Diffusion current and concentration gradient: Worked examples

This page is the drill hall for Diffusion current and concentration gradient. The parent note taught you the ideas. Here we hit every case the topic can throw at you — every sign, every degenerate input, every limiting shape — so that no exam scenario is new.

Before we start, one reminder of the two boxed results we lean on constantly (built in the parent):

The scenario matrix

Every diffusion problem you can be handed is built from a small set of independent choices. Enumerate them and we can guarantee coverage:

Cell	Carrier	Profile shape	Sign of slope $d n / d x$	What it tests
A	electrons	linear (falling)	negative	basic $J_{n}$ , sign of conventional current
B	holes	linear (falling)	negative	why $J_{p}$ sign differs from $J_{n}$
C	either	uniform ( $d n / d x = 0$ )	zero	the degenerate case — no current
D	electrons	linear rising	positive	flip the slope sign, check current flips
E	holes	exponential decay	negative	real-diode profile, differentiate $e^{- x / L}$
F	holes & electrons	Einstein $D \leftrightarrow μ$	zero	convert mobility to $D$ (no gradient at all)
G	both together	opposing gradients	mixed	electron/hole currents add
G2	both together	same-sign gradients	matched	electron/hole currents partially cancel
H	electrons	limiting shape	$\to \infty$	what happens as the region gets thin

The examples below hit each cell one by one. Each cell is labelled in its example title. Figure s03 (shown at the very end) is a strip of five thumbnail sketches — one carrier profile plus its current arrow for each of cells A, C, D, G and G2 — so you can see every case at a glance.

Example A — falling electrons, linear

Forecast: guess first — is the conventional current in $+ x$ or $- x$ ? (Electrons crowd on the left and diffuse rightward…)

Convert $x$ to cm. Why this step? Every $D$ is quoted in $cm^{2} / s$ , so lengths must be cm or the units clash. $2 μ m = 2 \times 1 0^{- 4} cm$ .
Slope. Why this step? Linear profile ⇒ one constant slope covers the whole region. $\frac{d n}{d x} = \frac{1 0 ^{16} - 1 0 ^{17}}{2 \times 1 0 ^{- 4}} = \frac{- 9 \times 1 0 ^{16}}{2 \times 1 0 ^{- 4}} = - 4.5 \times 1 0^{20} cm^{- 4}$
Apply $J_{n} = q D_{n} d n / d x$ . Why this step? This is the boxed electron law. $J_{n} = (1.6 \times 1 0^{- 19}) (25) (- 4.5 \times 1 0^{20}) = - 1800 A/cm^{2}$

Verify: Units: $C \cdot \frac{cm ^{2}}{s} \cdot \frac{1}{cm ^{4}} = \frac{C}{s \cdot cm ^{2}} = A/cm^{2}$ ✓. Sign check against our convention: $J_{n} < 0$ means current in $- x$ — and indeed electrons pile up on the left, diffuse in $+ x$ , and being negative that is conventional current in $- x$ . Matches. Forecast answer: $- x$ .

Figure s01 below draws exactly this: the blue density curve dropping left-to-right, the green arrow showing electrons physically diffusing in $+ x$ , and the red arrow showing the resulting conventional current pointing in $- x$ . Look at how the red and green arrows point opposite ways — that opposition is the whole reason $J_{n}$ carries a leading $+$ yet comes out negative here.

Figure — Diffusion current and concentration gradient

Example B — falling holes, linear

Forecast: the slope is identical to Cell A (negative). Will $J_{p}$ come out positive or negative? Guess before reading.

Slope (same as A). $\frac{d p}{d x} = - 4.5 \times 1 0^{20} cm^{- 4}$ . Why this step? Identical geometry — reuse it.
Apply $J_{p} = - q D_{p} d p / d x$ . Why this step? Holes use the minus law because their charge is $+ q$ . $J_{p} = - (1.6 \times 1 0^{- 19}) (12) (- 4.5 \times 1 0^{20}) = + 864 A/cm^{2}$

Verify: Two minus signs multiplied → plus, so by our convention current flows in $+ x$ . Physically: holes crowd on the left, diffuse in $+ x$ ; holes are positive, so conventional current is also $+ x$ → positive. ✓ This is the whole point of the sign difference: same physical drift as Cell A, opposite current sign.

Example C — uniform density (the degenerate case)

Forecast: there are enormously many electrons. Big current? Guess.

Slope. Why this step? Diffusion cares about the change, not the amount. Uniform ⇒ $\frac{d n}{d x} = 0$ .
Apply. $J_{n} = q D_{n} \cdot 0 = 0$ .

