ESE319 Introduction to Microelectronics
Class AB Output Stage
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Class AB amplifier Operation
Multisim Simulation - VTC
Class AB amplifier biasing
Widlar current source
Multisim Simulation - Biasing
2008 Kenneth R. Laker, updated 26Nov12 KRL
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ESE319 Introduction to Microelectronics
Class AB Operation
vI
VB
IQ
IQ (set by VB)
2008 Kenneth R. Laker, updated 26Nov12 KRL
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ESE319 Introduction to Microelectronics
Basic Class AB Amplifier Circuit
1. Bias QN and QP into slight conduction (fwd. act.)
when vI = 0: iN = iP.
i L =i N −i P
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!
2008 Kenneth R. Laker, updated 26Nov12 KRL
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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
2008 Kenneth R. Laker, updated 26Nov12 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, updated 26Nov12 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, updated 26Nov12 KRL
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
for DC vi = 0
V BB
v BEN =
−v O
2
V BB
v EBP =v O
2
vi +
i N =i P i L
Bias (QN & QP matched):
I N =I P =I Q =I S e
V BB
2V T
2008 Kenneth R. Laker, updated 26Nov12 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 2Q
for all vi
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
V BB
V BB
for vi 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
 
2008 Kenneth R. Laker, updated 26Nov12 KRL
BB
T
 
 
 
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation - cont.
from the previous slide
iN
iP
IQ
V T ln
V T ln
=2 V T ln
IS
IS
IS
 
i N =i P i L
V T ln
   
   
iN iP
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 2Q  or i N i P =I Q
Constant base voltage condition:
2008 Kenneth R. Laker, updated 26Nov12 KRL
v BEN v EBP=V BB => i N i P =I 2Q
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ESE319 Introduction to Microelectronics
Class AB Circuit Operation – VTC cont.
The constant base voltage condition i P i N =I Q2 where IQ is typically small.
For example let IQ = 1 mA 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 without the dead zone.
2008 Kenneth R. Laker, updated 26Nov12 KRL
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ESE319 Introduction to Microelectronics
Class AB Small-Signal Output Resistance
ac ground
i N =I S e
v BEN
vT
Instantaneous resistance for the
QN transistor - assume α  1:
CN
v BEN
VT
di N
ISe
iN
1
=
= =
dv BEN
VT
V T r eN
BN
vI = 0
EN
vO <=>
EP
BP
CP
ac ground
v I =0
R out =r eN∥r eP
2008 Kenneth R. Laker, updated 26Nov12 KRL
in
ip
Root
For the QP transistor:
di P
iP
1
= =
dv EBP V T r eP
Hence:
VT
VT
r eN =
and r eP =
iN
iP
vI > 0 V: iN > iP => R out ≈r eN
vI < 0 V: iP > iN => R out ≈r eP
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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
1 1

iN iP

iN
iP
i N iP


and
vO
i L = =i N −i P
RL
At iN = iP (the no-signal condition i.e. vO = 0 => iL = 0): i N =i P =I Q
VT
R out =
2IQ
So, for small signals, a small load current IQ flows => no dead-zone!
2008 Kenneth R. Laker, updated 26Nov12 KRL
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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.
2008 Kenneth R. Laker, updated 26Nov12 KRL
V o− peak
= 7.63 V
P Disp ≠ 0 when Vo-peak = 0
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ESE319 Introduction to Microelectronics
Class AB Amplifier Biasing
current mirror
D1
IQ
IQ
+
QN
VBB
D2
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
currents IQ set by matched R's:
2 V CC −1.4 V CC −0.7
I Q=
=
2R
R
or:
V CC −0.7
R=
IQ
Recall: With mirrors, the ambient temperature for all transistors needs to
be matched!
2008 Kenneth R. Laker, updated 26Nov12 KRL
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ESE319 Introduction to Microelectronics
Widlar Current Source
IN = bias current for Class AB amplifier NPN
R
IREF
VCC
IQ
IQ = IN
Q2 = QN
 
 
I REF
V BE1=V T ln
IS
IQ
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 IQ
IQ
V CC −V BE1 12V −0.7V
I REF =
=
=1 mA
V BE1=V BE2I Q R e
R
11.3 k 
I REF
Note: Pages 543-546 in Sedra & Smith Text.
I Q R e =V T ln
IQ
+
+
- VBE1 VBE2-
 
Note Re > 0 iff IQ < IREF
2008 Kenneth R. Laker, updated 26Nov12 KRL
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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
 
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
I REF
I Q R e =V T ln
IQ
If Re specified and IREF chosen
by the designer:
VT
I Q=
ln  I REF  −ln  I Q  

Re
Solve for IQ graphically.
2008 Kenneth R. Laker, updated 26Nov12 KRL
 
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 
R e =17.27 k 
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ESE319 Introduction to Microelectronics
Widlar Current Mirror Small-Signal Analysis
.≈.
Rout
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
2008 Kenneth R. Laker, updated 26Nov12 KRL
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
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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
I Q= I QN = I QP≈2 mA
R=2.8 kΩ
IREF
IQN
Q2
Q1
iN
iIiLLL
Q3
Amplitude: 0 Vp
Frequency: 1 kHz
Q4
R=2.8 kΩ
IREF
i L =i N −i P
Re=10 Ω R =100 Ω
L
Re=10 Ω
IQP
R=
V CC −V BE1 V CC −V EB3
=
≈2.8 k 
I REF
I REF
VT
I REF
Re =
ln
≈ 10
I QN
I QN
 
PNP Widlar current mirror
2008 Kenneth R. Laker, updated 26Nov12 KRL
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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.
2008 Kenneth R. Laker, updated 26Nov12 KRL
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