Class AB Output Stage


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1 Class AB Output Stage Class AB amplifier Operation Multisim Simulation  VTC Class AB amplifier biasing Widlar current source Multisim Simulation  Biasing 1
2 Class AB Operation v I V B (set by V B )
3 Basic Class AB Amplifier Operation 1. Bias Q N and Q P into slight conduction (fwd. act.) when v I = 0: i N = i P. i L =i N i P Ideally Q N and Q P are: a. Matched (unlikely with discrete transistors and challenging in IC). b. Operate at same ambient temperature. 3.For v i > 0: i N > i P i.e. Q N most cond. (like Class B). 4.For v i < 0: i P > i N i.e. Q P most cond. (like Class B). NOTE. This is basevoltage biasing with all its stability problems! 3
4 Class AB VTC Plot Ideally the two DC base voltage sources are matched and / / 0.7V = V. Ideally, zero crossover distortion 4
5 Amplitude: 0 V p Frequency: 1 khz Class AB VTC Simulation V CC Looks like Class A VTC / R Sig / R L V CC 5
6 Class AB VTC Simulation  cont. Amplitude: V p Frequency: 1 khz Crossover distortion =0.1V =0.5V =0.7V 6
7 Class AB Amplifier Operation  cont. for v i 0 v i + v BEN = v i v O v EBP =v O v i Output voltage for v i 0: for v i 0 v o =v i v BEN v o v i for v i 0 v o =v i v EBP v o v i i N =i P i L Baseto base voltage is constant! Bias (Q N & Q P matched): I N =I P = = e V T v BEN v EBP = Let us next show that i N i P = for all v i for all v i 7
8 i N = e ESE319 Introduction to Microelectronics Class AB Amplifier Operation  cont. for v i 0 v o =v i v BEN v BEN =v i v o for v i 0 v o =v i v EBP v EBP =v o v i Using the currents v BEN V T v BEN =V T ln i N v BEN v EBP = V T ln i N for all v i ADD Note for Class B = 0 v EBP V i P = e T v EBP =V T ln i P I S V I N =I P = = e T =V T ln I Q V T ln i P I =V T S ln I Q I for all v S i 8
9 Class AB Amplifier Operation  cont. i N =i P i L from the previous slide V T ln i N I V T S ln i P I =V T S ln I Q V T ln i N i P =V T ln ln i N i P ln =ln ln ln i N i P =ln or i N i P = Constant base voltage condition: v BEN v EBP = => i N i P = 9
10 Class AB Amplifier Operation VTC cont. The constant base voltage condition i P i N = For example let = 1 µa and i N = 10 ma. i P = I Q = =0.1 ma= 3 i N i N where is typically small. The Class AB circuit, over most of its input signal range, operates as if either the Q N or Q P transistor is conducting and the Q P or Q N transistor is cut off. For small values of v I both Q N and Q P conduct, and as v I is increased or decreased, the conduction of Q N or Q P dominates, respectively. Using this approximation we see that a class AB amplifier acts much like a class B amplifier; but with a much reduced dead zone. 10
11 Class AB Power Conversion Efficiency & Power Dissipation Similar to Class B P Disp P Disp(max) = 0.9 W Let V CC = 1 V and R L =100 P Disp max = V CC =0.9 W R L P Disp B = V o peak R L V CC 1 V o peak R L 0.0 W 0.7 V Accurate for small V opeak. = 7.63 V V o peak P Disp 0 when V opeak = 0 11
12 D1 D ESE319 Introduction to Microelectronics Class AB Amplifier Biasing current mirror +  QN QP =V CC R R V CC A straightforward biasing approach: D1 and D are diodeconnected transistors identical to QN and QP, respectively. They form mirrors with the quiescent currents set by matched R's: = V CC 1.4 R R= V CC 0.7 = V CC 0.7 R Recall: With mirrors, the ambient temperature for all transistors needs to be matched! or: 1
13 V CC I REF R V BE1 ESE319 Introduction to Microelectronics Widlar Current Source I N = bias current for Class AB amplifier NPN V BE R e = I N Q = QN emitter degeneration V BE1 =V T ln I REF V BE =V T ln I Q V BE1 =V BE R e V BE1 V BE = R e I REF = V CC V BE1 R = 1V 0.7V 11.3 k =1mA Note: Pages in Sedra & Smith Text. V BE1 V BE =V T ln I REF =V I T ln I REF Q R e =V BE1 V BE =V T ln I REF Note R e 0 iff I REF 13
14 V CC R I REF ESE319 Introduction to Microelectronics Widlar Current Source  cont. R e =V T ln I REF R= V CC V BE1 I REF R e 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 = 1 V 0.7V 10 ma R e = V T ln I REF I = 0.05V Q 10 A.=500 ln1000=17.7 k R=1.13 k =1.13 k ln 10 m A 10 A R e =17.7 k 14
15 Widlar Current Mirror SmallSignal Analysis.. R e R R e e R out r 1/ g m 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. g m r R e i x v x r o R out is greatly enhanced by adding emitter degeneration. R out = v x i x r o [ g m R e r ] g m R e r 1 15
16 Class AB Current Biasing Simulation Bias currents set at I REF and by R and emitter resistor(s) R e. I REF 4 ma =N =P ma NPN Widlar current mirror 6.31mA R=.8 kω I REF N Q1 i N Q R e =10 Ω R L =100 Ω i L =i N i P Amplitude: 0 V p Frequency: 1 khz R=.8 kω Q3 Q4 I REF P ii i L R e =10 Ω.3mA 39µA R= V V CC BE1 = V V CC EB3.8 k I REF I REF R e = V T N ln I REF I 9 QN PNP Widlar current mirror 16
17 Conclusions ADVANTAGE: Class AB operation improves on Class B linearity. Power conversion efficiency similar to Class B DISADVANTAGES: 1. Emitter resistors absorb output power.. Power dissipation for low signal levels higher than Class B. 3. Temperature matching will be needed more so. if emitter degeneration resistors are not used. 17
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