Electronic Circuit Analysis

Multistage Amplifiers 75
Therefore, for a CASCODE Transistor Configuration, its input Z is equal to that of a
single Common Emitter Transistor (hie)' Its Current Gain is equal to that of a single Common
Base Transistor (h fe ). Its output resistance is equal to that of a single Common Base
Transistor (hob)' The reverse voltage gain is very very small, i.e., there is no link between
V I (input voltage) and V 2 (output voltage). In otherwords, there is negligible internal feedback
in the case of, a CASCODE Transistor Circuit, acts like a single stage C.E. Transistor
(Since hIe and h fe are same) with negligible internal feedback (:. h re is very small) and very
small output conductance, (== hob) or large output resistance (== 2MO equal to that of a
Common Base Stage). The above values are correct, if we make the assumption that hob RL
< 0.1 or RL is < 200K. When the value of RL is < 200 K. This will not affect the values of
hI' hr' ho' h f of the CASCODE Transistor, since, the value of hr is very very small.
CASCODE Amplifier will have
I. Very Large Voltage Gain.
2. Large Current Gain (h je
).
3. Very High Output Resistance.
Example : 2.6
Find the voltage gains Ays' A YJ
and AY2 of the amplifier shown in Fig.2.35.
Assume
hie = 1 KO, h re
= 10-4,
and
hoe = 10 -8 A/V.
Fig. 2.35 Amplifier circuit Ex : 2.6