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ESE319 Introduction to Microelectronics
12008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Output Stage● Class AB amplifier Operation● Multisim Simulations - Operation● Class AB amplifier biasing● Multisim Simulations - Biasing
ESE319 Introduction to Microelectronics
22008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Operation
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker, updated 30Nov09 KRL
Basic Class AB Amplifier CircuitBias Q
N and Q
P into slight conduction
when vI = 0.
Ideally QN and Q
P are:
1. Matched (unlikely with discrete transistors and challenging in IC).
2. Operate at same ambient temperature.
NOTE. This is base-voltage biasing with all its stability problems!
iL=iN−iP
ESE319 Introduction to Microelectronics
42008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Circuit Operation
I N=I P=I Q= I S eV BB2V T
Output voltage:
for v i0 vo=v iV BB2
−vBEN ⇒vo≈vi
Base-to base voltage is constant!vBENvEBP=V BB
Using:
V T ln iNI S V T ln iPI S =2V T ln I QI S
for v i0 vo=v i−V BB2
vEBP⇒ vo≈v i
Also: iN=I S evBENvT iP=I S e
vEBPV T&
Bias (QN & Q
P matched):
iN=I S evBENvT iP=I S e
vEBPV T, I Q=I S e
V BB2V T&
iN=iPi L
vBEN=V BB2
−vO
vEBP=vOV BB2
for vi = 0
There is a corresponding conservation relation between i
N & i
P , i.e.
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Circuit Operation - cont.
V T ln iNI S V T ln iPI S =2V T ln I QI S V T ln iN iPI S2 =2V T ln I QI S
V T ln iN iP−V T ln I S2 =2V T ln I Q−2V T ln I S
ln iN iP =ln I Q2
iN iP=I Q2
iN=iPiL
Constant base voltage condition:
from the previous slide
vBENvEBP=V BB =>
or iN iP=I Q2
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Circuit Operation - cont.
iN iP=I Q2
Constant base voltage condition:
Kirchhoff's Current Law condition:
iN=iPiL⇒ iP=iN−iLCombining equations:
iN i N−iL=I Q2
Hence:i N2 −i N i L−I Q
2=0
iN=iPiL
iP iPiL=I Q2or
iP2i P i L−I Q
2=0orfor v
I > 0 V for v
I < 0 V
for vI > 0 V for v
I < 0 ViL=
vORL
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Circuit Operation - cont.
Observations:1. Since with constant base voltage V
BB:
, if iP increases by
then iN decreases by , and vice-versa.
2. For vO > 0, the load current i
L is supplied by the complementary emitter
followers QN and Q
P. As v
O increases, i
P decreases and for large, positive
vO i.e.
iN iP=I Q2
iP=iP 1 i iN=iN /1i
iP=iN−iL=iN−vORL
vOV CC−V CENsat⇒ iP0
iN=iPiL=iPvORL
hence
3. For vO < 0, the load current i
L supplied by the complementary emitter
followers QN and Q
P. As v
O decreases, i
N decreases and for large, negative
vO i.e. hence −vO−V CCV ECPsat⇒ iN 0
iN iP= I Q2iL=iN−iP
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Circuit Operation - cont.iP iN=I Q
2The constant base voltage condition
For example let IQ = 1 mA and iN = 10 mA.
iP=I Q2
iN=1⋅10−6
10⋅10−3=0.1mA= 1
100iN
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 vI both Q
N and Q
P conduct, and as v
I is increased or
decreased, the conduction of QN or Q
P dominates, respectively.
Using this approximation we see that a class AB amplifier acts much like a class B amplifier; but without the dead zone.
where IQ is typically small.
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB VTC Plot
Requires the two DC base voltage sources to be matched and VBB/2 = 0.7 V.
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB VTC Simulation
VBB/2
VBB/2
VCC
-VCC
RL
RSig
Amplitude: 20 Vp
Frequency: 1 kHz
ESE319 Introduction to Microelectronics
112008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB VTC Simulation - cont.
V BB2
=0.1V
V BB2
=0.5V
V BB2
=0.7V
Amplitude: 2 Vp
Frequency: 1 kHz
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Small-Signal Output Resistance Instantaneous resistance for theQ
N transistor - assume α 1 :
diNdvBEN
=I S e
vBENV T
V T=
iNV T
For the QP transistor:diP
dvEBP=
i PV T
Hence:reN=
V TiN
reP=V TiP
and
vI
ac ground
ac ground
BN
BP
EN
EP
CN
CP
Rout=reN∥rePfor v
I > 0 V: Rout≈r eN
for vI < 0 V: Rout≈r eP
iN=I S evBENvT
in
ipvI
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker, updated 30Nov09 KRL
Small-Signal Output Resistance - cont.The two emitter resistors are in parallel:
Rout=reN∣∣reP=
V T2
iN iPV TiN
V TiP
=V T
iN iP 1iN 1iP =
V TiNiP
At iN = iP (the no-signal condition i.e. vO = 0 => iL = 0): iN=iP=I Q
Rout=V T2 I Q
So, for small signals, a small load current IQ flows => no dead-zone!
iL=vORL
=iN−iPand
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB Amplifier BiasingA straightforward biasing approach: D1 and D2 are diode-connectedtransistors identical to QN and QP, respectively.They form mirrors with the quiescentcurrents IQ set by matched R's:
I Q=2V CC−1.42R
=V CC−0.7
R
R=V CC−0.7
I QRecall: With mirrors, the ambient temperature for all transistors needs to be matched!
or:
QN
QP
+
-
VBB
IQ
IQ
IQ
IQ
D2
D1
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker, updated 30Nov09 KRL
Widlar Current Source
I REF=V CC−V BE1
R=12V −0.7V
11.3k =1mA
V BE1=V T ln I REFI S
I O Re=V T ln I REFI Q
IREF
VCC
IO
IQ
VBE1
+ +- -VBE2
R
Re
emitter degeneration
Q2 = QNI
Q = I
N
Note: Pages 653-656 in Sedra & Smith Text.
