Einstein relation between mobility and diffusion: Worked examples

Before anything else, three plain-word reminders so every symbol below is earned:

Everything on this page is one master equation used in different directions:

The scenario matrix

Every problem this topic can pose falls into one of these cells. The examples below are labelled with the cell they cover, so together they leave no scenario unshown.

#	Cell (case class)	What is tricky about it	Example
A	Forward: $μ \to D$ (electrons)	plain multiply, get units right	Ex 1
B	Forward: $μ \to D$ (holes)	holes obey the same relation	Ex 2
C	Reverse: $D \to μ$	divide, don't multiply	Ex 3
D	Ratio $\to$ temperature	invert $V_{T} = k_{B} T / q$ to find $T$	Ex 4
E	Temperature scaling	$D / μ \propto T$ ; watch what stays fixed	Ex 5
F	Degenerate / limiting case	classical form fails; ratio $> V_{T}$	Ex 6
G	Unit trap ( $k T$ vs $k T / q$ )	forgetting the $/ q$	Ex 7
H	Real-world device word problem	pick the right $D$ , build a length	Ex 8
I	Exam twist: mixed $D_{n}, D_{p}$ ratio	$μ$ 's and $D$ 's ratios are equal	Ex 9

Constants used everywhere: $k_{B} = 1.38 \times 1 0^{- 23} J/K$ , $q = 1.6 \times 1 0^{- 19} C$ .

Figure below (alt text): a straight line through the origin on axes "mobility $μ$ " (horizontal) versus "diffusion coefficient $D$ " (vertical). The line has slope $V_{T} = 0.0259 V$ , and three worked points sit on it — Example 1 electrons at $μ = 1350, D = 35$ , Example 2 holes at $μ = 480, D = 12.4$ , and Example 3 (reverse) at $μ = 965, D = 25$ . Dashed guide lines show reading $D$ off from $μ$ and back.

Figure — Einstein relation between mobility and diffusion

The straight line in the figure is the Einstein relation $D = V_{T} μ$ : the horizontal axis is mobility $μ$ , the vertical axis is diffusion coefficient $D$ , and the slope of the line is exactly the thermal voltage $V_{T} = 0.0259 V$ . Every example is either "go up from the $μ$ -axis to the line, then across to read $D$ " (forward, Ex 1 and Ex 2, orange and plum dots), "start from the $D$ -axis and come back to read $μ$ " (reverse, Ex 3, the square marker), or "tilt the whole line by changing $T$ " (temperature). Keep this picture in your head — each example below points you back to the specific dot or motion on this line.

Worked examples

Cell A — Forward, electrons

Worked example Example 1 —

$D_{n}$ from $μ_{n}$ Silicon electrons at $300 K$ : $μ_{n} = 1350 cm^{2} / (V\cdots)$ . Find $D_{n}$ .

Forecast: will $D_{n}$ be bigger or smaller than $μ_{n}$ as a bare number? (Hint: $V_{T} \approx 0.026$ , so multiplying by it shrinks the number.)

Write the master equation: $D_{n} = V_{T} μ_{n}$ . Why this step? We know $μ$ and want $D$ — the forward direction, so multiply. On the figure this is the orange dot: go up from $μ = 1350$ on the horizontal axis to the teal line.
Insert $V_{T} = 0.0259 V$ (the $300 K$ value): $D_{n} = 0.0259 \times 1350$ . Why this step? At $300 K$ the thermal voltage is a fixed known number.
$D_{n} = 34.97 \approx 35.0 cm^{2} / s$ . Why this step? Just arithmetic — reading straight across from the line to the vertical $D$ -axis gives $35$ .

Verify: Units: $V \times \frac{cm ^{2}}{V\cdots} = \frac{cm ^{2}}{s}$ ✓ (volts cancel, exactly the units of $D$ ). Textbook Si value is $D_{n} \approx 35 cm^{2} / s$ ✓.

Cell B — Forward, holes

Worked example Example 2 —

$D_{p}$ for holes Silicon holes at $300 K$ : $μ_{p} = 480 cm^{2} / (V\cdots)$ . Find $D_{p}$ .

