Diffusion current and concentration gradient: Exercises

Throughout, these are our constants (memorise them — you will reuse them every problem):

Formula Working constants

First meet the symbols, then use them. $q$ is the size of one electron's charge (always positive — the sign of the carrier is handled separately). $k$ is the Boltzmann constant ( $1.38 \times 1 0^{- 23} J/K$ ): the conversion factor between temperature $T$ (in kelvin) and thermal energy, so $k T$ is "how much random jiggling energy" a carrier carries. With those in hand: $q = 1.6 \times 1 0^{- 19} C V_{T} = \frac{k T}{q} \approx 0.0259 V at 300 K$ $J_{n} = + q D_{n} \frac{d n}{d x} J_{p} = - q D_{p} \frac{d p}{d x} \frac{D}{μ} = V_{T}$ $n$ is electrons per cm³, $p$ is holes per cm³, $x$ is position in cm, and $\frac{d n}{d x}$ is the slope of the density-vs-position graph. $V_{T}$ (thermal voltage) turns thermal energy into a voltage — see Thermal voltage $V_T$.

Two more symbols enter here — meet them before you use them:

$D$ (written $D_{n}$ for electrons, $D_{p}$ for holes — the subscript is just $n$ or $p$ ) is the diffusion coefficient: how eagerly a carrier spreads out per unit of slope, measured in $cm^{2} / s$ . Bigger $D$ = faster spreading. See Fick's laws of diffusion.
$μ$ (written $μ_{n}$ for electrons, $μ_{p}$ for holes) is the mobility: how fast a carrier drifts per unit of electric field, in $cm^{2} / V\cdots$ . Bigger $μ$ = the carrier responds more briskly to a field. See Drift current and mobility.

Level 1 — Recognition

L1.1 — Which quantity drives diffusion?

A silicon bar has a uniform electron density $n = 5 \times 1 0^{16} cm^{- 3}$ everywhere, with no applied field. What is the diffusion current density $J_{n}$ ?

Recall Solution

What we ask: does a lot of carriers automatically mean current? Why the formula settles it: $J_{n} = q D_{n} \frac{d n}{d x}$ depends on the slope, not the count. Uniform density ⇒ the graph of $n$ vs $x$ is a flat horizontal line ⇒ slope $\frac{d n}{d x} = 0$ . $J_{n} = q D_{n} \cdot 0 = 0 A/cm^{2}$ Dense but flat = no diffusion. The carriers wander, but equal numbers cross each way.

L1.2 — Read a linear profile

Electrons drop linearly from $n = 8 \times 1 0^{16} cm^{- 3}$ at $x = 0$ to $n = 2 \times 1 0^{16} cm^{- 3}$ at $x = 3 μ m$ . Find the gradient $\frac{d n}{d x}$ .

Recall Solution

What we do: a straight line has constant slope $= \frac{rise}{run}$ . Why: "linear" is exactly the promise that $\frac{d n}{d x}$ is the same everywhere. Convert $3 μ m = 3 \times 1 0^{- 4} cm$ (because $1 μ m = 1 0^{- 4} cm$ ). $\frac{d n}{d x} = \frac{2 \times 1 0 ^{16} - 8 \times 1 0 ^{16}}{3 \times 1 0 ^{- 4}} = \frac{- 6 \times 1 0 ^{16}}{3 \times 1 0 ^{- 4}} = - 2 \times 1 0^{20} cm^{- 4}$ Negative slope: density falls as $x$ grows — matching the downhill yellow line in the figure below.

The figure below plots this exact profile: the yellow line runs from the $8 \times 1 0^{16}$ dot at $x = 0$ down to the $2 \times 1 0^{16}$ dot at $x = 3 μ m$ . The dashed white triangle marks the "rise $= - 6 \times 1 0^{16}$ " over "run $= 3 μ m$ " — that ratio is the slope $- 2 \times 1 0^{20} cm^{- 4}$ labelled in yellow. Notice the line never curves: linear means one constant slope everywhere.

Figure — Diffusion current and concentration gradient

Level 2 — Application

L2.1 — Electron diffusion current, plug and chug

Using the profile of L1.2 (so $\frac{d n}{d x} = - 2 \times 1 0^{20} cm^{- 4}$ ) with $D_{n} = 30 cm^{2} / s$ , find $J_{n}$ .

