University Physics IV · Molecular and Solid-State Physics · 12.9
The p–n Junction & Semiconductor Devices
Two doped slabs, one shared Fermi level, and a barrier neither of them chose. Build the depletion region from drift cancelling diffusion, read the built-in potential straight off the doping, then bias it — and the same junction becomes a rectifier, an LED and a solar cell, each fenced in by the same gap.
Build the model
Connect the measurement to the mechanism.
Push p-type silicon against n-type and neither piece changes chemically; what changes is that the two must now share one Fermi level. Holes diffuse one way, electrons the other, and every carrier that leaves strands an immobile dopant ion it can no longer screen — acceptors negative on the p side, donors positive on the n side. That exposed charge raises a field, the field drives a drift current back, and equilibrium is not a stalemate but an exact cancellation, separately for electrons and for holes.
Integrating that balance gives eVbi = kT ln(Nₐ Nd / nᵢ²): the barrier is set by how hard you doped and by nothing else, and it can never reach the gap. Now bias the junction. The barrier moves by exactly eV, multiplying the minority density at each depletion edge by exp(eV/kT); those injected minorities diffuse away and recombine, and the gradient they leave behind is the current.
That is the whole Shockley law, and its price is a stack of assumptions — abrupt doping, complete depletion, low injection, and nothing recombining inside the depletion layer. The last of those fails first, which is why a real diode climbs a decade of current in somewhere between 59 and 119 mV. Run the same junction backwards and it is a solar cell; run it hard forwards and it is an LED; both are fenced in by Eg.
- Simple definition
- A p–n junction is the boundary where p-type and n-type semiconductor meet: diffusion and drift cancel there at equilibrium, leaving a depletion region swept clear of mobile carriers and a built-in potential step fixed by the two doping levels.
- Example
- Silicon doped 1.0×10¹⁷ cm⁻³ acceptors against 1.0×10¹⁶ cm⁻³ donors has Vbi = 0.774 V at 300 K, spread over a depletion width of 0.332 µm — 91% of which lies inside the lightly doped n side.
10¹⁷ against 10¹⁶ cm⁻³ gives 0.774 V — set by doping alone, and always short of Eg/e = 1.12 V.
Nₐ, Nd, nᵢ in cm⁻³; kT/e = 25.85 mV at 300 K; silicon nᵢ = 1.0×10¹⁰ cm⁻³
0.332 µm here, 91% of it inside the lightly doped side; W ∝ √(Vbi − V) is the varactor's tuning law.
ε = 11.7ε₀ = 1.036×10⁻¹⁰ F m⁻¹ in silicon; W in m with N in m⁻³; V forward-positive
4.66 MV m⁻¹ at zero bias; silicon avalanches once Eₘₐₓ is driven toward 3×10⁷ V m-1.
Triangular profile peaking at the metallurgical junction; ε and xₙ as above
A decade of forward current costs η × 59.5 mV; in reverse the bracket floors at −1, capping I at −I₀.
η the ideality factor, between 1 and 2; kT/e = 25.85 mV at 300 K; I₀ the reverse saturation current
I₀ ∝ nᵢ² ∝ exp(−Eg/kT), so leakage roughly doubles per 10 K and germanium leaks far more than silicon.
A the junction area in cm², D in cm² s⁻¹, τ the minority lifetime in s, L the diffusion length in cm
GaAs at 1.42 eV emits at 873 nm; a silicon cell's 0.628 V keeps only 56% of its 1.12 V gap.
