ESE319 Introduction to Microelectronics
Class AB Output Stage
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Class AB amplifier Operation
Multisim Simulations - Operation
Class AB amplifier biasing
Multisim Simulations - Biasing
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Operation
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Basic Class AB Amplifier Circuit
Bias QN and QP into slight conduction
when vI = 0.
i L =i N −i P
Ideally QN and QP are:
1. Matched (unlikely with discrete
transistors).
2. Operate at same temperature.
NOTE. This is base-voltage biasing with all its stability problems!
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation
for vi = 0
Output voltage:
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
V BB
−v O
2
V BB
v EBP =v O −
2
v BEN =
Base-to base voltage is constant!
v BEN v EBP =V BB
v BEN
vT
i N =i P i L
Bias:
I N =I P =I Q = I S e
Also:
i N =I S e
Using: i N =I S e , i P =I S e
V BB
2VT
v BEN
vT
& i P =I S e
v EBP
VT
 
 
v EBP
VT
& I Q =I S e
 
iN
iP
IQ
V T ln
V T ln
=2 V T ln
IS
IS
IS
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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V BB
2V T
ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
i N =i P i L
 
   
   
iN
iP
IQ
V T ln
V T ln
=2 V T ln
IS
IS
IS
V T ln
iN iP
I 2S
IQ
=2V T ln
IS
V T ln i N i P −V T ln  I 2S =2V T ln  I Q −2 V T ln I S 
ln i N i P =ln  I 2Q 
Constant base voltage condition:
i N i P =I Q2
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
Constant base voltage condition:
i N i P =I Q2
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:
vO
i L=
RL
i N i N −i L =I Q2
or
i P i P i L =I 2Q
for vI > 0 V
for vI < 0 V
Hence:
2
2
i 2N −i N i L −I Q2 =0 or i P i P i L −I Q =0
for vI > 0 V
for vI < 0 V
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
i L =i N −i P
i N i P = I Q2
Observations:
1. Since with constant base voltage VBB: i N i P =I Q2 , if iP increases by
i P =i P 1 i then iN decreases by i N =i N /1i  , and vice-versa.
2. For vO > 0, the load current iL supplied by the complementary emitter
followers QN and QP. As vO increases, iP decreases and for large, positive
vO i.e. i =i −i =i − v O
hence v O  large⇒ i P  0
P
N
L
N
RL
3. For vO < 0, the load current iL supplied by the complementary emitter
followers QN and QP. As vO decreases, iN decreases and for large, negative
vO
vO i.e.
i N =i P i L =i P 
hence −v O −large ⇒i N  0
RL
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
I 2Q
where IQ is typically small.
Write the product equation as: i P =
iN
For example let IQ = 1 mA and iN = 10 mA.
1⋅10−6
1
i P=
=0.1 mA=
iN
−3
100
10⋅10
The Class AB circuit, over most of its input signal range, operates
as if the QN or QP transistor is conducting and the QP or QN 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.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB VTC Plot
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Amplitude: 20 Vp
Frequency: 1 kHz
Class AB VTC Simulation
VCC
VBB/2
RSig
VBB/2
RL
-VCC
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB VTC Simulation - cont.
Amplitude: 2 Vp
Frequency: 1 kHz
V BB
=0.1V
2
V BB
=0.5V
2
V BB
=0.7V
2
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Small-Signal Output Resistance
Instantaneous resistance for the
QN transistor - assume α  1:
ac ground
CN
v BEN
VT
di N
ISe
iN
=
=
dv BEN
VT
VT
BN
vI = 0 V
EN
<=>
EP
BP
CP
ac ground
R out =r eN ∥r eP
For the QP transistor:
di P
iP
=
dv EBP V T
Hence:
VT
VT
r eN =
and r eP =
iN
iP
for vI > 0 V: R out ≈r eN
for vI < 0 V: R out ≈r eP
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Small-Signal Output Resistance - cont.
The two emitter resistors are in parallel:
V T2
iN iP
VT
VT
R out =r eN ∣∣r eP =
=
=
VT VT
i N i P
i N i P

iN iP
iN
iP
iN iP
 
At iN = iP (the no-signal condition i.e. vO = 0 => iL = 0): i N =i P =I Q
VT
R out =
2 IQ
So, for small signals, a load current flows => no dead-zone!
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Amplifier Biasing
IQ
IQ
+
QN
VBB
IQ
QP
IQ
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:
2V CC −1.4 V CC −0.7
I Q=
=
2R
R
or:
V CC −0.7
R=
IQ
Recall: With mirrors, the device temperature for all transistors needs to
be matched!
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Widlar Current Source
IN = bias current for Class AB amplifier NPN
R
IREF
VCC
IO
IO = IN
Q2 = QN
 
