1. Emitter Followers

2. The common-collector or emitter follower amplifier
+VCC R1
ac ground
vout
vin R2 RE RL

3. T model of the emitter follower amplifier
re = RE RL re
re + re’
A = R1 R2
re’
vin
vout = iere re
vin= ie(re + re’)

4. p model of the emitter follower amplifier
zin(stage) = R1 R2 b(re + re’) R1 R2
b(re + re’) re
vout
vin

5. Applying Thevenin’s theorem:
The output side of a common-emitter amplifier
Applying Thevenin’s theorem: RC RL
vth RC RL
The output impedance is equal to RC.

6. T model of the emitter follower amplifier
Apply Thevenin’s theorem to point A:
vth
zout RL A RG R1 R2
re’ A RL RE

7. Output impedance of the emitter follower amplifier
re’ +
R1 R2 RG b ( )
zout = RE
The current gain of the amplifier steps down
the impedance of the base circuit. Thus, the
output impedance of this amplifier is small.

8. The dc load line IC in mA 100 mA 14 12 80 mA 10 60 mA 8 40 mA 6 Q 42 4 6 8 10 12 14 16 18
VCC
VCE in Volts
IC(sat) =
VCE(cutoff) = VCC RE
The ac load line has a higher slope:
re = RE RL

9. 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

10. Darlington connectionQ1 Q2
b = b1b2
Darlington transistor

11. Push-pull emitter follower
+VCC R1
When Q2 is on,
the capacitor
discharges.
When Q1 is on,
the capacitor
charges. Q1 R2
vout R3
vin RL Q2 R4

12. Class B push-pull emitter follower
ICQ = 0
VCEQ = VCC/2
MPP = VCC 1
zin(base) = b RL
PD(max) = MPP2/40RL (each transistor)
pout(max) = MPP2/8RL

13. Crossover distortion in class B
+VCC R1 Q1 R2
vout R3
vin RL Q2 R4

14. 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.

15. 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.

16. Diode bias VCC - 2VBE Ibias = 2R +VCC VCC R IC(sat) = 2RL IC(sat)
Iav =
IC(sat) p
Ibias
Idc(total) = ICQ + Iav
2VBE
Pdc(in) = VCCIdc(total) RL
vin
pout(max) =
VCC2
8RL R
pout
Pdc
x 100%
h =

17. Direct-coupled common emitter driver
+VCC R3 R4 R3 R1 Q2 Q3 RL
vin Q1 R2 R4

18. Two-stage negative feedback
R2 provides both dc and
ac negative feedback
to stabilize the bias
and the voltage gain.
+VCC R1 Q2 Q3 Q1
vin R2

19. Zener follower +VCC Vout = VZ - VBE Iout RS IB = bdc RZ zout = re’ +RL
Vout

20. Two-transistor voltage regulator
R3 + R4 R4
(VZ + VBE)
Vout =
+Vin
Vout Q2 R2 R1 R3 RL Q1 R4 VZ