MHT CET · Physics · Semiconductors
For a common emitter configuration, if ' \(\alpha\) ' and ' \(\beta\) ' have their usual meanings, the incorrect relation between ' \(\alpha\) ' and ' \(\beta\) ' is
- A \(\frac{1}{\alpha}=\frac{1}{\beta}+1\)
- B \(\alpha=\frac{\beta}{1-\beta}\)
- C \(\alpha=\frac{\beta}{1+\beta}\)
- D \(\frac{1}{\beta}=\frac{1}{\alpha}-1\)
Answer & Solution
Correct Answer
(B) \(\alpha=\frac{\beta}{1-\beta}\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{C}}} \text { and } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{E}}}\)
\(\therefore \frac{1}{\alpha}=\frac{\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{C}}} \Rightarrow \frac{1}{\alpha}=\frac{1}{\beta}+1 \)
\( \therefore \frac{1}{\alpha}-1=\frac{1}{\beta} \Rightarrow \alpha=\frac{\beta}{\beta+1}\)
\(\therefore \frac{1}{\alpha}=\frac{\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{C}}} \Rightarrow \frac{1}{\alpha}=\frac{1}{\beta}+1 \)
\( \therefore \frac{1}{\alpha}-1=\frac{1}{\beta} \Rightarrow \alpha=\frac{\beta}{\beta+1}\)
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