Class AB Output Stage - University of Pennsylvaniaese319/Lecture_Notes/Lec_22_ClassAB_Amplif… ·...

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker, updated 26Nov12 KRL

Class AB Output Stage Class AB amplifier Operation Multisim Simulation - VTC Class AB amplifier biasing Widlar current source Multisim Simulation - Biasing



Class AB Operation

IQ

IQ

VB

vI

(set by VB)



Basic Class AB Amplifier Circuit1. Bias Q

N and Q

P into slight conduction (fwd. act.)

when vI = 0: i

N = i

P.

2 Ideally QN and Q

P are:

a. Matched (unlikely with discrete transistors and challenging in IC).

b. Operate at same ambient temperature.

NOTE. This is base-voltage biasing with all its stability problems!

iL=iNiP

3.For vi > 0: i

N > i

P i.e. Q

N most cond. (like Class B).

4.For vi < 0: i

P > i

N i.e. Q

P most cond. (like Class B).



Class AB VTC Plot

Ideally the two DC base voltage sources are matched and VBB/2 = 0.7 V.V BB /20.7V



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 Circuit Operation - cont.

I N=I P=I Q=I S eV BB2V T

Output voltage for vi 0:

for vi0 vo=viV BB2

v BEN vovi

Base-to base voltage is constant!vBENv EBP=V BB

for vi0 vo=viV BB2

v EBP vov i

Bias (QN & Q

P matched):

iN=iPiL

vBEN=V BB2

vO

vEBP=vOV BB2

for DC vi = 0

+vi

for all vi

iN iP=I Q2

Let us next show thatfor all v

i




for vi0 vo=viV BB2

v BEN vBEN=vivoV BB2

for vi0 vo=viV BB2

v EBP vEBP=vov iV BB2

ADD

vBENv EBP=V BB for all vi

iN=I S evBENvT vBEN=V T ln iNI S iP=I S e

vEBPV T v EBP=V T ln iPI S

Using the currents

I N=I P=I Q=I S eV BB2V T V BB=2V T ln I QI S

V T ln iNI S V T ln iPI S =2V T ln I QI S for all vi

Note for Class B VBB

= 0




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

ln iN iP ln I S2 =2ln I Q2ln I S

ln iN iP=ln I Q2

iN iP=I Q2

iN=iPiL

Constant base voltage condition:

from the previous slide

vBENv EBP=V BB =>

or iN iP=I Q2



Class AB Circuit Operation VTC cont.iP iN=I Q

2The constant base voltage condition

For example let IQ = 1 mA and iN = 10 mA.

iP=I Q2

iN=1106

10103=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 Small-Signal Output Resistance Instantaneous resistance for theQ

N transistor - assume 1 :

diNdvBEN

=I S e

vBENV T

V T=

iNV T

= 1r eN

For the QP transistor:diP

dvEBP=

iPV T

=1

r ePHence:

reN=V TiN

reP=V TiP

andac ground

ac ground

BN

BP

EN

EP

CN

CP

Rout=r eNrePv

I > 0 V: i

N > i

P => RoutreN

RoutreP

iN=I S evBENvT

in

ip

vI = 0

Root

vO

v I=0

vI < 0 V: i

P > i

N =>



Small-Signal Output Resistance - cont.The two emitter resistors are in parallel:

Rout=r eNr eP=

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

=i NiPand



Class AB Power Conversion Efficiency & Power Dissipation Similar to Class B

Accurate for small Vo-peak

.

V o peak

Let VCC = 12 V and RL=100

= 7.63 V0.7 V

PDisp(max) = 0.29 W

0.20 W

PDispB=2

V opeakRL

V CC12

V opeak2

RL

P Disp

PDisp max=2V CC

2

2RL=0.29W

P Disp 0 when Vo-peak = 0



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 CC1.42R

=V CC0.7

R

R=V CC0.7

I QRecall: With mirrors, the ambient temperature for all transistors needs to be matched!

or:

QN

QP

+

-

VBB

IQ

IQ

IQ

IQ

D2

D1

current mirror



Widlar Current Source

I REF=V CCV BE1

R=12V 0.7V

11.3k =1mA

V BE1=V T ln I REFI S

I Q Re=V T ln I REFI Q

IREF

VCC

IO

IQ

VBE1

+ +- -VBE2

R

Re

emitter degeneration

Q2 = QNI

Q = I

N

Note: Pages 543-546 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 BE1V 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

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 chosen

by the designer:

Example Let IQ = 10 A & choose IREF = 10 mA, determine R and Re:

R=V CCV BE1

I REF=12V0.7V10mA

=1.13 k

Re=V TI Q

ln I REFI Q =0.025V10 A ln 10m A10A .=2500 ln 1000=17.27 k

R=1.13 k Re=17.27 k

IQ

Re



Widlar Current Mirror Small-Signal Analysis

r1/gm

..

i x=g m viro=gm vv xv

rov=rRe i x

i x=g m rRe i xvxro

rRe ixro

gm Rer1

Rout is greatly enhanced by adding emitter degeneration.

Rout=v xi xro[ gmRer]

.g m rRe ixvxro

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

Amplitude: 0 Vp

Frequency: 1 kHz

Re=10

Re=10

IREFR=2.8 k

R=2.8 k

iN

iL

I REF4mA

R=V CCV BE1

I REF=

V CCV EB3I REF

2.8 k

Re=V TI QN

ln I REFI QN 10

I Q= I QN= I QP2mA

iL=iNiPQ1

Q2

Q3

Q4



ConclusionsADVANTAGE:

Class AB operation improves on Class B linearity.Power conversion efficiency similar to Class B

DISADVANTAGES:1. Emitter resistors absorb output power.2. Power dissipation for low signal levels higher than Class B.3. Temperature matching will be needed more so. if emitter degeneration resistors are not used.

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