Zener As Voltage Regulator

8 MCQs9-step worked example
Source: NCERT Electronic DevicesOfficial key: NTA-verifiedLast updated: 7 Oct 2026

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In which condition is a Zener diode operated when it is used as a voltage regulator?
  1. A.Forward bias
  2. B.Reverse bias, in the breakdown region
  3. C.Reverse bias, below the breakdown voltage
  4. D.No bias applied
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Answer: B. B is correct: a Zener diode is designed to operate under reverse bias in the breakdown region and is used as a voltage regulator (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 485).

A is wrong: A is wrong because forward bias makes the Zener behave like an ordinary diode; regulation uses the breakdown region.

C is wrong: C is wrong because below the breakdown voltage the reverse current is tiny and the voltage across the diode is not held constant.

D is wrong: D is wrong because an unbiased diode is not connected to any supply and cannot regulate anything.

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Zener As Voltage Regulator, explained for NEET

The question first: a Zener diode of breakdown voltage 9.0 V is fed from a 15 V supply through a 300 Ω resistor. Raise the supply to 18 V. What happens to the voltage across the Zener? Nothing: it stays at 9.0 V. The extra 3.0 V lands across the resistor, and the current rises from 20 mA to 30 mA. That is the whole idea of regulation.

The concept. A Zener diode is designed to operate under reverse bias in the breakdown region and is used as a voltage regulator (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 485). After the breakdown voltage V_z, a large change in current can be produced by an almost insignificant change in the reverse voltage; the Zener voltage remains constant while the current varies over a wide range (page 485).

The circuit. The unregulated dc voltage is connected to the Zener diode through a series resistance R_s, so that the Zener is reverse biased. If the input rises, the current through R_s and the Zener rises, and the extra drop appears across R_s with no change across the Zener. If the input falls, the drop across R_s falls and the Zener voltage still does not change (page 486). NCERT's worked example picks the Zener current several times the load current, so the diode keeps regulating (page 486).

Where this sits. The 2026-27 reprint of Chapter 14 does not contain this section. The NTA NEET (UG) 2026 syllabus still lists "Zener diode as a voltage regulator", so the pre-2023 edition of the chapter is the NCERT text for it.

Bridge to NEET. Recent papers ask for the Zener's bias in a regulator (2021, answer: reverse), for the statement that it works under reverse bias in the breakdown region (2023), and for a resistor drop (2026: breakdown 3 V, input 5 V, the series resistor takes 5 − 3 = 2 V).

Watch-out: the Zener takes its own breakdown voltage and the series resistor takes the rest. A forward-biased Zener is just an ordinary diode and does not regulate.

How do you solve a Zener As Voltage Regulator question? A worked example

  1. 1

    Given

    An ideal Zener diode of breakdown voltage 4.0 V is reverse biased through a series resistor R_s = 500 Ω from a 9.0 V supply. There is no load.

  2. 2

    Required

    The drop across R_s and the current, and then the output voltage and the current if the supply rises to 11.0 V.

  3. 3

    Concept

    In breakdown the Zener voltage remains constant, so the series resistor takes whatever the supply has beyond V_z (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, pages 485–486).

  4. 4

    Formula

    V_R = V_in − V_z; I = V_R ÷ R_s; V_out = V_z.

  5. 5

    Substitution

    Supply 9.0 V: V_R = 9.0 − 4.0, I = V_R ÷ 500. Supply 11.0 V: V_R = 11.0 − 4.0, I = V_R ÷ 500.

  6. 6

    Calculation

    At 9.0 V: V_R = 5.0 V and I = 5.0 ÷ 500 = 10 mA. At 11.0 V: V_R = 7.0 V and I = 7.0 ÷ 500 = 14 mA. The output stays at 4.0 V in both cases. All values are given data; there are no exact constants.

  7. 7

    Final answer

    9.0 V supply: 5.0 V across R_s, 10 mA. 11.0 V supply: 7.0 V across R_s, 14 mA. The output stays at 4.0 V.

  8. 8

    Common trap

    Giving the Zener the extra voltage because the supply went up. In breakdown the Zener voltage stays fixed, so the whole 2.0 V increase lands on the resistor. The same reasoning gives 2 V in the 2026 item (5 V input, 3 V breakdown). The wrong option 3 V there is the drop across the Zener itself, not across the resistor.

  9. 9

    Similar NEET-style question

    A 7.5 V Zener is fed from 12 V through 900 Ω with no load. Find the current through the resistor. (Answer: V_R = 12 − 7.5 = 4.5 V, I = 4.5 ÷ 900 = 5.0 mA.)

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Can you answer these Zener As Voltage Regulator MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

In which condition is a Zener diode operated when it is used as a voltage regulator?

