p-n Junction

Semiconductor Electronics · CBSE

Theory

What a p-n junction is, and how it is made

A p-n junction is the basic building block of nearly every semiconductor device — diodes, transistors and the rest. Understanding it is the key to understanding all of them.
It is made from a single crystal, not from two pieces. A thin p-type silicon wafer is taken, and a precisely controlled small quantity of a pentavalent impurity is added to part of it, converting that part into n-Si. The wafer then contains a p-region, an n-region and a metallurgical junction between them.

Why two slabs cannot simply be joined (a standard 1- or 2-mark question)

No. Any slab, however flat it looks, has surface roughness far larger than the interatomic crystal spacing (about $2$ to $3$ Å), so continuous contact at the atomic level is impossible.
Whatever gaps remain make the interface a discontinuity for the flowing charge carriers, and the continuous single-crystal structure needed for a depletion region and a barrier potential never forms.
In practice, p-n junctions are grown as one continuous crystal, doped differently on the two sides.

The two processes: diffusion and drift

These two words are the marks. Name them first, then explain.

Diffusion

In the n-region the electron concentration is much higher than the hole concentration; in the p-region the hole concentration is much higher than the electron concentration. Across the junction there is therefore a steep concentration gradient for each carrier.
Holes diffuse from the p-side to the n-side, and electrons diffuse from the n-side to the p-side, each moving down its own concentration gradient. This motion constitutes the diffusion current across the junction.

What diffusion leaves behind

When an electron diffuses from n to p, it leaves behind an ionised donor on the n-side. That ionised donor carries a positive charge and is immobile, because it is bonded into the surrounding lattice. As electrons keep diffusing away, a layer of positive space charge builds up on the n-side of the junction.
Likewise, when a hole diffuses from p to n, it leaves behind an ionised acceptor on the p-side — a negative, immobile charge. As holes keep diffusing away, a layer of negative space charge builds up on the p-side.

Drift

These two space-charge layers set up an electric field across the junction, directed from the positive charge on the n-side towards the negative charge on the p-side.
Under this field, an electron that strays onto the p-side of the junction is pushed back to the n-side, and a hole that strays onto the n-side is pushed to the p-side. The motion of charge carriers under the influence of an electric field is called drift, and it constitutes a drift current, opposite in direction to the diffusion current.

The depletion region

The space-charge region on either side of the junction, taken together, is called the depletion region — so called because the electrons and holes that took part in the initial diffusion have depleted this region of its free charges. What is left there is only immobile ionised donors and acceptors.
Its thickness is of the order of one-tenth of a micrometre, about $10^{-7}$ m.
Diagram in words: a rectangle divided into a p-region on the left (small open circles for holes) and an n-region on the right (dots for electrons). Straddling the boundary is a narrow strip: on its p-side a column of $-$ signs (immobile acceptor ions), on its n-side a column of $+$ signs (immobile donor ions). Mark the strip "depletion region" and draw an arrow across it labelled $V_0$ , with the n-side at the higher potential.

The barrier potential

Because the n-region has lost electrons and the p-region has gained them, a potential difference develops across the junction. The n-material is positive relative to the p-material.
The polarity of this potential difference is such that it opposes any further flow of carriers across the junction, so it is called the barrier potential (or junction barrier, or potential barrier), written $V_0$ or $V_B$ . Its standard value is of the order of $0.1$ V to $0.3$ V.
The barrier is set up entirely by the layer of immobile charges in the depletion region; there is no battery involved.

Reaching equilibrium

Initially the diffusion current is large and the drift current is small.
As diffusion continues, the space-charge regions widen, the field across them strengthens, and the drift current grows.
This continues until the drift current becomes equal to the diffusion current. At that point the two currents are equal and opposite, and the p-n junction is formed.
In a p-n junction at equilibrium there is no net current. Carriers do keep crossing the junction in both directions; it is the two currents that cancel, not the crossings that cease.

Effect of bias on the depletion layer

Forward bias (p-side to the positive terminal, n-side to the negative): the applied voltage opposes the built-in barrier, so the effective barrier height falls to $V_0 - V$ and the width of the depletion layer decreases.
Reverse bias (n-side to the positive terminal, p-side to the negative): the applied voltage acts in the same sense as the barrier, so the effective barrier height rises to $V_0 + V$ and the width of the depletion layer increases.
Independently of bias: increasing the doping concentration decreases the depletion width. More donor and acceptor ions per unit volume means the same amount of space charge — and hence the same barrier — is built up over a shorter distance, so equilibrium is reached with a narrower depletion region and a stronger field inside it.
Consequence for resistance: forward bias narrows the depletion layer and lowers the barrier, so a small increase in forward voltage produces a large increase in current — a low resistance. Reverse bias widens the layer and raises the barrier, so even a large increase in reverse voltage produces only a very small (minority-carrier) current — a high resistance.