Recall Solution

What we do: apply $J_{n} = q D_{n} \frac{d n}{d x}$ directly. Why the $+$ sign: for electrons the formula already carries $+ q$ ; the physical direction is decided by the sign of the slope. $J_{n} = (1.6 \times 1 0^{- 19}) (30) (- 2 \times 1 0^{20}) = - 960 A/cm^{2}$ Reading the sign: electrons diffuse toward larger $x$ (high→low). Electrons are negative, so conventional current points toward $- x$ → the answer is negative. Magnitude $960 A/cm^{2}$ .

L2.2 — Hole diffusion current

Holes fall linearly from $p = 1 0^{17} cm^{- 3}$ at $x = 0$ to $p = 1 0^{16} cm^{- 3}$ at $x = 1 μ m$ , with $D_{p} = 12 cm^{2} / s$ . Find $J_{p}$ .

Recall Solution

Step 1 — gradient. $1 μ m = 1 0^{- 4} cm$ . $\frac{d p}{d x} = \frac{1 0 ^{16} - 1 0 ^{17}}{1 0 ^{- 4}} = \frac{- 9 \times 1 0 ^{16}}{1 0 ^{- 4}} = - 9 \times 1 0^{20} cm^{- 4}$ Step 2 — apply the hole formula $J_{p} = - q D_{p} \frac{d p}{d x}$ (note the leading minus, because holes are positive). $J_{p} = - (1.6 \times 1 0^{- 19}) (12) (- 9 \times 1 0^{20}) = + 1728 A/cm^{2}$ Reading the sign: holes diffuse toward $+ x$ , holes are positive, so conventional current is $+ x$ → positive. ✓

L2.3 — Einstein relation both ways

At 300 K an electron has $μ_{n} = 1350 cm^{2} / V\cdots$ . (a) Find $D_{n}$ . (b) If a hole has $D_{p} = 12 cm^{2} / s$ , find $μ_{p}$ .

Recall Solution

Why Einstein: the same random collisions set both drift ( $μ$ ) and diffusion ( $D$ ), so they must be locked together by $\frac{D}{μ} = V_{T}$ . See Einstein relation. (a) $D_{n} = μ_{n} V_{T} = 1350 \times 0.0259 = 34.97 cm^{2} / s$ . (b) $μ_{p} = \frac{D _{p}}{V _{T}} = \frac{12}{0.0259} = 463.3 cm^{2} / V\cdots$ .

Level 3 — Analysis

L3.1 — Exponential decay (the diode profile)

Excess holes injected into an n-region decay as $δ p (x) = δ p (0) e^{- x / L_{p}}$ with $δ p (0) = 2 \times 1 0^{14} cm^{- 3}$ , diffusion length $L_{p} = 4 μ m$ , and $D_{p} = 12 cm^{2} / s$ . Find $J_{p}$ at $x = 0$ and at $x = L_{p}$ .

Recall Solution

Step 1 — differentiate. Why an exponential's derivative is itself (scaled): $\frac{d}{d x} e^{- x / L_{p}} = - \frac{1}{L _{p}} e^{- x / L_{p}}$ . $\frac{d p}{d x} = - \frac{δ p ( 0 )}{L _{p}} e^{- x / L_{p}}$ Step 2 — at $x = 0$ ( $e^{0} = 1$ ), with $L_{p} = 4 \times 1 0^{- 4} cm$ : $\frac{d p}{d x}_{0} = - \frac{2 \times 1 0 ^{14}}{4 \times 1 0 ^{- 4}} = - 5 \times 1 0^{17} cm^{- 4}$ $J_{p} (0) = - q D_{p} \frac{d p}{d x} = - (1.6 \times 1 0^{- 19}) (12) (- 5 \times 1 0^{17}) = + 0.96 A/cm^{2}$ Step 3 — at $x = L_{p}$ the factor is $e^{- 1} = 0.3679$ , so everything scales by $e^{- 1}$ : $J_{p} (L_{p}) = 0.96 \times e^{- 1} = + 0.353 A/cm^{2}$ Interpretation: the current dies off with the same exponential shape as the carriers.

The figure below overlays two curves on the same board. The blue solid curve is the hole density $δ p (x)$ ; the pink dashed curve is $J_{p} (x)$ , drawn with the same $e^{- x / L_{p}}$ shape (rescaled to share the axis). The yellow dotted vertical line marks $x = L_{p}$ : read off that at exactly one diffusion length both curves have dropped to a factor $e^{- 1}$ of their start. The white arrow flags the steep left edge — where the slope is biggest, so the diffusion current is biggest.

