ESE319 Introduction to Microelectronics Class AB Output Stage ● ● ● ● ● Class AB amplifier Operation Multisim Simulation - VTC Class AB amplifier biasing Widlar current source Multisim Simulation - Biasing Kenneth R. Laker, updated 14Nov14 KRL 1 ESE319 Introduction to Microelectronics Class AB Operation vI VB IQ IQ (set by VB) Kenneth R. Laker, updated 14Nov14 KRL 2 ESE319 Introduction to Microelectronics Basic Class AB Amplifier Operation 1. Bias QN and QP into slight conduction (fwd. act.) when vI = 0: iN = iP. i L =i N −i P V BB 2 Ideally QN and QP are: a. Matched (unlikely with discrete transistors and challenging in IC). b. Operate at same ambient temperature. 3.For vi > 0: iN > iP i.e. QN most cond. (like Class B). 4.For vi < 0: iP > iN i.e. QP most cond. (like Class B). NOTE. This is base-voltage biasing with all its stability problems! Kenneth R. Laker, updated 14Nov14 KRL 3 ESE319 Introduction to Microelectronics Class AB VTC Plot Ideally the two DC base voltage sources are matched and V BB = 0.7 VV. / 2≈0.7 BB/2 Ideally, zero cross-over distortion Kenneth R. Laker, updated 14Nov14 KRL 4 ESE319 Introduction to Microelectronics Amplitude: 20 Vp Frequency: 1 kHz Class AB VTC Simulation VCC Looks like Class A VTC VBB/2 RSig VBB/2 RL -VCC Kenneth R. Laker, updated 14Nov14 KRL 5 ESE319 Introduction to Microelectronics Class AB VTC Simulation - cont. Amplitude: 2 Vp Frequency: 1 kHz Cross-over distortion V BB =0.1V 2 V BB =0.5V 2 V BB =0.7V 2 Kenneth R. Laker, updated 14Nov14 KRL 6 ESE319 Introduction to Microelectronics Class AB Amplifier Operation - cont. for vi ≠ 0 vi + V BB V BB v BEN = v i −v O 2 −V BB v EBP =v O − v i 2 i N =i P i L Bias (QN & QP matched): I N =I P = I Q = I S e V BB 2V T Kenneth R. Laker, updated 14Nov14 KRL Output voltage for vi ≠ 0: V BB for v i 0 v o =v i −v BEN ⇒ v o≈v i 2 V BB for v i 0 v o =v i − v EBP ⇒ v o≈v i 2 Base-to base voltage is constant! v BEN v EBP =V BB for all v i Let us next show that i N i P = I Q2 for all vi 7 ESE319 Introduction to Microelectronics Class AB Amplifier Operation - cont. V BB V BB for v i 0 v o=v i −v BEN ⇒ v BEN =v i −v o 2 2 V BB V BB for v i 0 v o=v i − v EBP ⇒ v EBP =v o−v i 2 2 v BEN v EBP =V BB Using the currents i N =I S e v BEN VT for all vi ADD Note for Class B VBB = 0 v EBP VT iP iN i P =I S e ⇒ v EBP =V T ln ⇒ v BEN =V T ln IS IS V IQ 2V I N =I P = I Q = I S e ⇒ V BB=2 V T ln IS iN iP IQ V T ln V T ln =2 V T ln for all vi IS IS IS Kenneth R. Laker, updated 14Nov14 KRL BB T 8 ESE319 Introduction to Microelectronics Class AB Amplifier Operation - cont. i N =i P i L from the previous slide iN iP IQ V T ln V T ln =2 V T ln IS IS IS V T ln iN i P I 2S IQ =2 V T ln IS ln i N i P −ln I 2S =2 ln I Q −2 ln I S 2 ln i N i P =ln I Q2 or i N i P = I Q Constant base voltage condition: Kenneth R. Laker, updated 14Nov14 KRL v BEN v EBP =V BB => i N i P = I Q2 9 ESE319 Introduction to Microelectronics Class AB Amplifier Operation – VTC cont. The constant base voltage condition i P i N = I Q2 where IQ is typically small. For example let IQ = 1 µA and iN = 10 mA. I 2Q 1⋅10−6 1 i P= = =0.1 mA= iN −3 i N 10⋅10 100 The Class AB circuit, over most of its input signal range, operates as if either the QN or QP transistor is conducting and the QP or QN transistor is cut off. For small values