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If $$\alpha $$ = 0.98, $${I_{{\text{CO}}}} = 6\mu {\text{A}}$$ and $${I_\beta } = 100\mu {\text{A}}$$ for a transistor, then the value of $${I_{\text{C}}}$$ will be
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{
& {I_{\text{C}}} = \frac{{{I_{{\text{CO}}}}}}{{1 - \alpha }} + \frac{\alpha }{{1 - \alpha }} \times {I_\beta } \cr
& = \frac{6}{{1 - 0.98}} + \frac{{0.98}}{{1 - 0.98}} \times 100 \cr
& = 5.2\,{\text{mA}} \cr} $$
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