107_25 - 1 Electrical Engineering Technology EET 107...

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Unformatted text preview: 1 Electrical Engineering Technology EET 107 Introduction to Circuit Analysis # 25 Professor Robert Herrick Purdue University © EET 107 - 25 Introduction to Circuit Analysis 2 Overview u Series-parallel model approach u Ladder circuit - laws approach u R-2R passive circuit u R-2R active circuit (op amp) Purdue University © EET 107 - 25 Introduction to Circuit Analysis 3 Overview Series-parallel Thevenin or Norton Model Approach Purdue University © EET 107 - 25 Introduction to Circuit Analysis 4 Series-parallel Circuit 2k 4k 4k 180V 4k 2k Iload Rload + Vload − ¿ Find the OPEN CIRCUIT voltage ¿ Find resistance the SHORT CIRCUIT current ¿ Find THEVENIN RESISTANCE Purdue University © EET 107 - 25 Introduction to Circuit Analysis 5 Ladder Network - VOC 0 mA 2k 4k + 4k 180V 4k + 60 V − 60 V 2k − + VOC − VOC = (4kΩ / 12kΩ) 180V = 60V Purdue University © EET 107 - 25 Introduction to Circuit Analysis 6 Ladder Network - ISC 4 k 18 mA 9 mA 2k ISC 4k 180V 2k 4k RT = 4kΩ + 2kΩ + 4kΩ = 10kΩ Isupply = 180V / 10kΩ = 18mA ISC = 18mA / 2 = 9 mA Purdue University © EET 107 - 25 Introduction to Circuit Analysis 7 Thevenin Resistance RTH = RN = ETH / IN RTH = 60V / 9mA = 6.67 kΩ 6.67 Purdue University © EET 107 - 25 Introduction to Circuit Analysis 8 Thevenin Resistance Or zero the supply and measure at the zero output terminals with an ohmmeter. ohmmeter Circuit E - replaced by short I - replaced by open Purdue University © EET 107 - 25 Ohmmeter RTH Introduction to Circuit Analysis 9 Ladder Network - VOC 0 mA 2k 4k 0V 4k 180V 4k 2k + Ohmmeter VOC RTH − RTH = 2kΩ + (4kΩ // 8kΩ) + 2kΩ = 6.67 kΩ Purdue University © EET 107 - 25 Introduction to Circuit Analysis 10 Circuit Models 6.67 k 9mA 60V Thevenin Purdue University © 6.67 k Norton EET 107 - 25 Introduction to Circuit Analysis 11 Circuit Model – Load 13.33kΩ Class – 2 minute quiz – pick your method. Find load current with Rload of 13.33 kΩ ¿Thevenin model with load, or ¿ Norton model with load, or ¿ Original circuit with load Purdue University © EET 107 - 25 Introduction to Circuit Analysis Ladder Network – Find IL 2k 4k 4k 180V 4k 2k 12 Iload Rload + Vload − Choose your method wisely. wisely Rload = 13.33 kΩ Purdue University © EET 107 - 25 Introduction to Circuit Analysis 13 Overview Ladder circuit Laws approach Purdue University © EET 107 - 25 Introduction to Circuit Analysis 14 Ladder Network 1Ω 27V + - 1Ω 1Ω 6Ω 6Ω 2Ω ¿ Series-parallel circuit ¿ Name - looks like a ladder Purdue University © EET 107 - 25 Introduction to Circuit Analysis 15 Ladder Network 1Ω 27V + - 1Ω 6Ω 1Ω 6Ω 2Ω Use basic laws to solve for all basic currents and voltages ! Purdue University © EET 107 - 25 Introduction to Circuit Analysis 16 Ladder - basic laws approach Steps 1. Find RT – use resistor reduction 2. Find IT - use Ohm’s Law 3. Find VR & IR - use basic laws Purdue University © EET 107 - 25 Introduction to Circuit Analysis 17 Ladder Network - resistor reduction 1Ω 27V 1Ω 6Ω + - 3Ω 6Ω 2Ω 2Ω 3Ω 2Ω 3Ω RT = 3Ω Purdue University © 1Ω Start here EET 107 - 25 Introduction to Circuit Analysis 18 Ladder Network - solve I & V ΙΤ 27V 9Α + - 1Ω +9V- Drop ? 