Electronic Circuit Analysis

High Frequency Transistor Circuits
105
3.7 Current Gain with Resistance Load :
fr == ~. hr. ==
Considering the load resistance Ru
21t (C e + C J
V b'e is the input voltage and is equal to VI
V ce is the output voltage and is equal to V 2
Vce
~==-
Vb'e
This circuit is still complicated for analysis.
Because, there are two time constants associated with the input and the other associated with
the output. The output time constant will be much smaller than the input time constant. So it can
be neglected.
K == Voltage gain. It will be » 1
gb'c (K; 1) :::: gb'c
gb'c < gce rb'c :::: 4 Mn, rce == 80 K (typical values)
So gb'c can be neglected in the equivalent circuit.
In a wide band amplifier RL will not exceed 2Kn, sincefH oc
RL
IfRL is smallf H is large.
fH == 21tC s (RC IIR L
)
Therefore gce can be neglected compared with RL
Therefore the output circuit consists of current generator gm V b'e feeding the load RL so the
Circuit simplifies as shown in Fig. 3.10.
I
I
B'
r-------~----~--_r.c
C
c
(l+g
m-'L
R )
Fig. 3.10 Simplified equivalent circuit
Vce
K == -V == - gm RL ; gm == 50 rnA \ V,
b'e
RL == 2Kn (typical values)
K==-lOO
So the maximum value is gb'c (1 - K) :::: 0.02595. So this can be neglected compared to
gb'e:::: 1 mAN.