Forward bias of a Zener is similar to ordinary diode. In reverse bias, current is negligible until reverse voltage reaches V_Z (Zener breakdown), where current rises sharply at almost constant voltage. This constant V_Z is exploited as a voltage regulator. Plot I_R vs V_R to identify V_Z.
-- NCERT Physics Lab Manual Class 12, Part 10, p. 125Experiment 17 Zener Diode Curves
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Experiment 17 Zener Diode Curves, explained for NEET
The recurring error in this experiment is treating the Zener's reverse branch like an ordinary diode's. On a rectifier diode the reverse branch is a flat, uninteresting microampere line; on a Zener it is the branch you came to measure. Everything of interest here happens in reverse bias.
Wire the Zener with its cathode to the positive terminal, always through a series resistance. Raise the reverse voltage in steps and record the reverse current. The plot (procedure in NCERT Physics Lab Manual Class 12, Part 10, page 125) stays almost on the voltage axis — currents of the order of microamperes — until a sharp knee. Past that knee the current climbs steeply while the voltage across the diode barely moves. The voltage at the knee is the reverse breakdown voltage, V_z.
That near-vertical segment is the point of the device: over a wide range of current the Zener holds its terminal voltage nearly fixed, which is why it serves as a voltage reference. The small residual slope of the segment, ΔV/ΔI, is its dynamic resistance — a few ohms for a good Zener, not zero.
Two practical consequences. First, the series resistance is not optional: without it, the current past breakdown is limited only by the supply, and the power rating is exceeded within a volt. Second, the knee is where reading precision matters. Stepping the supply in 1 V jumps places the knee anywhere inside a 1 V window; near the expected breakdown, step in 0.1 V and take several closely spaced points.
For NEET, the questions that come out of this experiment are circuit-level: read V_z off a curve or a table, size the series resistance for a stated Zener current, or split the resistor current between the Zener and a load.
Watch-out: breakdown is a normal, reversible operating state, not damage — provided the current stays limited.
Can you answer these Experiment 17 Zener Diode Curves 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
To obtain the breakdown region of a Zener diode's characteristic, the diode is connected in:
Show answer and why every option is right or wrong
Answer: C. C is correct — the breakdown region lies on the reverse branch, so the cathode is held at the higher potential, as the circuit in NCERT Physics Lab Manual Class 12, Part 10, page 125 shows.
Why A is wrong: A is wrong because forward bias gives only the ordinary conducting knee near a few tenths of a volt, where no breakdown appears.
Why B is wrong: B is wrong because at zero bias no measurable current flows and no point of the characteristic is traced.
Why D is wrong: D is wrong because an open circuit carries no current, so no reading of the characteristic can be recorded.
MCQ 2Easy RecallPractice
In the breakdown region of a Zener diode's reverse characteristic, as the reverse current is increased the voltage across the diode:
Show answer and why every option is right or wrong
Answer: A. A is correct — past the knee the curve is nearly vertical, so a large change in current produces only a small change in terminal voltage (NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why B is wrong: B is wrong because the terminal voltage settles near the breakdown value instead of collapsing to zero.
Why C is wrong: C is wrong because proportionality would describe an ohmic resistor, which is exactly what the breakdown segment is not.
Why D is wrong: D is wrong because the polarity of the applied bias stays the same throughout the reverse sweep.
MCQ 3Easy RecallPractice
In this experiment a resistance is always connected in series with the Zener diode. Its main purpose is to:
Show answer and why every option is right or wrong
Answer: D. D is correct — once breakdown begins the diode itself offers almost no resistance, so the series resistor is what fixes the current (NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why A is wrong: A is wrong because the breakdown voltage is set by the doping of the junction and no external resistor can shift it.
Why B is wrong: B is wrong because the series resistor is present for the reverse sweep, while the shape of the forward knee is a property of the junction.
Why C is wrong: C is wrong because the voltmeter sits across the diode and is not the component at risk; the diode is.
MCQ 4Direct ApplicationPractice
A Zener diode of breakdown voltage 6.0 V is to be run from a 10.0 V supply at a Zener current of 20.0 mA. The series resistance required is:
Show answer and why every option is right or wrong
Answer: B. B is correct — the resistor drops 10.0 V − 6.0 V = 4.0 V, so R = 4.0 V / 20.0 × 10⁻³ A = 2.0 × 10² Ω, the sizing step set out with the circuit in NCERT Physics Lab Manual Class 12, Part 10, page 125.
Why A is wrong: A is wrong because it uses a 2.0 V drop across the resistor, half the actual difference between supply and breakdown voltage.
Why C is wrong: C is wrong because it takes the resistor's drop as 6.0 V, which is the voltage the diode holds, not the resistor.
Why D is wrong: D is wrong because it divides the full 10.0 V supply by the current, ignoring the voltage held by the diode.
