Electronic Circuit Analysis

High Frequency ·Transistor Circuits
111
Example : 3.2
V T ·21tC e
= 1500 MHz
Giv~n the .follo~ing tr~sistor ~easurerne)'1lS mad~at Ic = 5 rnA ; V CE = 10V and at room temperature
h fe - 100 , ~e - 600n , [Aiel - 10 at-lo1vtHz C c - 3 pf.
Solution:
Find f~,
fT' Ce, rb'e and rbb,·
h fe
= 100, Aie = 10, atf= 10 MHz
10= FW
1 + [L) =
100
2
100xtOO = 100
f~ 10xlO
2
[~) = 100-1 =99
2
[{) = 99, f= 10 MHz for use,
I:. f~ = 1.005 MHz I
A = h fe · f~ = 100 x 1.005 MHz = 100.5 MHz
C =~. ~=1l
II 21tfT' V T