Formation of a PN junction: Visual walkthrough — Hardware deep dive

We will meet these tools as we go, and each time we will stop and ask why this tool and not another:

a gradient $\frac{d n}{d x}$ (a slope),
an electric field $E$ and its link to potential $V$ ,
the Einstein relation (borrowed from Einstein Relation),
and one integral — the only calculus move on the whole page.

Prerequisites you may want open: Doping of Semiconductors, Intrinsic vs Extrinsic Semiconductors, Diffusion and Drift Currents.

Step 1 — What "concentration" means, and the slope that lives on it

WHAT. A semiconductor is a crystal with mobile charge carriers sprinkled through it. On the n-side there are many free electrons; on the p-side there are many holes (empty electron seats). Concentration just means "how many per cubic centimetre." We write the electron concentration as $n$ and the hole concentration as $p$ . Units: $cm^{- 3}$ (things per cubic cm).

WHY. Before we can talk about flow, we need the quantity that flows. Carriers flow because there are more of them in one place than another — so the star of the show is not $n$ itself but how fast $n$ changes across space.

PICTURE. In the figure, the height of the coloured region is $n$ . It is tall on the n-side (crowded), short on the p-side (empty). The steepness of the ramp connecting them is the thing we will name next.

$n$ ::: number of free electrons per $cm^{3}$ at a point
$x$ ::: position along the crystal (the horizontal axis)
$\frac{d n}{d x}$ ::: the steepness of the concentration ramp at that point

Step 2 — Diffusion: the flow driven by that slope

WHAT. Random jiggling of carriers means more of them wander out of a crowded spot than wander in. Net result: carriers drift from high $n$ to low $n$ . We call the resulting electron current density $J_{n}^{diff}$ and write it proportional to the gradient: $J_{n}^{diff} = q D_{n} \frac{d n}{d x}$

WHY this form. The bigger the slope, the more lopsided the random wandering, the bigger the flow — so the current must be proportional to the gradient, not to $n$ itself. That is why $\frac{d n}{d x}$ appears and not $n$ .

PICTURE. The orange arrow rides down the ramp: electrons pour from the tall n-side toward the short p-side. (See Diffusion and Drift Currents for the full treatment.)

Step 3 — Diffusion uncovers fixed ions and builds a field

WHAT. Every electron that leaves the n-side abandons a fixed positive donor ion $N_{D}^{+}$ nailed into the lattice. Every hole that leaves the p-side abandons a fixed negative acceptor ion $N_{A}^{-}$ . A layer of $+$ charge (n-side) faces a layer of $-$ charge (p-side). Between two sheets of opposite charge lives an electric field $E$ .

WHY a field appears. Separated $+$ and $-$ charge always make a field pointing from $+$ to $-$ . Here that is n → p. This field is the crystal's automatic response — nobody plugged in a battery.

PICTURE. The violet arrows show $E$ pointing n→p. Notice it pushes electrons (negative) back toward n and holes back toward p: it fights the diffusion of Step 2.

$V$ ::: electric potential (potential energy per unit charge) at a point
$E$ ::: the electric field, the negative slope of $V$
why the minus sign ::: field points downhill but $d V / d x$ measures the uphill slope

Step 4 — Drift: the return current the field creates

WHAT. A field pushes charges. The electron current it drives is the drift current: $J_{n}^{drift} = q μ_{n} n E$

WHY this form. Drift current needs carriers present ( $n$ ) and a push ( $E$ ); double either and you double the flow, so both multiply in. The constant $μ_{n}$ (mobility) says how fast electrons slide per unit push.

WHY we need both currents. Diffusion (Step 2) shoves electrons n→p; drift (this step) shoves them back p→n. They point opposite ways. The junction is stable exactly when they cancel.

PICTURE. Two arrows of equal length pointing opposite ways — the tug-of-war that will define equilibrium.

Step 5 — Equilibrium: set total current to zero and simplify

WHAT. No battery is attached, so no net current can flow anywhere. Add the two pieces and demand they vanish: $J_{n} = q μ_{n} n E + q D_{n} \frac{d n}{d x} = 0$

WHY $= 0$ . A steady net current in an unconnected crystal would be free energy forever — forbidden. So drift $+$ diffusion $= 0$ at every point.

Now replace $E = - \frac{d V}{d x}$ (Step 3): $0 = - q μ_{n} n \frac{d V}{d x} + q D_{n} \frac{d n}{d x}$

WHY this swap. We ultimately want a voltage $V_{bi}$ , so we trade the field $E$ for the potential $V$ it comes from. Divide through by $q n$ and rearrange: $\frac{d V}{d x} = \frac{D _{n}}{μ _{n}} \frac{1}{n} \frac{d n}{d x}$

PICTURE. The balance beam: the diffusion arrow and the drift arrow are now the same length, and the net-current needle reads exactly zero.

Why divide by $q n$ ? ::: to isolate $\frac{d V}{d x}$ and expose the clean ratio $\frac{D _{n}}{μ _{n}}$
What does $J_{n} = 0$ physically mean? ::: drift exactly cancels diffusion, so no net electron flow

Step 6 — The Einstein relation turns the constant into a temperature

WHAT. The clump $\frac{D _{n}}{μ _{n}}$ is not a mystery. The Einstein Relation says: $\frac{D _{n}}{μ _{n}} = \frac{k T}{q} \equiv V_{T}$

WHY this tool and not another. Both $D_{n}$ (diffusion) and $μ_{n}$ (drift response) come from the same random thermal jiggling of electrons. Einstein's insight ties them together, and their ratio collapses to pure thermal energy $k T$ divided by charge $q$ — a voltage called the thermal voltage $V_{T}$ .

