PhysicsISC Class 12 · Electronic Devices

The p-n Junction Diode: Depletion Region, Biasing and V-I Characteristics

Updated 15 October 2026 · Fact-checked

A p-n junction forms when p-type and n-type semiconductors are joined. Electrons and holes diffuse across and recombine, leaving a depletion region of immobile ions whose field creates a potential barrier (about 0.7 V for silicon, 0.3 V for germanium). Forward bias lowers the barrier and current flows; reverse bias raises it and almost no current flows.

Understand The p-n Junction Diode and Its Characteristics

Formation of the junction. The n side has many free electrons; the p side has many holes. At the junction, electrons diffuse from n to p and holes from p to n (diffusion current), where they recombine. This leaves:

on the n side, a layer of positive donor ions, and on the p side, a layer of negative acceptor ions; these ions are fixed in the crystal (immobile);
a thin region with no mobile charge carriers: the depletion region (about a micrometre wide);
an electric field from the n side to the p side across it, which opposes further diffusion and pushes minority carriers across (drift current). In equilibrium, diffusion and drift balance.

The potential difference across the depletion region is the potential barrier $V_{0}$ : about 0.7 V for silicon and 0.3 V for germanium.

Forward bias. p side to the positive terminal, n side to the negative terminal. The applied voltage opposes the barrier: the barrier height falls to $(V_{0} - V)$ and the depletion region narrows. Once V exceeds the barrier (the knee or cut-in voltage), current rises sharply (milliamperes), carried by majority carriers.

Reverse bias. p side to negative, n side to positive. The barrier rises to $(V_{0} + V)$ and the depletion region widens. Only a tiny reverse saturation current (microamperes) flows, due to minority carriers; it is nearly independent of voltage but rises with temperature. At a large reverse voltage, breakdown occurs and the current rises suddenly.

V-I characteristic. Forward: almost no current until the knee, then a steep, nearly straight rise. Reverse: a very small, nearly constant current until breakdown. Draw V on the x-axis and I on the y-axis (mA forward, µA reverse); the curve does not start rising at the origin.

Resistance of a diode. It is a non-ohmic device.

Static resistance at a point: $R = \frac{V}{I}$ .
Dynamic (a.c.) resistance: $r_{d} = \frac{Δ V}{Δ I}$ , small in forward bias beyond the knee.

Simple circuits. With a diode modelled as a constant forward drop $V_{d}$ (0.7 V Si, 0.3 V Ge) and an open switch in reverse: $I = \frac{V - V _{d}}{R}$ for a series resistor R.

Special junction diodes (elementary ideas).

LED: forward-biased; electrons and holes recombine and emit light of energy about $E_{g}$ , so $λ \approx \frac{1240}{E _{g} ( eV )}$ nm. LEDs are efficient, long-lasting and save energy, reducing pollution and global warming compared with filament lamps.
Photodiode: reverse-biased; light creates electron-hole pairs in the depletion region, and the reverse current rises with light intensity. Used in light sensors.
Solar cell: an unbiased p-n junction of large area; light-generated carriers are separated by the junction field, producing an emf (about 0.5 to 0.6 V for silicon).

(The Zener diode, operated in reverse breakdown as a voltage regulator, is covered with rectifiers.)

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

Barrier potential: $V_{0} \approx 0.7 V (Si), 0.3 V (Ge)$; Knee voltage.
Forward-bias barrier: $V_{0} - V$; Lowered.
Reverse-bias barrier: $V_{0} + V$; Raised.
Series circuit current: $I = \frac{V - V _{d}}{R}$; Constant-drop model.
Static resistance: $R = \frac{V}{I}$; At one point.
Dynamic resistance: $r_{d} = \frac{Δ V}{Δ I}$; Slope of the curve.
LED wavelength: $λ \approx \frac{1240}{E _{g} ( eV )} nm$; Photon energy ≈ band gap.

How to solve The p-n Junction Diode and Its Characteristics questions

Decide whether each diode is forward or reverse biased; then apply Kirchhoff's laws with the diode as a fixed drop or an open switch.

1Find the bias: p side at higher potential than n side means forward bias.
2Forward: replace the diode by a drop of 0.7 V (Si) or 0.3 V (Ge), unless told it is ideal (0 V).
3Reverse: treat the diode as an open circuit (current ≈ 0).
4Apply KVL: supply = diode drops + IR; solve for I.
5Static R = V/I at a point; dynamic r = ΔV/ΔI over a small range.
6LED/photodiode/solar cell questions: state bias and principle in one line each.

The p-n Junction Diode and Its 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.

Choosing a resistor for an LED torch

Suppose an LED torch uses a 3 V battery and a white LED that needs 20 mA with a forward drop of about 2.0 V.

What series resistor is needed, and why can't the LED be connected directly?

Try it first, then show the solutionSolution

$R = \frac{3 - 2.0}{0.020} = 50$ Ω.
Without R, a small rise in voltage above the knee would make the current rise steeply and burn out the LED; R limits it.

Answer: 50 Ω; the resistor limits the current.

Memory hook: Every LED needs a bodyguard resistor.

