The p-n junction under bias

Set the doping on each side and the bias. Watch the depletion region, charge, field, potential and band diagram respond together, with Vbi, W, xp, xn, Emax and Cj worked out live.

N_A \text{ (p side)}

1×10¹⁶ cm⁻³

N_D \text{ (n side)}

1×10¹⁶ cm⁻³

V_a \text{ (bias, p relative to n)}

+0.00 V

V_{bi}

0.693 V

V_{bi}-V_a

0.693 V

W

0.423 µm

x_p

0.212 µm

x_n

0.212 µm

E_{max}

32.7 kV/cm

C_j/A

24.5 nF/cm²

All four plots share the same x axis (µm). The junction is at x = 0, p side on the left.

Charge density ρ/q (cm⁻³)

Electric field E (V/cm)

Potential φ (V)

Energy bands (eV, zero at the p-side Fermi level)

Current through the diode

Ideal diode: $I = I_s\left(e^{V_a/V_T}-1\right)$ . At the current bias, $I/I_s = 0$

What to notice

Charge neutrality: the two blocks in the charge plot have equal area, $N_A x_p = N_D x_n$ (here $2.12\times10^{11}$ cm⁻²).
The peak field sits at the junction: $E_{max} = qN_A x_p/\varepsilon = 32.7$ kV/cm, and the area under $E(x)$ is the voltage across the junction, $0.693$ V.
At zero bias the Fermi level is flat across the whole device: no net current flows.

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