Formation of a PN junction: Question bank — Hardware deep dive

These items lean on ideas from Doping of Semiconductors, Diffusion and Drift Currents, the Einstein Relation, and set up PN Junction under Bias and Diode I-V Characteristics.

Words before symbols — the ideas these traps assume

Before any symbol appears, here are the plain-word ideas the whole page leans on. Read these once and the traps become fair game.

Definition The four ideas you must own

Diffusion current — carriers randomly wander from where they are crowded to where they are sparse (like ink spreading in water). The moving charge this spreading carries is the diffusion current. Its strength grows with how steep the crowding gradient is. See Diffusion and Drift Currents.
Drift current — when an electric field is present, it drags charges along; that pushed charge is the drift current. Its strength grows with the field and with how many carriers are around to be pushed.
Mass-action law — in any semiconductor at temperature $T$ , the product of electron count and hole count is fixed: $n p = n_{i}^{2}$ . So making one kind of carrier plentiful forces the other kind to be rare (see Intrinsic vs Extrinsic Semiconductors).
Conventional current vs. electron flux — throughout this page "current" means conventional current, defined as the direction positive charge would move. Electrons are negative, so electron flux (which way the electrons physically go) points opposite to the conventional current they produce. Keep these two apart — a whole "Spot the error" trap lives on the difference.

Symbol reminder — read this first

Every symbol below is used bare in the traps; keep this card open until it's automatic.

Definition The symbols on this page

$k$ — Boltzmann's constant, the "how much energy per degree of temperature" number ( $1.38 \times 1 0^{- 23} J/K$ ).
$T$ — absolute temperature in kelvin (room temperature $\approx 300$ K).
$q$ — the size of one electron's charge ( $1.6 \times 1 0^{- 19}$ C), always written as a positive number.
$k T / q$ — the thermal voltage $V_{T}$ ; combines the three above into a voltage ( $\approx 0.026$ V at 300 K). Think of it as "how many volts one unit of random thermal jostling is worth."
$N_{A}$ — acceptor doping: number of p-type dopant atoms per cm³ (each becomes a fixed negative ion after giving away a hole).
$N_{D}$ — donor doping: number of n-type dopant atoms per cm³ (each becomes a fixed positive ion after giving away an electron).
$n_{i}$ — intrinsic carrier concentration: how many electron–hole pairs a pure (undoped) crystal makes by itself at temperature $T$ (see Intrinsic vs Extrinsic Semiconductors).
$n_{n}$ — the number of free electrons per cm³ on the n-side (the majority carriers there; $n_{n} \approx N_{D}$ ).
$n_{p}$ — the number of free electrons per cm³ on the p-side (the minority carriers there; by mass-action $n_{p} \approx n_{i}^{2} / N_{A}$ ).
$D_{n}$ — the electron diffusion coefficient: how fast electrons spread out per unit crowding gradient (bigger $D_{n}$ = they diffuse away faster).
$μ_{n}$ — the electron mobility: how fast electrons drift per unit electric field (bigger $μ_{n}$ = the field drags them faster).
$x_{p}, x_{n}$ — how far the depletion region reaches into the p-side and the n-side respectively.

Where $V_{bi}$ comes from (so the traps make sense)

The built-in potential formula is used all over this page, so here is the reasoning that produces it, with every symbol earned above.

Formula Why the balance produces a

logarithm — step by step 1. Set net electron current to zero. At equilibrium no net conventional current flows anywhere (else charge would pile up forever). For electrons this means their drift current and diffusion current are equal and opposite: $drift (field drags them) q μ_{n} n E + diffusion (crowding spreads them) q D_{n} \frac{d n}{d x} = 0.$ (Note: "electron current is zero" is shorthand for the net conventional current carried by electrons being zero — the two physical flows cancel.)

