1 / 21

MALVINO

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 ’.

lilli
Télécharger la présentation

MALVINO

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SIXTH EDITION MALVINO Electronic PRINCIPLES

  2. Emitter Followers Chapter 12

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

  4. re A = re + re’ re = RE RL Tmodel of the emitter follower amplifier re’ R2 R1 vin vout = iere re vin= ie(re + re’)

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

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

  7. zout A vth RL RL Tmodel of the emitter follower amplifier Apply Thevenin’s theorem to point A: re’ R2 RG R1 A RE

  8. 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.

  9. 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 The dc load line 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

  10. 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 = ICQreor VCEQ (whichever is smaller) • MPP = 2MP • When the Q point is at the center of the ac load line: ICQre= VCEQ

  11. Darlington connection Darlington transistor Q1 Q2 b = b1b2

  12. 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

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

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

  15. 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.

  16. Thermal runaway • 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.

  17. 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 ICQ@ Ibias Idc(total) = ICQ + Iav 2VBE Pdc(in) = VCCIdc(total) RL vin R

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

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

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

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

More Related