Doping is what makes any of this possible

The two sides of the junction must be doped, because an intrinsic semiconductor has very few carriers and a very small, temperature-dependent conductivity. It is only in doped material that one side has a large excess of holes and the other a large excess of electrons, so that diffusion, a depletion layer and a barrier potential can exist at all.
This is why the board's answer to "why is doping a necessity for making electronic devices" ends by saying that it is the doped material, not the intrinsic material, that is used to build p-n junctions, diodes and every device beyond them.

Standard results

Two processes: diffusion (majority carriers, down the concentration gradient) and drift (minority carriers, under the junction field).
Diffusion leaves immobile positive donor ions on the n-side and immobile negative acceptor ions on the p-side.
Depletion region: about $10^{-7}$ m thick, containing only immobile ions.
Barrier potential $V_0 \approx 0.1$ V to $0.3$ V, n-side at the higher potential.
At equilibrium: diffusion current = drift current, net current = zero.
Forward bias: barrier $V_0 - V$ , depletion layer narrows. Reverse bias: barrier $V_0 + V$ , depletion layer widens.
Heavier doping narrows the depletion layer.
Two slabs pressed together do not form a junction — atomic-scale roughness prevents continuous contact.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
(a) Explain the formation of a p-n junction. (b) Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get a p-n junction? Explain.
3 marksCBSE Class 12 Physics — SA · 2022C, U · Section C, 3 marksAnswer 1 ↓
2
Answer the following, giving reason: (a) The resistance of a p-n junction is low when it is forward biased and is high when it is reversed biased. (b) Doping of intrinsic semiconductors is a necessity for making electronic devices.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 2 ↓
3
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. OR Explain with the help of the diagram the formation of depletion region and barrier potential in a p-n junction. (2/3, AI 2016) OR Write briefly the important processes that occur during the formation of p-n junction. With the help of necessary diagrams, explain the term barrier potential. (Foreign 2015)
3 marksCBSE Class 12 Physics — SA · Delhi 2017, 2/3 AI 2016, Foreign 2015, U · Section C, 3 marksAnswer 3 ↓
4
(i) A germanium crystal is doped with antimony. With the help of an energy-band diagram, explain how the conductivity of the doped crystal is affected. (ii) Briefly explain the two processes involved in the formation of a p-n junction. (iii) What will be the effect of (1) forward biasing, and (2) reverse biasing be on the width of depletion layer in a p-n junction diode?
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 4 ↓
5
State briefly the processes involved in the formation of a p-n junction, explaining clearly how the depletion region is formed. OR Explain with the help of a diagram, how a depletion layer and barrier potential are formed in a junction diode. (3/5, Delhi 2014C)
5 marksCBSE Class 12 Physics — Long Answer · 2/5, Delhi 2014, 3/5 Delhi 2014C · Section E, 5 marksAnswer 5 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
(a) Two processes take place in the formation of a p-n junction: diffusion and drift. Since the p-side has a much higher hole concentration and the n-side a much higher electron concentration, holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side (down their respective concentration gradients), constituting a diffusion current across the junction. As electrons diffuse from the n-side, they leave behind ionised (positively charged) donor atoms near the junction on the n-side; as holes diffuse from the p-side, they leave behind ionised (negatively charged) acceptor atoms near the junction on the p-side. This narrow region of immobile, oppositely-charged ions on either side of the junction is called the depletion region, and it sets up an internal electric field (and hence a potential difference $V_0$ , of order 0.1–0.3 V, called the potential barrier) directed from the n-side to the p-side. This internal field drives majority carriers on either side back across the junction (a drift current) opposing further diffusion. At equilibrium, the diffusion current and the drift current become equal and opposite, and no net current flows — the p-n junction (junction diode) is then fully formed. (b) No. We cannot obtain a proper p-n junction simply by mechanically pressing together separate slabs of p-type and n-type semiconductor. This is because the surfaces of both crystals have some inherent roughness at the microscopic (atomic) level, so a truly intimate, continuous, atomic-scale contact between the two crystal lattices is not possible this way — any gap, however small, would prevent the required continuous single-crystal structure (and hence the depletion region and potential barrier) from forming properly. In practice, p-n junctions are grown as a single continuous crystal, doped differently on the two sides.
2.↑ back to the question