of vI both QN and QP conduct, and as vI is increased or decreased, the conduction of QN or QP 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. Kenneth R. Laker, updated 14Nov14 KRL 10 ESE319 Introduction to Microelectronics Class AB Power Conversion Efficiency & Power Dissipation Similar to Class B Let VCC = 12 V and R L =100 P Disp 2 P Disp max= 2 V CC 2 RL 2 =0.29 W 2 V o− peak 1 V o− peak P Disp−B = V CC − RL 2 RL PDisp(max) = 0.29 W 0.20 W 0.7 V Accurate for small Vo-peak. Kenneth R. Laker, updated 14Nov14 KRL V o− peak = 7.63 V P Disp ≠ 0 when Vo-peak = 0 11 ESE319 Introduction to Microelectronics Class AB Amplifier Biasing current mirror D1 D2 A straightforward biasing approach: IQ D1 and D2 are diode-connected IQ transistors identical to QN and QP, QN + respectively. They form mirrors with the quiescent VBB currents IQ set by matched R's: 2 V CC −1.4 V CC −0.7 QP I Q= = IQ IQ 2R R or: V CC −0.7 R= V BB=V CC −I Q R− I Q R−V CC IQ Recall: With mirrors, the ambient temperature for all transistors needs to be matched! Kenneth R. Laker, updated 14Nov14 KRL 12 ESE319 Introduction to Microelectronics Widlar Current Source IN = bias current for Class AB amplifier NPN R IREF VCC IQ IQ = IN IQ V BE2 =V T ln IS Q2 = QN + + - VBE1 VBE2IQ Re emitter degeneration V CC −V BE1 12V −0.7V I REF = = =1 mA R 11.3 k I REF V BE1=V T ln IS V BE1=V BE2 I Q R e ⇒V BE1−V BE2=I Q R e I REF I S I REF V BE1−V BE2 =V T ln =V T ln IS IQ IQ Note: Pages 543-546 in Sedra & Smith Text. I REF I Q R e =V BE1−V BE2=V T ln IQ Note Re ≥ 0 iff IQ ≤ IREF Kenneth R. Laker, updated 14Nov14 KRL 13 ESE319 Introduction to Microelectronics Widlar Current Source - cont. If IQ specified and IREF chosen by designer: R IREF VT I REF Re= ln IQ IQ IQ VCC IQ Re I REF I Q R e =V T ln IQ V CC −V BE1 R= I REF Example Let IQ = 10 µA & choose IREF = 10 mA, determine R and Re: V CC −V BE1 12 V −0.7 V R= = =1.13 k I REF 10 mA VT I REF 0.025 V 10 m A Re= ln = ln IQ IQ 10 A 10 A .=2500 ln 1000=17.27 k R=1.13 k Kenneth R. Laker, updated 14Nov14 KRL R e =17.27 k 14 ESE319 Introduction to Microelectronics Widlar Current Mirror Small-Signal Analysis .≈. Re Re v x −−v i x =g m v i ro =g m v ro v =−r ∥R e i x v x r ∥R e i x i x =−g m r ∥R e i x − ro ro vx .≈−g m r ∥R e i x ro Kenneth R. Laker, updated 14Nov14 KRL Re Rout r ≫1/ g m Rout is greatly enhanced by adding emitter degeneration. vx ⇒ R out = ≈r o [ g m R e∥r ] ix g m R e∥r ≫1 15 ESE319 Introduction to Microelectronics Class AB Current Biasing Simulation Bias currents set at IREF and IQ by R and emitter resistor(s) Re. NPN Widlar current mirror I REF ≈4 mA 6.312mA I Q = I QN = I QP ≈2 mA R=2.8 kΩ IREF IQN Q2 Q1 iN Re=10 Ω R =100 Ω L iIiLLL Q3 Amplitude: 0 Vp Frequency: 1 kHz Q4 R=2.8 kΩ IREF -39µA Re=10 Ω IQP 2.322mA i L =i N −i P V CC −V BE1 V CC −V EB3 R= = ≈2.8 k I REF I REF VT I REF R e= ln ≈ 9 I QN I QN PNP Widlar current mirror Kenneth R. Laker, updated 14Nov14 KRL 16 ESE319 Introduction to Microelectronics Conclusions ADVANTAGE: Class AB operation improves on Class B linearity. Power conversion efficiency similar to Class B DISADVANTAGES: 1. Emitter resistors absorb output power. 2. 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. Kenneth R. Laker, updated 14Nov14 KRL 17

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