JEE Mains Physics · Semiconductor Electronics

Semiconductors and the p-n Junction

Doping decides which carrier is in the majority; joining p-type to n-type builds a barrier, and the bias across it decides whether the junction conducts.

All Semiconductor Electronics notes JEE Mains Physics strategy

Why this matters

Thirty-one PYQs, twenty-eight of them multiple choice, and two from 2026. Eight are about carriers and doping: which dopant gives which type, where the Fermi level sits and what heat does to resistance. Thirteen are about the junction itself: the barrier and its field, forward and reverse bias, the I-V curve and Zener breakdown. Ten are about LEDs, photodiodes and solar cells. Most are statements to judge, so the facts below are worth learning exactly.

Concept 1 of 3: Intrinsic, n-type and p-type semiconductors

Pure silicon has very few free carriers, and every electron freed leaves a hole behind, so the two are equal. Doping adds atoms with one electron too many (pentavalent) or one too few (trivalent). The extra electrons, or the extra holes, then far outnumber the other kind. The crystal stays neutral, because each dopant ion carries the charge its carrier took away.

Definition

Intrinsic: pure Si or Ge, $n_e = n_h = n_i$ .
n-type: a pentavalent donor (P, As, Sb). Electrons are the majority carriers, holes the minority.
p-type: a trivalent acceptor (B, Al, Ga, In). Holes are the majority carriers, electrons the minority.
Mass action: in equilibrium $n_e n_h = n_i^{2}$ , doped or not. More of one carrier means fewer of the other.
A doped crystal is electrically neutral: the donor or acceptor ions balance the free carriers.
Heating frees more carriers, so $n_e$ rises steeply and resistivity falls. A semiconductor has a negative temperature coefficient of resistance; its resistivity curve falls towards zero but never reaches it.

Type	Dopant	Majority carriers	Fermi level	Net charge
Intrinsic (pure Si, Ge)	none	none: $n_e = n_h = n_i$	near the middle of the band gap	neutral
n-type	pentavalent donor: P, As, Sb	electrons	near the conduction band; rises with more doping	neutral
p-type	trivalent acceptor: B, Al, Ga, In	holes	near the valence band; falls with more doping	neutral
Metal	not doped	free electrons	inside the conduction band	neutral

Whatever the dopant, the crystal stays neutral and the product of the two carrier densities stays fixed.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q107Moderate

Example 1 · Semiconductor Electronics · Semiconductors and the p-n Junction

Statement I: When a Si sample is doped with Boron, it becomes

P

type and when doped by Arsenic it becomes

N

-type semi conductor such that

P

-type has excess holes and

N

-type has excess electrons. Statement II: When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter. In the light of above statements, choose the most appropriate answer from the options given below

Extra electrons do not make a negative crystal

An n-type crystal has more free electrons than holes, but every donor atom that gave an electron is left as a positive ion. The crystal as a whole is neutral. 'n-type has net negative charge' is false.

The product stays fixed, not the sum

Doping raises one carrier density and lowers the other so that n_e n_h = n_i² still holds. The minority density falls below n_i; it does not stay at n_i.

Resistivity falls with heat, but never to zero

A graph of a semiconductor's resistivity against temperature is a falling curve that flattens out. A straight line, a rising curve or one that touches zero is wrong.

Drill 7 more on intrinsic, n-type and p-type semiconductors

Concept 2 of 3: The p-n junction: barrier, bias and dynamic resistance

When p-type meets n-type, electrons and holes diffuse across and cancel near the joint. That leaves a thin depletion layer of fixed ions and a barrier potential across it, which stops further flow. With no battery the diffusion and drift currents balance and no net current flows. Forward bias lowers the barrier and current rises steeply; reverse bias raises it, and only a tiny minority-carrier current flows.