2. Swap field for potential. A field is just the downhill slope of voltage, $E = - \frac{d V}{d x}$ . Substituting turns the balance into a statement about $V$ and $n$ only: $q D_{n} \frac{d n}{d x} = q μ_{n} n \frac{d V}{d x} .$

3. Separate the variables. Divide by $q μ_{n} n$ so all the $V$ -stuff sits on one side and all the $n$ -stuff on the other: $\frac{d V}{d x} = \frac{D _{n}}{μ _{n}} \frac{1}{n} \frac{d n}{d x} .$

4. Use the Einstein Relation. It states $D_{n} / μ_{n} = k T / q$ — spreading-strength and dragging-strength share the same thermal origin. That converts the messy ratio into the thermal voltage: $d V = \frac{k T}{q} \frac{d n}{n} .$

5. Add up (integrate) from p-side to n-side. The right side is $\frac{k T}{q} \times (sum of d n / n)$ , and summing $d n / n$ across a change in $n$ is exactly what produces a logarithm of the ratio of end values — that is why the log appears: $V_{bi} = \frac{k T}{q} ln \frac{n _{n}}{n _{p}} = \frac{k T}{q} ln \frac{N _{A} N _{D}}{n _{i}^{2}} .$ Read the result as a story: the log measures how lopsided the electron populations are ( $n_{n} \approx N_{D}$ on the n-side vs. $n_{p} \approx n_{i}^{2} / N_{A}$ on the p-side by mass-action); the more lopsided, the taller the voltage hill needed to hold it in place. The prefactor $k T / q$ converts lopsidedness into volts. Full derivation with figures lives on the parent Formation of a PN junction.

True or false — justify

The p-side of an isolated (unconnected) p-type block carries a net positive charge because it is full of holes.

False. Each mobile hole is balanced by a fixed negative acceptor ion, so the whole block is electrically neutral; charge only appears near the junction after diffusion uncovers unbalanced ions.

A PN junction is made by pressing a p-type crystal against an n-type crystal.

False. It must be one continuous crystal whose doping changes from p to n; a physical contact between two crystals has surface states and a broken lattice, not a clean junction.

At equilibrium no carriers cross the junction at all.

False. Carriers keep crossing both ways constantly; it is the net current that is zero because diffusion crossing exactly balances drift crossing.

The depletion region is called "depleted" because it has no charge in it.

False. It is depleted of mobile carriers but is full of fixed ionic charge (negative acceptors on p, positive donors on n) — that charge is precisely what builds the field.

The built-in field points from the p-side to the n-side.

False. Field runs from + charge to − charge, i.e. from the positive donor ions on the n-side to the negative acceptor ions on the p-side, so it points n→p.

Doubling both dopings doubles the built-in potential.

False.

V_{bi} = \frac{k T}{q} ln \frac{N _{A} N _{D}}{n _{i}^{2}}

depends logarithmically on doping; doubling both only adds

\frac{k T}{q} ln 4 \approx 0.036

The built-in voltage

V_{bi}

appears on a voltmeter placed across the diode's leads.

False. The metal–semiconductor contact potentials cancel

V_{bi}

around the loop, so the external terminals read zero — otherwise you'd have a battery from nothing.

Raising temperature always increases

V_{bi}

because

k T / q

grows.

False. The prefactor

k T / q

grows, but

n_{i}^{2}

grows exponentially with temperature, shrinking the log term much faster; net effect is

V_{bi}

decreases with temperature.

The depletion region is always symmetric about the metallurgical junction.

False. Charge balance

N_{A} x_{p} = N_{D} x_{n}

forces the depletion width to extend farther into the lightly doped side, so it is symmetric only when

N_{A} = N_{D}

Spot the error

"Electrons diffuse n→p, so conventional diffusion current also flows n→p."

Wrong direction. Electrons carry negative charge, so electrons moving n→p make a conventional current p→n; conventional current is defined by positive-charge motion, always opposite to electron flux.

"The field is the cause of diffusion across the junction."

Reversed causation. Diffusion happens first (driven by concentration gradient) and creates the exposed ions and thus the field; the field then opposes further diffusion.

"Since holes are positive, the p-side develops positive ions."

