Electronic Circuit Analysis

High Fri/quency Transistor Circuits
125
IT : Frequency at which Current Gain = 1
roT = ro Ih fe ( ro)1
or
At
IT = Ilhfe(f)1
1= 20 MHz, h fe
= 4
Capacitance in Hybrid - 1t Model
IT = 4 x 20 = 80 MHz
C n
is in pf. If we take gm in Kn, C~ in pf, and roT in nano seconds.
0.2
--------;- - 3.16 x lO-12
2x7tx80xlO 6
lO6 xO.l _ 3.16 pf
3.14x80
10 5
= 251.20 - 3.16 pf
:::: lOO - 3.16 pf
= 96.84 pf.
Now we must convert it to the desired operating point of Ic = 5 rnA, V CE = 4V.
Cn::::C b
C n varies Iinerly with Ie-
Ie = 5 rnA, VCE '"" lOY,
5
= 96.84 x to = 48.42 pf
To convert C n to V CE = 4 V, instead of 10, we need to know the relation between 0) and V CEO
As it is not known we assume that C n
remains the same at V CE = 4V also.
rw at Ic = lO rnA, V CE = lOY = 30n
At
IC = 4 rnA