Definition

The barrier is about 0.7 V for silicon and 0.3 V for germanium. With no external battery the net current is zero.
The field in the depletion layer is about $E = V_b/d$ for a barrier $V_b$ across a width $d$ .
Charge balance: $N_A x_p = N_D x_n$ . The lightly doped side holds the wider part of the depletion layer.
An electron crossing from n to p climbs the barrier and loses kinetic energy $eV_b$ .
Forward bias: the p-side is at the higher potential. Compare the two potentials, not their signs: p at −5 V and n at −8 V is forward bias. The barrier and the depletion layer shrink; the diffusion (majority) current dominates.
Reverse bias: the n-side is higher. The depletion layer widens, and a small drift current of minority carriers flows, nearly independent of the voltage until breakdown.
Dynamic resistance $r = \Delta V/\Delta I$ , read from the I-V curve. It is smaller at higher forward currents, where the curve is steeper.
Zener breakdown needs both sides heavily doped: the depletion layer is thin and its field is strong. A Zener diode works in reverse bias; forward biased, it is an ordinary diode.
A multimeter shows a low resistance one way round and a high resistance the other way round for a good diode.

Barrier field, energy loss and dynamic resistance

E = \frac{V_b}{d}, \qquad \tfrac{1}{2}mv^{2} = \tfrac{1}{2}mu^{2} - eV_b, \qquad r = \frac{\Delta V}{\Delta I}

Worked example

An electron with kinetic energy 1.25 eV reaches a junction from the n-side. The barrier is 0.45 V. What kinetic energy does it have on the p-side, and by what factor has its speed changed?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q115Moderate

Example 2 · Semiconductor Electronics · Semiconductors and the p-n Junction

A potential barrier of

0.4\text{ }V

exists across a p-n junction. An electron enters the junction from the

n

-side with a speed of

6.0 \times10^{5}{\text{ }ms}^{- 1}

. The speed with which electron enters the

p

side will be

\frac{x}{3}\times10^{5}{\text{ }ms}^{- 1}

the value of

x

is (Given mass of electron

= 9 \times10^{- 31}\text{ }kg

, charge on electron

= 1.6 \times10^{- 19}C

Compare potentials, not signs

A diode is forward biased when its p-side is at the higher potential. p at −4 V and n at −9 V is forward biased even though both are negative; p at −4 V and n at 0 V is reverse biased.

No battery, no current

Joining p-type to n-type does not make a current flow round an external ammeter. Diffusion builds the barrier until the diffusion and drift currents cancel, and the ammeter reads zero.

The wider layer is on the lightly doped side

The charge uncovered on each side must be equal, so the side with fewer dopant atoms per volume must uncover a longer stretch. The heavily doped side has the thinner part of the depletion layer.

Subtract energy, then take the root

An electron crossing the barrier loses eV_b of kinetic energy, not a fixed amount of speed. Subtract in energy, then convert back to speed with a square root.

Drill 12 more on the p-n junction: barrier, bias and dynamic resistance

Concept 3 of 3: Special-purpose diodes and the bias each one uses

A photon of energy equal to the band gap is what an electron gives out when it drops across the gap, and what it needs to jump across it. An LED uses the first: forward bias pushes electrons and holes together and they give out light. A photodiode and a solar cell use the second: light makes electron-hole pairs at the junction. Which bias each device needs follows from what it is for.

Definition

Photon energy and wavelength: $\lambda\,(\text{nm}) = \dfrac{1240}{E\,(\text{eV})}$ , from $hc \approx 1240\ \text{eV nm}$ .
LED: heavily doped, forward biased; emits light of photon energy close to $E_g$ . Its light grows with current only up to a point. Visible light (400 to 700 nm) needs $E_g$ between about 1.8 eV and 3.1 eV.
Photodiode: reverse biased. The reverse current is tiny, so the extra carriers made by light change it by a large fraction, which makes it easy to detect. Only light with $\lambda < 1240/E_g$ nm is detected.
Solar cell: no external bias. A large junction area collects more light; it drives current through a load and works in the fourth quadrant of the I-V graph.
Zener diode: reverse biased at breakdown, holding the voltage across it constant.