Verify: $J_{n} = (1.6 \times 1 0^{- 19}) (25) (0) = 0 A/cm^{2}$ ✓. This is the trap warned about in the parent's [!mistake]: density $n$ does not drive diffusion — its slope does. Dense but flat = no diffusion current. In the s03 strip this is the flat middle panel with no current arrow at all.

Example D — rising electrons (slope flips sign)

Forecast: In Cell A the answer was $- 1800$ . Same numbers, mirror-image profile. Predict the sign now.

Slope. Why this step? Now the higher value is on the right, so the slope is positive. $\frac{d n}{d x} = \frac{1 0 ^{17} - 1 0 ^{16}}{2 \times 1 0 ^{- 4}} = + 4.5 \times 1 0^{20} cm^{- 4}$
Apply. $J_{n} = (1.6 \times 1 0^{- 19}) (25) (+ 4.5 \times 1 0^{20}) = + 1800 A/cm^{2}$ .

Verify: Exactly $-$ (Cell A) → $+$ : current now flows in $+ x$ . Physically electrons crowd on the right and diffuse in $- x$ ; negative charge moving $- x$ = conventional current in $+ x$ → positive. ✓ Every sign is self-consistent. In the s03 strip this is the rising-blue panel whose red current arrow points right, mirror-image of Cell A.

Example E — exponential decay (the real-diode profile)

Forecast: as we move deeper ( $x$ grows), does the current get bigger or smaller? Guess.

Differentiate the exponential. Why this step? The slope of a curve is its derivative; $\frac{d}{d x} e^{- x / L_{p}} = - \frac{1}{L _{p}} e^{- x / L_{p}}$ . $\frac{d p}{d x} = - \frac{δ p ( 0 )}{L _{p}} e^{- x / L_{p}}$
Evaluate at $x = 0$ ( $e^{0} = 1$ ). Why this step? We want the current entering the region. $\frac{d p}{d x}_{0} = - \frac{1 0 ^{14}}{5 \times 1 0 ^{- 4}} = - 2 \times 1 0^{17} cm^{- 4}$ $J_{p} (0) = - q D_{p} \frac{d p}{d x} = - (1.6 \times 1 0^{- 19}) (12) (- 2 \times 1 0^{17}) = + 0.384 A/cm^{2}$
Evaluate at $x = L_{p}$ ( $e^{- 1} = 0.368$ ). Why this step? To see the decay. $J_{p} (L_{p}) = 0.384 \times e^{- 1} = 0.384 \times 0.3679 = + 0.1413 A/cm^{2}$

Verify: Units as before → A/cm² ✓; both positive, so current runs in $+ x$ everywhere. Current shrinks by the factor $e^{- 1}$ over one diffusion length $L_{p}$ — exactly the decay rate of the profile itself, since both $δ p$ and its slope scale as $e^{- x / L_{p}}$ . Forecast answer: smaller deeper in. ✓ (See Fick's laws of diffusion for why the profile itself is exponential.)

Figure s02 plots this decay. The dashed gray line is the slope (steepest) at $x = 0$ ; the two marked dots at $x = 0$ and $x = L_{p}$ carry the current values we just computed. Notice the curve and its tangent both flatten as $x$ grows — that flattening is the current shrinking.

Example F — Einstein relation (no gradient at all)

Forecast: which one — $D_{n}$ or $D_{p}$ — comes out larger? Guess from the mobilities.

Thermal voltage. Why this step? The Einstein relation $D / μ = V_{T}$ needs Thermal voltage $V_T$. At 300 K, $V_{T} = k T / q = 0.02585 V$ .
Multiply. Why this step? $D = μ V_{T}$ . $D_{p} = 450 \times 0.02585 = 11.63 cm^{2} / s$ $D_{n} = 1350 \times 0.02585 = 34.90 cm^{2} / s$

Verify: Units: $\frac{cm ^{2}}{V\cdots} \cdot V = \frac{cm ^{2}}{s}$ ✓. Both $D$ and $μ$ come from the same collisions (Drift current and mobility), so higher mobility ⇒ higher $D$ . Forecast answer: $D_{n}$ larger. ✓

Example G — both carriers, opposing gradients (currents ADD)

Forecast: electrons and holes have opposite slopes here — will their currents add or fight? Guess.

