MHT CET · Physics · Semiconductors
In a study of transistor as an amplifier, the ratio of collector current to emitter current is \(0.98 .\) The collector current is \(3 \mathrm{~mA}\), then base current will be approximately
- A \(6 \mathrm{~mA}\)
- B \(60 \mathrm{~mA}\)
- C \(6 \mu \mathrm{A}\)
- D \(60 \mu \mathrm{A}\)
Answer & Solution
Correct Answer
(D) \(60 \mu \mathrm{A}\)
Step-by-step Solution
Detailed explanation
\(\frac{i_{c}}{i_{c}}=0.98 \quad i_{c}=3 \mathrm{~mA} \quad i_{b}=?\)
\(\frac{i_{c}}{i_{c}+i_{b}}=0.98=\frac{98}{100} \Rightarrow \frac{i_{c}+i_{b}}{i_{c}}=\frac{100}{98}\)
\(1+\frac{\mathrm{i}_{\mathrm{b}}}{\mathrm{i}_{\mathrm{c}}}=\frac{100}{98} \quad \therefore \frac{\mathrm{i}_{\mathrm{b}}}{\mathrm{i}_{\mathrm{c}}}=\frac{2}{98} \quad \therefore \mathrm{i}_{\mathrm{b}}=3 \times \frac{2}{98}\) \(=\frac{3}{49} \mathrm{~mA}\)
\(=\frac{3000}{49} \mu \mathrm{A}=61.22 \mu \mathrm{A} \approx 60 \mu \mathrm{A}\)
\(\frac{i_{c}}{i_{c}+i_{b}}=0.98=\frac{98}{100} \Rightarrow \frac{i_{c}+i_{b}}{i_{c}}=\frac{100}{98}\)
\(1+\frac{\mathrm{i}_{\mathrm{b}}}{\mathrm{i}_{\mathrm{c}}}=\frac{100}{98} \quad \therefore \frac{\mathrm{i}_{\mathrm{b}}}{\mathrm{i}_{\mathrm{c}}}=\frac{2}{98} \quad \therefore \mathrm{i}_{\mathrm{b}}=3 \times \frac{2}{98}\) \(=\frac{3}{49} \mathrm{~mA}\)
\(=\frac{3000}{49} \mu \mathrm{A}=61.22 \mu \mathrm{A} \approx 60 \mu \mathrm{A}\)
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