Ion implantation and diffusion: Visual walkthrough

Where we are going: the parent note handed you the implant profile $N (x) = N_{p} exp [- \frac{( x - R _{p} ) ^{2}}{2Δ R _{p}^{2}}]$ and the dose-to-peak link $N_{p} = Q / (2 π Δ R_{p})$ . Here we build both from a single physical picture — many ions, each stopping at a slightly random depth — using nothing but counting and one bell-shaped curve. Every symbol is earned before it appears.

Parent: Ion implantation and diffusion · prerequisites: Doping and PN Junctions, Diffusion Equation, Fick's Laws, Arrhenius Equation.

Step 1 — What we are actually firing at the wafer

WHAT. We define a single measuring stick, depth $x$ :

$x = 0$ is the top surface,
$x > 0$ points into the silicon,
$x$ is measured in centimetres (cm) — but the physics only cares about how deep, not units.

WHY. A transistor needs dopants placed at a controlled depth. So the one number that matters for each ion is: how deep did it stop? Everything on this page is a story about the depths where ions come to rest.

PICTURE. Look at the beam of orange darts coming down and burying themselves at various depths below the surface line.

Step 2 — Why the depths are not all the same (the histogram)

WHAT. Fire millions of ions and record every stopping depth. Chop the depth axis into thin bins and count how many ions landed in each bin. That count-vs-depth chart is a histogram.

WHY the histogram matters. The height of a bin at depth $x$ tells you how many dopant atoms per unit depth live there — which is exactly the dopant concentration $N (x)$ we care about (atoms per cm³, once we account for beam area). So:

$dopant concentration at depth x N (x) \propto height of the histogram bin (how many ions stopped near x)$

PICTURE. The scattered stopping depths pile up into a lumpy histogram — tall in the middle, short at the edges. That lump is the shape we are about to name.

Step 3 — Why the lump is a bell curve (the Gaussian)

WHAT. The idealised histogram, with infinitely many ions and infinitely thin bins, becomes a smooth curve called a Gaussian (a "bell curve"):

$N (x) = N_{p} exp [- \frac{( x - R _{p} ) ^{2}}{2 Δ R _{p}^{2}}]$

Let's read it term by term, right where each symbol sits:

Formula Every symbol, decoded

$x$ — the depth we are asking about (our measuring stick from Step 1).
$R_{p}$ — the projected range: the centre of the bell, i.e. the average stopping depth. This is where the exponent is zero and the curve is tallest.
$x - R_{p}$ — how far the depth $x$ is from the centre. Left of the peak this is negative, right of it positive — but it gets squared, so only the distance matters, not the side.
$Δ R_{p}$ — the straggle: the half-width of the bell, how spread out the stopping depths are. Small straggle = sharp spike; large straggle = wide smear.
$\frac{( x - R _{p} ) ^{2}}{2Δ R _{p}^{2}}$ — distance-from-centre measured in units of straggle, squared. At one straggle away it equals $\frac{1}{2}$ .
$exp [- (\dots)]$ — the exponential turns "far from centre" into "small height", and does it smoothly and symmetrically.
$N_{p}$ — the peak height, the concentration exactly at the centre $x = R_{p}$ (there $exp [0] = 1$ , so $N = N_{p}$ ).

WHY the minus and the square. The minus makes the curve decay as you move away from centre (deeper OR shallower). The square makes it symmetric — an ion is just as likely to overshoot as undershoot the average by the same amount.

PICTURE. The smooth bell, with $R_{p}$ marked at its peak and $Δ R_{p}$ marked as the horizontal distance from the peak to where the curve has dropped to about $61%$ of its height.

Step 4 — Why the peak is buried, not at the surface

WHAT. The peak sits at $x = R_{p} > 0$ . The surface concentration is

$N (0) = N_{p} exp [- \frac{R _{p}^{2}}{2Δ R _{p}^{2}}]$

which is smaller than $N_{p}$ because the exponent is now negative (the surface is a full $R_{p}$ away from the centre).

WHY it matters. This is the single biggest difference from thermal diffusion, where the maximum sits at the surface. Buried peaks are exactly what you want for a MOSFET channel or a threshold-adjust implant.

PICTURE. The bell shifted rightward off the surface line; the short red bar at $x = 0$ shows how little dopant reaches the very top compared to the tall bar at $x = R_{p}$ .

