Forward bias: plot I (mA) vs V (V) — exponential rise above the cut-in voltage (~0.3 V Ge, ~0.7 V Si). Reverse bias: plot I (μA) vs V (V) — small saturation reverse current until breakdown. Dynamic resistance r = ΔV/ΔI computed from slope. Connect ammeter in series, voltmeter in parallel.
-- NCERT Physics Lab Manual Class 12, Part 10, p. 120Experiment 16 p–n Diode Characteristics
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Experiment 16 p–n Diode Characteristics, explained for NEET
A junction diode does not have a resistance. That is the whole content of this experiment, and it is where readings go wrong: a student takes one voltmeter value, one milliammeter value, divides, and reports "the resistance of the diode". Two different operating points on the same curve give two different numbers, and neither is wrong — the characteristic is not a straight line through the origin.
The circuit, as set out in the NCERT Physics Lab Manual Class 12, Part 10, page 120, puts the voltmeter directly across the diode and the current meter in series with it, with a potential divider to vary the bias and a series resistance to protect the junction. Forward bias means the p-side to the positive terminal. Current stays negligible until the applied voltage reaches the knee — near 0.7 V for silicon, near 0.3 V for germanium — and then climbs steeply, so the meter for this branch is a milliammeter.
Reverse the supply and the picture inverts. Current drops to a few microamperes and stays almost flat however far the reverse voltage is pushed, until breakdown. A microammeter is required; a milliammeter reads zero throughout and the branch collapses onto the axis. For the same reason the two branches are plotted in the first and third quadrants with different current scales.
Two resistances then follow from the same curve. Static resistance is V/I at a point. Dynamic resistance is ΔV/ΔI over a small interval — the reciprocal slope. In the forward region past the knee the dynamic value is a few ohms while the static value is tens of ohms; in reverse it runs to tens of megohms.
Watch-out: NEET asks for the dynamic resistance far more often than the static one. Read which is wanted before dividing.
Can you answer these Experiment 16 p–n Diode Characteristics 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 tracing the reverse characteristic of a p-n junction diode, the current in the circuit is measured with:
Show answer and why every option is right or wrong
Answer: C. Option C. Below breakdown the reverse current is the saturation current, a few microamperes, so only a microammeter resolves it — the reverse branch of the circuit in the NCERT Physics Lab Manual Class 12, Part 10, page 120 uses one for exactly this reason.
Why A is wrong: A is incorrect: milliamperes is the scale of the forward branch, and a milliammeter reads a steady zero throughout the reverse branch, collapsing the curve onto the voltage axis.
Why B is wrong: B is incorrect: a voltmeter measures potential difference and is connected across the diode, never in series to read current.
Why D is wrong: D is incorrect: an ampere-range meter is coarser still than a milliammeter, and the reverse current sits roughly six orders of magnitude below its full-scale value.
MCQ 2Easy RecallPractice
The knee voltage of a silicon p-n junction diode in forward bias is approximately:
Show answer and why every option is right or wrong
Answer: A. Option A. Silicon junctions begin to conduct appreciably near 0.7 V, which is where the forward characteristic plotted in the NCERT Physics Lab Manual Class 12, Part 10, page 120 turns sharply upward.
Why B is wrong: B is incorrect: 0.3 V is the knee voltage of a germanium diode, and swapping the two materials' values is the standard slip in this question.
Why C is wrong: C is incorrect: 1.5 V is a dry-cell emf, not a junction property, and a silicon diode conducts heavily well before that voltage is reached.
Why D is wrong: D is incorrect: 5.0 V is a supply-level voltage, and a forward-biased silicon diode would be destroyed long before the voltage across it reached this value.
MCQ 3Easy RecallPractice
To forward-bias the diode in this experiment, the connection required is:
Show answer and why every option is right or wrong
Answer: D. Option D. Forward bias means the external supply opposes the barrier potential, which requires the p-region at the higher potential — the forward-branch connection described in the NCERT Physics Lab Manual Class 12, Part 10, page 120.
