Question:

In the diagram shown, the Zener diode has a reverse breakdown voltage of VZ . The current through the load resistance RL is IL The current through the Zener diode is
the Zener diode has a reverse breakdown voltage of VZ

Updated On: Apr 14, 2025
  • \(\frac{v_0-v_z}{R_s}\)
  • \(\frac{v_0-v_z}{R_L}\)
  • \(\frac{v_z}{R_L}\)
  • \((\frac{v_0-v_z}{R_S})-I_l\)
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The Correct Option is D

Approach Solution - 1

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) \( \left( \frac{V_0 - V_Z}{R_S} \right) - I_L \). 

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Approach Solution -2

In the given circuit, we have a Zener diode with reverse breakdown voltage \( V_Z \) connected in parallel with the load resistance \( R_L \). The supply voltage is \( V_0 \) and the series resistance is \( R_S \). The current through the load resistance is \( I_L \), and the current through the Zener diode is \( I_Z \). Using Kirchhoff's voltage law (KVL) for the loop with the series resistor \( R_S \), the voltage across \( R_S \) is: \[ V_0 - V_Z \] The current through \( R_S \) is: \[ I_S = \frac{V_0 - V_Z}{R_S} \] Since the total current from the source is split between the Zener diode and the load resistor, the current through the Zener diode \( I_Z \) is: \[ I_Z = I_S - I_L \] Substituting \( I_S \): \[ I_Z = \left( \frac{V_0 - V_Z}{R_S} \right) - I_L \] Thus, the current through the Zener diode is: \[ I_Z = \left( \frac{V_0 - V_Z}{R_S} \right) - I_L \]

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