A practical silicon diode with a cut-in voltage of 0.7 V is connected as
follows:
P-terminal (anode) → Ground (0V)
N-terminal (cathode) → +10V
Given that the current flowing through the diode is 1 μA, what is the DC
resistance in reverse biased diode?

Question

Question

This question was previously asked in

RRB JE ECE 22 Apr 2025 Shift 2 CBT 2 Official Paper

Answer 1

Concept:

The DC resistance of a reverse-bias diode is calculated using Ohm's Law:

\( R_{DC} = \frac{V_{reverse}}{I_R} \)

Where:
V_reverse = Reverse voltage across diode
I_R = Reverse leakage current

Given:

Calculation:

Determine Reverse Voltage:
\(V_{reverse} = V_{cathode} - V_{anode} = 10V - 0V = 10V\)

Calculate DC Resistance:

\( R_{DC} = \frac{10V}{1 \times 10^{-6}A} = 10 \times 10^6 \Omega = 10 M\Omega\)

Key Notes:
In reverse bias, the diode's resistance is very high (leakage current is small).

The cut-in voltage (0.7 V) is only relevant for forward bias and does not affect reverse-bias calculations.