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Analog Electronics

Q1: A certain transistor has adc of 0.98 and a collectors leakage current of 5 μA. If IE = 1 mA, the collector current will be

A 1.005 mA

B 0.985 mA

C 0.975 mA

D 0.955 mA

ANS:B - 0.985 mA

ie = ic + ib.

Ic = B ib (where B is beta).
Ib = ic/B.

B = adc/1 - adc (where adc is alpha).
B = 0.98/1 - 0.98 = 49.
Ic = ie * B/1 + B.

1 * 49/1 + 49 = 0.98 ma (where ma is mills ampere).



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