Why A is wrong: A is incorrect: that is the reverse-bias connection, which widens the depletion region and leaves only the microampere saturation current.
Why B is wrong: B is incorrect: the diode's whole usefulness is that the two orientations give completely different currents at the same applied voltage.
Why C is wrong: C is incorrect: shorting both ends to one terminal applies no potential difference across the junction, so no branch of the characteristic can be traced.
MCQ 4Direct ApplicationPractice
A resistance is deliberately left in series with the diode in the forward branch of the circuit. Its main purpose is to:
Show answer and why every option is right or wrong
Answer: B. Option B. Past the knee the characteristic is almost vertical, so a small extra supply voltage produces a large current jump; the series resistance carries that surplus voltage and protects the junction, as the forward-branch circuit in the NCERT Physics Lab Manual Class 12, Part 10, page 120 shows.
Why A is wrong: A is incorrect: a series resistance drops voltage rather than adding it, so it reduces the potential difference reaching the diode.
Why C is wrong: C is incorrect: the saturation current is set by the junction's minority-carrier supply and cannot be increased by series resistance; the reverse branch needs a microammeter instead.
Why D is wrong: D is incorrect: a current reading is converted to a voltage by measuring across a known resistance, which is not the job of a protective series resistor.
MCQ 5Direct ApplicationPractice
The forward and reverse branches of the characteristic are drawn on one sheet using different current scales. The reason is that:
Show answer and why every option is right or wrong
Answer: D. Option D. The two branches differ in current by about a thousandfold, so each needs its own scale if both are to show real structure — the plotting instruction given with the characteristic in the NCERT Physics Lab Manual Class 12, Part 10, page 120.
Why A is wrong: A is incorrect: reverse voltages are applied in volts, often several volts, so it is the current and not the voltage that changes scale between the branches.
Why B is wrong: B is incorrect: the quadrants are the other way round, forward bias (positive V, positive I) being the first quadrant and reverse bias the third.
Why C is wrong: C is incorrect: both branches are traced on the same diode, simply by reversing the supply connections.
MCQ 6Direct ApplicationPractice
On the forward characteristic of a silicon diode, the current is 10.0 mA at 0.70 V and 30.0 mA at 0.75 V. The dynamic resistance of the diode over this interval is:
Show answer and why every option is right or wrong
Answer: A. Option A. Dynamic resistance is ΔV/ΔI = (0.75 − 0.70) V ÷ (30.0 − 10.0) mA = 0.05 V ÷ 2.00 × 10⁻² A = 2.5 Ω, the reciprocal-slope definition used with this characteristic in the NCERT Physics Lab Manual Class 12, Part 10, page 120.
Why B is wrong: B is incorrect: 25 Ω is the static resistance V/I at the upper point (0.75 V ÷ 30.0 mA), not the slope of the curve between the two points.
Why C is wrong: C is incorrect: 70 Ω is the static resistance at the lower point (0.70 V ÷ 10.0 mA), a second reminder that a diode has no single V/I value.
Why D is wrong: D is incorrect: 0.40 is ΔI/ΔV in siemens, the conductance, and the resistance is its reciprocal.
MCQ 7CalculationPractice
For the same readings — 10.0 mA at 0.70 V and 30.0 mA at 0.75 V — the static and dynamic resistances at the operating point 0.75 V are related as:
Show answer and why every option is right or wrong
Answer: B. Option B. Static resistance is 0.75 V ÷ 30.0 mA = 25 Ω and dynamic resistance is 0.05 V ÷ 2.00 × 10⁻² A = 2.5 Ω; the factor of ten follows from the curvature of the characteristic shown in the NCERT Physics Lab Manual Class 12, Part 10, page 120.
Why A is wrong: A is incorrect: equality of the two would require the characteristic to be a straight line through the origin, which is exactly what a diode curve is not.
