Chapter 13 Small-Signal Modeling and Linear Amplification

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1 Chapter 13 Small-Signal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 13-1

2 Chapter Goals Understanding of concepts related to: Transistors as linear amplifiers dc and ac equivalent circuits Use of coupling and bypass capacitors and inductors to modify dc and ac equivalent circuits Concept of small-signal voltages and currents Small-signal models for diodes and transistors Identification of common-source and common-emitter amplifiers Amplifier characteristics such as voltage gain, input and output resistances, and linear signal range Rule-of-thumb estimates for the voltage gain of common-emitter and common-source amplifiers. 1/4/12 Chap 13-2

3 Introduction to Amplifiers Amplifiers usually use electronic devices operating in the Active Region A BJT is used as an amplifier when biased in the forward-active region The FET can be used as amplifier if operated in the pinch-off or saturation region In these regions, transistors can provide high voltage, current and power gains Bias is provided to stabilize the operating point (the Q-Point) in the desired region of operation Q-point also determines Small-signal parameters of transistor Voltage gain, input resistance, output resistance Maximum input and output signal amplitudes Power consumption 1/4/12 Chap 13-3

4 Transistor Amplifiers BJT Amplifier Concept The BJT is biased in the active region by dc voltage source V BE. Q-point is set at (I C, V CE ) = (1.5 ma, 5 V) with I B = 15 µa (β F = 100) Total base-emitter voltage is: v BE = V BE + v be Collector-emitter voltage is: v CE = 10 i C R C This is the load line equation. 1/4/12 Chap 13-4

5 Transistor Amplifiers BJT Amplifier (cont.) If changes in operating currents and voltages are small enough, then i C and v CE waveforms are undistorted replicas of the input signal. A small voltage change at the base causes a large voltage change at collector. Voltage gain is given by: 8 mv peak change in v BE gives 5 µa change in i B and 0.5 ma change in i C. 0.5 ma change in i C produces a 1.65 V change in v CE. A v = V ce V be = o o = o = 206 Minus sign indicates 180 o phase shift between th einput and output signals. 1/4/12 Chap 13-5

6 Transistor Amplifiers MOSFET Amplifier Concept A v = V ds V gs A v = 4 180o 1 0 o A v = 4.00 MOSFET is biased in active region by dc voltage source V GS. Q-point is set at (I D, V DS ) = (1.56 ma, 4.8 V) with V GS = 3.5 V Total gate-source voltage is: v GS = V GS + v gs 1 V p-p change in v GS yields 1.25 ma p-p change in i D and a 4 V p-p change in v DS 1/4/12 Chap 13-6

7 Transistor Amplifiers Coupling and Bypass Capacitors Capacitors are designed to provide negligible impedance at frequencies of interest and provide open circuits at dc. ac coupling through capacitors is used to inject ac input signal and extract output signal without disturbing Q-point C 1 and C 2 are low impedance coupling capacitors or dc blocking capacitors whose reactance at the signal frequency is designed to be negligible. C 3 is a bypass capacitor that provides a low impedance path for ac current from emitter to ground, thereby removing R E (required for good Q-point stability) from the circuit when ac signals are considered. 1/9/12 Chap 13-7

8 Transistor Amplifiers dc and ac Analysis Two Step Analysis dc analysis: Find dc equivalent circuit by replacing all capacitors by open circuits and inductors by short circuits. Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model. ac analysis: Find ac equivalent circuit by replacing all capacitors by short circuits, inductors by open circuits, dc voltage sources by ground connections and dc current sources by open circuits. Replace transistor by its small-signal model Use small-signal ac equivalent to analyze ac characteristics of amplifier. Combine end results of dc and ac analysis to yield total voltages and currents in the network. 1/9/12 Chap 13-8

9 Transistor Amplifiers dc Equivalent Circuit for BJT Amplifier All capacitors in the original amplifier circuit are replaced by open circuits, disconnecting v I, R I, and R 3 from circuit. 1/9/12 Chap 13-9

10 Transistor Amplifiers ac Equivalent Circuit for BJT Amplifier Capacitors are replaced by short circuits 1/9/12 Chap 13-10

11 Transistor Amplifiers dc and ac Equivalents for a MOSFET Amplifier Full circuit dc equivalent ac equivalent Simplified ac equivalent 01/10/12 Chap 13-11

12 Small-Signal Operation Diode Small-Signal Model The slope of the diode characteristic at the Q-point is called the diode conductance and is given by: Diode resistance is given by: r d = 1 g d 01/10/12 Chap 13-12

