In the common-base configuration, a transistor has current amplification factor 0.95 . If the transistor is…

In the common-base configuration, a transistor has current amplification factor 0.95 . If the transistor is used in common-emitter configuration and base current changes by $2 \mu \mathrm{A}$, then the change in the collector current is
  1. $19 \mu \mathrm{A}$
  2. $0.91 \mu \mathrm{A}$
  3. $1.9 \mu \mathrm{A}$
  4. $38 \mu \mathrm{A}$

Solution

Given, $\alpha=0.95$ So, $ \begin{aligned} \beta & =\frac{\alpha}{1-\alpha}=\frac{0.95}{1-0.95}=19 \\ & =\beta=\frac{\Delta I_C}{\Delta I_B}=19 \end{aligned} $ or $\Delta I_C=19 \times 2=38 \mu \mathrm{A}$

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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