SSC JE Electrical Engineering 2014 Set 1 Paper

Question : 195 of 200

Marks: +1, -0

In a CE (common emitter) transistor,

V_{CC} = 12 V

and the zero signal collector current is 1 mA.
Determine the operating point when collector load (

R_C

) is 6 kΩ.

Solution:

V_{CC}= V_{CC}-I_CR_C

= 12 -6× 1 ×10^{-3 }× 10^{3 }= 6 V

Therefore operating voltage is

Q = (6 V_1 mA)

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