PICTURE. A thermometer feeding the value $V_{T} \approx 0.0259$ V at room temperature into the equation, showing the constant is just "how warm it is."

Step 7 — Integrate across the junction to get $V_{bi}$

WHAT. Both sides of $d V = V_{T} \frac{d n}{n}$ are tiny changes. Sum them up (integrate) from deep in the p-side (potential $V_{p}$ , electron concentration $n_{p}$ ) to deep in the n-side (potential $V_{n}$ , concentration $n_{n}$ ): $\int_{V_{p}}^{V_{n}} d V = \frac{k T}{q} \int_{n_{p}}^{n_{n}} \frac{d n}{n}$

WHY an integral, and why $\int \frac{d n}{n} = ln$ . We only knew the slope $d V / d x$ ; to get the total voltage drop we must add up all the tiny slopes across the width — that is exactly what an integral does. And the integral of $\frac{1}{n}$ is the natural logarithm $ln n$ — that is where the $ln$ in the final formula is born, not by decree but by calculus.

$V_{n} - V_{p} = \frac{k T}{q} ln \frac{n _{n}}{n _{p}}$

Now put in the concentrations. On the n-side almost every donor gives an electron, so $n_{n} \approx N_{D}$ . On the p-side, minority electrons obey mass action $n p = n_{i}^{2}$ (from Intrinsic vs Extrinsic Semiconductors), giving $n_{p} \approx n_{i}^{2} / N_{A}$ . Therefore:

PICTURE. The potential $V$ climbs as a smooth log-shaped step from the p-side level up to the n-side level; the total height of the step is $V_{bi}$ .

Step 8 — Sanity, edge and degenerate cases

WHAT / WHY. A formula is only trustworthy if it behaves at the extremes. Walk the corners:

Case	Argument $\frac{N _{A} N _{D}}{n _{i}^{2}}$	What $V_{bi}$ does	Physical reading
Normal Si, $N_{A} = N_{D} = 1 0^{16}$	$\approx 4.4 \times 1 0^{11}$	$\approx 0.70$ V	the classic diode barrier
Double the doping	$\times 4$	$+ V_{T} ln 4 \approx + 0.036$ V	barrier grows only logarithmically — slowly
Very light doping $N_{A} N_{D} \to n_{i}^{2}$	$\to 1$	$ln 1 = 0 \Rightarrow V_{bi} \to 0$	no contrast between sides ⇒ no barrier
Undoped both sides	$= 1$	$0$ V	it is just one intrinsic crystal — no junction at all
Raise temperature $T ↑$	$n_{i}$ grows fast	$V_{bi}$ falls	more thermal carriers wash out the contrast

PICTURE. Two log-step profiles overlaid: a tall step for heavy doping, a vanishing flat line as $N_{A} N_{D} \to n_{i}^{2}$ . The degenerate case is literally no step.

Worked example — silicon, term by term

The one-picture summary

Everything at once: the concentration ramp (Step 1) drives diffusion (Step 2), which uncovers ions and builds a field (Step 3) that drives the opposing drift (Step 4); setting their sum to zero (Step 5), using Einstein (Step 6), and integrating (Step 7) delivers the log-shaped potential step whose height is $V_{bi}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture a hill of electrons: piled high on the n-side, almost empty on the p-side. Because they jiggle randomly, more roll down the pile than up — that rolling is diffusion. But each electron that rolls away leaves a nailed-down "+" ion behind on the n-side, and fills a seat that exposes a "−" ion on the p-side. Those two sheets of stuck charge make an invisible electric field that pushes the electrons back up the hill — that push is drift. The pile stops changing exactly when the pushing-back perfectly matches the rolling-down: no net flow. To find how tall the resulting voltage hill is, we notice the field is just the steepness of a potential slope, we borrow Einstein's fact that jiggle-diffusion and push-drift share the same thermal origin (so their ratio is just "how warm it is," $k T / q$ ), and we add up all the little slopes across the junction — that adding-up is an integral, and integrating $1/ n$ gives a logarithm. Out pops $V_{bi} = \frac{k T}{q} ln \frac{N _{A} N _{D}}{n _{i}^{2}}$ : a thermal rung times the amount of contrast between the crowded and empty sides. Crank the doping and the hill barely grows, because a logarithm is lazy; that is why silicon always turns on near $0.7$ volts.

Recall

Where does the $ln$ in $V_{bi}$ come from? ::: From integrating $\frac{d n}{n}$ across the junction — $\int \frac{d n}{n} = ln n$ . Why is the barrier only $\sim 0.7$ V even with heavy doping? ::: Because $V_{bi}$ depends on doping logarithmically, so it grows very slowly. What makes $V_{bi} \to 0$ ? ::: When $N_{A} N_{D} \to n_{i}^{2}$ (little/no doping) the log argument $\to 1$ and $ln 1 = 0$ .