Step 5 — Connecting the shape to the dose (area under the bell)

WHAT. The dose $Q$ (atoms per cm²) is the total area under the concentration curve — every atom must be somewhere:

$Q = \int_{- \infty}^{\infty} N (x) d x$

The Gaussian's area is a known result: $\int_{- \infty}^{\infty} N_{p} e^{- (x - R_{p})^{2} /2Δ R_{p}^{2}} d x = N_{p} Δ R_{p} 2 π$ . Reading it term by term:

$N_{p}$ — height of the bell,
$Δ R_{p}$ — its width,
$2 π \approx 2.5066$ — the fixed geometric constant for every Gaussian (area = height × width × $2 π$ ).

WHY this step. Height × width = area, roughly — and the exact Gaussian version fixes the proportionality constant. Setting that area equal to the known dose $Q$ solves for the one unknown, $N_{p}$ :

$N_{p} = \frac{Q}{2 π Δ R _{p}}$

Term by term: a bigger dose $Q$ raises the peak; a wider straggle $Δ R_{p}$ lowers it (same atoms smeared over more depth).

PICTURE. The shaded area under the bell, labelled "= total dose $Q$ ", with height $N_{p}$ and width $Δ R_{p}$ marked as the two levers that trade off against each other.

Step 6 — The degenerate cases (never leave a gap)

Case A — Zero straggle ( $Δ R_{p} \to 0$ ). All ions stop at exactly $R_{p}$ . The bell collapses to an infinitely tall, infinitely thin spike at $R_{p}$ . In the formula, $N_{p} = Q / (2 π Δ R_{p}) \to \infty$ , but the area stays $Q$ . This is the idealised "delta" implant.

Case B — Doubling the dose ( $Q \to 2 Q$ ). Same width, twice the height ( $N_{p}$ doubles). Shape unchanged, curve simply scaled up. Dose only controls height, energy controls centre.

Case C — Zero dose ( $Q = 0$ ). $N_{p} = 0$ : flat line, no dopant anywhere. The formula correctly gives nothing.

Case D — After the anneal. Heating to activate dopants also lets them wander (thermal diffusion). The bell keeps its centre $R_{p}$ but widens: the straggle grows

$Δ R_{p} ⟶ Δ R_{p}^{2} + 2 D t$

where $D$ is the diffusion coefficient (temperature-set via the Arrhenius law $D = D_{0} e^{- E_{a} / k T}$ ) and $t$ the anneal time. Term by term: $Δ R_{p}^{2}$ is the implant spread, $2 D t$ is the extra thermal spread added on top — variances add.

PICTURE. Four small panels: (A) the spike, (B) doubled height, (C) flat zero line, (D) a narrow bell widening into a broad one after anneal.

The one-picture summary

This single figure compresses the whole walkthrough: random ion stops → histogram → smooth Gaussian bell centred at $R_{p}$ , width $Δ R_{p}$ , whose enclosed area is the dose $Q$ , with the peak height pinned by $N_{p} = Q / (2 π Δ R_{p})$ .

Recall Feynman retelling (say it in plain words)

We shoot dopant darts straight down into silicon. Each dart plows through a chaotic hail of collisions and stops at some depth — never quite the same depth twice, because the collisions are random. Pile up millions of darts and the depths stack into a bell-shaped hill: tallest at the average depth $R_{p}$ (which is buried inside the wafer, because darts need room to stop), and fading smoothly to both sides. How wide the hill is, we call the straggle $Δ R_{p}$ ; how tall it is depends on how many darts we fired — the dose $Q$ . Since every dart lands somewhere, the area under the hill is exactly $Q$ ; and since a Gaussian's area is height × width × $2 π$ , the height must be $N_{p} = Q / (2 π Δ R_{p})$ . Squeeze the width to nothing and the hill becomes a spike; double the darts and the hill doubles in height; fire nothing and there's no hill. Finally, when we heat the wafer to wake the dopants up, they wander a bit and the hill quietly widens — its width becoming $Δ R_{p}^{2} + 2 D t$ .

Recall

Where does the implant peak sit, and why? ::: At $x = R_{p}$ inside the wafer, because ions need travel-distance to lose their energy — almost none stop right at the surface. Why is the profile Gaussian? ::: Each ion's depth is a sum of many small random collisions; sums of many small independent randoms pile into a bell curve (Central Limit Theorem). How is the peak height $N_{p}$ fixed? ::: Total area under the bell = dose $Q$ ; a Gaussian's area is $N_{p} Δ R_{p} 2 π$ , so $N_{p} = Q / (2 π Δ R_{p})$ . What does annealing do to the width? ::: Widens the straggle from $Δ R_{p}$ to $Δ R_{p}^{2} + 2 D t$ (thermal diffusion adds variance).