Class AB Output Stage

ESE319 Introduction to Microelectronics

Class AB Output Stage

●

Class AB amplifier Operation

●

Multisim Simulation - VTC

●

Class AB amplifier biasing

●

Widlar current source

●

Multisim Simulation - Biasing

ESE319 Introduction to Microelectronics

Class AB Operation

ESE319 Introduction to Microelectronics

Basic Class AB Amplifier Circuit

1. Bias Q_N and Q_P into slight conduction (fwd. act.) when v

I = 0: i_N = i_P.

2 Ideally Q_N 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!

i_L=i_N−i_P

3.For v_i > 0: i_N > i_P i.e. Q_N most cond. (like Class B).

4.For v_i < 0: i_P > i_N i.e. Q_P most cond. (like Class B).

ESE319 Introduction to Microelectronics

Class AB VTC Plot

Requires the two DC base voltage sources to be

matched and V

/2 = 0.7 V.

ESE319 Introduction to Microelectronics

Class AB VTC Simulation

V_BB/2

V_BB/2

V_CC

-V_CC

R_L R_Sig

Amplitude: 20 V_p Frequency: 1 kHz

ESE319 Introduction to Microelectronics

Class AB VTC Simulation - cont.

V_BB

2 =0.1V

V_BB

2 =0.5V

V_BB

2 =0.7V Amplitude: 2 V_p

Frequency: 1 kHz

ESE319 Introduction to Microelectronics

Class AB Circuit Operation - cont.

I_N=I P=IQ=I Se

V_BB 2V_T

Output voltage for v_i ≠ 0:

for v_i0v_o=v_iV _BB

2 −v_BEN ⇒ v_o≈v_i

Base-to base voltage is constant!

v_BENv_EBP=V _BB for v_i0v_o=v_i−V _BB

2 v_EBP⇒ v_o≈v_i

Bias (Q_N & Q_P matched):

i_N=i_Pi_L v_BEN=V_BB

2 −v_O v_EBP=v_OV _BB

2

for DC v_i = 0

v_i +

for all v

ESE319 Introduction to Microelectronics

Class AB Circuit Operation - cont.

for v_i0v_o=v_iV _BB

2 −v_BEN⇒ v_BEN=v_i−v_oV _BB 2 for v_i0v_o=v_i−V _BB

2 v_EBP⇒ v_EBP=v_o−v_iV _BB 2

ADD

v_BENv_EBP=V _BB

for all v

i_N=IS e

v_BEN v_T

⇒v_BEN=VT ln





^

^

Using the currents

I_N=I P=IQ=IS e

V_BB 2V_T

⇒V _BB=2V T ln





V _T ln











 ^{for all v}

Note for Class B V_BB = 0

ESE319 Introduction to Microelectronics

Class AB Circuit Operation - cont.

V _T ln













V_T ln





^

^

V _Tlni_N i_P−V_T ln I _S²=2V _Tln I_Q−2V _T ln I_S lni_Ni_P=ln I_Q² 

i_N i_P=I_Q²

i_N=i_Pi_L

Constant base voltage condition:

from the previous slide

v_BENv_EBP=V _BB

=>

or ⁱ_N ⁱ_P=I_Q²

ESE319 Introduction to Microelectronics

Class AB Circuit Operation – VTC cont.

i_Pi_N=I_Q² The constant base voltage condition

For example let I_Q = 1 mA and i_N = 10 mA.

i_P= I_Q²

i_N = 1⋅10⁻⁶

10⋅10⁻³=0.1 mA= 1

100 i_N

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 v_I both Q_N and Q_P conduct, and as v_I is increased or decreased, the conduction of Q_N 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 I_Q is typically small.

ESE319 Introduction to Microelectronics

Class AB Small-Signal Output Resistance

Instantaneous resistance for the Q_N transistor - assume α 1 :

di_N

dv_BEN = I _S e

v_BEN V_T

V _T = i_N V _T For the Q_P transistor:

di_P

dv_EBP = i_P V _T Hence:

r_eN=V_T

i_N r_eP=V_T i_P and

ac ground ac ground

<=>

BN

BP

EN EP CN

CP

R_out=r_eN∥r_eP ^vI > 0 V: i_N > i_P => R_out≈r_eN R ≈r i_N=IS e

v_BEN v_T

i_n i_p vI = 0

R_oot v_O

v_I=0

v < 0 V: i > i =>

ESE319 Introduction to Microelectronics

Small-Signal Output Resistance - cont.