13 Small-Signal Operation Diode Small-Signal Model (cont.) g d is small but non-zero for I D = 0 because slope of diode equation is nonzero at the origin. At the origin, the diode conductance and resistance are given by: g d = I S V T and r d = V T I S 01/10/12 Chap 13-13

14 Small-Signal Operation Definition of a Small-Signal Now let s determine the largest magnitude of v d that represents a small signal. V T i D = I S exp v D 1 V T I D +i d = I S exp V D 1 +I S exp v D V T I D +i d = I S exp V +v D d 1 v d V T Subtracting I D from both sides of the equation, i d =(I D + I S ) v d + V 1 T 2 2 v d V T v d V T V T v 2 d V T v 3 d V T For i d to be a linear function of signal voltage v d, v d << 2V T = 0.05 V or v d 5 mv. Thus v d 5 mv represents the requirement for small-signal (linear) operation of the diode i d =(I D + I S ) v d = g v i d d D = I D + g d v d V T 01/10/12 Chap 13-14

15 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) The bipolar transistor is assumed to be operating in the Forward-Active Region: i C I S exp v BE 1+ v CE and i B i C V T V A Using a two-port y-parameter network: i c = g m v be + g o v ce i b = g π v be + g r v ce The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. β F ( V CE V BE ) 01/10/12 Chap 13-15

16 Small-Signal Operation BJT Small-Signal Model (The Hybrid-Pi Model) i c = g m v be + g o v ce i b = g π v be + g r v ce i C I S exp v BE 1+ v CE i B i C β F V T V A β o is called the small-signal commonemitter current gain of the BJT. g m = i c = i C = I C v be v v ce=0 BE V Q-point T v be g o = i c v ce v be=0 g π = i b v be v ce=0 g r = i b v ce v be=0 v BE = i C v CE Q-point = i B v BE Q-point = i B v CE Q-point V T I C = V A +V T = I C β o V T = 0 01/10/12 Chap 13-16

17 BJT Small-Signal Operation Current Gain & Intrinsic Voltage Gain Intrinsic voltage gain is defined by: β o = g m r π = β F 1 β 1 I F C β F i C Q point β o > β F for i C < I M, and β o < β F for i C > I M. However, for simplicity β F and β o will be assumed to be equal µ f = g m r o = I C V T For V CE << V A µ f V A V T V A +V CE I C 40V A = V A +V CE V T µ f represents the maximum voltage gain an individual BJT can provide and does not change with operating point. 01/10/12 Chap 13-17

18 Small-Signal Operation BJT Hybrid-Pi Model - Summary Transconductance: g m = I C V T 40I C Input resistance: The hybrid-pi small-signal model is the intrinsic representation of the BJT. Small-signal parameters are controlled by the Q-point and are independent of geometry of the BJT r π = β o V T I C Output resistance: r o = V A +V CE I C = β o g m V A I C or β o = g m r π 01/10/12 Chap 13-18

19 BJT Small-Signal Operation Equivalent Forms of Small-Signal Model Voltage-controlled current source g mv be can be transformed into current-controlled current source, v be = i b r π g m v be = g m i b r π = β o i b i c = β o i b + v ce r o β o i b Basic relationship i c = βi b is useful in both dc and ac analysis when the BJT is in the forward-active region. 01/10/12 Chap 13-19

20 BJT Small-Signal Operation Small-Signal Definition i C = I S exp v BE = I S exp V BE + v be i C = I C + i c = I S exp V BE exp v be = I C exp v be V T I C + i c = I C 1+ v be + 1 V T 2 v be V T V T v i c = I be C v be + 1 v be V T 2 V T 6 V T 3 v be V T For linearity, i c should be proportional to v be with v be << 2V T or v be V. i C I C 1+ v be v = I C + I be C The change in i c that corresponds to small-signal operation is: V T V T i c I C = v be V T 0.005V 0.025V = V T V T = I C + g m v be V T 01/12/12 Chap 13-20

21 BJT Small-Signal Operation Small-Signal Model for pnp Transistor For the pnp transistor Signal current injected into base causes decrease in total collector current which is equivalent to increase in signal current entering collector. So the small-signal models for the npn and pnp devices are identical! 01/12/12 Chap 13-21

22 Common-Emitter Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. Assume that Q-point is already known. 01/12/12 Chap 13-22

23 Common-Emitter Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the base terminal Output signal appears at collector terminal Emitter is common to both input and output signals Thus circuit is termed a Common-Emitter (C-E) Amplifier. The terminal gain of the C-E amplifier is the gain from the base terminal to the collector terminal A vt CE = v c v b = g m R L R L = r o R C R 3 01/12/12 Chap 13-23

