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NEET PHYSICSMedium

A Zener diode, having breakdown voltage equal to 15 V15\text{ V}, is used in a voltage regulator circuit, as shown in the figure. The current through the diode is:

A

10 mA10\text{ mA}

B

15 mA15\text{ mA}

C

20 mA20\text{ mA}

D

5 mA5\text{ mA}

Step-by-Step Solution

  1. Analyze the Circuit: In a typical Zener voltage regulator circuit (from the corresponding NEET 2011 question), the input unregulated voltage is Vin=20 VV_{in} = 20\text{ V}, the series resistor is Rs=250 ΩR_s = 250\, \Omega, and the load resistor is RL=1 kΩR_L = 1\text{ 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, VL=VZ=15 VV_L = V_Z = 15\text{ V}.
  3. Calculate Load Current (ILI_L): Using Ohm's law, IL=VZRL=15 V1000 Ω=0.015 A=15 mAI_L = \frac{V_Z}{R_L} = \frac{15\text{ V}}{1000\, \Omega} = 0.015\text{ A} = 15\text{ mA}.
  4. Calculate Series Current (IsI_s): The voltage drop across the series resistor is Vs=Vin−VZ=20 V−15 V=5 VV_s = V_{in} - V_Z = 20\text{ V} - 15\text{ V} = 5\text{ V}. The current through it is Is=VsRs=5 V250 Ω=0.02 A=20 mAI_s = \frac{V_s}{R_s} = \frac{5\text{ V}}{250\, \Omega} = 0.02\text{ A} = 20\text{ mA}.
  5. Calculate Zener Current (IZI_Z): Applying Kirchhoff's Current Law at the junction, the total current IsI_s splits into IZI_Z and ILI_L. Therefore, IZ=Is−IL=20 mA−15 mA=5 mAI_Z = I_s - I_L = 20\text{ mA} - 15\text{ mA} = 5\text{ mA}.
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