Eg in eV and λ in nm; IL the photocurrent and I₀ the dark saturation current, in the same units
Equilibrium is drift cancelling diffusion, species by species
Bring the two slabs together and the concentration steps are enormous: holes sit at 10¹⁷ cm⁻³ on the p side against nᵢ²/Nd = 10⁴ cm⁻³ on the n side, a ratio of 1013. Diffusion starts at once, and every hole that crosses leaves behind an acceptor ion with nothing to screen it. The exposed acceptors are negative and the exposed donors positive, so a field builds pointing from n to p, and that field drives a drift current back against the diffusion. Equilibrium arrives when the two cancel — and the condition is stronger than 'no net current'. It is Jₚ = 0 and Jₙ = 0 separately, because a steady circulating current with nothing driving it would violate the second law. Writing Jₚ = eμₚ pE − eDₚ dp/dx = 0 and using the Einstein relation D/μ = kT/e turns the balance into E = (kT/e)(1/p)(dp/dx), which integrates in one line. Said the other way: the Fermi level is flat everywhere, and the bands bend to keep it flat.
Integrating the balance gives a barrier fixed by doping
Integrate E = (kT/e) d(ln p)/dx across the junction and the potential difference falls out as Vbi = (kT/e) ln(pₚ/pₙ), the ratio of hole densities on the two sides. Both are known from the doping alone: pₚ ≈ Nₐ, and on the n side mass action gives pₙ = nᵢ²/Nd. So eVbi = kT ln(Nₐ Nd/nᵢ²), with no material constant left in it but nᵢ. For silicon at 300 K, nᵢ = 1.0×10¹⁰ cm⁻³ and kT/e = 25.85 mV, so Nₐ = 10¹⁷ and Nd = 10¹⁶ cm⁻³ give Vbi = 0.02585 × ln(10¹³) = 0.774 V. Two consequences. The barrier grows only logarithmically: raising both dopings tenfold multiplies the argument by 100 and adds 0.119 V, not a factor of anything. And it can never reach Eg/e = 1.12 V, because pushing it there would drive both Fermi levels into their bands, where the Boltzmann statistics behind mass action have already failed.
The depletion approximation turns Poisson into a triangle
Assume the doping changes abruptly and that the transition region is swept completely clear of mobile carriers, so the charge density there is simply ±eN. Poisson's equation dE/dx = ρ/ε then integrates once into a field that rises linearly through the p side and falls linearly through the n side — a triangle peaking at the metallurgical junction — and again into a parabolic potential whose total drop is the area under that triangle. Overall neutrality forces Nₐ xₚ = Nd xₙ, so the depletion region eats into whichever side is doped less. With Nₐ = 10¹⁷ cm⁻³, Nd = 10¹⁶ cm⁻³ and ε = 11.7ε₀, W = √[(2ε/e)Vbi(1/Nₐ + 1/Nd)] = 0.332 µm, of which 0.302 µm lies on the n side, and Eₘₐₓ = 2Vbi/W = 4.66 MV m-1. Because W ∝ √(Vbi − V), reverse bias widens the layer and raises the field: that square root is the varactor's tuning law, and the rising peak field is what eventually avalanches.
Bias moves the barrier, injection gives the exponential
A forward bias V falls almost entirely across the depletion layer, since the neutral regions conduct comparatively well, so the barrier becomes e(Vbi − V). Boltzmann statistics then multiply the minority density at each depletion edge by exp(eV/kT) — the law of the junction, pₙ(xₙ) = (nᵢ²/Nd)exp(eV/kT). Those injected minorities are not consumed at the edge; they diffuse into the neutral region and recombine, dying away over a diffusion length L = √(Dτ). The current is the diffusion flux at that edge, eDΔp/L from each side, and that is where I₀ = eAnᵢ²[Dₚ/(Lₚ Nd) + Dₙ/(Lₙ Nₐ)] comes from. Notice what the depletion region does in this model: nothing but supply a boundary condition. Notice too where the temperature dependence really lives — I₀ ∝ nᵢ² ∝ exp(−Eg/kT) swamps the exp(eV/kT) factor, which is why a silicon diode held at fixed current needs about 2 mV less forward bias per kelvin of heating.