Why C is wrong: C is incorrect: the labels are swapped, since V/I at a point gives the larger number and ΔV/ΔI over the steep interval gives the smaller one.
Why D is wrong: D is incorrect: static resistance is well defined as V/I at a stated operating point, it simply changes from point to point.
MCQ 8CalculationPractice
On the reverse branch a student records 5.0 µA at 2.0 V reverse and 5.2 µA at 8.0 V reverse. These readings show that:
Show answer and why every option is right or wrong
Answer: C. Option C. The current changes by only 0.2 µA for a 6.0 V change, so ΔV/ΔI = 6.0 V ÷ 2.0 × 10⁻⁷ A ≈ 3 × 10⁷ Ω — the flat saturation region of the reverse branch in the NCERT Physics Lab Manual Class 12, Part 10, page 120.
Why A is wrong: A is incorrect: breakdown appears as a near-vertical rise in reverse current, the opposite of the flat 5.0 µA to 5.2 µA behaviour recorded here.
Why B is wrong: B is incorrect: the current rose by about 4 per cent, not by a factor of two, and reading a near-flat branch as proportional is the error this question targets.
Why D is wrong: D is incorrect: 4 × 10⁵ Ω is the static value V/I at the first point, and quoting it as an ohmic resistance ignores that the same ratio at 8.0 V gives 1.5 × 10⁶ Ω.
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How do you solve a Experiment 16 p–n Diode Characteristics question? A worked example
- 1
Given
Two points on the forward characteristic of a silicon diode, read from the same trace: V₁ = 0.60 V with I₁ = 2.0 mA, and V₂ = 0.70 V with I₂ = 22.0 mA. The voltmeter resolves 0.01 V and the milliammeter resolves 0.1 mA.
- 2
Required
The dynamic resistance of the diode between these two points, with the uncertainty the instruments impose on it.
- 3
Concept
Beyond the knee the characteristic is steep but not vertical. Its reciprocal slope over a short interval is the dynamic resistance — the quantity that matters when a small signal rides on a steady bias. A single V/I ratio is the static resistance and answers a different question.
- 4
Formula
r = ΔV / ΔI, with ΔV = V₂ − V₁ and ΔI = I₂ − I₁.
- 5
Substitution
ΔV = 0.70 V − 0.60 V = 0.10 V. ΔI = 22.0 mA − 2.0 mA = 20.0 mA = 2.00 × 10⁻² A. So r = 0.10 V ÷ (2.00 × 10⁻² A).
- 6
Calculation
r = 5.0 Ω. Each voltmeter reading carries ±0.01 V, so ΔV carries ±0.02 V — 20 per cent of 0.10 V. Each current reading carries ±0.1 mA, so ΔI carries ±0.2 mA — 1 per cent of 20.0 mA. The fractional errors add to about 21 per cent. The count of readings (two points, exact) is a counting integer and contributes nothing to the significant-figure count; only the measured voltages and currents limit the precision.
- 7
Final answer
r ≈ 5.0 Ω, uncertain by about 21 per cent, i.e. 5 ± 1 Ω. Quoting "5.00 Ω" would claim precision that a 0.01 V voltmeter cannot support over so small a voltage interval.
- 8
Common trap
Taking the two points too close together. The uncertainty in r is dominated by ΔV, so a 0.10 V interval measured with a 0.01 V voltmeter already costs 20 per cent. Spreading the points over a wider stretch of the steep region cuts that error, at the price of averaging over more curvature — which is why the interval is chosen deliberately rather than stumbled into.
- 9
Similar NEET-style question
A forward characteristic gives 5.0 mA at 0.65 V and 45.0 mA at 0.75 V. Find the dynamic resistance, and state whether it is larger or smaller than the static resistance at 0.75 V. (Answer: 2.5 Ω dynamic against 16.7 Ω static — smaller, as always on the steep part of the curve.)
What to remember before solving Experiment 16 p–n Diode Characteristics questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 16 p–n Diode Characteristics questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
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