Emitter Followers

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

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’)

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

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.

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

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.

The dc load line IC in mA 100 mA 14 12 80 mA 10 60 mA 8 40 mA 6 Q 4 2 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

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 b = b1b2 Darlington transistor

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

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 +VCC R1 Q1 R2 vout R3 vin RL Q2 R4

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

Direct-coupled common emitter driver +VCC AQ1 @ R3 R4 R3 R1 Q2 Q3 RL vin Q1 R2 R4

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

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

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