Emitter Followers

+VCC

RER2

R1

RL

vin

The common-collector or emitter follower amplifier

vout

ac ground

re’R2R1vin

T model of the emitter follower amplifier

vout = ierere

re = RE RL

vin= ie(re + re’)

re

re + re’

A =

reR2R1vin

p model of the emitter follower amplifier

vout(re + re’)

zin(stage) = R1 R2 (re + re’)

RC RL

RC

RLvth

The output side of a common-emitter amplifier

Applying Thevenin’s theorem:

The output impedance is equal to RC.

re’R2R1RG

T model of the emitter follower amplifier

RE RL

A

vth

zout

RL

A

Apply Thevenin’s theorem to point A:

Output impedance of the emitter follower amplifier

re’ +

R1 R2 RG

( )zout = RE

The current gain of the amplifier steps down the impedance of the base circuit. Thus, theoutput impedance of this amplifier is small.

0 2 4 6 8 10 12 14 16 18

2468

101214

VCE in Volts

IC in mA

20 A

0 A

100 A

80 A

60 A

40 A

IC(sat) =VCC

RE

The dc load line

VCE(cutoff) = VCC

Q

The ac load line has a higher slope:re = RE RL

Large signal operation• When the Q point is at the center of

the dc load line, the signal cannot use all of the ac load line without clipping.

• MPP < VCC

• MP = ICQre or VCEQ (whichever is smaller)

• MPP = 2MP• When the Q point is at the center of

the ac load line: ICQre = VCEQ

Darlington connection

Q1

Q2

=

Darlington transistor

Push-pull emitter follower

Q1

Q2

RLvout

R4

R1

vin

R3

R2

+VCC

When Q1 is on,the capacitor

charges.

When Q2 is on,the capacitordischarges.

Class B push-pull emitter follower

• ICQ = 0

• VCEQ = VCC/2

• MPP = VCC

• A @ 1

• zin(base) = b RL

• PD(max) = MPP2/40RL (each transistor)

• pout(max) = MPP2/8RL

Crossover distortion in class B

Q1

Q2

RLvout

R4

R1

vin

R3

R2

+VCC

Class AB• Crossover distortion is caused by

the barrier potential of the emitter diodes.

• ICQ must be increased to 1 to 5 percent of IC(sat) to eliminate crossover distortion.

• The new operating point is between class A and B but is much closer to B.

Thermal runaway• When temperature increases,

collector current increases.• More current produces more heat.• Compensating diodes that match

the VBE curves of the transistors are often used.

• Any increase in temperature reduces the bias developed across the diodes.

Diode bias

RL

R

R

vin

+VCC

2VBE

Ibias =VCC - 2VBE

2R

ICQ Ibias

IC(sat) =VCC

2RL

Iav =IC(sat)

Idc(total) = ICQ + Iav

Pdc(in) = VCCIdc(total)

pout(max) =VCC

2

8RL

pout

Pdc

x 100% =

RL

R1

vin

+VCC

R2

R3

R4

Direct-coupled common emitter driver

Q1

Q3

Q2

AQ1 R3

R4

vin

+VCC

R2

R1

Two-stage negative feedback

Q1

Q3

Q2

R2 provides both dc and ac negative feedback

to stabilize the bias

and the voltage gain.

+VCC

RL

RS

Zener follower

VoutVZ

Vout = VZ - VBE

IB =Iout

dc

zout = re’ +

dc

RZ

R1

Two-transistor voltage regulator

VZ

R2

Q1

Q2

R3

R4

RL

+Vin

Vout

Vout =R3 + R4

R4

(VZ + VBE)