Built-in potential of a junction: Foundations — Hardware deep dive

This is the toolbox page for the parent topic. Every symbol that appears in $V_{bi}$ is dismantled here, in the order you need to meet them.

0 — The stage: what is silicon carrying?

Before any symbol, picture the material. Silicon is a crystal — atoms locked in a grid. In a perfect cold crystal every electron is bonded and nothing moves, so nothing conducts. Warm it up, and a few electrons shake loose and roam free. Each freed electron leaves an empty bond behind — a "hole" — which behaves like a positive particle drifting the other way.

Figure — Built-in potential of a junction

Why the topic needs this: the built-in potential exists because these carriers move. No mobile carriers, no diffusion, no field, no $V_{bi}$ .

1 — Concentration: the symbol $n$ (and $p$ , and $n_{i}$ )

How many carriers are there? We never count them one by one — there are quintillions. Instead we count how crowded they are: how many per unit volume.

Now the special one, $n_{i}$ :

Why the topic needs $n_{i}$ : it is the "pure baseline" that the doped concentrations are compared against inside the logarithm. See Intrinsic carrier concentration for the full temperature story.

2 — Doping: the symbols $N_{D}$ and $N_{A}$

Pure silicon barely conducts. We deliberately pollute it with a trace of a different atom to flood it with one kind of carrier. That is doping.

Why the topic needs these: $N_{A}$ and $N_{D}$ are the two knobs the factory actually turns, so the final formula is written in terms of them rather than the abstract $n$ and $p$ .

3 — Mass-action law: why $n p = n_{i}^{2}$

Here is the quiet fact that lets us convert doping into the minority carrier count.

So on the n-side (electrons $\approx N_{D}$ ), the leftover holes are $p_{n} = n_{i}^{2} / N_{D}$ . On the p-side (holes $\approx N_{A}$ ), the leftover electrons are $n_{p} = n_{i}^{2} / N_{A}$ . That is precisely why the formula's denominator is $n_{i}^{2}$ : dividing the doping product by $n_{i}^{2}$ silently encodes the mass-action swap. See Mass-action law.

the absence of a bonded electron, behaving like a mobile positive charge.

Concentration

n

is measured in

carriers per cubic centimetre,

cm^{- 3}

n_{i}

means

the electron (= hole) concentration in pure silicon at a given temperature.

A donor after releasing its electron becomes

a fixed positive ion stuck in the lattice.

The mass-action law states

n p = n_{i}^{2}

in any silicon at equilibrium.

ln (x)

asks the question

e

to what power gives

x

Why a logarithm and not linear scaling

concentrations span many powers of ten;

ln

turns "10× bigger" into "one fixed step up," so doping has weak control over

V_{bi}

V_{T} = k T / q

at 300 K equals

about 25.85 mV.

Why

k T / q

must sit in front of the

ln

the

ln

is unitless, so the prefactor must carry the volts.

The electric field in the depletion region points

from the positive donor ions (n-side) toward the negative acceptor ions (p-side).

Connections

Parent topic (Hinglish)
Intrinsic carrier concentration · Mass-action law · Einstein relation
Depletion region width · Energy band diagrams · PN junction under bias · Diode I-V equation

Foundations — Built-in potential of a junction

0 — The stage: what is silicon carrying?

1 — Concentration: the symbol $n$ (and $p$ , and $n_{i}$ )

2 — Doping: the symbols $N_{D}$ and $N_{A}$

3 — Mass-action law: why $n p = n_{i}^{2}$

4 — The logarithm: what does $ln$ ask?

5 — Thermal voltage: the symbols $k$ , $T$ , $q$ , and $k T / q$

6 — Two supporting symbols: $E$ , $V$ , and the depletion region

The prerequisite map

Worked micro-example — assemble the pieces

Equipment checklist

Connections