Show answer and why every option is right or wrong

Answer: B. B is correct: a Zener diode is designed to operate under reverse bias in the breakdown region and is used as a voltage regulator (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 485).

Why A is wrong: A is wrong because forward bias makes the Zener behave like an ordinary diode; regulation uses the breakdown region.

Why C is wrong: C is wrong because below the breakdown voltage the reverse current is tiny and the voltage across the diode is not held constant.

Why D is wrong: D is wrong because an unbiased diode is not connected to any supply and cannot regulate anything.

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MCQ 2Easy RecallPractice

In NCERT's voltage-regulator circuit, the unregulated dc voltage is connected to the Zener diode through which element?

Show answer and why every option is right or wrong

Answer: D. D is correct: the unregulated dc voltage is connected to the Zener diode through a series resistance R_s, with the Zener reverse biased (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486).

Why A is wrong: A is wrong because a series capacitor blocks steady dc, and the circuit has no capacitor in the regulator path.

Why B is wrong: B is wrong because the circuit uses a resistor to take up the voltage difference, not an inductor.

Why C is wrong: C is wrong because a parallel resistor would not drop the surplus voltage; the surplus must fall across an element in series with the Zener.

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MCQ 3Easy RecallPractice

Beyond the breakdown voltage V_z, which quantity stays constant while the current through the Zener varies over a wide range?

Show answer and why every option is right or wrong

Answer: A. A is correct: after V_z, the Zener voltage remains constant even though the current through the diode varies over a wide range (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 485).

Why B is wrong: B is wrong because the Zener current changes when the input changes; that change is what the series resistor absorbs.

Why C is wrong: C is wrong because the current through R_s rises and falls with the input voltage.

Why D is wrong: D is wrong because the unregulated input is the quantity that fluctuates; regulation holds the output constant, not the input.

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MCQ 4Direct ApplicationPractice

A Zener diode with V_z = 9.0 V is reverse biased from a 15 V supply through R_s = 300 Ω, with no load. What is the current?

Show answer and why every option is right or wrong

Answer: C. C is correct: the Zener holds 9.0 V, so the drop across R_s is 15 − 9.0 = 6.0 V and I = 6.0 ÷ 300 = 20 mA (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486).

Why A is wrong: A is wrong because 15 ÷ 300 = 50 mA treats the whole supply as dropping across R_s, ignoring the 9.0 V held by the Zener.

Why B is wrong: B is wrong because 9.0 ÷ 300 = 30 mA uses the Zener voltage instead of the voltage left for R_s.

Why D is wrong: D is wrong because 24 ÷ 300 = 80 mA adds the Zener voltage to the supply instead of subtracting it.

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MCQ 5Direct ApplicationPractice

In the circuit of MCQ 4, the supply is raised from 15 V to 18 V. What are the voltage across the Zener and the drop across R_s now?

Show answer and why every option is right or wrong

Answer: B. B is correct: the Zener still holds 9.0 V, so the drop across R_s becomes 18 − 9.0 = 9.0 V and the current rises to 9.0 ÷ 300 = 30 mA (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486).

Why A is wrong: A is wrong because it puts the whole supply across the Zener; in breakdown its voltage stays at 9.0 V.

Why C is wrong: C is wrong because it leaves the drop across R_s at 6.0 V; the drop must rise by the 3.0 V of extra supply.

Why D is wrong: D is wrong because the Zener voltage cannot rise from 9.0 V to 12 V in breakdown.

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MCQ 6Direct ApplicationPractice

A Zener with V_z = 5.0 V is fed from 12 V through R_s = 500 Ω. A 1.0 kΩ load is connected across the Zener. What is the current through the Zener?

Show answer and why every option is right or wrong

Answer: D. D is correct: the current through R_s is (12 − 5.0) ÷ 500 = 14 mA, the load takes 5.0 ÷ 1000 = 5.0 mA, and the Zener carries the rest, 14 − 5.0 = 9.0 mA (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486).

Why A is wrong: A is wrong because 14 mA is the total current through R_s, before the load takes its share.

Why B is wrong: B is wrong because 5.0 mA is the load current, not the Zener current.

Why C is wrong: C is wrong because it adds the load current to the total instead of subtracting it.

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MCQ 7CalculationPractice

A Zener with V_z = 5.0 V is fed from 8.0 V through R_s = 300 Ω. Below what load resistance does the Zener current fall to zero, so that regulation is lost?

Show answer and why every option is right or wrong

Answer: A. A is correct: the current through R_s is (8.0 − 5.0) ÷ 300 = 10 mA. The load takes all of it when 5.0 ÷ R_L = 10 mA, i.e. R_L = 500 Ω; a smaller load needs more current than R_s can supply (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486, which requires the Zener current to stay well above the load current).