Electron piece. Why this step? Use the $+$ law. $J_{n} = (1.6 \times 1 0^{- 19}) (25) (- 4.5 \times 1 0^{20}) = - 1800 A/cm^{2}$
Hole piece. Why this step? Use the $-$ law. $J_{p} = - (1.6 \times 1 0^{- 19}) (12) (+ 3.0 \times 1 0^{20}) = - 576 A/cm^{2}$
Sum. Why this step? Total current is the sum of all carrier contributions. $J = - 1800 + (- 576) = - 2376 A/cm^{2}$

Verify: Both pieces came out negative even though the slopes had opposite signs — the sign flip in $J_{p}$ 's formula makes them add. So the total current here is $- 2376 A/cm^{2}$ (in $- x$ ), larger in magnitude than either alone. ✓ Forecast answer: they add. In the s03 strip this is the panel with two same-direction (red) current arrows stacked.

Example G2 — both carriers, same-sign gradients (currents PARTIALLY CANCEL)

Forecast: this time both slopes point the same way. Does that mean the currents help each other — or oppose? Guess (the surprise is the point).

Electron piece. Why this step? $+$ law. $J_{n} = (1.6 \times 1 0^{- 19}) (25) (- 4.5 \times 1 0^{20}) = - 1800 A/cm^{2}$
Hole piece. Why this step? $-$ law, and now $d p / d x$ is negative. $J_{p} = - (1.6 \times 1 0^{- 19}) (12) (- 4.5 \times 1 0^{20}) = + 864 A/cm^{2}$
Sum. Why this step? Add the contributions. $J = - 1800 + (+ 864) = - 936 A/cm^{2}$

Verify: Now the two pieces have opposite signs and partially cancel: $∣ - 936 ∣ < ∣ - 1800 ∣$ . ✓ The lesson is the mirror image of Cell G — same-sign gradients make the currents oppose, because the charge-sign flip in $J_{p}$ turns matched slopes into opposed currents. Electrons and holes both diffuse the same physical way ( $+ x$ ), but their opposite charge sends their conventional currents opposite ways, so they subtract. In the s03 strip this is the final panel: two red arrows pointing against each other, the net one shorter.

Example H — the thin-region limit

Forecast: if we squeeze the same concentration drop into a thinner slab, does the current grow, shrink, or stay put?

General expression. Why this step? Slope $= \frac{- Δ n}{W}$ , so $J_{n} = q D_{n} \frac{- Δ n}{W}$ . $J_{n} = - (1.6 \times 1 0^{- 19}) (25) (9 \times 1 0^{16}) \cdot \frac{1}{W} = - \frac{3.6 \times 1 0 ^{- 1}}{W} (with W in cm)$
$W = 2 μ m = 2 \times 1 0^{- 4}$ cm. Why this step? Plug in the first width to get a concrete number to compare against. $J_{n} = - 0.36/ (2 \times 1 0^{- 4}) = - 1800 A/cm^{2}$ .
$W = 0.2 μ m = 2 \times 1 0^{- 5}$ cm. Why this step? Shrink the width by a factor of 10 and see how the current responds. $J_{n} = - 0.36/ (2 \times 1 0^{- 5}) = - 18000 A/cm^{2}$ .
Limit $W \to 0$ . Why this step? Push the shrinking to its extreme; $J_{n} \propto 1/ W \to - \infty$ : an infinitely steep gradient gives unbounded current.

Verify: Shrinking the width $\times 10$ multiplied the current magnitude $\times 10$ ✓ ( $- 1800 \to - 18000$ ). Physically the slope is what matters: pack the same drop into a thinner slab and the slope — hence the current — blows up. Forecast answer: grows without bound. (In reality other physics like Continuity equation recombination and finite carrier supply cap it, but the diffusion formula itself is unbounded.)

The whole matrix at a glance

The strip below sketches five key cells side by side — profile (blue) on top, resulting conventional-current arrow (red) below. Reading it left to right is the fastest way to internalise how the current arrow tracks the slope and the carrier's charge sign.

Recall

Recall Which cell breaks each intuition? (hide and answer)

Uniform dense bar gives what current? ::: Zero — slope is zero (Cell C). Same falling profile: why do $J_{n}$ and $J_{p}$ differ in sign? ::: The $\pm q$ charge flips the current sign (Cells A vs B). Opposite carrier slopes — do the currents add or cancel? ::: They add (Cell G). Same-sign carrier slopes — do the currents add or cancel? ::: They partially cancel (Cell G2). A positive $J$ means current flows which way? ::: In $+ x$ (to the right), by our convention. Squeeze the same $Δ n$ into half the width — current does what? ::: Doubles; $J \propto 1/ W$ (Cell H).