24 Common-Emitter Amplifiers Input Resistance and Signal Source Gain (β o +1)R E Define R ib as the input resistance looking into the base of the transistor: R ib = v b i b = r π The input resistance presented to v i is: R in = R I + R B R ib = R I + R B r π The signal source voltage gain is: A v CE = v o v i = v o v b v b v i = A vt CE R B r π R I + R B r π 01/12/12 Chap 13-24

25 Common-Emitter Amplifiers Rule of Thumb Design Estimate A CE CE R v = A B r π CE vt A vt R I + R B r π A CE vt = g m R L R L = r o R C R 3 Typically: r o >> R C and R 3 >> R C A v CE g m R C = 40I C R C I C R C represents the voltage dropped across collector resistor R C A typical design point is I C R C = V CC 3 A v CE 40 V CC 3 =13.3V CC To help account for all the approximations and have a number that is easy to remember, our "rule-of-thumb" estimate for the voltage gain becomes A v CE 10V CC 01/13/12 Chap 13-25

26 Common-Emitter Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-emitter amplifier with a specified Q-point Given data: β F = 100, V A = 75 V, Q-point is (0.245 ma, 3.39 V) Assumptions: Transistor is in active region, β O = β F. Signals are low enough to be considered small signals. Room temperature. Analysis: g m = 40I C = mA r o = V A +V CE I C = ( ) = 9.80 ms r π = β o = 100 =10.2 kω g m 9.8mS 75V V 0.245mA = 320 kω R B = R 1 R 2 =160kΩ 300kΩ =104 kω R L = r o R C R L = 320kΩ 22kΩ 100kΩ =17.1 kω R B r π =104kΩ 10.2kΩ = 9.29 kω 7/20/10 Chap 13-26

27 Common-Emitter Amplifiers Voltage Gain Example (cont.) Analysis (cont): R A v = g m R B r π L R I + R B R in = R I + R B r π r π =10.3 kω = 9.8mS 17.1kΩ ( ) 9.29kΩ = kΩ+ 9.29kΩ R v be = v B r π i R I + R B r v be 0.005V v i 5mV 10.3kΩ = 5.54 mv π 9.29kΩ ( ) = 151 Check the rule-of-thumb estimate: A v CE ( ) = 120 (ballpark estimate) ( ) = V What is the maximum amplitude of the output signal: v o 5.54mV 151 1/13/12 Chap 13-27

28 Common-Emitter Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 5-mV, 10-kHz sine wave. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion. The input is near our small-signal limit for linear operation. 1/13/12 Chap 13-28

29 Common-Emitter Amplifiers Dual Supply Operation - Example Analysis: To find the Q-point, the dc equivalent circuit is constructed I B +V BE +(β F +1)I B ( )= 5 I B =3.71 µa I C =65I B =241 µa Problem: Find voltage gain, input and output resistances for the circuit above Given data: β F = 65, V A = 50 V Assumptions: Active-region operation, V BE = 0.7 V, small signal operating conditions. I E =66I B =245 µa I C V CE ( )I E ( 5)=0 V CE =3.67 V 1/28/12 Chap 13-29

30 Common-Emitter Amplifiers Dual Supply Operation - Example (cont.) Next we construct the ac equivalent and simplify it kω R in = R B r π =6.31 kω R out = R C r o = 9.57 kω ( ) A v CE = v o v i = g m R out R 3 R in = 84.0 R I + R in Gain Estimate: A CE v 10( V CC +V EE ) = 100 1/28/12 Chap 13-30

31 Small-Signal Operation MOSFET Small-Signal Model g π = i g v gs vds =0 g r = i g v ds v ds =0 = i G v GS Q-point = i G v DS Q-point Using a two-port y-parameter network, i d g m = vgs vds=0 i D = vgs Q-point i g = g π v gs + g r v ds i d = g m v gs + g o v ds The port variables can represent either time-varying part of total voltages and currents or small changes in them away from Q-point values. g o = i d v ds v ds =0 I G = 0 = i D v DS Q-point I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) 1/17/12 Chap 13-31

32 Small-Signal Operation MOSFET Small-Signal Model (cont.) I G = 0 I D = K n ( 2 V V GS TN ) 2 ( 1+ λv DS ) g π = i G v GS Q-point g r = i G v DS Q-point = 0 = 0 i g = g π v gs + g r v ds i d = g m v gs + g o v ds g m = i D v GS Q-point g o = i D v DS Q-point = K n ( 2 V V GS TN )( 1+ λv DS ) = = λ K n ( 2 V GS V TN ) 2 = λi D = 1+ λv DS 2I D V GS V TN I D 1 λ +V DS 1/17/12 Chap 13-32