The ideality factor is where the ideal law confesses
The Shockley law needs an abrupt junction, complete depletion, low injection, negligible series resistance, and no generation or recombination inside the depletion layer. That last assumption fails first. A Shockley–Read–Hall centre in the depletion region consumes one electron and one hole together, and both densities there scale as exp(eV/2kT), so the recombination current goes as exp(eV/2kT) — an ideality factor of 2. It dominates at low forward bias, where the diffusion current is still tiny, so a real log I–V is steep near the origin, settles toward η = 1 higher up, then bends over as series resistance takes hold. A diode drawing 1.00 mA at 0.600 V and 10.0 mA at 0.680 V needs 80 mV per decade against the 59.5 mV that η = 1 would give, so η = 1.34. Reverse bias tells the same story: depletion-layer generation current is proportional to W ∝ √(Vbi + VR), so the reverse saturation current is not saturated at all.
One junction, three devices, all fenced in by the gap
Forward bias floods each side with minority carriers, and where they recombine radiatively the photon carries hν ≈ Eg. That is the LED, and it works only in a direct-gap material where the transition needs no phonon to conserve momentum — silicon's indirect gap is why it makes a superb diode and a hopeless emitter. Reverse the flow and the same junction is a photodiode or a solar cell: a photon above Eg makes a pair, the built-in field sweeps it out, and the illuminated characteristic is the Shockley curve shifted down by the photocurrent, I = I₀[exp(eV/ηkT) − 1] − IL. Setting I = 0 gives Voc = (ηkT/e)ln(IL/I₀ + 1), bounded by Eg/e because I₀ itself carries exp(−Eg/kT). Silicon under one sun reaches 0.628 V against its 1.12 V gap. Widening the gap buys volts but discards every photon below it, and that trade is what puts the single-junction optimum near 1.3 eV.
Change one variable at a time
Make the relationship visible.
Raise the donor slider toward 10¹⁷ and the long right flank pulls back while the short left flank grows — charge balance forces Nₐ xₚ = Nd xₙ. Then sweep the bias down to −0.6 V: width and peak field both climb as the square root of the junction drop, which is the varactor.
BUILT-IN Vbi0.774 V
JUNCTION DROP0.774 V
DEPLETION WIDTH0.332 µm
PEAK FIELD4.66 MV/m
Live interpretationBUILT-IN Vbi: 0.774 V. JUNCTION DROP: 0.774 V. DEPLETION WIDTH: 0.332 µm. PEAK FIELD: 4.66 MV/m
Catch the common trap
Explain before calculating.
A silicon diode at 300 K gives a straight forward log I–V line whose current rises by one decade for every 119 mV of applied bias. What does that slope say?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA silicon p–n junction at 300 K is doped with Nₐ = 1.0×10¹⁷ cm⁻³ acceptors on one side and Nd = 1.0×10¹⁶ cm⁻³ donors on the other. Take nᵢ = 1.0×10¹⁰ cm⁻³ and kT/e = 25.85 mV. Find the built-in potential, then work out how heavily you would have to dope to push it up to the 1.12 eV gap.
- Both dopants are fully ionised at 300 K, so far from the junction p ≈ Nₐ = 1.0×10¹⁷ cm⁻³ on the p side and n ≈ Nd = 1.0×10¹⁶ cm⁻³ on the n side.
- Feed those into eVbi = kT ln(Nₐ Nd/nᵢ²). The argument is (1.0×10¹⁷)(1.0×10¹⁶)/(1.0×10¹⁰)² = 10³³/10²⁰ = 1.0×1013.
- ln(1.0×10¹³) = 13 ln10 = 13 × 2.3026 = 29.93, so Vbi = 0.02585 V × 29.93 = 0.774 V.
- To reach 1.12 V the argument would have to be exp(1.12/0.02585) = exp(43.3) = 6.6×10¹⁸, i.e. Nₐ Nd = 6.6×10³⁸ cm⁻⁶, about 2.6×10¹⁹ cm⁻³ on each side. Both sides would then be degenerate and the Boltzmann statistics behind mass action — and behind this formula — would already have failed.