1Ω 6Ω 1Ω 6Ω 2Ω 3Ω IT = 27V / 3Ω = 9Α Ohm’s Law Ohm’s V1Ω = 9A * 1Ω = 9V Ohm’s Law Purdue University © EET 107 - 25 Introduction to Circuit Analysis 19 Ladder Network - solve I & V 9Α 27V + - Drop ? 1Ω 1Ω +9V- + 18V - 6Ω 3Α 1Ω 6Ω 2Ω Current ? V6Ω = 27V - 9V = 18V KVL KVL I6Ω = 18V / 6Ω = 3Α Ohm’s Law Ohm’s Purdue University © EET 107 - 25 Introduction to Circuit Analysis 20 Ladder Network - solve I & V 9Α 1Ω 27V +9V+ - Ι ? 6Α 1Ω V? + 18V - +6V- 6Ω 3Α I1Ω = 9A - 3A = 6A V1Ω = 6A * 1Ω = 6V Purdue University © EET 107 - 25 1Ω 6Ω 2Ω KCL KCL Ohm’s Law Introduction to Circuit Analysis 21 Ladder Network - solve I & V 9Α 1Ω 27V +9V+ - 6Α 1Ω + 18V - 1Ω 4Α +6V- + 6Ω 3Α 12V - +4V- + 6Ω 8V 2Α - 2Ω Continue finding the next best I or V ! Purdue University © EET 107 - 25 Introduction to Circuit Analysis 22 Overview R-2R Passive Purdue University © EET 107 - 25 Introduction to Circuit Analysis 23 R-2R Passive Circuit 16 mA R R R A 2R I3 I2 2R 2R I1 2R I0 2R A R 2R R 2R R Start here RT = R Purdue University © 2R R EET 107 - 25 Introduction to Circuit Analysis 24 R-2R Passive Circuit 16 mA A 2R 8 mA R 8 mA I3 4 mAR 4 mA I2 2R R 2R I1 2R I0 2R A 2R 2R 2R I3 = ½ (16mA) = 8 mA I2 = ½ (8mA) = 4 mA Purdue University © EET 107 - 25 Etc. Introduction to Circuit Analysis 25 Overview R-2R Active Purdue University © EET 107 - 25 Introduction to Circuit Analysis 26 R-2R Active DAC Circuit DAC - Digital to analog converter 1 è 2 è 3 è 4 è 3 è 2 è 1 CD translates to analog sound signal Purdue University © EET 107 - 25 Introduction to Circuit Analysis 27 R-2R Active Circuit 16 mA R R R 2R 2R 2R 2R 8 mA 4 mA 2 mA 1 mA SW3 I3 SW2 I2 I3 8 mA SW1 I1 I32 12 mA 2R Esupply SW0 VOUT I0 I321 I3210 14 mA 15 mA −Esupply 1k SW# - digital bits: all SW closed =11112 Find I3210 Purdue University © EET 107 - 25 Introduction to Circuit Analysis R-2R Active Circuit – Find Vout 16 mA R 2R R SW3 I3 R 2R 2R SW2 I2 2R SW1 I1 2R Esupply SW0 I32 I321 V1k = 15 mA × 1kΩ = 15V 15 Vout = 0V − 15V = −15V Purdue University © −15V VOUT I0 15 mA I3 28 EET 107 - 25 I3210 0V −Esupply 1k + 15V − Introduction to Circuit Analysis 29 R-2R Active Circuit 16 mA R R R 2R 2R 2R 2R 8 mA 0 mA 2 mA 0 mA SW3 I3 SW2 I2 I3 8 mA SW1 I1 I32 8 mA 2R Esupply SW0 −10V VOUT I0 I321 10 mA 10 mA −Esupply I3210 Suppose we changed the switch 1k + 10V − positions to 1010, what would happen? Purdue University © EET 107 - 25 Introduction to Circuit Analysis 30 Overview u Series-parallel model approach uLadder circuit - laws approach uR-2R passive circuit u R-2R active circuit (op amp) Purdue University © EET 107 - 25 Introduction to Circuit Analysis ...
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