L3.2 — Total current when a field is also present

At a point, electrons have $n = 1 0^{16} cm^{- 3}$ , $\frac{d n}{d x} = + 1 \times 1 0^{20} cm^{- 4}$ , $μ_{n} = 1000 cm^{2} / V\cdots$ , $D_{n} = 25.9 cm^{2} / s$ , and a field $E = + 500 V/cm$ points in $+ x$ . Find the total electron current density.

Recall Solution

Why add two terms: total electron current = drift + diffusion (from Drift current and mobility plus our formula). They are independent physical mechanisms that superpose. $J_{n}^{drift} = q n μ_{n} E = (1.6 \times 1 0^{- 19}) (1 0^{16}) (1000) (500) = + 800 A/cm^{2}$ $J_{n}^{diff} = q D_{n} \frac{d n}{d x} = (1.6 \times 1 0^{- 19}) (25.9) (1 0^{20}) = + 414.4 A/cm^{2}$ $J_{n}^{total} = 800 + 414.4 = + 1214.4 A/cm^{2}$ Both push conventional current toward $+ x$ here, so they add. (Had the slope been negative they could partly cancel.)

Level 4 — Synthesis

L4.1 — Derive the Einstein relation from equilibrium

Inside an unbiased region, the total electron current is zero in equilibrium. Using $n (x) = n_{0} e^{q V (x) / k T}$ (Boltzmann) and $E = - \frac{d V}{d x}$ , show that $\frac{D _{n}}{μ _{n}} = \frac{k T}{q}$ .

Recall Solution

Reminder of the symbols: $k$ is the Boltzmann constant (energy-per-kelvin, defined in the Working-constants box), $T$ the temperature in kelvin, and $V (x)$ the electric potential. So $k T$ is thermal energy and $\frac{k T}{q} = V_{T}$ . What we set up: equilibrium means drift + diffusion cancel: $J_{n} = q n μ_{n} E + q D_{n} \frac{d n}{d x} = 0.$ Why Boltzmann: in thermal equilibrium the carrier density follows the potential energy exactly as $n = n_{0} e^{q V / k T}$ . Differentiate it: $\frac{d n}{d x} = n_{0} e^{q V / k T} \cdot \frac{q}{k T} \frac{d V}{d x} = n \cdot \frac{q}{k T} \frac{d V}{d x} .$ Substitute, and replace $\frac{d V}{d x} = - E$ : $q n μ_{n} E + q D_{n} (n \frac{q}{k T} (- E)) = 0.$ State the assumption before dividing: in a doped region the electron density is nonzero ( $n \neq = 0$ ), and at any point where a gradient (hence a built-in field) exists we have $E \neq = 0$ ; the elementary charge is $q \neq = 0$ . Therefore the common factor $q n E$ is nonzero and we may legally divide the whole equation by it: $μ_{n} - D_{n} \frac{q}{k T} = 0 \Rightarrow \frac{D _{n}}{μ _{n}} = \frac{k T}{q} = V_{T} .$ The two mechanisms lock together because they spring from the same thermal statistics.

L4.2 — Steady-state carrier profile from the continuity equation

In an n-region with no field, excess holes obey the steady-state diffusion equation (a form of the Continuity equation): $D_{p} \frac{d ^{2} δ p}{d x ^{2}} = \frac{δ p}{τ _{p}},$ where $τ_{p}$ is the hole lifetime. Show that $δ p (x) = δ p (0) e^{- x / L_{p}}$ solves it, and find $L_{p}$ in terms of $D_{p}, τ_{p}$ .

Recall Solution

Why guess an exponential: this is a linear, constant-coefficient differential equation — the coefficients $D_{p}$ and $τ_{p}$ do not depend on $x$ . For such equations the derivative of a function must return the same function (only rescaled) so that both sides can match term-for-term; the exponential $e^{λ x}$ is the unique function whose derivative is a constant times itself. That is why the exponential ansatz is the natural — indeed the only — elementary trial form here. What we do: plug the trial exponential in and demand both sides match. First derivative — the chain rule pulls down one factor of $- \frac{1}{L _{p}}$ : $\frac{d}{d x} [δ p (0) e^{- x / L_{p}}] = - \frac{1}{L _{p}} δ p (0) e^{- x / L_{p}} .$ Second derivative — differentiate again; the chain rule pulls down another $- \frac{1}{L _{p}}$ , and the two minus signs multiply to a plus, leaving $+ \frac{1}{L _{p}^{2}}$ : $\frac{d ^{2}}{d x ^{2}} [δ p (0) e^{- x / L_{p}}] = (- \frac{1}{L _{p}}) (- \frac{1}{L _{p}}) δ p (0) e^{- x / L_{p}} = \frac{1}{L _{p}^{2}} δ p (0) e^{- x / L_{p}} = \frac{δ p}{L _{p}^{2}} .$ Why this matters: each derivative of a decaying exponential just multiplies by $- 1/ L_{p}$ , so an even number of derivatives always comes back positive — that is what lets the exponential balance a positive right-hand side. Substitute into the PDE: $D_{p} \cdot \frac{δ p}{L _{p}^{2}} = \frac{δ p}{τ _{p}} \Rightarrow \frac{D _{p}}{L _{p}^{2}} = \frac{1}{τ _{p}} .$ $L_{p} = D_{p} τ_{p}$ Numeric check: if $D_{p} = 12 cm^{2} / s$ and $τ_{p} = 1 \times 1 0^{- 6} s$ , then $L_{p} = 12 \times 1 0^{- 6} = 3.464 \times 1 0^{- 3} cm = 34.6 μ m .$ $L_{p}$ is literally the average distance a hole diffuses before it recombines — the natural length scale of the whole problem.

