Question bank — Einstein relation between mobility and diffusion

False. They ride on the same scattering time $\tau$ and effective mass $m^{\ast }$ (see Figure 2); fixing one and the temperature fixes the other through $D=\mu k_{B}T/q$.

True. $D/\mu =k_{B}T/q\propto T$, so hotter carriers jiggle harder and $D$ climbs even if the field-response $\mu$ is held fixed.

False. It is derived at zero net current in thermal equilibrium (no battery) — the balanced arrows of Figure 1 — where drift and diffusion cancel; that cancellation is exactly what pins the ratio.

False. It has units of volts. $k_{B}T$ is energy, but dividing by charge $q$ turns energy-per-charge into volts, matching $D/\mu$.

False. The same random collisions cause both (Figure 2) — drift is that jiggling biased by a field, diffusion is the same jiggling unbiased. That shared origin is why $\tau$ and $m^{\ast }$ cancel.

False. It assumes non-degenerate (Boltzmann) statistics. In heavily doped/degenerate material you need the generalized form, and $D/\mu$ then exceeds $k_{B}T/q$.

True. Both give $D/\mu =k_{B}T/q$; the ratio is universal, even though ${\mu }_n\ne {\mu }_p$ and $D_n\ne D_p$ individually.

True (in the non-degenerate regime). $D=(k_{B}T/q)\mu$, so with $T$ fixed, $D$ scales linearly with $\mu$.

False. $\tau$ (and $m^{\ast }$) cancel in the ratio; the scattering physics affects drift and diffusion identically, so it drops out entirely.

They forgot to divide by $q$. $k_{B}T\approx 0.0259\text{eV}$ is an energy; the thermal voltage is $k_{B}T/q\approx 0.0259\text{V}$. An answer in eV cannot be a voltage.

A sign slip. Electron charge is $-q$ and particle flux is $-D_{n}dn/dx$; their product gives $+q{D}_{n}dn/dx$. Two minus signs make a plus. See Diffusion current and Fick's law.

No. Hole charge is $+q$ and hole flux is $-D_{p}dp/dx$, so $J_{p,\text{diff}}=(+q)(-D_{p}dp/dx)=-q{D}_{p}dp/dx$. The hole current carries a minus sign — the opposite of the electron case, because holes are positive.

The diffusion term. $J_n=0$ means drift ($qn{\mu }_{n}E$) is balanced by diffusion ($q{D}_{n}dn/dx$) — the two arrows of Figure 1 — not that $E=0$. The field is opposed by a self-built-up concentration gradient.

The relation constrains the ratio, not $\mu$ alone. In real silicon $\mu$ actually drops with $T$ (more phonon scattering); the ratio still rises because it is set by the same $T$ that appears on top.

The equilibrium carrier distribution (Boltzmann): $n=n_0\exp (q\psi /k_{B}T)$. That physical input, not Fick's law, is what injects the $k_{B}T/q$. See Boltzmann distribution in semiconductors.

The field is the negative slope of potential, $E=-d\psi /dx$. Dropping the minus sign flips the balance and can give a spurious negative ratio.

Because $D\sim v_{th}^2\tau /2\sim (k_{B}T/m^{\ast })\tau$ and $\mu =q\tau /m^{\ast }$ share the same $\tau$ and $m^{\ast }$ (Figure 2). Dividing cancels both, leaving only $k_{B}T/q$.

From the equipartition theorem: each motion direction carries thermal energy $\frac{1}{2}{k}_{B}T=\frac{1}{2}{m}^{\ast }{v}_{th}^2$, so $v_{th}^2=k_{B}T/m^{\ast }$. That's how temperature $T$ physically enters $D$.

In a 1D random walk, a carrier steps distance $\ell =v_{th}\tau$ left or right with equal chance; the standard 1D diffusion result is $D={\ell }^2/(2\tau )=v_{th}^2\tau /2$ — the $2$ counts the two directions. In 3D the same argument spreads motion over three axes and gives $D=v_{th}^2\tau /3$. Either way the $\tau$ and $m^{\ast }$ (via $v_{th}^2$) cancel against $\mu$, so the geometric prefactor never touches the $k_{B}T/q$ ratio.

Equilibrium forces zero net current (Figure 1), so drift and diffusion must cancel exactly. That exact cancellation is a hard constraint that squeezes out the ratio; under bias there's no such clean condition. See Continuity equation and carrier transport.

$D/\mu$ must come out in volts. $k_{B}T$ is energy (joules); one factor of $1/q$ converts energy to energy-per-charge, i.e. volts. Any other power would break the units.

Higher $T$ means carriers carry more thermal energy $\sim k_{B}T$ and jiggle faster ($v_{th}^2=k_{B}T/m^{\ast }$ climbs), so random spreading (governed by $D$) intensifies — captured by $D/\mu \propto T$.

The cancellation of $\tau$ and $m^{\ast }$ happens for each species separately, and the leftover $k_{B}T/q$ contains no species-specific quantity. So the ratio is identical even when ${\mu }_n\ne {\mu }_p$.

In the non-degenerate limit the Fermi–Dirac distribution reduces to Maxwell–Boltzmann, giving an exponential occupation of energy states — valid when carrier levels sit well below the Fermi level.

$V_T=k_{B}T/q\to 0$, so $D\to 0$: no thermal jiggle means no diffusion. Drift ability $\mu$ can remain finite, so the ratio collapses even if $\mu$ doesn't.

The generalized Einstein relation using Fermi–Dirac statistics; $D/\mu$ becomes larger than $k_{B}T/q$ because the exclusion of filled states adds an extra "pressure" driving diffusion. See Degenerate semiconductors & generalized Einstein relation.

The relation $D/\mu =k_{B}T/q$ still holds as a material/temperature property. A zero gradient just means the diffusion current is zero at that instant, not that $D$ or the ratio is zero.

Above. As statistics start to become degenerate, $D/\mu$ rises above $k_{B}T/q$; it never dips below in the standard picture.

Yes. As long as statistics stay non-degenerate, the ratio depends only on $T$; combining scattering channels changes $\tau$ (hence $\mu$ and $D$ individually) but not $D/\mu$.

The equilibrium derivation no longer strictly applies; carriers may have an effective temperature above the lattice, so $D/\mu$ uses that elevated $T$ and can exceed the equilibrium $k_{B}T/q$.