PhysicsCBSE Class 12 · Semiconductor Electronics: Materials, Devices and Simple Circuits

p-n Junction and Diode Characteristics Class 12: Bias and I-V Curve

Updated 15 October 2026 · Fact-checked

When p-type and n-type regions meet, majority carriers diffuse across and recombine, leaving a depletion layer of fixed ions and a barrier potential (about 0.7 V for silicon). Forward bias lowers the barrier and a large current flows above the knee voltage; reverse bias raises it, allowing only a tiny current until breakdown.

Fixed ions in the depletion layer create the barrier; the diode conducts only in forward bias above about 0.7 V and gives tiny current in reverse.

Illustration drawn for QuestionBuddy.

Understand p-n Junction and Diode Characteristics

A diode lets current through one way and blocks it the other way. All of that comes from what happens at the boundary between p-type and n-type silicon: the p-n junction.

Formation of the junction. Take a crystal that is p-type on one side and n-type on the other.

Diffusion: electrons are plentiful in the n-side and holes in the p-side, so electrons diffuse into the p-side and holes into the n-side, producing a diffusion current.
Depletion layer: diffusing electrons and holes recombine near the junction. The n-side near the junction is left with uncovered positive donor ions, and the p-side with negative acceptor ions. These ions are fixed in the lattice. This thin region, empty of free carriers, is the depletion region (about a micrometre wide or less).
Barrier potential: the fixed charges set up an electric field from the n-side to the p-side and a potential difference across the junction, the barrier potential (about 0.7 V for silicon and 0.3 V for germanium). It opposes further diffusion of majority carriers.
Drift: the same field sweeps minority carriers across the junction, producing a drift current opposite to the diffusion current. In equilibrium the two are equal and the net current is zero.

Forward bias. Connect the p-side to the positive terminal of a battery. The applied voltage opposes the barrier potential, so the effective barrier falls and the depletion layer becomes narrower. Majority carriers cross easily and a large current flows. On the I-V graph the current is tiny until the applied voltage nears the barrier (the knee or threshold voltage, about 0.7 V for Si), and then rises sharply, almost exponentially.

Reverse bias. Connect the p-side to the negative terminal. The applied voltage adds to the barrier, so the depletion layer becomes wider and majority-carrier diffusion stops. Only the drift of minority carriers remains: a very small reverse saturation current (microamperes for Si) that hardly changes with voltage, because the number of minority carriers is set by temperature. At a large enough reverse voltage (breakdown voltage) the reverse current suddenly increases sharply.

I-V characteristic and resistance. The forward and reverse curves together show strongly non-ohmic, one-way behaviour. The dynamic resistance at a point is $r_{d} = \frac{Δ V}{Δ I}$ : low (a few ohms) in forward bias above the knee, very high in reverse bias. An ideal diode is treated as a closed switch (zero resistance) in forward bias and an open switch (infinite resistance) in reverse bias. A real silicon diode in forward bias drops about 0.7 V.

Using a multimeter. In its diode-test setting, a multimeter shows about 0.6 to 0.7 V for a good silicon diode connected one way and 'open' (no reading) the other way; this identifies the p- and n-terminals.

p-n Junction and Diode Characteristics: formulas and key facts

Barrier potential (approximate): $V_{0} \approx 0.7 V (Si), 0.3 V (Ge)$; Depends on doping and temperature.
Dynamic resistance: $r_{d} = \frac{Δ V}{Δ I}$; Low in forward bias, very high in reverse bias.
Current in a series circuit with a Si diode: $I = \frac{V - 0.7}{R}$; Forward bias, above the knee.
Field in the depletion layer (estimate): $E \approx \frac{V _{0}}{d}$; d = depletion width.
Bias effect on depletion width: $forward: d ↓, reverse: d ↑$; Qualitative.

How to solve p-n Junction and Diode Characteristics questions

Use this method for p-n junction and diode circuit questions.

