Built-in potential of a junction

WHY does a potential appear at all?

HOW to derive it from first principles

We derive it two equivalent ways. Both start from the same law.

Step 1 — At equilibrium, net current = 0

For electrons, current has two parts: $J_n = \underbrace{q\mu_n n E}_{\text{drift}} + \underbrace{qD_n \frac{dn}{dx}}_{\text{diffusion}} = 0$

Why this step? Equilibrium (no battery) means no net current anywhere, so the two mechanisms must exactly balance.

Step 2 — Use the Einstein relation

The Einstein relation links diffusion and mobility: $D_n = \frac{kT}{q}\mu_n$

Why? Both diffusion and drift come from the same random thermal motion of carriers, so they can't be independent — Einstein's relation ties them together.

Substitute and cancel $\mu_n$ : $q\mu_n n E + q\frac{kT}{q}\mu_n \frac{dn}{dx} = 0 \;\Rightarrow\; E = -\frac{kT}{q}\frac{1}{n}\frac{dn}{dx}$

Step 3 — Turn $E$ into a potential

Since $E = -\dfrac{dV}{dx}$ : $-\frac{dV}{dx} = -\frac{kT}{q}\frac{1}{n}\frac{dn}{dx} \;\Rightarrow\; dV = \frac{kT}{q}\frac{dn}{n}$

Why? We want a voltage (integratable), and $E$ is the slope of voltage. This cleverly converts a position integral into a much easier integral over $n$ .

Step 4 — Integrate across the junction

Integrate from the p-side (where $n = n_p$ ) to the n-side (where $n = n_n$ ): $V_{bi} = V_n - V_p = \frac{kT}{q}\int_{n_p}^{n_n}\frac{dn}{n} = \frac{kT}{q}\ln\!\left(\frac{n_n}{n_p}\right)$

Step 5 — Plug in doping values

For non-degenerate, fully-ionized doping:

n-side majority electrons: $n_n \approx N_D$
p-side minority electrons (mass-action law $np = n_i^2$ ): $n_p = \dfrac{n_i^2}{N_A}$

$\boxed{V_{bi} = \frac{kT}{q}\ln\!\left(\frac{n_n}{n_p}\right) = \frac{kT}{q}\ln\!\left(\frac{N_D \cdot N_A}{n_i^2}\right)}$

Why this step? We replaced abstract carrier densities with quantities the fab actually controls: the doping levels.

Figure — Built-in potential of a junction

Common Mistakes (Steel-manned)

Q: If both sides are doped at $10^{16}$ (Si), forecast $V_{bi}$ : closer to 0.3, 0.7, or 1.1 V? Verify: $\ln(10^{32}/10^{20})=\ln(10^{12})=27.6$ ; $\times0.02585 = 0.714$ V → ~0.7 V. Textbook silicon value ✓.

Flashcards

What physical process is stopped at equilibrium by the built-in field?

The net diffusion of carriers — drift exactly cancels diffusion so net current = 0.

Write the built-in potential formula.

V_{bi} = \frac{kT}{q}\ln\!\left(\frac{N_A N_D}{n_i^2}\right)

Which relation links

D_n

and

\mu_n

in the derivation?

The Einstein relation,

D_n = (kT/q)\mu_n

Why can't you extract energy from

V_{bi}

with a voltmeter?

Contact potentials at the probe junctions cancel it; total loop voltage = 0.

How much does

V_{bi}

change per 10× increase in doping?

About

+60

mV (since

kT/q \cdot \ln 10 \approx 59.6

mV).

Why is Ge's

V_{bi}

lower than Si's?

Ge has a much larger

n_i

, shrinking the ratio

N_AN_D/n_i^2

and hence the log.

What is

kT/q

at 300 K?

About 25.85 mV (the thermal voltage

V_T

What happens to

V_{bi}

as temperature rises?

It decreases (~

-2

mV/°C for Si) because

n_i

grows fast, shrinking the log ratio.

On which side do positive donor ions get exposed?

The n-side of the depletion region.

Recall Feynman: explain to a 12-year-old

Imagine two crowded rooms joined by a door. One room is packed with blue balls (electrons), the other with red balls (holes). Open the door and balls spill into the emptier room. But each ball that leaves its home leaves behind a "sticky spot" (a charged ion) that pulls balls back. Very quickly, a "wall of pull" builds up at the doorway strong enough to stop any more balls from crossing. The strength of that wall, measured as a voltage, is the built-in potential. Nobody plugged in a battery — the balls built the wall themselves just by wanting to spread out.

Connections

Depletion region width — $V_{bi}$ sets how wide the carrier-free zone is.
Mass-action law — $np = n_i^2$ , used to find $n_p$ in the derivation.
Einstein relation — the bridge between drift and diffusion.
PN junction under bias — external voltage adds/subtracts from $V_{bi}$ .
Diode I-V equation — the same $kT/q$ thermal voltage governs the exponential.
Intrinsic carrier concentration — temperature dependence of $n_i$ drives $V_{bi}(T)$ .
Energy band diagrams — $qV_{bi}$ = the band bending across the junction.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab hum p-type aur n-type silicon ko jodte hain, to n-side me electrons bahut zyada hote hain aur p-side me holes. Nature ko concentration gradient pasand nahi — isliye electrons n se p taraf diffuse karte hain. Lekin jaise hi electron apni jagah chhodta hai, wahan ek fixed positive donor ion khula reh jata hai, aur p-side pe negative acceptor ion. Yeh charged ions ek electric field bana dete hain jo aur carriers ko rukne ke liye bolta hai. Is field ki wajah se jo voltage difference banta hai, usko hi hum built-in potential $V_{bi}$ kehte hain — bina kisi battery ke, khud ba khud.

Formula yaad rakho: $V_{bi} = \frac{kT}{q}\ln(N_A N_D / n_i^2)$ . Yahan $kT/q$ ko thermal voltage kehte hain, jo room temperature (300 K) pe lagbhag 25.85 mV hota hai. Derivation ka core idea simple hai: equilibrium pe net current zero hota hai, matlab drift aur diffusion ek dusre ko exactly cancel karte hain. Einstein relation ( $D_n = (kT/q)\mu_n$ ) use karke integrate karo, to yeh log wala formula nikal aata hai.

Ek important baat — yeh log hai, isliye doping 10 guna badha do to $V_{bi}$ sirf ~60 mV badhta hai. Bahut weak control hai doping ka. Aur silicon me $V_{bi}$ ~0.7 V aata hai, germanium me ~0.3 V, kyunki germanium ka $n_i$ bada hota hai. Ek common galti: socho mat ki voltmeter laga ke $V_{bi}$ measure kar loge aur free energy mil jayegi — contact potentials cancel kar dete hain, reading zero aati hai. $V_{bi}$ ek internal quantity hai, EMF nahi.

WHAT is it?

WHY does a potential appear at all?

HOW to derive it from first principles

Step 1 — At equilibrium, net current = 0

Step 2 — Use the Einstein relation

Step 3 — Turn $E$ into a potential

Step 4 — Integrate across the junction

Step 5 — Plug in doping values

Worked Examples

Common Mistakes (Steel-manned)

Forecast-then-Verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Doping & PN Junctions

Connections