Einstein relation between mobility and diffusion

HOW do we derive it? (from first principles)

WHY this derivation works: In thermal equilibrium with no external wires, the net current must be zero everywhere. Drift and diffusion currents must cancel exactly. Forcing that cancellation, plus the known equilibrium carrier distribution, squeezes out the $D/\mu$ ratio.

Step 1 — Write both currents (electrons). $J_n = \underbrace{q\,n\,\mu_n E}_{\text{drift}} + \underbrace{q D_n \frac{dn}{dx}}_{\text{diffusion}}$

Why this step? Total electron current is drift + diffusion. Note the diffusion term is $+qD_n\,dn/dx$ because electron charge $-q$ times flux $-D_n\,dn/dx$ gives $+qD_n\,dn/dx$ .

Step 2 — Equilibrium condition. In equilibrium $J_n = 0$ : $q n \mu_n E = -q D_n \frac{dn}{dx}$

Why this step? No battery, no net current — the two microscopic drives balance.

Step 3 — Relate field to potential. $E = -\dfrac{d\psi}{dx}$ , so: $n\mu_n \frac{d\psi}{dx} = D_n \frac{dn}{dx}$

Why this step? Field is the negative slope of electrostatic potential; this lets us plug in the equilibrium $n(\psi)$ .

Step 4 — Use the equilibrium carrier distribution. In equilibrium the electron density follows Boltzmann statistics in the potential energy $-q\psi$ : $n(x) = n_0 \exp\!\left(\frac{q\psi(x)}{k_BT}\right) \;\Rightarrow\; \frac{dn}{dx} = n\cdot\frac{q}{k_BT}\frac{d\psi}{dx}$

Why this step? This is the physical input. At temperature $T$ , carriers occupy higher-potential-energy states less often, exponentially. The gradient in $n$ is caused entirely by the potential.

Step 5 — Substitute and cancel. $n\mu_n \frac{d\psi}{dx} = D_n \cdot n\frac{q}{k_BT}\frac{d\psi}{dx}$ Cancel $n\,\dfrac{d\psi}{dx}$ from both sides: $\boxed{\frac{D_n}{\mu_n} = \frac{k_BT}{q}}$

Why this step? Everything position-dependent cancels — the ratio is a pure constant set only by temperature. The identical argument for holes gives $D_p/\mu_p = k_BT/q$ .

Figure — Einstein relation between mobility and diffusion

WHY it's true, intuitively (kinetic view)

Worked examples

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine kids in a playground. If a teacher blows a wind (a "field"), kids drift the same way — that's drift/mobility. Even with no wind, kids running around randomly slowly spread from a crowded corner to empty space — that's diffusion. The same running legs cause both! So a kid who spreads out fast (big $D$ ) is also a kid the wind pushes easily (big $\mu$ ). How hot/energetic the kids are ( $k_BT$ ) sets how strong the random spreading is compared to the wind push. That's the whole secret.

Active recall

Einstein relation formula

D/\mu = k_BT/q

(thermal voltage

V_T

)

What condition is used to derive it

Zero net current in thermal equilibrium — drift and diffusion currents cancel

Value of thermal voltage

k_BT/q

at 300 K

≈ 25.9 mV (~0.026 V)

Physical input in the derivation (Step 4)

Boltzmann distribution:

n = n_0\exp(q\psi/k_BT)

Why is

D/\mu

independent of scattering details

Same

\tau

and

m^*

govern both drift and diffusion, so they cancel in the ratio

Express

D_n

in terms of

\mu_n

D_n = (k_BT/q)\,\mu_n

Common unit error to avoid

Using

kT

instead of

kT/q

;

D/\mu

must be in volts

When does the classical Einstein relation FAIL

Degenerate (heavily doped) semiconductors — need Fermi–Dirac; then

D/\mu > k_BT/q

D_n

for Si electrons (

\mu_n=1350

) at 300 K

0.0259\times1350 \approx 35\ \text{cm}^2/\text{s}

How does

D/\mu

scale with temperature

Linearly,

D/\mu \propto T

Connections

Drift current and mobility
Diffusion current and Fick's law
Boltzmann distribution in semiconductors
Thermal voltage V_T
Continuity equation and carrier transport
Degenerate semiconductors & generalized Einstein relation
Fermi-Dirac vs Maxwell-Boltzmann statistics

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, semiconductor ke andar carriers do wajah se chalte hain. Ek to drift — jab electric field lagta hai to field unko push karta hai, aur ye kitni aasani se hota hai wo mobility $\mu$ batati hai. Doosra hai diffusion — jahan carriers zyada crowded hain wahan se kam wale region me apne aap random thermal motion se phail jaate hain, aur uski strength diffusion coefficient $D$ deti hai. Dono alag lagte hain, par asli baat ye hai ki dono ke peeche same collisions (scattering) hain. Isliye $\mu$ aur $D$ independent nahi ho sakte — inka rishta fixed hai.

Ye rishta hai Einstein relation: $D/\mu = k_BT/q$ . Right side ko thermal voltage $V_T$ kehte hain, jo 300 K par lagbhag 26 mV hoti hai. Derivation ka trick simple hai: equilibrium me (koi battery nahi) net current zero hona chahiye, matlab drift current aur diffusion current exactly cancel. Ismein Boltzmann distribution $n = n_0 e^{q\psi/kT}$ daalo, sab position-dependent cheezein cancel ho jaati hain, aur bachta hai sirf $k_BT/q$ .

Practical faayda: agar aapko mobility pata hai, to bas usko $0.0259$ se multiply karo aur $D$ mil gaya. Jaise Si electrons ka $\mu=1350$ ho to $D \approx 35\ \text{cm}^2/\text{s}$ . Yaad rakho — $kT$ nahi, $kT/q$ lena hai warna units galat ho jayenge. Aur ye classical formula sirf non-degenerate (halka doped) material me chalta hai; bahut zyada doping par Fermi-Dirac wala generalized version use karna padta hai.

Einstein relation between mobility and diffusion

HOW do we derive it? (from first principles)

WHY it's true, intuitively (kinetic view)

Worked examples

Common mistakes

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Band Theory & Carrier Physics

Connections