Visual walkthrough — Diffusion current and concentration gradient

Nothing is assumed. If you have never seen a slope, a derivative, or the letter $n$ used for "how crowded", start at line one — we define each thing the moment it appears.

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Step 1 — Picture "concentration" as crowdedness

WHAT. We are counting particles. Call the number of carriers packed into each tiny box of space the concentration, written $n(x)$. The little $x$ means "at position $x$ along a line", and $n$ is just a headcount per unit volume (units: particles per ${\text{cm}}^3$).

WHY. Before anything can flow, we must have a way to say "here it is crowded, there it is empty." $n(x)$ is exactly that: a number that changes as you walk along $x$.

PICTURE. On the left the dots are packed tight (large $n$); on the right they thin out (small $n$). The height of the purple curve above each point is the crowdedness there.

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Step 2 — Turn "crowdedness changing" into a slope

WHAT. We measure how fast the crowd thins out as we move right. That rate of change is the gradient, written $\frac{dn}{dx}$.

WHY. Notice: if $n$ is the same everywhere (flat curve), nothing special happens — no reason for a net flow. What drives flow is the change in $n$ across space. So the important quantity is not $n$ itself but its steepness.

Read the symbol literally:

• $dn$ = a tiny change in the headcount,
• $dx$ = a tiny step you take to the right,
• $\frac{dn}{dx}$ = (change in crowd) ÷ (step size) = the slope of the purple curve.

Why a derivative and not just subtraction? Because we want the slope at a single point, not averaged over a big region. A derivative is the tool that asks "how steep is the curve right here, in the limit of an infinitesimally small step?" — exactly the local steepness a particle at that spot actually feels.

PICTURE. The orange tangent line touches the curve; its tilt is $\frac{dn}{dx}$. Going downhill (crowd shrinking as $x$ grows) means the slope is negative.

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Step 3 — The hopping model: how a single carrier actually moves

WHAT. Zoom into one particle. It doesn't glide — it hops. Every collision time $\tau$ it jumps left or right by one mean free path $\ell$, with a 50/50 coin flip on direction.

WHY. This is the honest microscopic truth: at room temperature carriers rattle around from thermal energy. We introduce two named quantities:

• $\ell$ (mean free path) = the typical distance travelled before bumping into something,
• $\tau$ (collision time) = the typical time between bumps.

From these we define a speed: $v_{th}=\underset{\text{distance per hop}}{\underbrace{\ell }}/\underset{\text{time per hop}}{\underbrace{\tau }}$ Read it as "distance over time" — that is what speed always means. $v_{th}$ is the thermal speed, the pace of the random jiggle.

PICTURE. One carrier's zig-zag path: equal-length steps $\ell$, direction decided by a coin at each dot.

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Step 4 — Count who crosses a dividing plane

WHAT. Plant an imaginary vertical plane at position $x_0$. Ask: in one collision time, how many carriers cross it going right, and how many going left?

WHY. A "current" is a net crossing rate at some plane. So we must literally tally crossings. Only carriers within one hop $\ell$ of the plane can reach it in time $\tau$; by the coin-flip symmetry, exactly half of those aim at the plane.

Split space into two slabs hugging the plane:

• Left slab, centred at $x_0-\frac{\ell }{2}$: sends carriers rightward.
• Right slab, centred at $x_0+\frac{\ell }{2}$: sends carriers leftward.

Rightward crossers (per unit area, per unit time): $F_{\to }=\underset{\text{half aim right}}{\underbrace{\frac{1}{2}}}\underset{\text{how crowded the left slab is}}{\underbrace{n\left(x_0-\frac{\ell }{2}\right)}}\underset{\text{how fast they go}}{\underbrace{v_{th}}}$

Leftward crossers: $F_{\leftarrow }=\frac{1}{2}\underset{\text{how crowded the right slab is}}{\underbrace{n\left(x_0+\frac{\ell }{2}\right)}}{v}_{th}$

PICTURE. Two shaded slabs on either side of the plane; magenta arrows for rightward crossers (thick, because the left is crowded) and thinner violet arrows for leftward crossers.