Why B is wrong: B is wrong because 300 Ω is the series resistor, not the load that would absorb the full 10 mA at 5.0 V.

Why C is wrong: C is wrong because 8.0 ÷ 10 mA = 800 Ω uses the supply voltage; the load sits at the Zener voltage, 5.0 V.

Why D is wrong: D is wrong because 50 Ω would need 5.0 ÷ 50 = 100 mA, ten times the 10 mA available.

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MCQ 8CalculationPractice

A Zener with V_z = 6.0 V is fed through R_s = 400 Ω, with no load. The input varies between 12 V and 14 V. What is the range of the Zener current, and what is the output voltage?

Show answer and why every option is right or wrong

Answer: C. C is correct: at 12 V the current is (12 − 6.0) ÷ 400 = 15 mA and at 14 V it is (14 − 6.0) ÷ 400 = 20 mA, while the voltage across the Zener stays at 6.0 V (NCERT Class 12 Physics (pre-2023 edition), Chapter 14, page 486).

Why A is wrong: A is wrong because it lets the output follow the input; the regulator holds the output at 6.0 V.

Why B is wrong: B is wrong because 45 mA and 50 mA come from (12 + 6.0) ÷ 400 and (14 + 6.0) ÷ 400, adding the Zener voltage instead of subtracting it.

Why D is wrong: D is wrong because the current changes when the input changes: 15 mA at 12 V but 20 mA at 14 V.

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What to remember before solving Zener As Voltage Regulator questions

5 NCERT lines
Key Fact

Zener diode as a voltage regulator

In the NEET syllabus; removed from current NCERT.

The unregulated dc voltage (filtered rectifier output) is connected to the Zener diode through a series resistance R_s so that the Zener diode is reverse biased. If the input voltage rises (or falls), the current through R_s and the Zener diode rises (or falls), changing the drop across R_s but not the voltage across the Zener diode, because in the breakdown region the Zener voltage stays constant. The Zener diode thus gives a constant output voltage.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 486
Key Fact

Zener I-V characteristic: beyond V_z a large current change for almost no voltage change; used to regulate

In the NEET syllabus; removed from current NCERT.

The I-V characteristics of a Zener diode is shown in Fig. 14.21(b). It is seen that when the applied reverse bias voltage(V) reaches the breakdown voltage (Vz) of the Zener diode, there is a large change in the current. Note that after the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant, even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating supply voltages so that they are constant.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 485
Key Fact

Breakdown at V_z: internal field emission (field ionisation), field of order 10^6 V/m

In the NEET syllabus; removed from current NCERT.

When the reverse bias voltage V = Vz, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side. These electrons account for high current observed at the breakdown. The emission of electrons from the host atoms due to the high electric field is known as internal field emission or field ionisation. The electric field required for field ionisation is of the order of 106 V/m.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 485
Key Fact

Zener voltage regulator: series resistance R_s, Zener reverse biased; input changes appear across R_s, not across the Zener

In the NEET syllabus; removed from current NCERT.

The unregulated dc voltage (filtered output of a rectifier) is connected to the Zener diode through a series resistance Rs such that the Zener diode is reverse biased. If the input voltage increases, the current through Rs and Zener diode also increases. This increases the voltage drop across Rs without any change in the voltage across the Zener diode. This is because in the breakdown region, Zener voltage remains constant even though the current through the Zener diode changes. Similarly, if the input voltage decreases, the current through Rs and Zener diode also decreases. The voltage drop across Rs decreases without any change in the voltage across the Zener diode. Thus any increase/ decrease in the input voltage results in, increase/ decrease of the voltage drop across Rs without any change in voltage across the Zener diode. Thus the Zener diode acts as a voltage regulator.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 486
Key Fact

Worked example 14.5: choose the Zener current several times the load current; R_s from the drop across it

In the NEET syllabus; removed from current NCERT.

The value of RS should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., IZ = 20 mA. The total current through RS is, therefore, 24 mA. The voltage drop across RS is 10.0 – 6.0 = 4.0 V. This gives RS = 4.0V/(24 × 10–3) A = 167 Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate.

-- NCERT Class 12 Physics (pre-2023 edition), Chapter 14, p. 486

Zener As Voltage Regulator: NEET previous year questions (PYQs) with answers

17 questions in Electronic Devices

No question in our NEET 2020–2025 set targets this topic directly.

All 17 past-paper questions from Electronic Devices →

How does NEET ask about Zener As Voltage Regulator?

1 recurring pattern from past papers

More in Electronic Devices: 3 exam traps and mistakes · 3 formulas · 2 question patterns from its other lessons.

Sources

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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