zeyeθ znmax Zxgrid Zygrid zθstep zt0de zzmin zeyeφ zpltstrt Statistics Zxmin Zymin Zxres Ztmin Ztmaxde Zmax zeyeϕ zpltstep ü ý Σxy corr medx1 medy2 nStat regEq(x)* Sy ∆tbl SysData Ok Exp* Σx Σy maxX medx2 medy3 q1 seedl R2 tblInput σx σy maxY medx3 minX q3 seed2 Table Data/Matrix Miscellaneous Solver Σx2 Σy2 medStat medyl minY regCoef Sx tblStart C1-c99 Main Eqn* 18 Errornum ...
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