350 Solved Electrical Engineering Problems: For the FE/PE Exams in Electrical Engineering

2.9: Common collector amplifier

2.9 Common collector amplifier

The common collector BJT circuit shown in Figure 2.9 has the following hybrid parameters. Input impedance, h ie, equals 2 10 3 ohms. Voltage feedback ratio, h re, equals 4 10 ?4. Current gain, h fe, equals 50. Output admittance, h oe, equals 20 10 ?6 siemens. Calculate the (a) Z in, (b) Z out, (c) A i, and (d) A v respectively.


Figure 2.9: Common collector amplifier

Solution:

Since h oeR E ? 0.1

  1. Z in = [h ie + (1 + h fe)R E] R 1 R 2

    Z in = [2 10 3 + (51)5.1 10 3] (10 10 3) (90 10 3)

    Z in = (27.1 10 4) (9 10 3) = 8,700 ohms

  2. Z out = h ie / (1 + h fe) R E = (2 10 3 /51) (5.1 10 3)

    Z out = 39 (5.1 10 3) = 39 ohms

  3. A i = (1 + h fe) = 51

  4. A v = 1 ? (h ie / Z in) = 1 ? (2 10 3 / 8,700) = 0.77

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