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

Before we start, one shared picture so every symbol below is already earned. Look at the sketch:

Figure — Built-in potential of a junction

Definition The cast of characters (read alongside the figure)

$V_{bi}$ = the built-in potential. Sign convention pinned down here: $V_{bi} \equiv V_{n} - V_{p}$ , the potential of the neutral n-side minus that of the neutral p-side. In the figure this is the height of the hill going left (p) → right (n); with this convention $V_{bi} > 0$ for a normal junction.
$N_{A}$ , $N_{D}$ = acceptor (p-side) and donor (n-side) doping densities — how many dopant atoms per cm³ the fab put in.
$n_{i}$ = intrinsic carrier concentration — how many electron–hole pairs pure silicon makes on its own from thermal shaking.
$k T / q$ = the thermal voltage $V_{T} \approx 25.85$ mV at 300 K — the "energy of thermal jiggling" expressed as a voltage.
$E$ = the electric field inside the depletion region. Orientation convention: field points from the positive donor ions (n-side) toward the negative acceptor ions (p-side), i.e. n→p (the red arrow in the figure). It relates to potential by $E = - d V / d x$ — field is the downhill slope of voltage.
$E_{g}$ = the bandgap energy of the semiconductor — the energy gap (in electron-volts) an electron must gain to jump from the bound valence band into the mobile conduction band. Dividing by charge gives the bandgap voltage $E_{g} / q$ (~1.12 V for Si, ~0.66 V for Ge). It is the ceiling no built-in potential can reach.
$V_{bi} = \frac{k T}{q} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$

Spot the error

Each line quotes a plausible-sounding statement; the answer names the flaw.

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

The ratio is upside down. Doping exceeds

n_{i}^{2}

, so the argument must be

> 1

(giving positive

V_{bi}

); the flipped version gives a negative number.

"Since

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

is the minority electron density on the p-side, I'll use

N_{A}

for

n_{n}

too."

Confusion of sides:

n_{n} \approx N_{D}

(majority electrons on the n-side), while

N_{A}

sets the p-side. Mixing them scrambles which doping goes where.

"The Einstein relation

D_{n} = μ_{n} k T / q

was assumed just for convenience."

Not convenience — it's physically required. Drift and diffusion arise from the same thermal motion, so their coefficients must be linked; without it the derivation cannot cancel

μ_{n}

and close.

"I integrated

E

over position to get

V_{bi}

, which needs the exact field profile."

The clever move was rewriting

d V = (k T / q) d n / n

(the relation in the [!formula] box) so we integrate over carrier density, not position — the messy field shape cancels out entirely.

V_{bi}

is the diode's turn-on voltage, so a Si diode conducts once you apply 0.7 V equal to

V_{bi}

V_{bi}

and the turn-on voltage are related but not identical. Turn-on is where forward current becomes appreciable on the Diode I-V equation; you never fully reach

V_{bi}

in normal forward operation.

"Because current is zero at equilibrium, the carriers must be sitting still."

Wrong picture — carriers are in vigorous thermal motion and cross constantly. Zero is the balance of drift against diffusion, not stillness.

Why questions

Why does a potential appear even though no battery is connected?

Because the concentration difference drives diffusion; the departing carriers expose fixed ions whose field builds a potential hill — the system pays for spreading out with an internal voltage.

Why is

V_{bi}

a logarithm of the doping rather than linear in it?

It comes from integrating

d n / n

across the junction;

\int d n / n = ln n

, so the ratio of carrier densities enters as a natural log.

Why does the formula use

n_{i}^{2}

and not

n_{i}

Because Mass-action law gives the minority electron density

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

; substituting that into

ln (n_{n} / n_{p})

produces

ln (N_{A} N_{D} / n_{i}^{2})

Why does

V_{bi}

decrease as the junction heats up?

n_{i}

climbs exponentially with temperature, so the ratio

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

collapses and the log shrinks faster than the

k T / q

prefactor grows.

Why can't you extract useful energy from

V_{bi}

Any measuring loop introduces contact potentials that exactly cancel it; the built-in voltage is an equilibrium bookkeeping quantity, not a source of free work.

Why is the field said to "stop" diffusion rather than reverse it?

The field grows only until its drift push exactly matches the diffusion pull; at that balance point net flow is zero — it opposes, it doesn't over-power.

Why does using

n_{n} \approx N_{D}

assume full ionization and non-degeneracy?

We assume every donor gives up its electron (full ionization) and that densities stay below the conduction-band limit (non-degenerate) so majority electron count equals dopant count.

Edge cases

What is

V_{bi}

for a slab of intrinsic silicon touching intrinsic silicon?

Zero — with no doping difference,

N_{A} N_{D} \to n_{i}^{2}

, the ratio is 1,

ln 1 = 0

; no concentration gradient, no built-in field.

What happens to

V_{bi}

n_{i} \to 0

(a hypothetical perfect insulator)?

The ratio

N_{A} N_{D} / n_{i}^{2} \to \infty

, so

V_{bi}

diverges — physically capped by the bandgap voltage

E_{g} / q

; wide-gap materials indeed show larger built-in potentials.

If one side is doped extremely lightly (nearly intrinsic), does

V_{bi}

vanish?

No — even a modest doping on one side keeps the product

N_{A} N_{D} ≫ n_{i}^{2}

, so

V_{bi}

stays a few tenths of a volt; it only vanishes when both sides approach intrinsic.

What limits

V_{bi}

from above as doping is cranked to its maximum?

The bandgap voltage

E_{g} / q

(~1.12 V for Si); near it the material becomes degenerate, the simple log formula fails, and the Fermi level enters the bands (see Energy band diagrams).

Under forward bias, does the internal built-in potential change?

The net potential barrier drops by the applied voltage, but the material's intrinsic

V_{bi}

(the equilibrium value) is a fixed reference — bias adds on top of it (see PN junction under bias).

At absolute zero, what does the formula predict and why is it suspect?

With

k T / q \to 0

the prefactor kills

V_{bi}

to 0, but

n_{i} \to 0

too and freeze-out defeats "full ionization"; the non-degenerate formula simply stops applying at very low

T

Does swapping which side is p and which is n flip the sign of

V_{bi}

The magnitude is identical (the product

N_{A} N_{D}

is symmetric); only the reference direction of "n-side minus p-side" flips sign — the physical hill is the same height.

Recall One-line summary of every trap

$V_{bi}$ is (1) internal not extractable, (2) logarithmic not linear in doping, (3) falling with temperature, (4) larger for larger bandgap / smaller $n_{i}$ , and (5) zero only when the two sides are indistinguishable.

Connections

Built-in potential of a junction — parent
Mass-action law, Einstein relation, Intrinsic carrier concentration — tools used above
PN junction under bias, Diode I-V equation, Depletion region width, Energy band diagrams — where these ideas go next

Question bank — Built-in potential of a junction

True or false — justify