33 Small-Signal Operation MOSFET Small-Signal Model - Summary Transconductance: g m = 2I D = 2K n I D V GS V TN Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite. Small-signal parameters are controlled by the Q-point. For the same operating point, MOSFET has lower transconductance and an output resistance that is similar to the BJT. Output resistance: r o = 1 g o = 1+λV DS λi D 1 λi D Amplification factor for λv DS <<1: µ f =g m r o = 1+λV DS λi D 1 λ 2K n I D 1/17/12 Chap 13-33

34 MOSFET Small-Signal Operation Small-Signal Definition Assume λv DS <<1. Then i D K n ( 2 v V GS TN ) 2 for v DS v GS V TN For v GS = V GS + v gs, i D = I D +i d = K n 2 2 ( ) + v gs ( v GS V TN ) 2 + 2v gs V GS V TN 2 ( ) + v gs i d = K n 2 2v gs V GS V TN For linearity, i d should be proportional to v gs and we require v 2 gs << 2v gs V GS V TN or v gs << 2( V GS V TN ) v gs 0.2( V GS V TN ) ( ) Since the MOSFET can be biased with (V GS - V TN ) equal to several volts, it can handle much larger values of v gs than corresponding the values of v be for the BJT. The change in drain current that corresponds to small-signal operation is: i d = g m 2 v gs I D I D V GS V TN ( ) 0.2 V GS V TN i d 0.4 I D 1/17/12 Chap 13-34

35 MOSFET Small-Signal Operation Body Effect in Four-terminal MOSFETs Drain current depends on threshold voltage which in turn depends on v SB. Back-gate transconductance is: g mb = i D = i D v BS v Q-point SB Q-point i D g mb = VTN V TN v SB Q-point = ( g m η)=g m η 0 < η < 3 is called the back-gate transconductance parameter. The bulk terminal is a reverse-biased diode. Hence, no conductance from the bulk terminal to other terminals. 01/18/12 Chap 13-35

36 MOSFET Small-Signal Operation Small-Signal Model for PMOS Transistor For a PMOS transistor v SG = V GG v gg i D = I D i d Positive signal voltage v gg reduces sourcegate voltage of the PMOS transistor causing decrease in total current exiting the drain, equivalent to an increase in the signal current entering the drain. The NMOS and PMOS small-signal models are the same! 01/18/12 Chap 13-36

37 Common-Source Amplifiers Small-Signal Analysis - ac Equivalent Circuit ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. 01/19/12 Chap 13-37

38 Common-Source Amplifiers Small-Signal Equivalent Circuit Input voltage is applied to the gate terminal Output signal appears at the drain terminal Source is common to both input and output signals Thus circuit is termed a Common-Source (C-S) Amplifier. The terminal gain of the C-S amplifier is the gain from the gate terminal to the drain terminal A vt CE = v d v g = g m R L R L = r o R D R 3 01/19/12 Chap 13-38

39 Common-Source Amplifiers Input Resistance and Signal-Source Gain Define R ig as the input resistance looking into the base of the transistor. R vg ig = = R G ii R in is the resistance presented to v i. R in = R I + R G The signal source voltage gain is: A v CS = v o v i = v o v g v g v i CS = A vt A CS R v = g m R G L R I + R G R G R I + R G 01/12/12 Chap 13-39

40 Common-Source Amplifiers Rule of Thumb Design Estimate R A CS v = g m R G L A vt R I + R G CS A CS vt = g m R L R L = r o R D R 3 I Typically: r o >> R D and R 3 >> R D A CS v g m R D = D R D V GS V TN 2 I D R D represents the voltage dropped across drain resistor R D A typical design point is I D R D = V DD 2 with V GS V TN =1 V A v CS V DD Our rule-of-thumb estimate for the C-S amplifier: the voltage gain equals the power supply voltage. Note that this is 10 times smaller than that for the BJT! 01/19/12 Chap 13-40

41 Common-Source Amplifiers Voltage Gain Example Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-source amplifier with a specified Q-point Given data: Κ n = 0.50 ma/v 2, V TN = 1 V, λ = V -1, Q-point is (0.241 ma, 3.81 V) Assumptions: Transistor is in the active region. Signals are low enough to be considered small signals. Analysis: g m = 2K n I D ( 1+ λv DS ) = ms r o = λ 1 +V DS I D = 328 kω R G = R 1 R 2 = 892 kω R L = r o R D R 3 =17.1 kω 1/25/12 Chap 13-41

