Built-in potential of a junction: Visual walkthrough

Before any symbol appears, here is the whole cast of characters, in plain words:

Everything below is built from just these ideas.

Step 1 — Two crowds, one door

WHY we start here: a potential is a difference between two places. So the very first thing to draw is the imbalance that will drive everything — a big pile of electrons on one side, almost none on the other.

PICTURE:

Figure — Built-in potential of a junction

Look at the density of cyan dots: dense on the left, sparse on the right. That lopsidedness has a name.

Step 2 — Diffusion: crowds spread out

WHY this tool: we need to describe flow caused purely by crowding, with no pushing force. Diffusion is exactly that — the flow a concentration gradient creates all by itself. The measured law is: flow $\propto$ steepness of the crowd.

PICTURE:

Step 3 — Exposed ions build a field

WHY this matters: those bare ions are charge that cannot move. Positive charge on the left, negative on the right — that is exactly what makes an electric field.

PICTURE:

The field's job: it pushes electrons back toward the n-side — opposing the diffusion of Step 2. Flow caused by a field has its own name: drift.

Step 4 — Equilibrium: the tug-of-war ties

WHY this is the key move: "no battery ⇒ no net current" is the single physical fact that pins down $V_{bi}$ . It converts a hard dynamics problem into one algebra equation.

PICTURE:

Solve for the field by moving one term across and dividing by $q μ_{n} n$ : $E = - \frac{D _{n}}{μ _{n}} \frac{1}{n} \frac{d n}{d x}$ Notice the ratio $\frac{D _{n}}{μ _{n}}$ appeared on its own. That ratio is not free — it is fixed by physics, which is the next step.

Step 5 — The Einstein relation replaces $D_{n} / μ_{n}$

WHY this tool and not another: we have an unknown ratio $D_{n} / μ_{n}$ stuck in our equation. Einstein's relation is the one law that evaluates exactly that ratio — it is the perfect key for this lock.

PICTURE:

Substitute into Step 4's field: $E = - \frac{k T}{q} \frac{1}{n} \frac{d n}{d x}$ The messy microscopic coefficients are gone; only the thermal voltage remains.

Step 6 — Turn the field into a voltage

WHY: a field lives at every point and is hard to add up directly across the messy junction. A voltage difference only cares about the two endpoints. So we trade the field for the thing we can integrate cleanly.

PICTURE:

Set the two expressions for $E$ equal: $- \frac{d V}{d x} = - \frac{k T}{q} \frac{1}{n} \frac{d n}{d x}$ The $d x$ cancels on both sides (multiply through by $d x$ ), leaving a much friendlier relation between $d V$ and $d n$ : $d V = \frac{k T}{q} \frac{d n}{n}$

Step 7 — Add up across the junction

WHY: the sum of all the little $d V$ steps across the whole depletion region is the built-in potential, by definition.

PICTURE:

The integral $\int \frac{d n}{n}$ is exactly the one whose answer is a logarithm (the $ln$ is defined as the area under $1/ n$ ): $V_{bi} = V_{n} - V_{p} = \frac{k T}{q} \int_{n_{p}}^{n_{n}} \frac{d n}{n} = \frac{k T}{q} ln (\frac{n _{n}}{n _{p}})$

Step 8 — Replace carrier counts with doping we control

WHY: to make the formula usable, every symbol must be something an engineer can dial in.

PICTURE:

On the n-side, essentially every donor gives an electron: $n_{n} \approx N_{D}$ .
On the p-side, electrons are rare; use the Mass-action law (restated below) to get $n_{p} = \frac{n _{i}^{2}}{N _{A}}$ .

Substitute: $\frac{n _{n}}{n _{p}} = \frac{N _{D}}{n _{i}^{2} / N _{A}} = \frac{N _{A} N _{D}}{n _{i}^{2}}$

Step 9 — Where the derivation breaks: the degenerate limit

WHY this deserves its own step: the boxed formula is only as trustworthy as its assumptions. A reader must know the fence around it, not just the field inside.