Forecast: Holes are positive and heavier (lower $μ$ ). Does the relation change form for holes? (No — same $V_{T}$ .)

Same master equation, hole version: $D_{p} = V_{T} μ_{p}$ . Why this step? The parent derivation gave $D_{p} / μ_{p} = k_{B} T / q$ too — the sign of the charge cancels in the ratio, so holes obey the identical rule. On the figure this is the plum dot: same teal line, just a point lower down at $μ = 480$ .
$D_{p} = 0.0259 \times 480 = 12.43 cm^{2} / s$ . Why this step? Multiply — forward direction again.

Verify: $D_{p} \approx 12.4 cm^{2} / s$ matches textbook Si holes ( $\approx 12$ ). Sanity: holes drift slower ( $μ_{p} < μ_{n}$ ), so they also diffuse slower ( $D_{p} < D_{n}$ ) — same collision physics for both ✓. Both dots sit on the same line, confirming one $V_{T}$ for both carriers.

Cell C — Reverse direction

Worked example Example 3 —

$μ$ from $D$ A measurement gives $D_{n} = 25 cm^{2} / s$ at $300 K$ . What is $μ_{n}$ ?

Forecast: We now go backwards. Will the number get bigger or smaller? (Dividing by $0.026$ makes it much bigger.)

Rearrange master equation: $μ_{n} = \frac{D _{n}}{V _{T}}$ . Why this step? We know $D$ , want $μ$ — divide, don't multiply. This is the single most common flip-up. On the figure this is the square marker: start on the vertical $D$ -axis at $25$ and read back down to the horizontal axis.
$μ_{n} = \frac{25}{0.0259} = 965.3 cm^{2} / (V\cdots)$ . Why this step? We substitute the known $D_{n} = 25$ and the $300 K$ thermal voltage $V_{T} = 0.0259 V$ so that only $μ_{n}$ is left; dividing the two undoes the forward multiplication, giving the mobility that produces this measured diffusion coefficient.

Verify: Units: $\frac{cm ^{2} / s}{V} = \frac{cm ^{2}}{V\cdots}$ ✓ — exactly mobility units. Cross-check forward: $0.0259 \times 965.3 = 25.0$ ✓.

Cell D — Ratio gives temperature

Worked example Example 4 — Find

$T$ from a measured ratio An experiment reports $D / μ = 0.0345 V$ . Assuming non-degenerate material, what temperature is the sample at?

Forecast: $0.0345 V$ is bigger than the $300 K$ value $0.0259 V$ . Since $V_{T} \propto T$ , the sample must be hotter than $300 K$ .

Set $\frac{D}{μ} = V_{T} = \frac{k _{B} T}{q}$ . Why this step? The measured ratio is the thermal voltage (non-degenerate assumption). On the figure this ratio is the slope of the line — a steeper slope means a hotter sample.
Solve for $T$ : $T = \frac{q}{k _{B}} \cdot \frac{D}{μ} = \frac{q V _{T}}{k _{B}}$ . Why this step? Invert $V_{T} = k_{B} T / q$ ; the only unknown left is $T$ . Here $V_{T} = 0.0345 V$ is the sample's own thermal voltage, not the $300 K$ number.
$T = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 0.0345 )}{1.38 \times 1 0 ^{- 23}} = 400.0 K$ . Why this step? Numbers in.

Verify: $400 K > 300 K$ , matching the "hotter" forecast ✓. Cross-check: $V_{T} (400) = \frac{k _{B} \cdot 400}{q} = 0.0345 V$ ✓ (see Boltzmann distribution in semiconductors for why the exponent scales with $T$ ).

Cell E — Temperature scaling

Worked example Example 5 — How

$D$ changes with $T$ Take Si electrons with $D_{n} = 35 cm^{2} / s$ at $300 K$ . Pretend $μ$ is fixed. Find $D_{n}$ at $250 K$ .

Forecast: Cooler sample $\Rightarrow$ less thermal jiggling $\Rightarrow$ smaller $D$ .

