Small Signal Model. S. Sivasubramani EE101- Small Signal - Diode

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1 Small Signal Model i v

2 Small Signal Model i I D i d i D v d v D v V D

3 Small Signal Model -Mathematical Analysis V D - DC value v d - ac signal v D - Total signal (DC ac signal) Diode current and voltage are related by i d = I s (exp( v d ) 1) Let us apply V D, I D = I s (exp( V D ) 1) I D I s (exp( V D )) (1)

4 Small Signal Model -Mathematical Analysis Let us apply v D, sum of DC and ac signal, v D = V D v d The current i D = I s (exp( v D ) 1) i D I s (exp( v D )) i D I s (exp( V D v d )) i D I s exp( V D ) exp( v d )

5 Contd.. From (1) i D = I D exp( v d ) If v d <<, it can be written using Taylor s series i D = I D (1 v d ) I D i d = I D I Dv d i d = I D v d The small signal voltage and current related by linear relationship like resistor. i d = g d v d where g d = I D and is a small signal conductance in.

6 Small Signal Model It can also be written as v d = r d i d where r d =. r d is a small signal resistance in Ω. I D It can also be obtained [ ] did r d = 1/ dv D v D =V D For small signal (v d < V T ) v D v d r d i D i d

7 Small Signal Model - Various Elements Resistor R i D v D R i d v d Voltage Source - Independent V i D i d Current Source - Independent I v D v d

8 Test Draw the small signal equivalent circuit at V D = 1 V, I D = 1 ma. The element N has the following i v relationship. i D = v 2 D R R v i sin(ωt) v i sin(ωt) v d r N v D N i d V I i D r N = 1 1 = di D 2 Ω vd =V dv D D

9 Small Signal Model - Analysis If an input ac signal is superimposed on DC, use the following steps to solve for voltage and current in any nonlinear circuit. 1 Calculate DC operating point without ac signal using any one of the nonlinear techniques such as analytical method or graphical method. 2 Draw small signal equivalent circuit. 3 As it is a linear circuit, use any linear techniques to solve for voltage/current. 4 Add DC operating point and small signal voltage current to find the response of total signal (DC small signal). Keep in mind that this method will work till v d <.

10 Example-I 1 kω 1 sin(ωt) V v D 5 V i D

11 PSPICE Result of Total Signal * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch Date/Time run: 09/15/14 17:12:36 (A) Diode_Analysis (active) 660.0mV Temperature: mV 637.5mV 630.0mV 0s 50ms 100ms V(R1:2,D1:2) Time Date: September 15, 2014 Page 1 Time: 17:14:25

12 Small Signal Analysis Approach 5 V 1 Find operating point. (Use any nonlinear technique) 1 kω V D I D V D = V; I D = ma 2 Obtain small signal circuit 1 sin(ωt) V 3 Add them 1 kω v d i d r d v D = sin(ωt) V r d = I D = = Ω v d = 1 r d R r d = V

13 PSPICE of Total and Small Signal Analysis * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch Date/Time run: 09/16/14 21:27:22 (A) Diode_Analysis (active) 660.0mV Temperature: mV 637.5mV 630.0mV 0s 50ms 100ms V(x) V(y)V(z) Time Date: September 16, 2014 Page 1 Time: 21:32:21

14 Example-II 1 kω 5 sin(ωt) V v D 5 V i D

15 PSPICE Result of Total Signal * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch Date/Time run: 09/18/14 14:34:35 (A) Diode_Analysis (active) 800mV Temperature: mV 0V 0s 50ms 100ms V(x) Time Date: September 18, 2014 Page 1 Time: 14:35:33

16 Small Signal Analysis Approach 1 DC operating point (Already done.) 2 Get the small signal circuit 5 sin(ωt) V 1 kω 3 Add operating point and small signal output v d i d r d r d = I D = = Ω v d = 5 r d R r d = V v D = sin(ωt) V As v d >, Small signal approach failed here... For large variation from DC operating point, any one of the nonlinear techniques should be used.

17 PSPICE of Total and Small Signal Analysis * C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch Date/Time run: 09/18/14 14:34:35 (A) Diode_Analysis (active) 800mV Temperature: mV 0V 0s 50ms 100ms V(x) V(y)V(z) Time Date: September 18, 2014 Page 1 Time: 14:36:37

Bipolar Junction Transistor (BJT) - Introduction

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