1Decide the bias: p-side at higher potential than n-side means forward bias.
2Forward-biased real Si diode: subtract about 0.7 V; ideal diode: treat as a wire.
3Reverse-biased diode: treat as an open switch (current ≈ 0, unless breakdown).
4Redraw the circuit with diodes replaced by wires or gaps, then use Ohm's law and Kirchhoff's rules.
5For dynamic resistance from a graph or table, take two nearby points and find ΔV/ΔI.
6For explanations, mention diffusion, recombination, depletion layer, barrier potential and drift in that order.

p-n Junction and Diode Characteristics in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Reverse-polarity protection

Suppose a 9 V battery powers a small gadget (equivalent to a 300 Ω load) through a silicon diode in series, added so that the gadget is not damaged if the battery is put in the wrong way round.

(a) Find the current when the battery is connected correctly. (b) What happens if the battery is reversed? (c) What is the cost of this protection?

Try it first, then show the solutionSolution

Forward bias: I = (9 − 0.7)/300 ≈ 0.0277 A ≈ 27.7 mA.
Reversed: the diode is reverse biased and passes only a tiny reverse current, so the gadget is protected.
About 0.7 V is always lost across the diode, wasting a little energy and slightly reducing the voltage available.

Answer: (a) about 27.7 mA; (b) almost no current flows; (c) about 0.7 V lost across the diode.

Memory hook: A diode is a one-way valve with a 0.7 V toll.

Testing a diode

Imagine checking an unmarked diode with a multimeter on its diode-test setting, which passes about 1 mA. Connected one way it reads 0.62 V; reversed, it shows 'OL' (open).

(a) Which way round was it forward biased? (b) Estimate its static resistance in the forward direction at this current. (c) Why is the reverse reading 'open'?

Try it first, then show the solutionSolution

The 0.62 V reading occurs when the meter's positive lead is on the p-side (anode): forward bias.
R = V/I = 0.62/10⁻³ = 620 Ω at this low current (it would be much lower at larger currents).
In reverse bias the depletion layer widens and only a tiny minority-carrier current flows, too small for the meter to register.

Answer: (a) positive lead on the anode (p-side); (b) about 620 Ω at 1 mA; (c) the reverse current is negligible.

Memory hook: About 0.6 to 0.7 V one way, open the other: a healthy silicon diode.

Tricks to remember p-n Junction and Diode Characteristics

Short memory aids. Say each one aloud once, then use it on a question the same day.

P to Positive is Forward
p-side to the positive terminal: forward bias.
Use it when: Identifying bias.
Forward squeezes, reverse widens
Forward bias narrows the depletion layer; reverse bias widens it.
Use it when: Depletion width questions.
Majority diffuse, minority drift
Diffusion current is by majority carriers, drift current by minority carriers.
Use it when: Explaining formation.
0.7 for Si, 0.3 for Ge
Typical barrier (and knee) voltages.
Use it when: Circuit numericals.

p-n Junction and Diode Characteristics: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Dynamic resistance from readings · 2 marks · Very short answerBoard level

In the forward characteristic of a silicon diode, the current rises from 10 mA to 30 mA when the voltage rises from 0.7 V to 0.8 V. Find the dynamic resistance in this range.

Try it first, then show the solutionSolution

ΔV = 0.1 V; ΔI = 20 mA = 0.02 A. (1)
r_d = 0.1/0.02 = 5 Ω. (1)

Answer: 5 Ω.

Example 2: Formation of the depletion layer · 5 marks · Long answerBoard level

Explain, with a diagram, how a depletion layer and a barrier potential are formed at a p-n junction. Name the two currents across the junction and state their condition in equilibrium.

Try it first, then show the solutionSolution

Electrons (majority in n) diffuse into the p-side and holes (majority in p) diffuse into the n-side, because of the difference in concentration: diffusion current. (1)
Near the junction they recombine, leaving uncovered positive donor ions on the n-side and negative acceptor ions on the p-side; this region without free carriers is the depletion layer. (1)
The fixed charges create an electric field from n to p and a potential difference, the barrier potential (≈ 0.7 V for Si), which opposes further diffusion. (1)
The field drives minority carriers across the junction: drift current, opposite to the diffusion current. (1)
In equilibrium, diffusion and drift currents are equal and opposite, so the net current is zero. Diagram: p and n regions, depletion layer with − ions on p-side and + ions on n-side, field arrow from n to p. (1)

Answer: Diffusion and recombination leave fixed ions (depletion layer) whose field forms the barrier; diffusion current (majority) equals drift current (minority) in equilibrium.