AnswerVbi = 0.774 V, fixed by the doping alone. The 1.12 V gap is a ceiling this expression can approach only where its own assumptions break down.
MediumA silicon diode at 300 K carries 1.00 mA at 0.600 V and 10.0 mA at 0.680 V, both far above its reverse saturation current. Find the ideality factor and I₀, and check whether the −1 in the Shockley law makes any difference at 0.600 V.
- Well above a few kT/e the −1 is negligible, so I ≈ I₀ exp(eV/ηkT) and ln I is linear in V with slope e/ηkT.
- One decade of current costs ΔV = η(kT/e)ln10 = η × 0.02585 × 2.3026 = η × 59.5 mV, while the measured decade costs 0.680 − 0.600 = 80 mV.
- So η = 80/59.5 = 1.344, and ηkT/e = 1.344 × 25.85 mV = 34.7 mV.
- Invert at the lower point: I₀ = I exp(−eV/ηkT) = 1.00×10⁻³ × exp(−0.600/0.0347) = 1.00×10⁻³ × exp(−17.27) = 3.1×10⁻¹¹ A.
- At 0.600 V the exponential is exp(17.27) = 3.2×10⁷, so dropping the −1 shifts I by three parts in 108. It matters only within a few tens of millivolts of zero bias, and in reverse, where it caps the current at −I₀.
Answerη = 1.34 and I₀ = 3.1×10⁻¹¹ A, about 31 pA. An η between 1 and 2 says the current is a mixture of diffusion, which gives 1, and depletion-layer recombination, which gives 2.
HardTwo solar cells under one sun (100 mW cm⁻²) at 300 K, both with η = 1. Cell A is silicon: Eg = 1.12 eV, Jsc = 35 mA cm⁻², J₀ = 1.0×10⁻¹² A cm⁻², fill factor 0.80. Cell B has Eg = 1.90 eV, so its saturation current falls by exp(−0.78 eV/kT) to J₀ = 1.0×10⁻¹² × exp(−0.78/0.02585) A cm⁻², but the wider gap leaves only Jsc = 12 mA cm⁻²; take its fill factor as 0.83. Find both open-circuit voltages and both efficiencies, and say which gap wins.
- At open circuit the photocurrent is exactly balanced by the dark diode current: JL = J₀[exp(eVoc/ηkT) − 1], so Voc = (ηkT/e)ln(JL/J₀ + 1).
- Cell A: JL/J₀ = 3.5×10⁻²/1.0×10⁻¹² = 3.5×10¹⁰, and ln(3.5×10¹⁰) = 1.253 + 23.026 = 24.28, so Voc = 0.02585 × 24.28 = 0.628 V.
- Cell B's dark current: exp(−0.78/0.02585) = exp(−30.17) = 7.9×10⁻¹⁴, so J₀ = 7.9×10⁻²⁶ A cm⁻² — thirteen orders of magnitude below silicon's.
- Cell B: JL/J₀ = 1.2×10⁻²/7.9×10⁻²⁶ = 1.52×10²³, ln = 0.420 + 52.96 = 53.38, so Voc = 0.02585 × 53.38 = 1.380 V. It keeps 73% of its gap voltage against silicon's 56%.
- Maximum power is FF × Voc × Jsc. Cell A: 0.80 × 0.628 V × 35 mA cm⁻² = 17.6 mW cm-2. Cell B: 0.83 × 1.380 V × 12 mA cm⁻² = 13.7 mW cm-2.
- Against 100 mW cm⁻² incident that is 17.6% against 13.7%. Cell B more than doubles the voltage, 1.380/0.628 = 2.20, but keeps only 12/35 = 34% of the current, and the current loses.
AnswerCell A: Voc = 0.628 V, efficiency 17.6%. Cell B: Voc = 1.380 V, efficiency 13.7%. Silicon wins — a 2.2× voltage gain does not repay throwing away two thirds of the photocurrent, which is why the single-junction optimum sits near 1.3 eV.