The two emitter resistors are in parallel:

R_out=r_eN∣∣r_eP=

V _T² i_N i_P V_T

i_N V _T i_P

= V _T

i_N i_P





At i_N = i_P (the no-signal condition i.e. v

O = 0 => i

L = 0): ⁱ_N=i_P=I_Q R_out= V_T

2 I_Q

So, for small signals, a small load current I_Q flows => no dead-zone!

i_L= v_O

R_L=i_N−i_P and

ESE319 Introduction to Microelectronics

Class AB Power Conversion Efficiency &

Power Dissipation Similar to Class B

Accurate for small V .

V_{o− peak}

Let V_CC = 12 V and ^R_L=100

= 7.63 V 0.7 V

P_Disp(max) = 0.29 W

0.20 W

P_Disp−B= 2



V_{o− peak}

R_L V_CC−1 2

V_o²_{− peak} R_L

P_Disp

P_{Dispmax}=2 V_CC²

²R_L=0.29 W

P ≠ 0 when V = 0

ESE319 Introduction to Microelectronics

Class AB Amplifier Biasing

A straightforward biasing approach:

D1 and D2 are diode-connected transistors identical to QN and QP, respectively.

They form mirrors with the quiescent currents I_Q set by matched R's:

I_Q= 2 V_CC−1.4

2R =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 +

- V_BB I_Q

I_Q I_Q

I_Q

D2 D1

current mirror

ESE319 Introduction to Microelectronics

Widlar Current Source

I _REF=V_CC−V BE1

R =12V −0.7V

11.3 k  =1mA

V _BE1=V_T ln





I_QR_e=V _T ln





I_REF V_CC

I_O I_Q

V+_BE1 + - V_BE2- R

R_e emitter degeneration

Q2 = QN I_Q = I_N

Note: Pages 543-546 in Sedra & Smith Text.

I_N = bias current for Class AB amplifier NPN

Note R > 0 iff I < I V _BE2=V _T ln





V _BE1=V _BE2I_Q R_e V _BE1−V _BE2=V _T ln I_REF

I_S

I_S

I_Q =V _T ln I _REF I_Q 

ESE319 Introduction to Microelectronics

Widlar Current Source - cont.

I_QR_e=V _T ln





I_Q=V _T

R_e



^

^

^

^ 

I_REFR

I_Q V_CC

Solve for I

Q graphically.

If IQ specified and I_REF chosen by designer:

R_e=V_T

I_Q ln





If Re specified and I_REF chosen by the designer:

Example Let I_Q = 10 µA & choose I_REF = 10 mA, determine R and R_e:

R=V _CC−V _BE1

I _REF =12 V −0.7V

10 mA =1.13 k  R_e=V_T

I_Q ln





.=2500 ln1000=17.27 k 

R=1.13 k  R_e=17.27 k 

I_Q

R_e

ESE319 Introduction to Microelectronics

Widlar Current Mirror Small-Signal Analysis

r_≫1/ g_m

.≈.

i_x=g_mv_i_ro=g_mv_ v_x−−v_ r_o

v_=−r_∥R_ei_x i_x=−g_mr_∥R_ei_x v_x

r_o − r_∥R_ei_x r_o

g R ∥r ≫1

R_out is greatly enhanced by adding emitter degeneration.

⇒R_out=v_x

i_x =ro1gmR_e∥r_

.≈−g_mr_∥R_ei_x v_x r

R_out

ESE319 Introduction to Microelectronics

Quick Review

i_L=i_N−i_P

i_N i_P=I_Q² v_BENv_EBP=V _BB

=>

QN

QP +

- V_BB I_Q

I_Q I_Q

I_Q

D2 D1

current mirror

Widlar Current Mirror

R_e=V_T

I_Q ln





ESE319 Introduction to Microelectronics

i_L

Class AB Current Biasing Simulation

Bias currents set at I

and I

by R and emitter resistor(s) R

e

.

I_REF I_QN

I_QP I_L

NPN Widlar current mirror

R_L=100 Ω

Amplitude: 0 V_p Frequency: 1 kHz

R_e=10 Ω

R_e=10 Ω I_REF

R=2.8 kΩ R=2.8 kΩ

i_N

i_L I_REF≈4 mA

R=V_CC−V _BE1

I_REF =V_CC−V_EB3

I_REF ≈2.8 k  R_e= V_T

I_QN ln





I_Q=IQN=IQP≈2 mA

i_L=i_N−i_P

Q1

Q2

Q3 Q4

ESE319 Introduction to Microelectronics

Conclusions

ADVANTAGE:

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.