MCQ 5Direct ApplicationPractice
A reverse sweep on a Zener diode gives: 2.0 V → 0.001 mA; 4.0 V → 0.002 mA; 5.0 V → 0.005 mA; 5.6 V → 0.50 mA; 5.7 V → 8.0 mA; 5.8 V → 20.0 mA. The reverse breakdown voltage is:
Show answer and why every option is right or wrong
Answer: D. D is correct — the breakdown voltage is the knee, the first reading at which the current leaves the microampere floor and begins to climb steeply, which here is 5.6 V (NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why A is wrong: A is wrong because 2.0 V sits on the flat leakage part of the curve, where the current is still about a microampere.
Why B is wrong: B is wrong because at 4.0 V the current has barely doubled from its 2.0 V value, so no turn has begun.
Why C is wrong: C is wrong because 5.0 V is the last point still on the leakage floor, below the turn.
MCQ 6Direct ApplicationPractice
Beyond breakdown, a Zener diode reads 6.10 V at a reverse current of 5.0 mA and 6.20 V at 25.0 mA. Its dynamic resistance in this region is:
Show answer and why every option is right or wrong
Answer: A. A is correct — dynamic resistance is the slope of the breakdown segment, ΔV/ΔI = 0.10 V / 20.0 × 10⁻³ A = 5.0 Ω (breakdown-region reading practice, NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why B is wrong: B is wrong because 20 is the current change in milliamperes, reported as though it were a resistance.
Why C is wrong: C is wrong because it uses a voltage change of 1.0 V instead of the measured 0.10 V.
Why D is wrong: D is wrong because 6.20 V ÷ 25.0 mA ≈ 2.5 × 10² Ω is the static ratio V/I at one point, not the slope of the breakdown segment.
MCQ 7CalculationPractice
A 6.0 V Zener diode is fed from a 12.0 V supply through a series resistance of 2.0 × 10² Ω, and a load connected across the diode draws 10.0 mA. The current through the Zener diode is:
Show answer and why every option is right or wrong
Answer: C. C is correct — the resistor carries (12.0 − 6.0) V / 2.0 × 10² Ω = 30.0 mA, and the load takes 10.0 mA of it, leaving 20.0 mA through the diode (NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why A is wrong: A is wrong because 10.0 mA is the load current; the diode carries what remains of the resistor current.
Why B is wrong: B is wrong because it halves the resistor current instead of subtracting the stated load current.
Why D is wrong: D is wrong because 30.0 mA is the total current through the series resistor, before the load branch takes its share.
MCQ 8Concept TrapPractice
A student sweeps the reverse voltage in steps of 1.0 V and reports the breakdown voltage as "between 5 V and 6 V". The change that most improves this determination is to:
Show answer and why every option is right or wrong
Answer: B. B is correct — the knee is a feature narrower than the step size, so only finer steps across that region can locate it (NCERT Physics Lab Manual Class 12, Part 10, page 125).
Why A is wrong: A is wrong because a larger current past the knee says nothing about where the knee lies, and it risks exceeding the diode's power rating.
Why C is wrong: C is wrong because a microammeter would go off-scale past breakdown; the limitation is the spacing of the voltage settings, not the current range.
Why D is wrong: D is wrong because averaging repeats of the same coarse settings reproduces the same 1.0 V uncertainty every time.
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How do you solve a Experiment 17 Zener Diode Curves question? A worked example
- 1
Given.
Supply voltage V_s = 12.0 V. Zener breakdown voltage V_z = 6.0 V. Required Zener current I_z = 25.0 mA = 2.50 × 10⁻² A.
- 2
Required.
The series resistance R that sets this operating point, and the power dissipated in the diode.
- 3
Concept.
Past breakdown the Zener holds its terminal voltage near V_z, so the remainder of the supply voltage appears across the series resistor, and that resistor alone fixes the current.
- 4
Formula.
R = (V_s − V_z) / I_z, from the loop equation V_s = I_z R + V_z. Diode power P = V_z I_z.
- 5
Substitution.
R = (12.0 V − 6.0 V) / 2.50 × 10⁻² A.
- 6
Calculation.
R = 6.0 / 2.50 × 10⁻² = 2.40 × 10² Ω. P = 6.0 × 2.50 × 10⁻² = 0.15 W. The integers in the loop equation (one resistor, one diode) are exact counting numbers and do not contribute to the significant-figure count; the two significant figures come from the measured 6.0 V difference.
- 7
Final answer.
R = 2.4 × 10² Ω, with the diode dissipating 0.15 W — comfortably inside a 0.5 W rating.
- 8
Common trap.
Dividing the full 12.0 V by the current, giving 4.8 × 10² Ω. That treats the Zener as a short circuit and halves the actual Zener current in the built circuit.
- 9
Similar NEET-style question.
The same diode is run from a 9.0 V supply through 1.5 × 10² Ω, with a load drawing 5.0 mA across the diode. Find the Zener current. (Resistor current = 3.0 V / 1.5 × 10² Ω = 20.0 mA; Zener current = 15.0 mA.)
What to remember before solving Experiment 17 Zener Diode Curves questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 17 Zener Diode Curves questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
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