PICTURE:

Recall The ordinary (non-degenerate) edge cases — all safe

These stay inside the valid regime and are handled correctly by the boxed formula:

Equal doping $N_{A} = N_{D} = N$ : $V_{bi} = V_{T} ln (N^{2} / n_{i}^{2}) = 2 V_{T} ln (N / n_{i})$ — symmetric, finite.
Raise doping 10×: ratio grows 10×, but $ln$ turns that into $+ V_{T} ln 10 \approx + 60$ mV. Weak control (Step 7's promise).
Raise temperature: $n_{i} \propto T^{3/2} e^{- E_{g} /2 k T}$ grows fast, so the ratio shrinks and $V_{bi}$ falls (~ $- 2$ mV/°C for Si) — the rising $V_{T}$ prefactor cannot win.
Undoped limit $N_{A}, N_{D} \to n_{i}$ : ratio $\to 1$ , $ln 1 = 0$ , so $V_{bi} \to 0$ . Correct — no junction, no built-in potential.
Add a battery: equilibrium breaks, $J_{n} \neq = 0$ , and the whole $J_{n} = 0$ derivation stops — that regime is PN junction under bias leading to the Diode I-V equation.

The link to bands: $q V_{bi}$ is exactly how far the Energy band diagrams bend across the junction — the potential hill of height $V_{bi}$ in volts becomes an energy step of height $q V_{bi}$ in joules (or electron-volts), which is precisely the offset between the n-side and p-side band edges at equilibrium. That is the same $V_{bi}$ we just derived, now read off a band diagram instead of a potential plot.

The one-picture summary

Recall Feynman retelling of the whole walkthrough

Two crowded rooms share a door — n on the left, p on the right (that's our $x$ -axis, pointing right). Room N is packed with cyan balls; room P is nearly empty of them. Open the door — balls spill toward the empty side just because there's more room there (that's diffusion, Step 2). But each ball that leaves abandons a sticky charged spot in the wall (an ion, Step 3), and those spots build up an invisible wall of pull — an electric field that yanks balls back (drift). Balls keep crossing until the pull-back exactly matches the spill-out: a standstill with no net flow, $J_{n} = 0$ (Step 4). Because both the spilling and the pulling come from the same warmth of the room, their ratio is fixed — that's Einstein's rule, and it hands us a clean "thermal voltage" (Step 5). We measure the wall not as a force but as a height — a voltage — and add up the tiny climbs across the door (Steps 6–7). The adding-up of a $1/ n$ field is always a logarithm — that's why cranking the crowd harder barely raises the wall. Finally we swap "how many balls" for "how hard we doped" (Step 8), and out pops $V_{bi} = V_{T} ln (N_{A} N_{D} / n_{i}^{2})$ — trustworthy as long as we didn't dope so hard the room became degenerate (Step 9). Nobody plugged in a battery; the balls built the wall themselves, just by wanting to spread out.

Recall Quick self-test

Where does the logarithm come from, physically? ::: The field goes as $1/ n$ , and integrating $1/ n$ across the junction gives $ln$ . Why is $V_{bi}$ zero for undoped silicon? ::: Then $N_{A}, N_{D} \to n_{i}$ , the ratio $\to 1$ , and $ln 1 = 0$ — no imbalance, no potential. Which single assumption lets us set $J_{n} = 0$ ? ::: No external battery, so no net current can persist at equilibrium. What does Einstein's relation eliminate from the field equation? ::: The unknown ratio $D_{n} / μ_{n}$ , replacing it with $k T / q$ . Which two steps break for a degenerate (very heavily doped) semiconductor? ::: The mass-action law (Step 8) and the simple Einstein relation (Step 5) — both assumed Boltzmann statistics.

Connections

Built-in potential of a junction — the parent result derived here
Einstein relation · Mass-action law · Intrinsic carrier concentration
Depletion region width · Energy band diagrams
PN junction under bias · Diode I-V equation