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MCQEasyJEE 2023Zener Diode & Voltage Regulation

JEE Physics 2023 Question with Solution

A Zener diode of power rating 1.6 W1.6 \, \text{W} is used as a voltage regulator. If the Zener diode has a breakdown voltage of 8 V8 \, \text{V} and it has to regulate a voltage fluctuating between 3 V3 \, \text{V} and 10 V10 \, \text{V}, what is the value of resistance RsR_s for safe operation of the diode?

Circuit of a Zener diode voltage regulator showing series resistor Rs, unregulated voltage input, Zener diode shunt branch, and regulated voltage output terminals.
  • A

    13.3 Ω13.3\,\Omega

  • B

    13 Ω13\,\Omega

  • C

    10 Ω10\,\Omega

  • D

    12 Ω12\,\Omega

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Power rating of Zener diode = 1.6 W1.6 \, \text{W}, breakdown voltage = 8 V8 \, \text{V}, maximum input voltage = 10 V10 \, \text{V}.

Find: The value of series resistance RsR_s for safe operation.

For safe operation, use the maximum input voltage so that the Zener current does not exceed the permitted value.

The total current is given by

It=PtVtI_t = \frac{P_t}{V_t}

Substituting the values,

It=1.68=0.2 AI_t = \frac{1.6}{8} = 0.2 \, \text{A}

The voltage drop across RsR_s is

ΔV=10−8=2 V\Delta V = 10 - 8 = 2 \, \text{V}

Using Ohm’s law,

Rs=ΔVItR_s = \frac{\Delta V}{I_t}

Substituting the values,

Rs=20.2R_s = \frac{2}{0.2}

So,

Rs=10 ΩR_s = 10 \, \Omega

Therefore, the series resistance is 10 Ω10 \, \Omega. The solution states the correct option is A, but the computed value matches option C in the given options.

Answer Discrepancy Note

The working in the solution gives Rs=10 ΩR_s = 10 \, \Omega. However, the same the solution labels the correct option as A, while the listed options show 10 Ω10 \, \Omega as option C. Since the numerical derivation is consistent, the defensible answer from the options is C.

Common mistakes

  • Using the minimum input voltage 3 V3 \, \text{V} to design RsR_s is incorrect because Zener safety is decided by the maximum current condition, which occurs at the maximum input voltage. Use 10 V10 \, \text{V} for safe-design calculation.

  • Taking the full input voltage across RsR_s is wrong because the Zener diode already maintains 8 V8 \, \text{V} across itself in breakdown. The resistor only drops the excess voltage, so use 10−8=2 V10 - 8 = 2 \, \text{V}.

  • Using the power formula incorrectly can lead to a wrong current. Since P=VIP = VI, the maximum safe current is I=PVI = \frac{P}{V}, not I=PVI = PV or any other rearrangement. Here it is 1.68=0.2 A\frac{1.6}{8} = 0.2 \, \text{A}.

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