class ab output stage - penn engineeringese319/lecture_notes/lec_22...ese319 introduction to...
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![Page 1: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/1.jpg)
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
![Page 2: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/2.jpg)
ESE319 Introduction to Microelectronics
22008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB Operation
![Page 3: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/3.jpg)
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
![Page 4: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/4.jpg)
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
![Page 5: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/5.jpg)
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:
![Page 6: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/6.jpg)
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
![Page 7: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/7.jpg)
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
![Page 8: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/8.jpg)
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.
![Page 9: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/9.jpg)
ESE319 Introduction to Microelectronics
92008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08 KRL
Class AB VTC Plot
![Page 10: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/10.jpg)
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
![Page 11: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/11.jpg)
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
![Page 12: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/12.jpg)
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
![Page 13: Class AB Output Stage - Penn Engineeringese319/Lecture_Notes/Lec_22...ESE319 Introduction to Microelectronics 1 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 11Nov08](https://reader030.vdocuments.net/reader030/viewer/2022040204/5ea6ed3529ce151b3642d927/html5/thumbnails/13.jpg)
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!
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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
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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
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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
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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.
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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
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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
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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.
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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
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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
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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.