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Step 5 — Subtract: the net flux appears

WHAT. Net rightward flux $F$ = (rightward crossers) − (leftward crossers).

WHY. "Net" always means the winner minus the loser. Whatever is left over after cancellation is the real, observable flow.

$F=F_{\to }-F_{\leftarrow }=\frac{1}{2}{v}_{th}[\underset{\text{left slab count}}{\underbrace{n\left(x_0-\frac{\ell }{2}\right)}}-\underset{\text{right slab count}}{\underbrace{n\left(x_0+\frac{\ell }{2}\right)}}]$

Look at the bracket: it is (more crowded slab) − (less crowded slab). If the left is denser, the bracket is positive → net flow rightward. If both slabs are equally crowded, the bracket is zero → no net flow, even though carriers are hopping like crazy. That is the whole secret in one line.

PICTURE. The two arrow-bundles from Step 4 placed head-to-head; the leftover magenta arrows (rightward) are the net flux $F$.

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Step 6 — Taylor expand: convert the count-difference into a slope

WHAT. We rewrite $n(x_0\pm \frac{\ell }{2})$ in terms of the value at $x_0$ and the slope there.

WHY. The bracket in Step 5 is a difference of two nearby crowd values. A difference of nearby values is what a slope measures. The Taylor expansion is the tool that says exactly this: for a small step $\pm \frac{\ell }{2}$, $n\left(x_0\pm \frac{\ell }{2}\right)\approx \underset{\text{value here}}{\underbrace{n(x_0)}}\pm \underset{\text{step}\times \text{slope}}{\underbrace{\frac{\ell }{2}\frac{dn}{dx}}}$ The picture: start at the curve's height $n(x_0)$, then walk half a hop and rise/fall by (how far you walked) × (how steep it is). We use Taylor because it turns "the function a tiny bit away" into "the function here plus slope times distance" — precisely the ingredients we have.

Subtract the two: $n\left(x_0-\frac{\ell }{2}\right)-n\left(x_0+\frac{\ell }{2}\right)=(n_0-\frac{\ell }{2}{n}^{\prime })-(n_0+\frac{\ell }{2}{n}^{\prime })=-\ell \frac{dn}{dx}$ The $n(x_0)$ terms cancel — proof that the absolute crowd level does not matter, only the slope survives.

PICTURE. The tangent line from Step 2 with the two slab-points marked at $x_0\pm \frac{\ell }{2}$; the vertical gap between them is $-\ell \frac{dn}{dx}$.

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Step 7 — Fick's first law falls out

WHAT. Plug the subtracted result back into Step 5.

$F=\frac{1}{2}{v}_{th}(-\ell \frac{dn}{dx})=\underset{\text{call this }-D}{\underbrace{-\frac{1}{2}{v}_{th}\ell }}\frac{dn}{dx}$

WHY. The clump of constants $\frac{1}{2}{v}_{th}\ell$ appears in front of the slope every time; we give it one name, the diffusion coefficient $D=\frac{1}{2}{v}_{th}\ell$, so we never write it out again.

$\boxed{F=-D\frac{dn}{dx}}$

Term by term:

• $F$ = net particle flux (carriers per area per second),
• $-$ = flow goes downhill, from crowded to sparse,
• $D$ = how vigorously they diffuse (bigger $v_{th}$ or $\ell$ → bigger $D$), units ${\text{cm}}^2/\text{s}$,
• $\frac{dn}{dx}$ = the steepness that drives it.

PICTURE. The finished law: a downhill purple curve, an orange gradient arrow, and a magenta flux arrow $F$ pointing the opposite way (downhill).

