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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 :

  1. A \(250 \mathrm{~k} \Omega\)
  2. B \(250 \Omega\)
  3. C \(400 \Omega\)
  4. D \(400 \mathrm{~k} \Omega\)
Verified Solution

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…