A \(p\)-\(n\) junction is designed to withstand current up to a maximum in 10 mA. A resistor \(R = 200\ \Omega\) is connected in series with it. When forward biased the diode has a potential drop of 0.5 V. The maximum voltage of the battery required to forward bias the diode is
Show Hint
- 2.5 V
- 1.5 V
- 2 V
- 10 V
The Correct Option is A
Solution and Explanation
Total voltage = voltage across resistor + voltage across diode.
Step 2: Detailed Explanation:
\(V_{resistor} = I_{max} \times R = 10 \times 10^{-3} \times 200 = 2\) V
\(V_{total} = V_{resistor} + V_{diode} = 2 + 0.5 = 2.5\) V
Step 3: Final Answer:
Maximum voltage required \(= \mathbf{2.5}\) V.
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