Level 5 — Mastery

L5.1 — Full diode saturation-current-style build

An abrupt one-sided junction injects holes into a long n-region. Given $δ p (0) = 1 \times 1 0^{13} cm^{- 3}$ , $D_{p} = 12 cm^{2} / s$ , $τ_{p} = 2 \times 1 0^{- 6} s$ , find (a) $L_{p}$ , (b) the hole diffusion current density $J_{p}$ at the injection edge $x = 0$ , and (c) explain in one line what physical current this feeds in a PN junction diode.

Recall Solution

(a) Diffusion length (from L4.2): $L_{p} = D_{p} τ_{p} = 12 \times 2 \times 1 0^{- 6} = 2.4 \times 1 0^{- 5}$ . $L_{p} = 4.899 \times 1 0^{- 3} cm = 48.99 μ m .$ (b) Gradient of $δ p (x) = δ p (0) e^{- x / L_{p}}$ at $x = 0$ ( $e^{0} = 1$ ): $\frac{d p}{d x}_{0} = - \frac{δ p ( 0 )}{L _{p}} = - \frac{1 0 ^{13}}{4.899 \times 1 0 ^{- 3}} = - 2.041 \times 1 0^{15} cm^{- 4} .$ Apply the hole formula: $J_{p} = - q D_{p} \frac{d p}{d x} = - (1.6 \times 1 0^{- 19}) (12) (- 2.041 \times 1 0^{15}) = + 3.92 \times 1 0^{- 3} A/cm^{2} .$ (c) This diffusion current at the junction edge is exactly the hole part of the diode's forward/saturation current — the diode's whole $I$ – $V$ law is diffusion current in disguise. The steeper the injected profile (small $L_{p}$ ), the bigger the current.

L5.2 — Reverse-engineer a profile from a measured current

You measure a constant electron diffusion current $J_{n} = + 50 A/cm^{2}$ across a $2 μ m$ slab with $D_{n} = 25 cm^{2} / s$ . What must the profile $n (x)$ look like, and what is $Δ n$ across the slab?

Recall Solution

Why constant $J$ forces a straight line: $J_{n} = q D_{n} \frac{d n}{d x}$ is constant only if $\frac{d n}{d x}$ is constant ⇒ $n (x)$ is linear. Solve for the slope: $\frac{d n}{d x} = \frac{J _{n}}{q D _{n}} = \frac{50}{( 1.6 \times 1 0 ^{- 19} ) ( 25 )} = 1.25 \times 1 0^{19} cm^{- 4} .$ Positive slope ⇒ density rises with $x$ (electrons diffuse toward $- x$ , giving $+ x$ conventional current ✓). Change across $2 μ m = 2 \times 1 0^{- 4} cm$ : $Δ n = \frac{d n}{d x} \cdot Δ x = 1.25 \times 1 0^{19} \times 2 \times 1 0^{- 4} = 2.5 \times 1 0^{15} cm^{- 3} .$ A linear ramp of about $2.5 \times 1 0^{15} cm^{- 3}$ over the slab produces exactly the measured 50 A/cm².

Active recall

Recall One-line answers

Uniform dense region, no field — diffusion current? ::: Zero — slope $d n / d x = 0$ . Why convert µm to cm first? ::: So $J$ comes out in A/cm² (units must match $D$ and $n$ ). Diffusion length in terms of $D, τ$ ? ::: $L_{p} = D_{p} τ_{p}$ . Which way is conventional current for diffusing electrons down $+ x$ ? ::: Toward $- x$ (opposite to electron motion).