Question

In the diagram shown, the Zener diode has a reverse breakdown voltage of V_Z . The current through the load resistance R_L is I_L The current through the Zener diode is

• A. \(\frac{v_0-v_z}{R_s}\)
• B. \(\frac{v_0-v_z}{R_L}\)
• C. \(\frac{v_z}{R_L}\)
• D. \((\frac{v_0-v_z}{R_S})-I_l\)

Answer 1

The Correct Option is D

In the given circuit, the Zener diode has a reverse breakdown voltage \( V_Z \). The current through the load resistance \( R_L \) is \( I_L \). The Zener diode is in breakdown, and the voltage across it is constant at \( V_Z \). The total voltage across the series combination of \( R_S \) and the Zener diode is \( V_0 \). The current through the series combination is given by: \[ I = \frac{V_0 - V_Z}{R_S} \] The current through the Zener diode \( I_Z \) is the difference between the total current \( I \) and the current through the load resistance \( I_L \), because both the Zener diode and the load resistance are in parallel: \[ I_Z = I - I_L = \frac{V_0 - V_Z}{R_S} - I_L \] Thus, the current through the Zener diode is: \[ I_Z = \left( \frac{V_0 - V_Z}{R_S} \right) - I_L \] Therefore, the correct answer is (D) \( \left( \frac{V_0 - V_Z}{R_S} \right) - I_L \).