Electronic Circuit Analysis

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Electronic Circuit Analysis
Therefore) total input current,
Ii = I, + Ib = 4 + 100 = 104 IlA.
10 is the current through the 1 Kn load.
1
-h = 100 Kn is very large compared with R4 and R L . Therefore, all the current on the
oe
output side, h fe Ib gets divided between R4 and RL only.
Therefore;current through RL is 1 0
,
= 6.78 rnA.
Therefore ,current amplification,
I
A=-..2...
I I.
I
6.78xI0- 3
----;- =65.
104 x 10- 6
Vo =-10' RL
= (- 6.78 x 10 3 ) x (10 3 )
=-6.78V
Because, the direction of 10 is taken as entering into the circuit. But actually 10 flows down,
because Vo is measured with respect to ground.
= - 33.9
Negative sign indicates that there is phase shift of 180 0 between input and output voltages,
i.e. as base voltage goes more positive, (it is NPN transistor), the collector voltage goes
more negative.