class ab output stageese319/lecture_notes/... · basic class ab amplifier circuit bias q n and q p...

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



Class AB Operation



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



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.



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




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




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



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.



Class AB VTC Plot

Requires the two DC base voltage sources to be matched and VBB/2 = 0.7 V.



Class AB VTC Simulation

VBB/2

VBB/2

VCC

-VCC

RL

RSig

Amplitude: 20 Vp

Frequency: 1 kHz



Class AB VTC Simulation - cont.

V BB2

=0.1V

V BB2

=0.5V

V BB2

=0.7V

Amplitude: 2 Vp

Frequency: 1 kHz



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



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



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



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



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



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



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



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



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.



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



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



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.

class ab output stageese319/lecture_notes/... · basic class ab amplifier circuit bias q n and q p...

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