Use DC analysis techniques to find the DC voltages at the base, collector, and e

Question

Use DC analysis techniques to find the DC voltages at the base, collector, and emitter of the transistor. The voltages are referenced to circuit ground. What is VCE,Q and IC,Q, the quiescent (dc) values for the operating collector-emitter voltage and the collector current? Assume that current gain beta = 100. Using the small-signal model for the transistor shown in Figure 1. draw an equivalent small signal circuit model and find the expected values of Av,Ai, Zin and Zout. Again, assume beta = 100. The Av = vout/vin, Ai = iout/iin, Zin = vin/iin, Zout = vout/iz Recall that to calculate the output impedance, the input voltage source must be set to zero and the output terminals must be driven with a test current source of value iz into the collector connection. Small-signal model of a bipolar junction transistor. The small-signal parameters can be computed by using the dc operating point values: gm = IC/VT, rx = beta/gm.IC is the dc value of the collector current and VT is the thermal voltage which is approximately 26mV at room temperature. Common-collector amplifier with coupling capacitors Cc- Component values are as follows: Rs = 3.3k ohm(used to measure iin), R1 = 82k ohm, R2 = 68k ohm, RE = 3.9k ohm, RL = 5k Ohm (variable), Cc = 5 mu F, Vcc = 20V.

Explanation / Answer

DC analysis
Voltage at the base of the transistor VB = VCCR2 / (R1 + R2) = (20)(68000) / (82000+68000) = 9.06V
Applying KVL at the input section VB = VBE + VE VE = 9.06 - 0.7 = 8.36V
IE = VE / RE = 8.36 / 3900 = 2.14mA IC ˜ IE = 2.14mA
VCE = VCC - VE = 20 - 8.36 = 11.64V
AC analysis
re =VT / IE = 26mV / 2.14mA = 12.15
AV ˜ RE / RE + re = 3900/(3900+12.15) = 0.996 or rac = 3.9k || 5k = 2.2k AV = rac / (rac + re) = 2200/(2200+12.15) = 0.994
Ai = 1 + = 1 + 100 = 101
Input impedance RB = 82k || 68k = 37.2k Zin = RB || (re + rac) Zin = 37200 || (100 (12.15+2200) = 37200 || 221215 = 31.8k
Output impedance Zout = re + (Rs || RB ) / Zout = 12.15 + (3300 ||2200) /100 = 25.35
VE = 9.06 - 0.7 = 8.36V
IE = VE / RE = 8.36 / 3900 = 2.14mA IC ˜ IE = 2.14mA
VCE = VCC - VE = 20 - 8.36 = 11.64V
AC analysis
re =VT / IE = 26mV / 2.14mA = 12.15
AV ˜ RE / RE + re = 3900/(3900+12.15) = 0.996 or rac = 3.9k || 5k = 2.2k AV = rac / (rac + re) = 2200/(2200+12.15) = 0.994
Ai = 1 + = 1 + 100 = 101
Input impedance RB = 82k || 68k = 37.2k Zin = RB || (re + rac) Zin = 37200 || (100 (12.15+2200) = 37200 || 221215 = 31.8k
Output impedance Zout = re + (Rs || RB ) / Zout = 12.15 + (3300 ||2200) /100 = 25.35

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