Wrong sign. The p-side is doped with acceptors; when holes leave, they uncover fixed negative acceptor ions (

N_{A}^{-}

). Positive donor ions appear on the n-side.

"At equilibrium diffusion current is zero."

Not zero individually. Diffusion current is large and nonzero; it is exactly cancelled by an equal and opposite drift current, giving zero net current.

"The barrier stops forming once all mobile carriers have recombined everywhere."

It stops far earlier. The field grows only until drift balances diffusion at the junction; the bulk regions keep their mobile carriers — only the thin depletion zone is emptied.

"We used the Einstein relation just to simplify algebra."

It's physics, not cosmetics.

D_{n} / μ_{n} = k T / q

links how strongly a carrier diffuses to how strongly a field drags it, both set by thermal energy — that shared origin is why drift and diffusion can balance at one clean potential.

Why questions

Why does the depletion region stop growing instead of eating the whole crystal?

As it widens the ionic dipole and its field grow, and the drift force from that field rises until it exactly opposes diffusion; at that balance the driving force for net crossing vanishes.

Why is the built-in potential barely changed by doping while diode turn-on is roughly fixed at 0.7 V?

Because

V_{bi}

depends on doping only through a logarithm, so even large doping changes shift it by tens of millivolts — hence silicon barriers cluster near 0.6–0.7 V.

Why must the total charge on the p-side of the depletion region equal that on the n-side?

The depletion region as a whole is neutral (it came from neutral material), so exposed negative acceptor charge must match exposed positive donor charge:

N_{A} x_{p} = N_{D} x_{n}

Why can't

V_{bi}

act as a free battery to run a motor?

It's an equilibrium potential; the contact potentials at the wires exactly cancel it, giving zero net loop EMF. A nonzero one would let you extract work forever, violating the second law.

Why does the field point specifically from the n-side toward the p-side?

The uncovered ions form a dipole with positive donors on n and negative acceptors on p, and field lines run from positive to negative charge, hence n→p — conveniently the direction that pushes electrons back to n and holes back to p.

Why is

n_{p}

(electrons on the p-side) taken as

n_{i}^{2} / N_{A}

in the

V_{bi}

formula?

On the p-side holes dominate at

p \approx N_{A}

, and mass-action

n p = n_{i}^{2}

fixes the minority electrons at

n_{p} = n_{i}^{2} / N_{A}

Edge cases

What happens to the depletion width as one side is doped extremely heavily (a "one-sided" junction)?

The heavily doped side needs almost no width to supply its charge, so the depletion region lies almost entirely in the lightly doped side.

If a region were doped so lightly it were nearly intrinsic (

N_{D} \to n_{i}

), what happens to

V_{bi}

The ratio

N_{A} N_{D} / n_{i}^{2}

shrinks toward

N_{A} / n_{i}

, and if both sides approached intrinsic the log tends to zero — no doping difference, no built-in barrier.

What is the built-in potential of a junction between two identically doped n-type regions?

Zero. There is no p-to-n transition, no net concentration gradient of the same carrier across a boundary, so nothing to diffuse and unbalance — no junction forms.

As temperature approaches absolute zero, what happens to the equilibrium picture?

Carriers freeze onto their dopants (

n_{i} \to 0

), diffusion effectively stops, and the simple thermal-voltage derivation (

V_{T} = k T / q \to 0

) breaks down — the equilibrium argument assumes thermally mobile carriers.

If diffusion somehow overshot and made the field too strong, what would happen?

The excess field would drift carriers back faster than they diffuse out, refilling the edge of the depletion region until drift and diffusion re-balance — the equilibrium is self-correcting.

Recall One-line self-test

Cover everything: can you state (a) which ions sit on each side, (b) which way the field points, (c) why net current is zero, and (d) why $V_{bi}$ isn't a usable battery — each in one sentence with a because? Check ::: (a) negative acceptors on p, positive donors on n; (b) field points n→p; (c) drift cancels diffusion; (d) contact potentials cancel it around the loop.