How are the built-in potential and depletion width calculated?

For an abrupt, fully ionized junction the built-in potential is V_bi = (kT/q)·ln(N_A·N_D/n_i²), about 0.72 V for silicon at 300 K with 10¹⁶ cm⁻³ on both sides. The depletion width is W = √(2ε(V_bi − V)/q · (1/N_A + 1/N_D)), so it widens under reverse bias and shrinks under forward bias. Charge neutrality requires N_A·x_p = N_D·x_n, so the depletion region reaches mostly into the more lightly doped side.

Why do the quasi-Fermi levels split in the band diagram?

In equilibrium there is a single, flat Fermi level. With a bias V applied, the Fermi levels in the n and p regions differ by qV, and across the depletion region electrons and holes each keep the level of their own side: the electron and hole quasi-Fermi levels E_Fn and E_Fp. Their separation sets np = n_i²·e^{qV/kT}, far above n_i² under forward bias (driving injection current) and far below it under reverse bias.

Why does a diode with series resistance need an implicit solution?

With series resistance the junction sees V_d = V − I·R_s, while the current depends on V_d through I = I_s(e^{V_d/nV_T} − 1), so I appears on both sides. There is no elementary closed form, although it can be written with the Lambert W function. The tool bisects on V_d instead: the junction current rises monotonically with V_d, so there is exactly one solution, and reverse breakdown fits in the same scheme. At high current the curve straightens out with slope 1/R_s.

How do LED colour and forward voltage relate to the band gap?

An LED emits photons with an energy close to the band gap: λ = hc/E_g ≈ 1240 nm·eV / E_g. Red AlGaInP LEDs (about 1.9–2.0 eV) therefore need about 2 V, blue InGaN (about 2.7 eV) about 3 V, and UV LEDs more. The forward voltage is usually a bit above E_g/q because of series resistance and contact drops. Silicon and germanium have indirect gaps and barely emit light, which is why LEDs use direct-gap materials such as GaAs and GaN.

What limits a solar cell’s fill factor and efficiency?

The fill factor FF = P_max/(V_oc·I_sc) measures how square the I–V curve is. Series resistance makes the knee sag near the maximum power point and a low shunt resistance tilts the flat part, and both lower FF. Heating raises J₀ steeply, as T³e^{−E_g/kT}, so a silicon cell loses about 2 mV of V_oc per °C and its efficiency drops. The theoretical limit for a single junction, the Shockley–Queisser limit, is about 33.7 %.

🔆

🔆PN Junction & Diode Simulator - Band Diagram, Depletion Width, Diode I–V, LED, Solar Cell

Choose silicon, germanium or gallium arsenide, set the doping on each side, the temperature and the bias, and the tool draws the band bending, quasi-Fermi levels, space charge, field, potential and carrier densities using the abrupt-junction depletion approximation, along with the built-in potential, depletion width, peak field and junction capacitance. The diode page solves the Shockley equation with series resistance and reverse breakdown on linear and log axes. The LED page turns a band gap into wavelength and colour, and the solar-cell page uses the single-diode model to find V_oc, I_sc, fill factor and efficiency.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

Loading tool...

How to use Semiconductors & PN Junctions

1On the PN junction page, pick a material and drag N_A, N_D, temperature and bias to watch the band diagram and profiles change
2On the diode page, choose a 1N4148, Schottky, Zener or LED preset, adjust I_s, n, R_s, breakdown voltage and temperature, and drag the operating point to read current and dynamic resistance
3On the LED page, drag the band gap or wavelength to see the colour and typical forward voltages, and size a current-limiting resistor
4On the solar-cell page, choose a cell type and adjust irradiance, temperature, series and shunt resistance to read V_oc, I_sc, FF, the maximum power point and efficiency

FAQ

How are the built-in potential and depletion width calculated?: For an abrupt, fully ionized junction the built-in potential is V_bi = (kT/q)·ln(N_A·N_D/n_i²), about 0.72 V for silicon at 300 K with 10¹⁶ cm⁻³ on both sides. The depletion width is W = √(2ε(V_bi − V)/q · (1/N_A + 1/N_D)), so it widens under reverse bias and shrinks under forward bias. Charge neutrality requires N_A·x_p = N_D·x_n, so the depletion region reaches mostly into the more lightly doped side.
Why do the quasi-Fermi levels split in the band diagram?: In equilibrium there is a single, flat Fermi level. With a bias V applied, the Fermi levels in the n and p regions differ by qV, and across the depletion region electrons and holes each keep the level of their own side: the electron and hole quasi-Fermi levels E_Fn and E_Fp. Their separation sets np = n_i²·e^{qV/kT}, far above n_i² under forward bias (driving injection current) and far below it under reverse bias.
Why does a diode with series resistance need an implicit solution?: With series resistance the junction sees V_d = V − I·R_s, while the current depends on V_d through I = I_s(e^{V_d/nV_T} − 1), so I appears on both sides. There is no elementary closed form, although it can be written with the Lambert W function. The tool bisects on V_d instead: the junction current rises monotonically with V_d, so there is exactly one solution, and reverse breakdown fits in the same scheme. At high current the curve straightens out with slope 1/R_s.
How do LED colour and forward voltage relate to the band gap?: An LED emits photons with an energy close to the band gap: λ = hc/E_g ≈ 1240 nm·eV / E_g. Red AlGaInP LEDs (about 1.9–2.0 eV) therefore need about 2 V, blue InGaN (about 2.7 eV) about 3 V, and UV LEDs more. The forward voltage is usually a bit above E_g/q because of series resistance and contact drops. Silicon and germanium have indirect gaps and barely emit light, which is why LEDs use direct-gap materials such as GaAs and GaN.
What limits a solar cell’s fill factor and efficiency?: The fill factor FF = P_max/(V_oc·I_sc) measures how square the I–V curve is. Series resistance makes the knee sag near the maximum power point and a low shunt resistance tilts the flat part, and both lower FF. Heating raises J₀ steeply, as T³e^{−E_g/kT}, so a silicon cell loses about 2 mV of V_oc per °C and its efficiency drops. The theoretical limit for a single junction, the Shockley–Queisser limit, is about 33.7 %.

Features

Si, Ge and GaAs parameters with Varshni band gap and temperature-dependent intrinsic carrier density
Abrupt PN junction band diagram: E_c, E_v, E_i and quasi-Fermi levels splitting under forward or reverse bias
Space charge, electric field, potential and log-scale n and p profiles with depletion edges marked
Shockley diode with series resistance and reverse breakdown solved implicitly, plus dynamic resistance and temperature coefficient
Band gap ↔ wavelength ↔ colour for LEDs, a table of common LEDs and a current-limiting resistor calculator
Single-diode solar cell: I–V and P–V curves, maximum power point, fill factor, efficiency and temperature effects

Related tools

View all 196 Physics Tools →