### Class AB Amplifier cont.

```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
Kenneth R. Laker, updated 14Nov14 KRL
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ESE319 Introduction to Microelectronics
Class AB Operation
vI
VB
IQ
IQ (set by VB)
Kenneth R. Laker, updated 14Nov14 KRL
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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
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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
Ideally, zero cross-over
distortion
Kenneth R. Laker, updated 14Nov14 KRL
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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
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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
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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
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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
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
 
 
 
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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
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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
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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.
Kenneth R. Laker, updated 14Nov14 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
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
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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
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
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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
 
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 
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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
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
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
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ESE319 Introduction to Microelectronics
Conclusions