Example 3: Diode in a series circuit · 3 marks · Short answerHarder board level

A silicon diode (knee voltage 0.7 V) is connected in series with a 1 kΩ resistor across a 5 V battery. Find the current when the diode is (a) forward biased and (b) reverse biased. (c) What is the potential difference across the resistor in each case?

Try it first, then show the solutionSolution

(a) I = (5 − 0.7)/1000 = 4.3 mA. (1)
(b) Practically zero (only a tiny reverse saturation current). (1)
(c) Forward: 4.3 V across the resistor; reverse: about 0 V across it (nearly all 5 V appears across the diode). (1)

Answer: (a) 4.3 mA; (b) about zero; (c) 4.3 V and about 0 V.

Example 4: Field in the depletion layer · 2 marks · Very short answerHarder board level

The barrier potential of a silicon junction is 0.7 V and the depletion layer is 0.5 μm wide. Estimate the electric field in it. How does the width change under forward and reverse bias?

Try it first, then show the solutionSolution

E ≈ V/d = 0.7/(0.5 × 10⁻⁶) = 1.4 × 10⁶ V/m. (1)
Forward bias narrows the layer; reverse bias widens it. (1)

Answer: About 1.4 × 10⁶ V/m; narrower in forward, wider in reverse bias.

Example 5: Ideal diodes in a network · 5 marks · Long answerHOTS / competency

A 6 V battery is connected through a 2 Ω resistor to two parallel branches. Branch 1 has an ideal diode D₁ in series with 4 Ω, pointing so that it is forward biased; branch 2 has an ideal diode D₂ in series with 4 Ω, pointing the opposite way.
(a) Find the current from the battery and in each branch. (2)
(b) Find the currents if the battery is reversed. (2)
(c) Find the battery current if both diodes were forward biased. (1)

Try it first, then show the solutionSolution

(a) D₁ conducts (a short), D₂ is reverse biased (open). Circuit: 2 Ω + 4 Ω in series. (1)
I = 6/6 = 1 A, all through branch 1; branch 2 carries zero. (1)
(b) Now D₂ is forward biased and D₁ reverse biased: I = 6/(2 + 4) = 1 A, in the opposite direction, all through branch 2. (1)
Branch 1 carries zero. (1)
(c) Both branches conduct: 4 Ω ∥ 4 Ω = 2 Ω; I = 6/(2 + 2) = 1.5 A. (1)

Answer: (a) 1 A through branch 1, none in branch 2; (b) 1 A through branch 2, none in branch 1; (c) 1.5 A.

Example 6: Reading an I-V characteristic · 4 marks · Case-basedHarder board level

Read the passage and answer.

The I-V curve of a silicon diode shows almost no current up to 0.6 V in forward bias, then a sharp rise. In reverse bias the current stays at about 1 μA up to −50 V, then increases suddenly.
(a) What is the knee voltage? (1)
(b) Why is the reverse current almost constant? (1)
(c) What happens at −50 V, and what is this voltage called? (1)
(d) Compare the dynamic resistance in forward and reverse bias. (1)

Try it first, then show the solutionSolution

(a) About 0.6 to 0.7 V, where the forward current begins to rise sharply. (1)
(b) It is due to minority carriers, whose number depends on temperature, not on the applied voltage. (1)
(c) The reverse current rises sharply: breakdown; −50 V is the breakdown voltage. (1)
(d) Forward (above the knee): very low, a few ohms; reverse (before breakdown): very high, of the order of megaohms. (1)

Answer: (a) about 0.6 to 0.7 V; (b) minority-carrier current set by temperature; (c) breakdown at the breakdown voltage; (d) low forward, very high reverse.

Where students lose marks in p-n Junction and Diode Characteristics

The slips examiners see most often, and what to write instead.

Mistake: Saying the depletion layer contains free charges.
Confusing ions with carriers.
Do this: It contains fixed (immobile) ions and no free carriers.
Usually costs: 1 mark.
Mistake: Reversing the direction of the field at the junction.
Not checking which side has positive ions.
Do this: Positive ions on the n-side, negative on the p-side: field from n to p.
Usually costs: 1 mark.
Mistake: Drawing the reverse characteristic with the same scale as the forward one.
The reverse current is microamperes and the forward current milliamperes.
Do this: Use mA for forward and μA for reverse current on the graph and label them.
Usually costs: 1 mark of the graph.
Mistake: Forgetting the 0.7 V drop for a real silicon diode.
Treating it as ideal when the question gives a knee voltage.
Do this: Use I = (V − 0.7)/R unless told the diode is ideal.
Usually costs: The answer mark.