Device	Bias in use	Doping and junction	What it does
Rectifier diode	forward to conduct, reverse to block	moderate doping	lets current through one way only
Zener diode	reverse, at breakdown	both sides heavily doped; thin depletion layer	holds the voltage across it constant
LED	forward	heavily doped	electrons and holes recombine and give out light of photon energy about $E_g$
Photodiode	reverse	junction close to the surface so light reaches it	light makes electron-hole pairs and raises the reverse current
Solar cell	no external bias	large junction area, thin top layer	light produces an emf; works in the fourth quadrant of the I-V graph

The LED is the only one forward biased in use; the photodiode and Zener work in reverse, and the solar cell needs no battery at all.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q5Moderate

Example 3 · Semiconductor Electronics · Semiconductors and the p-n Junction

A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is

1.42eV

. The wavelength of light emitted from the LED is:

A photodiode is reverse biased

Forward biased, a photodiode carries a large majority current that light hardly changes. It is used in reverse bias, where light changes the small minority current by a large fraction.

Use eV with 1240, or joules with hc

λ in nm = 1240 ÷ E in eV. Dividing 1240 by an energy in joules, or hc in joule metres by an energy in eV, gives an answer off by a factor of about 10¹⁹.

A solar cell needs a large area

A solar cell's junction area is made large to collect as much light as possible, and it has no battery. A photodiode has the small junction and the reverse bias.

Drill 9 more on special-purpose diodes and the bias each one uses

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

The p-n junction: barrier, bias and dynamic resistance
Barrier field, energy loss and dynamic resistance
$E = \frac{V_b}{d}, \qquad \tfrac{1}{2}mv^{2} = \tfrac{1}{2}mu^{2} - eV_b, \qquad r = \frac{\Delta V}{\Delta I}$

Reference tables (2)

Intrinsic, n-type and p-type semiconductors4 rows

Type	Dopant	Majority carriers	Fermi level	Net charge
Intrinsic (pure Si, Ge)	none	none: $n_e = n_h = n_i$	near the middle of the band gap	neutral
n-type	pentavalent donor: P, As, Sb	electrons	near the conduction band; rises with more doping	neutral
p-type	trivalent acceptor: B, Al, Ga, In	holes	near the valence band; falls with more doping	neutral
Metal	not doped	free electrons	inside the conduction band	neutral

Whatever the dopant, the crystal stays neutral and the product of the two carrier densities stays fixed.

Special-purpose diodes and the bias each one uses5 rows

Device	Bias in use	Doping and junction	What it does
Rectifier diode	forward to conduct, reverse to block	moderate doping	lets current through one way only
Zener diode	reverse, at breakdown	both sides heavily doped; thin depletion layer	holds the voltage across it constant
LED	forward	heavily doped	electrons and holes recombine and give out light of photon energy about $E_g$
Photodiode	reverse	junction close to the surface so light reaches it	light makes electron-hole pairs and raises the reverse current
Solar cell	no external bias	large junction area, thin top layer	light produces an emf; works in the fourth quadrant of the I-V graph

The LED is the only one forward biased in use; the photodiode and Zener work in reverse, and the solar cell needs no battery at all.

Watch out for (10)

Extra electrons do not make a negative crystal
→ Intrinsic, n-type and p-type semiconductors
The product stays fixed, not the sum
→ Intrinsic, n-type and p-type semiconductors
Resistivity falls with heat, but never to zero
→ Intrinsic, n-type and p-type semiconductors
Compare potentials, not signs
→ The p-n junction: barrier, bias and dynamic resistance
No battery, no current
→ The p-n junction: barrier, bias and dynamic resistance
The wider layer is on the lightly doped side
→ The p-n junction: barrier, bias and dynamic resistance
Subtract energy, then take the root
→ The p-n junction: barrier, bias and dynamic resistance
A photodiode is reverse biased
→ Special-purpose diodes and the bias each one uses
Use eV with 1240, or joules with hc
→ Special-purpose diodes and the bias each one uses
A solar cell needs a large area
→ Special-purpose diodes and the bias each one uses

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