COMEDK · Physics · 30. Semiconductors
The reverse current in the semiconductor diode changes from \(20 \mu \mathrm{A}\) to \(40 \mu \mathrm{A}\) when the reverse potential is changed from \(10 \mathrm{~V}\) to \(15 \mathrm{~V}\), then the reverse resistance of the junction diode will be :
- A \(250 \mathrm{~k} \Omega\)
- B \(250 \Omega\)
- C \(400 \Omega\)
- D \(400 \mathrm{~k} \Omega\)
Answer & Solution
Correct Answer
(A) \(250 \mathrm{~k} \Omega\)
Step-by-step Solution
Detailed explanation
The reverse resistance \(R\) of a junction diode is defined as the ratio of the change in reverse potential \(\Delta V\) to the change in reverse current \(\Delta I\). Given: Initial potential \(V_1 = 10 \text{ V}\) Final potential \(V_2 = 15 \text{ V}\) Initial current…
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