42 Common-Source Amplifiers Voltage Gain Example (cont.) g m = ms r o = 328 kω R G = 892 kω R L =17.1 kω R A CS v = g m R G 892kΩ L = 0.503mS ( 17.1kΩ) = R I + R G 1kΩ+892kΩ R R in = R I + R G = 893 kω v gs = v G R i v G i 0.2 V GS V TN R I + R G R I + R G V GS V TN 2I D K n = V v i V ( ) 893kΩ ( ) = 8.59 ( ) = V 892kΩ Check the rule-of-thumb estimate: A v CS V DD = 12 V (ballpark estimate) 1/25/12 Chap 13-42

43 Common-Source Amplifiers Voltage Gain Example (cont.) Simulation Results: The graph below presents the output voltage for an input voltage that is a 0.15-V, 10-kHz sine wave. The expected output voltage amplitude is vo = 8.59(0.15) = 1.29 V. Note that although the sine wave at first looks good, the positive and negative peak amplitudes are different indicating the presence of distortion, and the amplitude is actually larger than expected. The input is near our small-signal limit for linear operation. 1/25/12 Chap 13-43

44 C-E and C-S Amplifiers Output Resistance 01/19/12 Chap 13-44

45 C-E and C-S Amplifiers Output Resistance (cont.) Apply test source v x and find i x (with v i = 0) v be = 0 g m v be = 0 R out = v x i x = R C r o R out R C for r o >> R C v gs = 0 g m v gs = 0 R out = v x i x = R D r o R out R D for r o >> R D For comparable bias points, output resistances of C-S and C-E amplifiers are similar. 01/19/12 Chap 13-45

46 JFET Small-Signal Operation Small-Signal Model g g = 1 r g = i G v GS Q-point = I G + I SG V T 0 for I G = 0 i D = I DSS 1 v GS V P for v DS v GS V P i G = I SG exp v GS 1 V T 2 ( 1+ λv DS ) g m = i D = v GS Q-point g o = 1 r o = i D v DS Q-point 2I D 2 I DSS V 2 GS V P V GS V P V P = I D 1 λ +V DS ( ) 01/18/12 Microelectronic Circuit Design Chap 13-46

47 JFET Small-Signal Operation Small-Signal Model (cont.) For small-signal operation, the input signal limit is: v gs <0.2( V GS V P ) Since JFET is normally operated with gate junction reverse-biased, I G = I SG r g = The amplification factor is given by: 1 µ f = g m r o = 2 λ +V DS 2 V GS V λv P P I DSS I D 01/18/12 Microelectronic Circuit Design Chap 13-47

48 Common-Source Amplifiers JFET Example Analysis: Dc equivalent circuit is constructed. Ι G = 0, Ι S = Ι D. V GS = 2000I D V GS = ( )( )1 V GS ( 1) Choose V GS less negative than V P. 2 Problem: Find voltage gain, input and output resistances. Given data: Ι DSS = 1 ma, V P = -1V, λ = 0.02 V -1 Assumptions: Pinch-off region of operation. V GS = 0.5 V I D = ma 12= 27,000I D +V DS +2000I S V DS = 4.75V 1/28/12 Microelectronic Circuit Design Chap 13-48

49 Common-Source Amplifier JFET Example (cont.) Next we construct the ac equivalent and simplify it. 1/28/12 Microelectronic Circuit Design Chap 13-49

50 BJT and FET Small-Signal Model Summary 01/18/12 Chap 13-50

51 Common-Emitter/Common-Source Amplifiers Summary 1/25/11 Chap 13-51

52 Amplifier Power Dissipation Static power dissipation in amplifiers is found from their dc equivalent circuits. (a) Total power dissipated in the C-B and E-B junctions is: P D = V CE I C + V BE I B Total power supplied is: P S = V CC I C + V EE I E where V CE = V CB + V BE (b) Total power dissipated in the transistor is: P D = V DS I D + V GS I G = V DS I D Total power supplied is: P S = V DD I D +V DD2 /(R 1 +R 2 ) 1/28/12 Chap 13-52

53 Amplifier Signal Range v CE = V CE V m sinωt where V m is the output signal. Active region operation requires v CE v BE So: V m V CE V BE Also: v Rc ( t) = I C R C V m sinωt 0 ( ) V m min I C R C, V CE V BE Similarly for MOSFETs and JFETs: V M min I D R D,(V DS (V GS V TN )) V M min I D R D,(V DS (V GS V P )) 1/28/12 Chap 13-53

54 End of Chapter 13 1/28/12 Chap 13-54

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