Electronic Circuit Analysis

14 Electronic Circuit Analysis
Example : 1.1
For the circuit shown in Fig.I.18 estimate AI' A y
, Ri and Ro using reasonable approximations. The
h-parameters for the transistor are given as
h fe
= 100 hie = 2000 n hre is negligible and hqe = 10- 5 mhos (U) .
Ib = 100 J.LA..
Solution:
Fig. 1.18 CE Amplifier Circuit
At the test frequency capacitive reactances can be neglected. V cc point is at ground because the AC
potenth:~l at V cc = o. So it is at ground. R) is connected between base and ground forAC. Therefore,
R) II R 2 • R4 is connected between collector and ground. So R4 is in parallel with IIhoe in the output.
The A.C. equivalent circuit in terms of h-parameters of the transistor is shown in Fig. l.l9.
-+Ib
\rr-.....,---i'---. B
Fig. 1.19 Equivalent Circuit
The voltage source ~e Vee is not shown since, h re is negligible. At the test frequency of the
input signal ,the capacitors C) and C 2
can be regarded as short circuits. So they are not shown in the
AC equivalent circuit. The emitter is at ground potential. Because Xc is also negligible, all the
AC passes through C 3
• Therefore, emitter is at ground potential and {his circuit is in Common
Emitter Configuration.