 
I REF
V BE1=V T ln
IS
IO
V BE2=V T ln
IS
emitter
IO Re
degeneration
I REF I S
I REF
V BE1−V BE2=V T ln 
=V T ln 

IS IO
IO
V CC −V BE1 12V −0.7 V
I REF =
=
=1 mA
V BE1 =V BE2I O R e
R
11.3k 
I REF
Note: Pages 653-656 in Sedra & Smith Text.
I O R e =V T ln
IO
+
+
- VBE1 VBE2-
 
Note Re > 0 iff IO < IREF
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Widlar Current Source - cont.
R
If IO specified (and IREF chosen by
VT
I REF
designer):
Re=
ln
IO
IO
 
IO
IREF
VCC
IO
Re
 
I REF
I O R e =V T ln
IO
If Re specified and IREF given:
VT
I O=
ln  I REF  −ln  I O  

Re
Solve for IO graphically.
Example Let IO = 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
0.025V
10 m A
Re=
ln
=
ln

IO
IO
10 A
10  A
.=2500 ln 1000=17.27 k 
R=1.13 k 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
R e =17.27 k 
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ESE319 Introduction to Microelectronics
Widlar Current Mirror Small-Signal Analysis
.≈.
v x −−v  
i x =g m v i ro=g m v  
ro
r  ≫1/ g m
Rout is greatly enhanced by
adding emitter degeneration.
v  =−r ∥Re i x
v x r ∥R e i x
vx
i x =−g m r ∥R e i x  −
⇒ R out = = Re∥r  1g m r o 
ro
ro
ix
g m r o ≫1
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB Current Biasing Simulation
Bias currents set at IREF and IO by R and emitter resistor(s) Re.
I REF ≈4 mA
NPN Widlar current mirror
I O ≈2 mA
R=2.8 kΩ
IREF
iN
ION
Re=10 Ω R =100 Ω
L
iIiLLL
Re=10 Ω
Amplitude: 0 Vp
Frequency: 1 kHz
R=2.8 kΩ
IREF
IOP
V CC −V BE1
R=
I REF
VT
I REF
Re=
ln
IO
IO
 
PNP Widlar current mirror
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB IREF = 4 mA Current Bias Simulation
Amplitude: 0 Vp
Frequency: 1 kHz
NPN Widlar current mirror
4.031m
2.329m
Quiescent Power
Dissipation vI = 0 V:
P Disp =P D−av =76.31 mW 75.53 mW
.=151.84 mW
2.270m
4.025m
PNP Widlar current mirror
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB 4 mA Current Bias Simulation - cont.
Amplitude: 12 Vp
Frequency: 1 kHz
For linear operation: - 9 V < vO < +9 V
VTC has no dead-zone.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB VTC Limits
Linear VTC for vI-max ≤ vI ≥ 0 =>
1. Q2, Q3 forward-active
iREF
Q2
Q3
vI
Q4
vO
iP
Q1
Q2 ≠ Saturation =>
v CE2=V CC −i N Re −v O ≥ V CE2sat
where v O =−i N Re v I
v CE2=V CC −v I ≥ V CE2sat
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 
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
Class AB 4 mA Current Bias Simulation - cont.
3.324m
I REFN
V BQ3
PD+av
0.018m
309 mW
I ON
V BQ3
VI
2.697m
PD+av
PL-av
VO
PLav
I OP
PD-av
1.776u
4.734m
PD-av
NOTE:
1. Linear operation up to VI = 9 V:
PDav = 946mW, PLav = 510mW =>
P Lav 510 mW
=
=
=0.540.785
P Dav 946 mW
2. At VI = 10 V: VBEQ3 = VBN – VI = -0.1V
NPN Q3 is cutoff for VI ≥ 9.5 V.
3. By symmetry PNP Q4 is cutoff
for VI ≤ -9.5V .
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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ESE319 Introduction to Microelectronics
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.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
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