ESE319 Introduction to Microelectronics
12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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 (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Operation
ESE319 Introduction to Microelectronics
32008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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).
2. Operate at same temperature.
NOTE. This is base-voltage biasing with all its stability problems!
iL=iN−iP
ESE319 Introduction to Microelectronics
42008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Circuit Operation
I N=I P=I Q= I S eV BB
2V T
Output voltage:for v i0 vo=v i
V BB
2−vBEN ⇒vo≈vi
Base-to base voltage is constant!vBENvEBP=V BB
Using:
V T ln iN
I S V T ln iP
I S =2V T ln I Q
I S
for v i0 vo=v i−V BB
2vEBP⇒ vo≈v i
iN=iPi L
vBEN=V BB
2−vO
vEBP=vO−V BB
2
Also: iN=I S evBEN
vT iP=I S evEBP
V T&
Bias:
iN=I S evBEN
vT iP=I S evEBP
V T, I Q=I S eV BB
2V T&
for vi = 0
ESE319 Introduction to Microelectronics
52008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Circuit Operation - cont.
V T ln iN
I S V T ln iP
I S =2V T ln I Q
I S V T ln i N iP
I S2 =2V T ln I Q
I S V T ln iN iP−V T ln I S
2 =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:
ESE319 Introduction to Microelectronics
62008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Circuit Operation - cont.
iN iP=I Q2
Constant base voltage condition:
Kirchhoff's Current Law condition:
iN=iPiL⇒ iP=iN−iL
Combining equations:iN iN−iL=I Q
2
Hence:iN2 −i N iL−I Q
2=0
iN=iPiL
iP iPiL=I Q2or
iP2iP i L−I Q
2=0orfor v
I > 0 V for v
I < 0 V
for vI > 0 V for v
I < 0 ViL=
vO
RL
ESE319 Introduction to Microelectronics
72008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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 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 i N=iN /1 i
iP=iN−iL=iN−vO
RLvO large⇒ iP0
i N=iPiL=iPvO
RL
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−large⇒ iN 0
iN iP= I Q2iL=iN−iP
ESE319 Introduction to Microelectronics
82008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Circuit Operation - cont.
iP=I Q2
iNWrite the product equation as:
For example let IQ = 1 mA and iN = 10 mA.
iP=1⋅10−6
10⋅10−3=0.1mA= 1
100iN
The Class AB circuit, over most of its input signal range, operates as if the Q
N or Q
P transistor is conducting and the Q
P or Q
N 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.
where IQ is typically small.
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB VTC Plot
ESE319 Introduction to Microelectronics
102008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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 (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB VTC Simulation - cont.
V BB
2=0.1V
V BB
2=0.5V
V BB
2=0.7V
Amplitude: 2 Vp
Frequency: 1 kHz
ESE319 Introduction to Microelectronics
122008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Small-Signal Output Resistance Instantaneous resistance for theQ
N transistor - assume α 1 :
diN
dvBEN=
I S evBEN
V T
V T=
iN
V T
For the QP transistor:diP
dvEBP=
iP
V T
Hence:reN=
V T
iNreP=
V T
iPand
vI = 0 V
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
ESE319 Introduction to Microelectronics
132008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Small-Signal Output Resistance - cont.The two emitter resistors are in parallel:
Rout=reN∣∣reP=
V T2
iN iP
V T
iN
V T
iP
=V T
iN iP i NiP
iN iP =
V T
i NiP
At iN = iP (the no-signal condition i.e. vO = 0 => i
L = 0): iN=iP=I Q
Rout=V T
2 I Q
So, for small signals, a load current flows => no dead-zone!
ESE319 Introduction to Microelectronics
142008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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 quiescentcurrent set by R:
I Q=2V CC−1.42R
=V CC−0.7
Ror:
R=V CC−0.7
I Q
Recall: With mirrors, the device temperature for all transistors needs to be matched!
QN
QP
+
-
VBB
IQ
IQ
IQ
IQ
ESE319 Introduction to Microelectronics
152008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Widlar Current Source
I REF=V CC−V BE1
R=12V −0.7V
11.3k =1mA
V BE1=V T ln I REF
I S
I O Re=V T ln I REF
I O
IREF
VCC
IO
IO
VBE1
+ +- -V
BE2
R
Re
emitter degeneration
Q2 = QNI
O = I
N
Note: Pages 653-656 in Sedra & Smith Text.
IN = bias current for Class AB amplifier NPN
Note Re > 0 iff IO < IREF
V BE2=V T ln I O
I S
V BE1=V BE2I O Re
V BE1−V BE2=V T ln I REF
I S
I S
I O=V T ln
I REF
I O
ESE319 Introduction to Microelectronics
162008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Widlar Current Source - cont.
I O Re=V T ln I REF
I O I O=
V T
Reln I REF −ln I O
RI
REF
IO
IO R
e
VCC
Solve for IO graphically.
If IO specified (and IREF chosen by
designer): Re=V T
I Oln I REF
I O
If Re specified and IREF given:
Example Let IO = 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 T
I Oln I REF
I O =0.025V10 A
ln 10m A10A
.=2500 ln 1000=17.27k R=1.13k Re=17.27k
ESE319 Introduction to Microelectronics
172008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Widlar Current Mirror Small-Signal Analysis
r≫1/gm
.≈.
i x=g m viro=gm vv x−−v
r o
v=−r∥Re i x
i x=−gm r∥Re i xvx
r o−
r∥Re i x
ro⇒ Rout=
vx
i x=Re∥r1g m ro
gm ro≫1
Rout is greatly enhanced by adding emitter degeneration.
ESE319 Introduction to Microelectronics
182008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
iL
Class AB Current Biasing SimulationBias currents set at I
REF and I
O by R and emitter resistor(s) R
e.
IREF I
ON
IOP
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
Re=V T
I Oln I REF
I O
I O≈2mA
ESE319 Introduction to Microelectronics
192008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB IREF = 4 mA Current Bias Simulation
Quiescent Power Dissipation v
I = 0 V:
PDisp=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 (based on P. V. Lopresti 2006) updated 11Nov08 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 (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB VTC Limits
vCE2=V CC−iN Re−vO ≥V CE2sat
Q2 ≠ Saturation =>
where
2. Q3 ≠ Cutoff =>vBQ3≥ v I0.7V
vO=−iN Rev I
vCE2=V CC−v I ≥V CE2sat
v I ≤V CC−V CE2sat=v I−max1
vBQ3=V CC−iREF R
v I ≤V CC−iREF R−0.7V=v I−max2
v I−max=max v I−max1 , v I−max2
iP
Q2
Q3
Q1
Q4vI
iREF
vO
Linear VTC for vI-max ≤ vI ≥ 0 => 1. Q2, Q3 forward-active
ESE319 Introduction to Microelectronics
222008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
VI V
O
PD+av
PD-av
PL-av
PD+av
PLav
PD-av
I REFN I ON
I OP
NOTE:1. Linear operation up to V
I = 9 V:
PDav
= 946mW, PLav
= 510mW =>
=PLav
PDav=510mW946mW
=0.540.785
2. At VI = 10 V: V
BEQ3 = V
BN – V
I = -0.1V
NPN Q3 is cutoff for VI ≥ 9.5 V.
3. By symmetry PNP Q4 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 BQ3
V BQ3
ESE319 Introduction to Microelectronics
232008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 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.
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