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# MALVINO - PowerPoint PPT Presentation

SIXTH EDITION. MALVINO. Electronic. PRINCIPLES. Emitter Followers. Chapter 12. v out. The common-collector or emitter follower amplifier. +V CC. R 1. ac ground. v in. R 2. R E. R L. r e. A =. r e + r e ’. r e = R E R L. T model of the emitter follower amplifier. r e ’.

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Presentation Transcript

MALVINO

Electronic

PRINCIPLES

Chapter 12

vout

The common-collector or emitter follower amplifier

+VCC

R1

ac ground

vin

R2

RE

RL

re

A =

re + re’

re = RE RL

Tmodel of the emitter follower amplifier

re’

R2

R1

vin

vout = iere

re

vin= ie(re + re’)

zin(stage) = R1 R2 b(re + re’)

p model of the emitter follower amplifier

re

b(re + re’)

vout

R2

R1

vin

RC

vth

RL

The output side of a common-emitter amplifier

Applying Thevenin’s theorem:

RC

RL

The output impedance is equal to RC.

zout

A

vth

RL

RL

Tmodel of the emitter follower amplifier

Apply Thevenin’s theorem to point A:

re’

R2

RG

R1

A

RE

re’+

)

(

zout = RE

R1 R2 RG

b

Output impedance of the emitter follower amplifier

The current gain of the amplifier steps down

the impedance of the base circuit. Thus, the

output impedance of this amplifier is small.

100 mA

80 mA

60 mA

40 mA

20 mA

0 mA

re = RE RL

The ac load line has a higher slope:

VCC

VCE(cutoff) = VCC

IC(sat) =

RE

14

12

10

IC in mA

8

6

Q

4

2

6

16

2

4

10

12

0

14

18

8

VCE in Volts

• 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 = ICQreor VCEQ (whichever is smaller)

• MPP = 2MP

• When the Q point is at the center of the ac load line: ICQre= VCEQ

Darlington transistor

Q1

Q2

b = b1b2

When Q2 is on,

the capacitor

discharges.

vout

Push-pull emitter follower

+VCC

R1

When Q1 is on,

the capacitor

charges.

Q1

R2

R3

vin

RL

Q2

R4

• ICQ = 0

• VCEQ = VCC/2

• MPP = VCC

• A @ 1

• zin(base) = bRL

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

• pout(max) = MPP2/8RL

vout

Crossover distortion in class B

+VCC

R1

Q1

R2

R3

vin

RL

Q2

R4

• 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.

• When temperature increases, collector current increases.

• More current produces more heat.

• Compensatingdiodes that match the VBE curves of the transistors are often used.

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

Ibias =

VCC - 2VBE

IC(sat) =

VCC

2R

2RL

Iav =

IC(sat)

p

pout

VCC2

x 100%

h =

Pdc

pout(max) =

8RL

+VCC

R

Diode bias

[email protected] Ibias

Idc(total) = ICQ + Iav

2VBE

Pdc(in) = VCCIdc(total)

RL

vin

R

R3

R4

Q1

Direct-coupled common emitter driver

+VCC

R3

R1

Q2

Q3

RL

vin

R2

R4

+VCC

R1

R2 provides both dc and

ac negative feedback

to stabilize the bias

and the voltage gain.

Q2

Q3

Q1

vin

R2

RZ

IB =

zout = re’ +

bdc

Iout

bdc

Zener follower

+VCC

Vout = VZ - VBE

RS

VZ

Vout

RL

Vout

R3 + R4

(VZ + VBE)

R4

Two-transistor

voltage regulator

Vout =

+Vin

Q2

R2

R1

R3

RL

Q1

R4

VZ