# Class AB Output Stage

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1 Class AB Output Stage Class AB amplifier Operation Multisim Simulations - Operation Class AB amplifier biasing Multisim Simulations - Biasing 1

2 Class AB Operation 2

3 Basic Class AB Amplifier Circuit Bias Q N and Q P into slight conduction when v I = 0. i L =i N i P Ideally Q N and Q P are: 1. Matched (unlikely with discrete transistors). 2. Operate at same temperature. NOTE. This is base-voltage biasing with all its stability problems! 3

4 Class AB Circuit Operation for v i = 0 v BEN = V BB 2 v O v EBP =v O V BB 2 Output voltage: for v i 0 v o =v i V BB 2 v BEN v o v i for v i 0 v o =v i V BB 2 v EBP v o v i Base-to base voltage is constant! v BEN v EBP =V BB Bias: I N =I P =I Q = e Also: i N = e V BB 2 V T v BEN v T & i N =i P i L i P = e v EBP V T Using: V T ln i N i N = e v BEN v T, i P = e & I Q = e V T ln i P v EBP V T =2V T ln I Q V BB 2V T 4

5 Class AB Circuit Operation - cont. i N =i P i L V T ln i N V T ln i P =2V T ln I Q V T ln i N i P 2 =2V T ln I Q V T ln i N i P V T ln 2 =2V T ln I Q 2V T ln ln i N i P =ln I Q 2 Constant base voltage condition: i N i P =I Q 2 5

6 Class AB Circuit Operation - cont. Constant base voltage condition: 2 i N i P =I Q i N =i P i L Kirchhoff's Current Law condition: i N =i P i L i P =i N i L Combining equations: i L = v O R L i N i N i L =I Q 2 for v I > 0 V Hence: i 2 N i N i L I 2 Q =0 or or 2 i P i P i L =I Q for v I < 0 V i 2 P i P i L I 2 Q =0 for v I > 0 V for v I < 0 V 6

7 Class AB Circuit Operation - cont. 2 i L =i N i P i N i P =I Q Observations: 2 1. Since with constant base voltage V BB : i N i P =I Q, if i P increases by i P =i P 1i then i N decreases by i N =i N /1i, and vice-versa. 2. For v O > 0, the load current i L supplied by the complementary emitter followers Q N and Q P. As v O increases, i P decreases and for large, positive v O i.e. i P =i N i L =i N v O hence v R O large i P 0 L 3. For v O < 0, the load current i L supplied by the complementary emitter followers Q N and Q P. As v O decreases, i N decreases and for large, negative v O i.e. i N =i P i L =i P v O R L hence v O large i N 0 7

8 Class AB Circuit Operation - cont. i P = I 2 Q Write the product equation as: where I i Q is typically small. N For example let I Q = 1 ma and i N = 10 ma. i P = =0.1 ma= i N The Class AB circuit, over most of its input signal range, operates as if the Q N or Q P transistor is conducting and the Q P or Q N transistor is cut off. Using this approximation we see that a class AB amplifier acts much like a class B amplifier; but without the dead zone. 8

9 Class AB VTC Plot 9

10 Amplitude: 20 V p Frequency: 1 khz Class AB VTC Simulation V CC V BB /2 R Sig V BB /2 R L -V CC 10

11 Class AB VTC Simulation - cont. Amplitude: 2 V p Frequency: 1 khz V BB 2 =0.1V V BB 2 =0.5V V BB 2 =0.7V 11

12 v I = 0 V ESE319 Introduction to Microelectronics Class AB Small-Signal Output Resistance ac ground BN BP ac ground CN EN EP CP <=> R out =r en r ep for v I > 0 V: R out r en for v I < 0 V: R out r ep Instantaneous resistance for the Q N transistor - assume α 1: di N = e dv BEN V T v BEN V T For the Q P transistor: di P = i P dv EBP V T Hence: r en = V T i N and = i N V T r ep = V T i P 12

13 Small-Signal Output Resistance - cont. The two emitter resistors are in parallel: V T 2 R out =r en r ep = i N i P V T i N V T i P = V T i N i P i Ni P i N i P V T = i N i P At i N = i P (the no-signal condition i.e. v O = 0 => i L = 0): i N =i P =I Q R out = V T 2 I Q So, for small signals, a load current flows => no dead-zone! 13