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Step 8 — From particle flux to electric current (both carrier signs)

WHAT. Particles carry charge, so a particle flux is a current. Multiply by the charge each carrier carries. Let $q=1.6\times 10^{-19}$ C be the size of the elementary charge (a positive number).

WHY. Current density $J$ = (charge per carrier) × (particle flux). This is definitional. The subtlety is the sign of the charge, and it differs for the two carrier types.

Electrons carry $-q$: $J_n=(-q)F_n=(-q)(-D_n\frac{dn}{dx})=\boxed{+q{D}_n\frac{dn}{dx}}$

Holes carry $+q$: $J_p=(+q)F_p=(+q)(-D_p\frac{dp}{dx})=\boxed{-q{D}_p\frac{dp}{dx}}$

The physical motion is identical (both slide downhill), but an electron drifting right is conventional current flowing left — that flips $J_n$'s sign, giving the leftover $+$. Holes keep the $-$.

PICTURE. Same downhill profile, drawn twice: electrons (magenta) move right but their current arrow points left; holes (orange) move right and their current arrow also points right.

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Step 9 — Edge and degenerate cases (never get surprised)

WHAT / WHY / PICTURE, four scenarios the formula must survive:

1. Flat profile, $\frac{dn}{dx}=0$. Crowd is uniform. Carriers still hop furiously, but equal numbers cross each way → $F=0$, $J=0$. Dense does not mean current.
2. Uphill profile, $\frac{dn}{dx}>0$ (crowd grows to the right). The minus sign sends $F$ leftward — flow still runs downhill, now toward $-x$. The formula handles it automatically.
3. Steeper slope, same average $n$. Double the steepness → double $F$. The height of the curve is irrelevant; only its tilt scales the flow.
4. Exponential decay $\delta p(x)=\delta p(0)e^{-x/L_p}$ (the real diode case). Its slope is $-\frac{\delta p(0)}{L_p}{e}^{-x/L_p}$ — steepest at $x=0$, so the diffusion current is largest right at the junction edge and fades with $x$.

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The one-picture summary

Everything on one canvas: crowd curve $n(x)$ (violet), its slope $\frac{dn}{dx}$ (orange tangent), the microscopic hops, the net flux $F=-Ddn/dx$ (magenta), and the two current boxes $J_n=+q{D}_n{n}^{\prime }$, $J_p=-q{D}_p{p}^{\prime }$.

Recall Feynman retelling of the whole walkthrough

Picture a hallway crammed with kids on the left and nearly empty on the right. Nobody is told to move; every kid just steps randomly left or right, like flipping a coin. But because the left is packed, more kids happen to spill rightward across the middle than come back — so the crowd slowly evens out. That leftover rightward spill is the flux $F$. It only exists when the crowding changes across the hall: if kids were spread evenly everywhere, equal numbers would cross each way and nothing would flow — even in a jam-packed but uniform hall. The steeper the crowding difference (the slope $dn/dx$), the stronger the spill, and $D$ just measures how energetically the kids jiggle. Finally, pin a charged badge on each kid: the spill becomes an electric current. Electrons wear negative badges, so their rightward spill counts as current pointing left (the $+$ in $J_n$ after the double flip); holes wear positive badges and keep their sign (the $-$ in $J_p$). That's the entire story: random hops + a crowding slope = a clean directed current.

Recall Quick checks

What must be nonzero for diffusion current? ::: The gradient $dn/dx$ — the slope, not the value of $n$. Where did the minus sign in Fick's law come from? ::: Subtracting the two slab counts gives $-\ell dn/dx$; flow runs downhill. Why does the $n(x_0)$ term vanish in Step 6? ::: It cancels in the subtraction, proving only the slope survives. Why is $J_n$ positive but $J_p$ negative for the same downhill flow? ::: Electron charge $-q$ flips the sign; hole charge $+q$ keeps it.

See also: Drift current and mobility · Einstein relation · Continuity equation · PN junction diode · Thermal voltage $V_T$