Doping makes two kinds of silicon
Pure silicon has very few free carriers, about 10¹⁰ cm⁻³ at room temperature. Doping changes that. Adding acceptor atoms (boron) gives P-type silicon, full of mobile holes; adding donor atoms (phosphorus) gives N-type, full of free electrons. Each side stays electrically neutral: every hole is balanced by a fixed negative acceptor ion, every electron by a fixed positive donor ion.
The minority carriers on each side follow the mass-action law, n·p = ni², so a side doped to 10¹⁶ cm⁻³ has only 10⁴ cm⁻³ of the other carrier.
n · p = ni² P side: p ≈ Na, n ≈ ni²/Na N side: n ≈ Nd, p ≈ ni²/Nd
The depletion region and built-in potential
Where the two meet, electrons diffuse into P and holes into N, and they recombine. They leave behind the fixed ions: a layer of negative charge on the P side and positive charge on the N side. That charge sets up an electric field, pointing from N to P, which pushes back on further diffusion. The balance is reached at the built-in potential Vbi.
The region swept clear of mobile carriers is the depletion region. Its width shrinks with heavier doping, and it reaches mostly into the more lightly doped side, because the charge on both sides must balance.
Vbi = VT · ln(Na·Nd / ni²), VT = kT/q ≈ 25.9 mV W = √( 2ε(Vbi − Va)/q · (1/Na + 1/Nd) ) xp·Na = xn·Nd, Emax = q·Na·xp / ε
Forward and reverse bias
Forward bias (P positive) works against the built-in field: the barrier falls to Vbi − Va, the depletion region narrows, and carriers pour across. The current rises exponentially with voltage. Reverse bias adds to the barrier: the depletion region widens, the field grows, and only a tiny leakage of thermally generated minority carriers flows. This one-way behaviour is the diode.
I = Is · (exp(Va / VT) − 1)
Reading the band diagram
The band diagram plots electron energy across the junction. The conduction band Ec and valence band Ev bend by q(Vbi − Va): the hill an electron must climb to cross from N to P. With no bias the Fermi level EF is flat; under bias it splits by qVa between the two sides.
Worked example (the simulator’s default values)
- Silicon, Na = Nd = 10¹⁶ cm⁻³, ni = 10¹⁰ cm⁻³, T = 300 K.
- Vbi = 0.0259 · ln(10³² / 10²⁰) = 0.714 V.
- At zero bias W = 0.430 µm, split equally (0.215 µm each side); Emax = 33.2 kV/cm.
- Forward 0.5 V: W shrinks to 0.235 µm and the current grows by a factor of about 2.5 × 10⁸.
- Reverse 5 V: W grows to 1.22 µm and Emax to 94 kV/cm, but only leakage current flows.
Common mistakes
- Thinking the depletion region holds no charge. It holds no mobile carriers, but it is full of fixed ions; that charge makes the field.
- Expecting the depletion region to sit equally on both sides. It extends mostly into the lightly doped side.
- Applying forward bias above Vbi. The barrier never vanishes: long before Va reaches Vbi the current is so large that the resistance of the silicon and wires takes over.
Common questions
What is the depletion region in a P-N junction?
A thin layer around the junction where electrons and holes have diffused across and recombined, leaving only fixed ions. Their charge creates an electric field and the built-in potential that stop further diffusion.
How do you calculate the built-in potential?
Vbi = (kT/q)·ln(Na·Nd/ni²). For silicon with Na = Nd = 10¹⁶ cm⁻³ at 300 K, Vbi ≈ 0.71 V.
What happens to the depletion region under reverse bias?
It widens, in proportion to the square root of (Vbi − Va), and the electric field grows. Only a small leakage current of minority carriers flows.