14 I Q I Q + V BB - ESE319 Introduction to Microelectronics Class AB Amplifier Biasing I Q QN QP I Q A straightforward biasing approach: D1 and D2 are diode-connected transistors identical to QN and QP, respectively. They form mirrors with the quiescent current set by R: or: I Q = 2V CC 1.4 2R R= V CC 0.7 I Q = V CC 0.7 R Recall: With mirrors, the device temperature for all transistors needs to be matched! 14

15 V CC I REF R V BE1 ESE319 Introduction to Microelectronics Widlar Current Source I N = bias current for Class AB amplifier NPN I REF = V CC V BE1 12V 0.7V = R 11.3k =1mA V BE2 R e = I N Q2 = QN emitter degeneration Note: Pages in Sedra & Smith Text. V BE1 =V T ln I REF V BE2 =V T ln I O V BE1 V BE2 =V T ln I REF V BE1 =V BE2 R e R e =V T ln I REF Note R e > 0 iff < I REF =V T ln I REF 15

16 V CC ESE319 Introduction to Microelectronics Widlar Current Source - cont. R e =V T ln I REF = V T I REF R If R e specified and I REF given: R e ln I REF ln R e Solve for graphically. If specified (and I REF chosen by designer): R e = V T ln I REF Example Let = 10 µa & choose I REF = 10 ma, determine R and R e : R= V CC V BE1 I REF R e = V T ln I REF 12V 0.7V = 10 ma = 0.025V 10 A.=2500 ln1000=17.27 k R=1.13k =1.13 k ln 10 m A 10 A R e =17.27 k 16

17 Widlar Current Mirror Small-Signal Analysis.. i x =g m v i ro =g m v v x v r o v = r R e i x i x = g m r R e i x v x r o r R e i x r o r 1/ g m R out is greatly enhanced by adding emitter degeneration. R out = v x i x = R e r 1g m r o g m r o 1 17

18 Class AB Current Biasing Simulation Bias currents set at I REF and by R and emitter resistor(s) R e. I NPN Widlar current mirror REF 4 ma 2 ma R=2.8 kω I REF N i N R e =10 Ω R L =100 Ω Amplitude: 0 V p Frequency: 1 khz ii i L R e =10 Ω R= V CC V BE1 I REF R=2.8 kω I REF P R e = V T ln I REF PNP Widlar current mirror 18

19 Class AB I REF = 4 ma Current Bias Simulation Amplitude: 0 V p Frequency: 1 khz NPN Widlar current mirror 4.031m 2.329m Quiescent Power Dissipation v I = 0 V: P Disp =P D av =76.31 mw mw.= mw 4.025m 2.270m PNP Widlar current mirror 19

20 Class AB 4 ma Current Bias Simulation - cont. Amplitude: 12 V p Frequency: 1 khz For linear operation: - 9 V < v O < +9 V VTC has no dead-zone. 20

21 Q3 i REF Class AB VTC Limits Q2 Linear VTC for v I-max v I 0 => 1. Q2, Q3 forward-active Q2 Saturation => v CE2 =V CC i N R e v O V CE2sat where v O = i N R e v I v CE2 =V CC v I V CE2sat v I Q4 i P v O Q1 v I V CC V CE2sat =v I max1 2. Q3 Cutoff => v BQ3 v I 0.7V v BQ3 =V CC i REF R v I V CC i REF R 0.7 V =v I max2 v I max =max v I max1, v I max2 21

22 Class AB 4 ma Current Bias Simulation - cont. P D+av V BQ m I REFN 0.018m N 309 mw P D+av V BQ3 P L-av V I 4.734m 2.697m P 1.776u VO P D-av P D-av P Lav NOTE: 1. Linear operation up to V I = 9 V: P Dav = 946mW, P Lav = 510mW => = P Lav 510 mw = P Dav 946mW = At V I = 10 V: V BEQ3 = V BN V I = -0.1V NPN Q3 is cutoff for V I 9.5 V. 3. By symmetry PNP Q4 is cutoff for V I -9.5V. 22

23 Conclusions ADVANTAGE: Class AB operation improves on Class B linearity. DISADVANTAGES: 1. Emitter resistors absorb output power. 2. Power Conversion Efficiency is less than Class B. 3. Temperature matching will be needed more so. if emitter resistors are not used. TRADEOFFS: Tradeoffs - involving bias current - between power efficiency, power dissipation and output signal swing need to be addressed. 23

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