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AP EAMCET · PHYSICS · Semiconductors

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. A \(19 \mu \mathrm{A}\)
  2. B \(0.91 \mu \mathrm{A}\)
  3. C \(1.9 \mu \mathrm{A}\)
  4. D \(38 \mu \mathrm{A}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(38 \mu \mathrm{A}\)

Step-by-step Solution

Detailed explanation

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 \(\Delta I_C=19 \times 2=38 \mu \mathrm{A}\)