Reverse-polarity protection

Imagine a child putting the batteries of a toy in the wrong way round. A diode in series with the battery protects the circuit.

Explain how.

Try it first, then show the solutionSolution

Correct polarity: the diode is forward biased and conducts, dropping about 0.7 V.
Wrong polarity: the diode is reverse biased and blocks current (only microamperes flow), so the electronics are not damaged.

Answer: The diode conducts only one way, so a reversed battery cannot drive current.

Memory hook: A diode is a one-way door.

Tricks to remember The p-n Junction Diode and Its Characteristics

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

P to positive
Forward bias: P to Positive, N to Negative.
Use it when: Bias questions (2024 Q17).
Forward squeezes, reverse widens
Forward bias narrows the depletion region; reverse bias widens it.
Use it when: Depletion-layer questions.
Seven and three
Knee: Si 0.7 V, Ge 0.3 V.
Use it when: Numericals and graphs.
Immobile ions, no carriers
Depletion region = fixed ions, no free carriers.
Use it when: Definitions.

The p-n Junction Diode and Its Characteristics: solved examples

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

Example 1: Reverse bias and the depletion region · 1 mark · MCQBoard level

When a p-n junction is reverse biased, the depletion region:
(a) becomes narrower
(b) becomes wider
(c) disappears
(d) is unchanged

Try it first, then show the solutionSolution

The applied voltage adds to the barrier, pulling carriers away from the junction.

Answer: (b).

Example 2: A germanium diode circuit · 2 marks · Short answerBoard level

A germanium diode (forward drop 0.3 V) is connected in series with a 1.5 kΩ resistor across a 9 V battery, forward biased. Find the current and the voltage across the resistor.

Try it first, then show the solutionSolution

$I = \frac{9 - 0.3}{1500} = 5.8 \times 1 0^{- 3}$ A = 5.8 mA. (1)
Voltage across resistor = 9 − 0.3 = 8.7 V. (1)

Answer: 5.8 mA; 8.7 V.

Example 3: Formation of the depletion region · 3 marks · Long answerHarder board level

Explain how the depletion region and the potential barrier are formed at a p-n junction. Define each term.

Try it first, then show the solutionSolution

Electrons diffuse from n to p and holes from p to n, recombining near the junction. (1)
This leaves immobile positive ions on the n side and negative ions on the p side: the depletion region is the narrow region at the junction with no mobile charge carriers, only immobile ions. (1)
These ions set up a field from n to p; the potential difference across the depletion region, which opposes further diffusion of majority carriers, is the potential barrier (≈ 0.7 V Si, 0.3 V Ge). (1)

Answer: Diffusion and recombination leave fixed ions; their field gives the barrier.

Example 4: Static and dynamic resistance · 2 marks · Short answerHarder board level

In the forward characteristic of a diode, the current is 20 mA at 0.72 V and rises from 12 mA to 28 mA as the voltage rises from 0.70 V to 0.74 V. Find the static resistance at 0.72 V and the dynamic resistance in this range.

Try it first, then show the solutionSolution

Static: $R = \frac{0.72}{0.020} = 36$ Ω. (1)
Dynamic: $r_{d} = \frac{0.74 - 0.70}{0.028 - 0.012} = \frac{0.04}{0.016} = 2.5$ Ω. (1)

Answer: 36 Ω; 2.5 Ω.

Example 5: Two branches, two diodes · 3 marks · Short answerHOTS / competency

A 6 V battery feeds two parallel branches. Branch 1: a silicon diode, forward biased, in series with 200 Ω. Branch 2: a silicon diode, reverse biased, in series with 300 Ω. Find the current in each branch and from the battery.

Try it first, then show the solutionSolution

Branch 1: $I_{1} = \frac{6 - 0.7}{200} = 26.5$ mA. (1)
Branch 2: the diode is reverse biased, so $I_{2} \approx 0$ . (1)
Battery current ≈ 26.5 mA. (1)

Answer: 26.5 mA, ≈ 0, 26.5 mA.

Example 6: Lighting with LEDs · 4 marks · Case-basedHOTS / competency

Suppose you are designing an indicator using a blue LED of wavelength 450 nm on a 5 V supply. The LED needs 20 mA and drops 2.0 V when lit.
(i) Estimate the band gap of the LED material. (1 mark)
(ii) Find the series resistor needed. (1 mark)
(iii) Find the power used by the LED. (1 mark)
(iv) Why are LEDs important for the environment? (1 mark)

Try it first, then show the solutionSolution

(i) $E_{g} \approx \frac{1240}{450} \approx 2.76$ eV.
(ii) $R = \frac{5 - 2.0}{0.020} = 150$ Ω.
(iii) $P = 2.0 \times 0.020 = 0.04$ W = 40 mW.
(iv) They convert electrical energy to light efficiently and last long, saving energy and so reducing fuel burnt in power stations, pollution and global warming.

Answer: (i) ≈ 2.76 eV; (ii) 150 Ω; (iii) 40 mW; (iv) energy saving.

