A special diode with 0 Ω forward resistance and 1 kΩ reverse resistance is connected in series with a 1 kΩ load resistor. A square wave input is applied with:
Maximum voltage (Vmax) = 4 V
Minimum voltage (Vmin) = −4 V
What is the output voltage across the load when the diode is in reverse bias?

Concept:

This problem involves a special diode with asymmetric resistance characteristics:

• Forward resistance = 0 Ω → behaves like a short circuit in forward bias
• Reverse resistance = 1 kΩ → behaves like a 1 kΩ resistor in reverse bias

The diode is in series with a 1 kΩ load resistor, and a square wave input is applied with: \( V_{\text{max}} = +4~\text{V} \) and \( V_{\text{min}} = -4~\text{V} \).

Case: When the Diode is in Reverse Bias

In reverse bias, diode has a 1 kΩ resistance and is in series with the 1 kΩ load resistor. The total resistance in the circuit becomes:

\( R_{\text{total}} = 1~\text{k}\Omega + 1~\text{k}\Omega = 2~\text{k}\Omega \)

Input voltage during reverse bias = -4 V

Using the voltage division rule:

\( V_{\text{out}} = V_{\text{in}} \cdot \frac{R_{\text{load}}}{R_{\text{total}}} = -4 \cdot \frac{1}{2} = -2~\text{V} \)

Conclusion:

Correct Answer: Option 1) -2 V