Built-in voltage

The built-in voltage $V_0$ (also called the built-in potential or the junction potential barrier) is the stable potential difference that develops across the Depletion region of a PN junction in thermal equilibrium, with no external voltage applied. It is the energy hill a carrier must climb to cross the junction.

Where it comes from

When p-type and n-type silicon meet, carriers diffuse across the junction and leave behind fixed ionised dopant charge — positive donors on the n-side, negative acceptors on the p-side (see Depletion region). That charge separation produces an electric field, and integrating that field across the depletion region gives a potential difference: $V_0$. The field grows until the drift current it drives exactly cancels the diffusion current (Drift and diffusion current); $V_0$ is the equilibrium value of that potential.

The formula

$V_0=V_T\ln \frac{N_A{N}_D}{n_i^2}$

where:

• $V_T=kT/q$ is the Thermal voltage, about $25\text{mV}$ at room temperature on the convention used in this course ($k$ = Boltzmann constant, $T$ = absolute temperature, $q$ = electron charge). A slightly different rounding gives $26\text{mV}$; this course uses $25\text{mV}$ throughout.
• $N_A$ is the acceptor (p-side) doping concentration.
• $N_D$ is the donor (n-side) doping concentration.
• $n_i$ is the Intrinsic carrier concentration ($\approx 1.5\times 10^{10}{\text{cm}}^{-3}$ for silicon).

The structure is intuitive: $V_0$ grows logarithmically with how strongly each side is doped. More doping means a steeper concentration step across the junction, so a bigger built-in potential is needed to hold back the larger diffusion tendency. The $n_i^2$ in the denominator ties it back to the Mass-action law.

Worked example. Silicon with $N_A=10^{17}{\text{cm}}^{-3}$, $N_D=10^{16}{\text{cm}}^{-3}$, $n_i=1.5\times 10^{10}{\text{cm}}^{-3}$, $V_T=25\text{mV}$:

$\frac{N_A{N}_D}{n_i^2}=\frac{10^{17}\times 10^{16}}{(1.5\times 10^{10}{)}^2}=\frac{10^{33}}{2.25\times 10^{20}}\approx 4.4\times 10^{12}$

$V_0=0.025\times \ln (4.4\times 10^{12})=0.025\times 29.1\approx 0.73\text{V}$

which is the familiar "$\approx 0.7\text{V}$" for a silicon junction.

[Background from general knowledge, not the source PDF]

Real but not measurable

$V_0$ is genuinely present inside the device: a carrier crossing from the n-side to the p-side must climb an energy barrier of height $q{V}_0$, and that barrier is exactly what biasing raises or lowers to switch a diode off or on. But you cannot read $V_0$ with a voltmeter. To probe it you must attach metal leads, and each metal–semiconductor contact develops its own contact potential. Going around the loop, those contact potentials sum to exactly $-V_0$, cancelling it. So $V_0$ does no work on an external circuit at equilibrium (otherwise you would have a free perpetual current), yet it is the very barrier that Forward bias lowers (to $V_0-V$) and Reverse bias raises (to $V_0+V$) to control current flow.