Ratio trick: with $μ$ fixed, $\frac{D ( T _{2} )}{D ( T _{1} )} = \frac{V _{T} ( T _{2} )}{V _{T} ( T _{1} )} = \frac{T _{2}}{T _{1}}$ . Why this step? $D = V_{T} μ$ and $V_{T} \propto T$ (that is exactly why $V_{T}$ is temperature-dependent), so with $μ$ constant the $μ$ cancels — only the temperature ratio survives. On the figure, changing $T$ tilts the whole line: cooling flattens its slope.
$D_{n} (250) = 35 \times \frac{250}{300} = 29.17 cm^{2} / s$ . Why this step? Multiply by the temperature ratio.

Verify: $29.17 < 35$ ✓ (cooler = weaker diffusion, matches forecast). Ratio $250/300 = 0.833$ , and $35 \times 0.833 = 29.17$ ✓.

Cell F — Degenerate / limiting case (where it breaks)

First, one plain-word definition so no new symbol sneaks in undefined:

Definition The degeneracy factor

$g$ (in plain words) When a semiconductor is so heavily doped that carriers are packed shoulder-to-shoulder, the Pauli exclusion principle forces them to obey Fermi–Dirac statistics instead of the simple Boltzmann exponential used in the parent's Step 4. The correction is captured by a single dimensionless number we will call the degeneracy factor $g \geq 1$ : $\frac{D}{μ} = g V_{T}, g \geq 1.$

In lightly doped (non-degenerate) material $g = 1$ and we recover the classic $D / μ = V_{T}$ .
The more degenerate the material, the larger $g$ grows above $1$ . The exact value of $g$ comes from a ratio of two Fermi–Dirac integrals, but for this page you are simply handed $g$ as a number — you never compute the integrals yourself. See Degenerate semiconductors & generalized Einstein relation and Fermi-Dirac vs Maxwell-Boltzmann statistics for where $g$ comes from.

Worked example Example 6 — Heavily doped: classical form fails

A heavily doped $n^{+}$ sample at $300 K$ is given a degeneracy factor $g = 1.6$ . Compute the actual $D / μ$ and compare to the naive $25.9 mV$ .

Forecast: Does the ratio go up or down when carriers are packed so tightly they must obey the Pauli exclusion principle? (Up — since $g \geq 1$ .)

Naive (Boltzmann) value: $V_{T} = 0.0259 V$ (the $300 K$ thermal voltage). Why this step? This is what the classical Einstein relation would predict, i.e. the $g = 1$ case — the ordinary teal line of the figure.
Generalized relation: $\frac{D}{μ} = g V_{T}$ with the handed value $g = 1.6$ . Why this step? Degenerate carriers occupy states up to the Fermi level; the Boltzmann exponential in the parent's Step 4 is no longer valid, so the ratio picks up the factor $g > 1$ defined above. On the figure this is a steeper line than the drawn one — the slope is $1.6 \times$ larger.
Actual $\frac{D}{μ} = 1.6 \times 0.0259 = 0.04144 V = 41.4 mV$ . Why this step? Scale the classical value by the degeneracy factor.

Verify: $41.4 mV > 25.9 mV$ ✓ — the generalized ratio always exceeds $k_{B} T / q$ in degenerate material, exactly the parent's warning "Einstein relation always holds → NO." This is the limiting/degenerate cell: the one place the boxed formula must not be used blindly.

Cell G — Unit trap

Worked example Example 7 —

$k T$ vs $k T / q$ A student writes $D / μ = k T = (1.38 \times 1 0^{- 23}) (300) = 4.14 \times 1 0^{- 21}$ and is baffled. What went wrong, and what is the right value?

Forecast: $D / μ$ must be a voltage (volts). Is $4.14 \times 1 0^{- 21}$ in volts? (No — it's joules, an energy.)

Compute the energy: $k_{B} T = (1.38 \times 1 0^{- 23}) (300) = 4.14 \times 1 0^{- 21} J$ . Why this step? This is the thermal energy, not a voltage.
Divide by charge to get volts: $V_{T} = \frac{k _{B} T}{q} = \frac{4.14 \times 1 0 ^{- 21}}{1.6 \times 1 0 ^{- 19}}$ . Why this step? $D / μ$ has units of volts, and voltage = energy per charge, so you must divide by $q$ .
$V_{T} = 0.0259 V = 25.9 mV$ . Why this step? Correct value — this is the slope the figure's line actually has.

