A transistor having $\alpha=0.8$ is connected in common emitter configuration. When the base current changes…

A transistor having $\alpha=0.8$ is connected in common emitter configuration. When the base current changes by $6 \mathrm{~mA}$, the change in collector current is
  1. $12 \mathrm{~mA}$
  2. $1.5 \mathrm{~mA}$
  3. $24 \mathrm{~mA}$
  4. $0.66 \mathrm{~mA}$

Solution

A transistor having $\alpha=0.8$ is connected in common emitter configuration. When the base current changes by 6 mA , then the change in collector current is 24 mA . Given, $\alpha=0.8$ Change In base current, $\Delta \mathrm{l}_{\mathrm{b}}=6 \mathrm{~mA}$ For common emitter configuration, $\beta=\frac{\alpha}{1-\alpha}=\frac{0.8}{1-0.8}=4$ Also, $\beta=\frac{\Delta \mathrm{l}_{\mathrm{c}}}{\Delta \mathrm{l}_{\mathrm{b}}}$ $\begin{aligned} & 4=\frac{\Delta \mathrm{l}_{\mathrm{c}}}{6 \mathrm{~mA}} \\ & \Delta \mathrm{l}_{\mathrm{c}}=24 \mathrm{~mA} \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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