Exam tips

The p-n junction and its characteristics appear in seven of the eight 2023-2026 board papers in our bank (for example 2023 55/5/1 Q32(a), 2024 55/5/1 Q28, 2025 55/1/1 Q28 and 2026 55/1/1 Q26).
In the formation answer, use the sequence: diffusion, recombination, depletion layer, barrier, drift, equilibrium.
Draw both bias circuits and the full I-V graph with knee voltage and breakdown voltage marked.
In circuit problems, state which diodes conduct before calculating.
Special-purpose diodes (Zener, LED, photodiode, solar cell) are not in the 2026-27 syllabus.

Practice questions on p-n Junction and Diode Characteristics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The depletion region of a p-n junction contains
(a) free electrons only (b) free holes only (c) immobile ions (d) both free electrons and holes
Hint
Depleted of carriers.
Answer
(c) immobile ions.
Q2 · 1 mark · Very short answerBoard level
How does the width of the depletion layer change in (a) forward bias and (b) reverse bias?
Hint
Applied voltage opposes or adds to the barrier.
Answer
(a) It decreases; (b) it increases.
Q3 · 1 mark · Very short answerHarder board level
In a forward-biased diode, the current changes by 25 mA when the voltage changes by 0.1 V. Find the dynamic resistance.
Hint
ΔV/ΔI.
Answer
r_d = 0.1/0.025 = 4 Ω.
Q4 · 1 mark · Very short answerHarder board level
A silicon diode is connected in series with a 100 Ω resistor and a 3 V battery in forward bias. Find the current.
Hint
Subtract 0.7 V.
Answer
I = (3 − 0.7)/100 = 23 mA.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Below breakdown, the reverse current of a p-n junction diode is almost independent of the reverse voltage.
Reason (R): The reverse current is due to minority carriers, whose number is fixed by temperature.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What carries reverse current?
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
Explain why a current, though very small, flows in a reverse-biased p-n junction, and why it increases if the diode is heated.
Hint
Minority carriers and temperature.
Answer
The field across the junction sweeps minority carriers (electrons in p, holes in n) across it, giving a small drift current. Heating breaks more bonds, creating more electron-hole pairs, so there are more minority carriers and the reverse current increases.
Q7 · 2 marks · Very short answerHOTS / competency
Why is the barrier potential of a p-n junction not measured with a voltmeter across the diode's terminals?
Hint
Think about the metal contacts.
Answer
Contact potentials at the metal-semiconductor junctions at the two ends exactly cancel the barrier potential in equilibrium, so no net voltage appears at the terminals; otherwise an unbiased diode would act as a battery and could deliver energy for nothing.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Barrier potential of a silicon junction?Show answer
About 0.7 V.
2.What is the depletion layer?Show answer
A region near the junction with no free carriers, only fixed ions.
3.Effect of forward bias on the barrier?Show answer
It is reduced.
4.What carries reverse saturation current?Show answer
Minority carriers.
5.Dynamic resistance?Show answer
r_d = ΔV/ΔI.
6.Direction of the field in the depletion layer?Show answer
From the n-side to the p-side.

p-n Junction and Diode Characteristics: frequently asked questions

What is the depletion region?

A thin region at the junction from which free electrons and holes have been removed by diffusion and recombination, leaving only fixed positive ions (n-side) and negative ions (p-side).

Why does current rise sharply only after about 0.7 V in a silicon diode?

Until the applied forward voltage nearly cancels the barrier potential, few majority carriers can cross the junction. Once it does, the current rises steeply.

What is breakdown in a p-n junction?

At a high enough reverse voltage, the reverse current increases suddenly and sharply. The voltage at which this happens is the breakdown voltage; ordinary diodes can be damaged if operated there.

What is an ideal diode?

A diode with zero resistance in forward bias and infinite resistance in reverse bias, i.e. a perfect one-way switch with no voltage drop.

Are Zener diodes and LEDs in the 2026-27 syllabus?

No. Special-purpose p-n junction diodes are not to be assessed; the only application of the junction diode in the syllabus is the rectifier.