The base current in a transistor circuit changes from $45 \mu \mathrm{A}$ to $140 \mu \mathrm{A}$.…

The base current in a transistor circuit changes from $45 \mu \mathrm{A}$ to $140 \mu \mathrm{A}$. Accordingly, the collector current changes from $0.2 \mathrm{~mA}$ to $0.400 \mathrm{~mA}$. The gain in current is
  1. $9.5$
  2. $1$
  3. $40$
  4. $20$

Solution

$\begin{aligned} & \text { Current gain } \beta=\frac{\Delta i_c}{\Delta i_b} \\ & \Delta i_c=(4-0.2) \mathrm{mA}=3.8 \times 10^{-3} \mathrm{~A} \\ & \Delta i_b=(140-45) \mu \mathrm{A}=95 \times 10^{-6} \mathrm{~A} \\ & \therefore \quad \beta=\frac{3.8 \times 10^{-3}}{95 \times 10^{-5}}=40 \\ & \end{aligned}$

Asked in: AP EAMCET 2013

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