IN = bias current for Class AB amplifier NPN
Note Re > 0 iff IQ < IREF
V BE2=V T ln I QI S
V BE1=V BE2I Q Re
V BE1−V BE2=V T ln I REFI S
I SI Q
=V T ln I REFI Q
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker, updated 30Nov09 KRL
Widlar Current Source - cont.
I Q Re=V T ln I REFI Q I Q=
V TRe
ln I REF −ln I Q
RI
REF
IQ
IQ Re
VCC
Solve for IQ graphically.
If IQ specified (and IREF chosen by
designer): Re=V TI Q
ln I REFI Q
If Re specified and IREF given:
Example Let IQ = 10 µA & choose IREF = 10 mA, determine R and Re:
R=V CC−V BE1
I REF=12V−0.7V
10mA=1.13 k
Re=V TI Q
ln I REFI Q =0.025V10 A ln 10m A10A .=2500 ln 1000=17.27 k
R=1.13k Re=17.27 k
ESE319 Introduction to Microelectronics
172008 Kenneth R. Laker, updated 30Nov09 KRL
Widlar Current Mirror Small-Signal Analysis
r≫1/gm
.≈.
i x=g m viro=gm vv x−−v
r ov=−r∥Re i x
i x=−gm r∥Re i xvxro−
r∥Re i xro
gm ro≫1
Rout is greatly enhanced by adding emitter degeneration.
⇒Rout=v xi x=Re∥r1gm ro
.≈−g m r∥Re i xv xr o
Rout
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker, updated 30Nov09 KRL
iL
Class AB Current Biasing SimulationBias currents set at I
REF and I
Q by R and emitter resistor(s) R
e.
IREF IQN
IQP
IL
PNP Widlar current mirror
NPN Widlar current mirror
RL=100 Ω
Amplitude: 0 Vp
Frequency: 1 kHz
Re=10 Ω
Re=10 Ω
IREFR=2.8 kΩ
R=2.8 kΩ
iN
iL
I REF≈4mA
R=V CC−V BE1
I REF=
V CC−V EB3I REF
≈2.8k
Re=V TI QN
ln I REFI QN ≈10
I Q= I QN= I QP≈2mA
iL=iN−iPQ1
Q2
Q3
Q4
ESE319 Introduction to Microelectronics
192008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB IREF = 4 mA Current Bias Simulation
Quiescent Power Dissipation vI = 0 V:
P Disp=P D−av=76.31mW75.53mW
.=151.84mW
NPN Widlar current mirror
PNP Widlar current mirror
Amplitude: 0 Vp
Frequency: 1 kHz 4.031m
2.329m
4.025m2.270m
ESE319 Introduction to Microelectronics
202008 Kenneth R. Laker, updated 30Nov09 KRL
For linear operation: - 9 V < vO < +9 V
VTC has no dead-zone.
Amplitude: 12 Vp
Frequency: 1 kHz
Class AB 4 mA Current Bias Simulation - cont.
ESE319 Introduction to Microelectronics
212008 Kenneth R. Laker, updated 30Nov09 KRL
Class AB VTC Limits
vCE2=V CC−iN Re−vO ≥V CE2satQ2 ≠ Saturation =>
where
2. Q1 ≠ Cutoff =>vBQ1≥ v I0.7V
vO=−iN Rev IvCE2=V CC−v I ≥V CE2satv I ≤V CC−V CE2sat=v I−max1
vBQ1=V CC−iREF Rv I ≤V CC−iREF R−0.7V=v I−max2v I−max=max v I−max1 , v I−max2
iP
Q2
Q1
Q4
Q3vI
iREF
vO
Linear VTC for vI-max ≤ vI ≥ 0 => 1. Q1, Q2 forward-active
vO=−iN ReV EB1V BE2v I
IQ
ESE319 Introduction to Microelectronics
222008 Kenneth R. Laker, updated 30Nov09 KRL
VI V
O
PD+av
PD-av
PL-av
PD+av
PLav
PD-av
I REFN I QN
I QP
NOTE:1. Linear operation up to V
I = 9 V:
PDav
= 946mW, PLav
= 510mW =>
=PLavPDav
=510mW946mW
=0.540.785
2. At VI = 10 V: V
BEQ3 = V
BQ3 – V
I = -0.1V
NPN Q1 is cutoff for VI ≥ 9.5 V.
3. By symmetry PNP Q3 is cutoff for V
I ≤ -9.5V .
3.324m0.018m
2.697m
4.734m1.776u
Class AB 4 mA Current Bias Simulation - cont.
309 mW
V BQ1
V BQ3
9 9.7VQ2
Q1
Q3Q4
IQN
IQP V BQ1
ESE319 Introduction to Microelectronics
232008 Kenneth R. Laker, updated 30Nov09 KRL
ConclusionsADVANTAGE:
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 powerefficiency, power dissipation and output signal swingneed to be addressed.