Verify: In electron-volts, $k_{B} T = 0.0259 eV$ ; dividing an energy in eV by "one electron charge" gives the same number in volts, $0.0259 V$ ✓. The missing $/ q$ was the whole error.

Cell H — Real-world device word problem

Before this example, one new symbol must be earned:

Worked example Example 8 — Diffusion length in a transistor base

In an npn transistor base, minority electrons at $300 K$ have $μ_{n} = 1000 cm^{2} / (V\cdots)$ and lifetime $τ = 1 0^{- 7} s$ . The diffusion length is $L_{n} = D_{n} τ$ . Find $L_{n}$ in micrometres.

Forecast: Diffusion lengths in transistor bases are typically a few to tens of micrometres. Watch this land in that range.

Get $D_{n}$ via Einstein: $D_{n} = V_{T} μ_{n} = 0.0259 \times 1000 = 25.9 cm^{2} / s$ . Why this step? The problem gives $μ$ , but the length needs $D$ — Einstein bridges them (the forward move on the figure's line).
Diffusion length: $L_{n} = D_{n} τ = 25.9 \times 1 0^{- 7} cm$ . Why this step? $L_{n} = D τ$ is how far a carrier random-walks during its lifetime $τ$ before recombining (from the Continuity equation and carrier transport). Units: $\frac{cm ^{2}}{s} \cdot s = cm$ .
$L_{n} = 2.59 \times 1 0^{- 6} = 1.609 \times 1 0^{- 3} cm = 16.1 μ m$ . Why this step? Convert: $1 cm = 1 0^{4} μ m$ , so $1.609 \times 1 0^{- 3} cm = 16.09 μ m$ .

Verify: $16 μ m$ is a physically sensible base-diffusion length ✓. Units checked at every step ✓. Without the Einstein relation we could not have turned the measured $μ$ into the $D$ this formula needs.

Cell I — Exam twist: mixed ratios

Worked example Example 9 — Ratio of ratios

Given $D_{n} = 35$ , $D_{p} = 12$ , $μ_{n} = 1350 cm^{2} / (V\cdots)$ (all at $300 K$ ), find $μ_{p}$ without using $V_{T}$ directly.

Forecast: Since $D / μ$ is the same $V_{T}$ for electrons and holes, the four quantities are linked by one proportion.

Equal ratios: $\frac{D _{n}}{μ _{n}} = \frac{D _{p}}{μ _{p}}$ (both equal $V_{T}$ ). Why this step? The parent proved $D_{n} / μ_{n} = D_{p} / μ_{p} = k_{B} T / q$ — identical for both carrier types. On the figure, both carriers live on the same line, so their $D / μ$ slopes match.
Solve for $μ_{p}$ : $μ_{p} = μ_{n} \cdot \frac{D _{p}}{D _{n}} = 1350 \times \frac{12}{35}$ . Why this step? Cross-multiply the equal-ratio equation; $V_{T}$ cancels entirely.
$μ_{p} = 462.9 cm^{2} / (V\cdots)$ . Why this step? Arithmetic.

Verify: Cross-check via $V_{T}$ : $μ_{p} = D_{p} / V_{T} = 12/0.0259 = 463.3$ ✓ (matches within rounding). Sensible: $μ_{p} < μ_{n}$ , holes are less mobile ✓.

Recall Which cell is this? (self-quiz — reveal answers)

You are told $D / μ = 0.031 V$ and asked for $T$ . Answer ::: Cell D — invert $V_{T} = k_{B} T / q$ to get $T = q V_{T} / k_{B} \approx 359 K$ .

You are told $μ_{p}$ and asked for $D_{p}$ . Answer ::: Cell B — forward, holes: multiply by $V_{T}$ .

You are told the material is $n^{++}$ degenerate. Answer ::: Cell F — use the generalized ratio $g V_{T}$ with $g > 1$ , NOT the boxed formula.

Your answer for $D / μ$ came out as $\sim 1 0^{- 21}$ . Answer ::: Cell G — you used $k T$ (joules) instead of $k T / q$ (volts). Divide by $q$ .

Connections

2.1.10 Einstein relation between mobility and diffusion (Hinglish)
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