(a) In forward biasing, the applied (forward) voltage opposes the internal potential barrier, so the effective barrier height is reduced and the depletion region narrows. A small increase in forward voltage then produces a large increase in forward current — this behaviour corresponds to a LOW resistance in forward bias. In reverse biasing, the applied voltage instead adds to (supports) the potential barrier, increasing its height and widening the depletion region; even a large increase in reverse voltage produces only a very small increase in (minority-carrier) reverse current — this behaviour corresponds to a HIGH resistance in reverse bias. (b) Intrinsic (pure) semiconductors have very small electrical conductivity at room temperature, since they contain only the small number of electron-hole pairs generated by thermal excitation. This tiny, temperature-sensitive conductivity is not controllable or reliable enough to build practical electronic devices. Doping (adding a small, controlled amount of trivalent or pentavalent impurity) increases the conductivity to a much greater, controllable extent, and this doped (extrinsic) material — not the intrinsic material — is what is actually used to build p-n junctions, diodes, transistors and other electronic devices.
3.↑ back to the question
The two processes are diffusion and drift. Diagram: a p-region (majority holes, shown as small circles, on the left) and an n-region (majority free electrons, shown as dots, on the right) are joined at a junction. Near the junction, a narrow strip on the p-side is shown with immobile negative ( $-$ ) ions and a narrow strip on the n-side with immobile positive ( $+$ ) ions — together these two strips make up the depletion region, and the potential difference $V_0$ across them (n-side at higher potential) is shown by an arrow pointing from n to p. When the p-n junction is formed, free electrons from the n-side diffuse across to the p-side (filling in holes there), and holes from the p-side diffuse across to the n-side. This leaves a layer of uncompensated positive ionised-donor charge on the n-side of the junction and a layer of uncompensated negative ionised-acceptor charge on the p-side. This layer of immobile opposite charges (the depletion layer) forms within a very short time of the junction being made, and quickly grows to a point where it prevents any further net movement of charge carriers across the junction — it sets up a potential difference $V_0$ (of order 0.1 to 0.3 V) across the junction, called the potential barrier or junction barrier, directed so as to oppose further diffusion.
4.↑ back to the question
(i) Germanium is an intrinsic semiconductor with its own band gap $E_g\approx0.72$ eV. Antimony (Sb) is a pentavalent impurity, so doping Ge with it produces an n-type extrinsic semiconductor, with electrons as the majority charge carriers. Energy band diagram: conduction band (top) and valence band (bottom) separated by Ge's own band gap of $0.72$ eV; a donor energy level, contributed by the antimony atoms, is drawn just below the conduction band, separated from it by only about $0.01$ eV. Because this donor level lies so close to the conduction band, at ordinary temperatures nearly all the donor atoms are thermally ionised, releasing their extra (fifth) electron directly into the conduction band. This supplies a large number of additional free electrons — far more than the intrinsic thermal generation alone would — so the electrical conductivity of the doped crystal is greatly increased compared to pure germanium. (ii) Diffusion: due to the concentration gradient across the junction, holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side, constituting a diffusion current. Drift: the immobile ions left behind near the junction set up an internal electric field (the potential barrier), which drives the minority carriers on each side across the junction in the opposite sense, constituting a drift current; at equilibrium these two currents balance. (iii) (1) Forward biasing: the applied forward voltage opposes the potential barrier, reducing the effective barrier height, so the width of the depletion layer DECREASES. (2) Reverse biasing: the applied reverse voltage supports (adds to) the potential barrier, increasing the effective barrier height, so the width of the depletion layer INCREASES.
5.↑ back to the question
Two processes take place in the formation of a p-n junction: diffusion and drift. Diagram: p-region (holes, small circles) on the left joined to n-region (electrons, dots) on the right; a narrow strip of immobile negative ions on the p-side of the junction and a narrow strip of immobile positive ions on the n-side, with the potential $V_0$ (n-side higher) shown across this strip. When the p-n junction is formed, free electrons from the n-side diffuse over to the p-side, filling in holes there; holes from the p-side similarly diffuse over to the n-side. This leaves a layer of uncompensated (immobile) positive charge on the n-side and a layer of uncompensated (immobile) negative charge on the p-side, right at the junction. This layer of oppositely-charged immobile ions is the DEPLETION REGION — so called because it is depleted of the mobile charge carriers (free electrons and holes) that would normally be present there. It grows within a very short time of the junction forming, until it is wide enough that its internal electric field prevents any further net diffusion of carriers across the junction. The potential difference $V_0$ (typically 0.1 to 0.3 V) that this layer of charge sets up across the junction is called the barrier potential (or junction barrier) — it is this barrier that a forward-bias voltage must overcome before appreciable current can flow.