Class AB Output Stage - Penn Engineering · Class AB Amplifier Operation – VTC cont. i P i N =I Q...

ESE319 Introduction to Microelectronics

1Kenneth R. Laker, updated 14Nov14 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 Operation1. 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=iN−iP

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).

V BB



Class AB VTC Plot

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

Ideally, zero cross-over distortion



Class AB VTC Simulation

VBB/2

VBB/2

VCC

-VCC

RL

RSig

Amplitude: 20 Vp

Frequency: 1 kHz

Looks like Class A VTC



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

Cross-over distortion



Class AB Amplifier Operation - cont.

I N=I P= I Q= I S eV BB

2V T

Output voltage for vi ≠ 0:

for vi0 vo=viV BB

2−v BEN ⇒ vo≈vi

Base-to base voltage is constant!vBENvEBP=V BB

for vi0 vo=vi−V BB

2v EBP⇒ vo≈v i

Bias (QN & Q

P matched):

iN=iPiL

vBEN=V BB

2v i−vO

vEBP=vO−−V BB

2v i

for vi ≠ 0

+vi

for all vi

iN iP= I Q2

Let us next show thatfor all v

i

V BB




for v i0 vo=v iV BB

2−vBEN ⇒vBEN=vi−vo

V BB

2

for v i0 vo=v i−V BB

2vEBP⇒ vEBP=vo−v i

V BB

2

ADD

vBENv EBP=V BB for all vi

iN=I S evBEN

V T ⇒vBEN=V T ln iN

I S iP=I S evEBP

V T ⇒vEBP=V T ln iP

I S Using the currents

I N=I P= I Q= I S eV BB

2V T ⇒V BB=2V T ln I Q

I S V T ln iN

I S V T ln iP

I S =2V T ln I Q

I S for all vi

Note for Class B VBB

= 0




V T ln iN

I S V T ln iP

I S =2V T ln I Q

I S V T ln iN iP

I S2 =2V T ln I Q

I S ln iN iP −ln I S

2 =2ln I Q−2ln 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



Class AB Amplifier Operation – VTC cont.iP iN= I Q

2The constant base voltage condition

For example let IQ = 1 µA 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 with a much reduced dead zone.

where IQ is typically small.



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

PDisp−B=2

V o−peak

RLV CC−

12

V o−peak2

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 CC−1.42R

=V CC−0.7

R

R=V CC−0.7

I Q

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

or:

QN

QP

+

-

VBB

IQ

IQ

IQ

IQ

D2

D1

current mirror

V BB=V CC−I Q R− I Q R−V CC



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 Q Re=V BE1−V BE2=V T ln I REF

I Q

IREF

VCC

IQ

IQ

VBE1

+ +- -V

BE2

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 Q

I S V BE1=V BE2I Q Re⇒V BE1−V BE2=I Q Re

V BE1−V BE2=V T ln I REF

I S

I S

I Q=V T ln

I REF

I Q



Widlar Current Source - cont.

I Q Re=V T ln I REF

I Q

RI

REF

IQ

VCC

If IQ specified and IREF chosen by designer:

Re=V T

I Qln I REF

I Q Example Let IQ = 10 µA & choose IREF = 10 mA, determine R and Re:

R=V CC−V BE1

I REF=12V−0.7V10mA

=1.13 k

Re=V T

I Qln I REF

I Q =0.025V10 A

ln 10m A10A

.=2500 ln 1000=17.27 k R=1.13 k Re=17.27 k

IQ

Re

R=V CC−V BE1

I REF



Widlar Current Mirror Small-Signal Analysis

r≫1/gm

.≈.

i x=g m viro=gm vv x−−v

rov=−r∥Re i x

i x=−g m r∥Re i xv x

ro−

r∥Re i x

ro

gm Re∥r≫1

Rout is greatly enhanced by adding emitter degeneration.

⇒Rout=v x

i x≈ro[ gmRe∥r]

.≈−gm r∥Re i xv x

r o

Rout

Re ReRe



iL

Class AB Current Biasing SimulationBias currents set at I

REF and I

Q by R and emitter resistor(s) R

e.

IREF I

QN

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

R=2.8 kΩ

R=2.8 kΩ

iN

iL

I REF≈4mA

R=V CC−V BE1

I REF=

V CC−V EB3

I REF≈2.8k

Re=V T

I QNln I REF

I QN ≈9

I Q= I QN= I QP≈2mA

iL=iN−iPQ1

Q2

Q3

Q4

6.312mA

IREF

-39µA

2.322mA



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.

Class AB Output Stage - Penn Engineering · Class AB Amplifier Operation – VTC cont. i P i N =I Q...

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