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

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.
Solution:
Since h oeR E ? 0.1
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
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
A i = (1 + h fe) = 51
A v = 1 ? (h ie / Z in) = 1 ? (2 10 3 / 8,700) = 0.77