Electronics and Communication Engineering - Analog Electronics - Discussion

3. 

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

Answer: Option B

Explanation:

No answer description available for this question.

S Naveena said: (Jun 12, 2016)  
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).

Rajashekar Balya said: (Dec 28, 2017)  
Here, a=Ic÷ Ie.

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