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14- Diodes and Transistors 2

# 14- Diodes and Transistors 2 - Non-Linear Circuits II Load...

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Non-Linear Circuits II: Load Line nalysis Diodes Square Law Analysis, Diodes, Square Law MOSFET Common Source Amp © Fred Terry all, 008 Fall, 2008 1

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Simple Example: 2 Linear Resistors In Series ' L L I V R Ohm s Law = 10 -3 Example V cc =10V R S =15K Ω R L =10K Ω ' LL SS S SC C L IV R O h m s L a w VV V K V L = =− 1 x 10 I L I S KCL Is Satisfied At Intersection . LS L Plot I and I vs V 0.5 t (amps) R s + V s +- I L I S -0.5 0 Curren V CC R L V L - -10 -8 -6 -4 -2 0 2 4 6 8 10 -1 2 V L (volts)
Simple Example: 2 Linear Resistors In Series ' L L I V R Ohm s Law = 10 -3 Example V cc =10V R S =15K Ω R L =10K Ω ' LL SS S SC C L IV R O h m s L a w VV V K V L = =− 0.8 0.9 1 x 10 V cc /R L . LS L Plot I and I vs V .5 0.6 0.7 V cc /R S I L I S R s + V s +- I L I S 0.3 0.4 0.5 V CC R L V L - 0 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 3 V L (volts)

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esistively Loaded Exponential Diode Resistively Loaded Exponential Diode 3.5 4 x 10 -3 Diode Circuit with R D =0 Ω diode current Load Resistor Current R L =10K Ω Load Resistor Current R L =5K Ω q Vk T I e 2.5 3 t (Amps) ( ) 1 25.85 300 A DS RC CA II kT q mV atT K VV V K V L = ≈= =− 1 1.5 2 Curren t ( ) LL L C L IV R V VR == 0 1 2 3 4 5 6 7 8 9 10 0 0.5 Diode Voltage V (volts) 4 a
oom In To Estimate Solutions Zoom In To Estimate Solutions x 10 -3 Diode Circuit with R D =0 Ω iode current 6 1.8 2 diode current Load Resistor Current R L =10K Ω Load Resistor Current R L =5K Ω 1.2 1.4 1.6 t (Amps) 0.8 1 Curren t 0.2 0.4 0.6 5 0.4 0.5 0.6 0.7 0.8 0.9 1 Diode Voltage V a (volts)

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oom In To Estimate Solutions Zoom In To Estimate Solutions 9 x 10 -3 Diode Circuit with R D =0 Ω 1.7 1.8 1.9 iode current .4 1.5 1.6 ent (Amps) diode current Load Resistor Current R L =10K Ω Load Resistor Current R L =5K Ω 1.2 1.3 1.4 Curr e 0.9 1 1.1 6 0.635 0.64 0.645 0.65 0.655 0.66 0.665 0.67 0.675 Diode Voltage V a (volts)
umerical Solution Numerical Solution t h e r m LD VV II R I e = ( ) () 1 At LL S LC C A VR VVV = =− 1 h e r m CC A S Ie R Seek such that: 0 t h e r m L A V V V V R I e = ( ) ( ) 10 In the presence of imperfect data, seek such that: error A CC A L S A fV V ⎡⎤ −− ⎣⎦ 7 2 is minimized error A

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xample Example • 2V 60Hz Cos wave into ½ Wave Rectifier Circuit • Solution Using Matlab fzero function – This is a bisection based zero solver ther Approaches Use Newton’sMe thod – Other Approaches Use Newton s Method – Many Other Nonlinear Regression Methods Exist (Newton-Raphsen, Levenberg Marquardt, Genetic lgorithms etc ) Algorithms, etc.) – A Course Like Math 471/571 Is Good for Many Engineers and Scientists Dealing With Real Data atlab and Similar Codes Will Often Do the Hard – Matlab and Similar Codes Will Often Do the Hard Work, but Understanding the Limitations in the Methods is often Important 8
xample Matlab Code clear all; close all; format long; 60; Example Matlab Code f=60; t=-1/f:1e-4/f:1/f; vcc=2*cos(2*pi*f*t); x=vcc(1)-0.7; RL=10e3; IS=1e-14; vtherm=25.85e-3; r kk=1:length(vcc) for kk=1:length(vcc) x0=x; % line below finds the zero of the function diode_err using Va as the % variable, x0 as the starting guess, and all other variables, vcc, RL, IS, % etc. are treated as KNOWN parameters. The syntax below is the Matlab

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14- Diodes and Transistors 2 - Non-Linear Circuits II Load...

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