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

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

Mistake: Missing key words in definitions
The ISC 2024 examiners (Q8) saw depletion-region definitions without "immobile ions" or "no mobile charge carriers", and barrier definitions like "obstructs flow".
Do this: Use the full definitions: no mobile carriers, immobile ions; p.d. that opposes diffusion of majority carriers.
Usually costs: 1 mark each.
Mistake: Incomplete forward-bias statement
In 2024 Q17 candidates wrote "p connected to positive" only, or described reverse bias.
Do this: p region to the positive terminal and n region to the negative terminal.
Usually costs: 1 mark.
Mistake: Graph from the origin
Also 2024 Q17: curve rising from the origin, axes swapped or unlabelled, cut-in voltage not marked.
Do this: Flat until ≈ 0.7 V (Si) or 0.3 V (Ge), then steep; V on x-axis, I on y-axis.
Usually costs: 1 mark.
Mistake: Reversing p and n carriers
The 2025 examiners (Q20) saw majority carriers of p and n types swapped.
Do this: p: holes majority; n: electrons majority.
Usually costs: 1 mark.

Exam tips

Draw a battery, resistor and diode symbol with the arrow from p to n for forward bias.
In V-I graphs use mA for forward and µA for reverse current and mark the knee.
Treat diodes as 0.7 V (Si) drops unless told "ideal".
Give one line each for LED, photodiode and solar cell: bias plus principle.

Practice questions on The p-n Junction Diode and Its 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 knee (cut-in) voltage of a silicon diode is about:
(a) 0.1 V
(b) 0.3 V
(c) 0.7 V
(d) 1.5 V
Hint
Germanium is 0.3 V.
Answer
(c).
Q2 · 1 mark · Short answerBoard level
Define the potential barrier of a p-n junction.
Hint
Across the depletion region.
Answer
The potential difference developed across the depletion region, due to the immobile ions, which opposes the diffusion of majority carriers across the junction.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The reverse current of a p-n junction is very small and nearly independent of the reverse voltage (before breakdown).
Reason (R): It is carried by minority carriers, whose number is fixed by temperature, not by the voltage.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Minority carriers.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
A silicon diode in series with a 2.2 kΩ resistor is forward biased by a 12 V supply. Find the current.
Hint
12 − 0.7.
Answer
≈ 5.14 mA.
Q5 · 3 marks · Short answerHarder board level
State the bias used and the working principle of (a) an LED, (b) a photodiode and (c) a solar cell.
Hint
Forward, reverse, none.
Answer
(a) Forward: recombination emits light. (b) Reverse: light creates carriers, increasing reverse current. (c) No bias: light-generated carriers separated by the junction field give an emf.
Q6 · 2 marks · Short answerHOTS / competency
A 1.5 V cell drives a diode in series with 100 Ω. Find the current with (a) a silicon diode, (b) a germanium diode.
Hint
Different knee voltages.
Answer
(a) 8 mA; (b) 12 mA.
Q7 · 2 marks · Short answerHOTS / competency
Two silicon diodes in series, both forward biased, are connected with a 330 Ω resistor across 5 V. Find the current.
Hint
Two drops of 0.7 V.
Answer
$\frac{5 - 1.4}{330} \approx 10.9$ mA.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student records a diode's forward characteristic: V = 0.2, 0.4, 0.6, 0.7, 0.8 V gives I = 0, 0.1, 2, 10, 40 mA.
(i) How must the diode be connected for these readings? (1 mark)
(ii) Estimate the cut-in voltage. (1 mark)
(iii) Find the static resistance at 0.8 V. (1 mark)
(iv) Find the dynamic resistance between 0.7 V and 0.8 V. (1 mark)
Hint
p to +.
Answer
(i) p side to the positive terminal, n side to the negative. (ii) About 0.6 to 0.7 V (silicon). (iii) 20 Ω. (iv) ≈ 3.3 Ω.

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.Depletion region?Show answer
The region at the junction with no mobile carriers, only immobile ions.
2.Effect of forward bias on the barrier?Show answer
It is lowered (to $V_{0} - V$ ).
3.Knee voltage of germanium?Show answer
About 0.3 V.
4.Bias for a photodiode?Show answer
Reverse bias.
5.Dynamic resistance formula?Show answer
$\frac{Δ V}{Δ I}$ .

The p-n Junction Diode and Its Characteristics: frequently asked questions

Why does current not flow in forward bias until about 0.7 V?

The applied voltage must first overcome the barrier potential before majority carriers can cross the junction in large numbers.

Why does the reverse current increase with temperature?

Heat breaks more covalent bonds, creating more minority carriers, which carry the reverse current.

Is a diode ohmic?

No. Its current is not proportional to voltage; its resistance changes with voltage and direction.

What does the ISC 2027 syllabus need here?

The junction diode and its symbol, depletion region and potential barrier, forward and reverse bias, V-I characteristics and numericals, and elementary ideas of LED, photodiode and solar cell, including why LEDs save energy.

How was this asked recently?

ISC 2024 Q8 (2 marks) defined the depletion region and potential barrier; ISC 2024 Q17 (3 marks) asked forward-bias connections and the forward I-V graph.