A Zener diode, having breakdown voltage equal to $15 ext{ V}$, is used in a voltage regulator circuit, as sho

Question

A Zener diode, having breakdown voltage equal to $15	ext{ V}$, is used in a voltage regulator circuit, as shown in the figure. The current through the diode is:

Accepted Answer

1. **Analyze the Circuit:** In a typical Zener voltage regulator circuit (from the corresponding NEET 2011 question), the input unregulated voltage is $V_{in} = 20	ext{ V}$, the series resistor is $R_s = 250\, \Omega$, and the load resistor is $R_L = 1	ext{ k}\Omega$.
2. **Voltage Across Load:** Since the Zener diode is in parallel with the load resistor and is in breakdown region, the voltage across the load is equal to the Zener voltage, $V_L = V_Z = 15	ext{ V}$.
3. **Calculate Load Current ($I_L$):** Using Ohm's law, $I_L = \frac{V_Z}{R_L} = \frac{15	ext{ V}}{1000\, \Omega} = 0.015	ext{ A} = 15	ext{ mA}$.
4. **Calculate Series Current ($I_s$):** The voltage drop across the series resistor is $V_s = V_{in} - V_Z = 20	ext{ V} - 15	ext{ V} = 5	ext{ V}$. The current through it is $I_s = \frac{V_s}{R_s} = \frac{5	ext{ V}}{250\, \Omega} = 0.02	ext{ A} = 20	ext{ mA}$.
5. **Calculate Zener Current ($I_Z$):** Applying Kirchhoff's Current Law at the junction, the total current $I_s$ splits into $I_Z$ and $I_L$. Therefore, $I_Z = I_s - I_L = 20	ext{ mA} - 15	ext{ mA} = 5	ext{ mA}$.

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