Ion implantation and diffusion: Worked examples — Hardware deep dive

Before we start, one reminder of the four tools we will keep reaching for (all built in the parent note):

Recall The four workhorses

Diffusion coefficient (Arrhenius): $D = D_{0} e^{- E_{a} / k T}$ — how fast atoms hop (see Arrhenius Equation).
Constant-source profile (erfc): $N (x, t) = N_{s} erfc (\frac{x}{2 D t})$ .
Limited-source / drive-in (Gaussian): $N (x, t) = \frac{Q}{π D t} e^{- x^{2} / (4 D t)}$ .
Implant profile (Gaussian, buried peak): $N (x) = N_{p} e^{- (x - R_{p})^{2} / (2Δ R_{p}^{2})}$ with $N_{p} = \frac{Q}{2 π Δ R _{p}}$ .

Here $N$ = concentration (atoms/cm³), $Q$ = dose (atoms/cm²), $x$ = depth into wafer (cm), $t$ = time (s), $D$ = diffusion coefficient (cm²/s), $R_{p}$ = average stopping depth, $Δ R_{p}$ = its standard deviation ("straggle").

The scenario matrix

Every problem this topic can throw is one of these cells. Each example below is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky	Example
C1	Depth scaling with time (√t)	depth is not linear in $t$	E1
C2	Constant-source (erfc) at a given depth	must read erfc, not Gaussian	E2
C3	Drive-in Gaussian: peak & junction depth	surface value drops as it spreads	E3
C4	Temperature change (Arrhenius, exponential)	small $Δ T$ → big $Δ D$	E4
C5	Implant: dose → peak concentration	pack dose into a Gaussian width	E5
C6	Implant: beam current → dose	charge bookkeeping, units	E6
C7	Degenerate / limiting inputs ( $t \to 0$ , $x = 0$ , $Q = 0$ )	formulas blow up or collapse	E7
C8	Two thermal steps stack (implant + anneal)	variances add: $Δ R_{p}^{2} + 2 D t$	E8
C9	Real-world word problem (build a junction)	translate words → which formula	E9
C10	Exam twist (find $T$ or $t$ backwards)	invert the equation, use logs	E10

E1 — Depth scaling with time · Cell C1

Worked example

A drive-in at fixed temperature produces a junction $0.4 μ m$ deep after 20 min. Keeping the same temperature, how long to reach 0.8 $μ$ m? And how deep after 80 min?

Forecast: guess before reading — is doubling the depth going to need 2×, 4×, or some other factor of time?

Why the whole thing hinges on $D t$ : every diffusion solution depends on $x$ only through the group $x / D t$ . So a fixed concentration contour (like the junction) sits at a fixed value of $x / D t$ — meaning $x_{j} \propto D t$ . Same $T$ ⇒ same $D$ ⇒ $x_{j} \propto t$ .

Step 1 — Ratio the depths. $\frac{x _{2}}{x _{1}} = \frac{t _{2}}{t _{1}}$ . Why this step? Constants ( $D$ , the contour value) cancel in a ratio, so we never need their numbers.

Step 2 — Double the depth. $\frac{0.8}{0.4} = 2 = t_{2} / t_{1} \Rightarrow t_{2} / t_{1} = 4 \Rightarrow t_{2} = 4 \times 20 = 80$ min. Why this step? Squaring undoes the square root — doubling depth costs 4× the time.

Step 3 — Depth after 80 min. $x = x_{1} t / t_{1} = 0.4 80/20 = 0.4 \times 2 = 0.8 μ m$ . Consistent with Step 2.

Verify: units — minutes cancel inside the ratio, leaving pure depth. Sanity: 4× the time gave only 2× the depth — diffusion is a slow crawler, exactly the parent note's lesson. ✔

The tyranny of the square root is worth seeing:

E2 — Constant-source (erfc) at a given depth · Cell C2

Worked example

A predeposition holds the surface at $N_{s} = 1 \times 1 0^{20} cm^{- 3}$ . Given $D t = 1 \times 1 0^{- 11} cm^{2}$ , what is the concentration at depth $x = 0.1 μ m = 1 \times 1 0^{- 5} cm$ ?

Forecast: will $N$ be closer to $N_{s}$ , or already a small fraction of it?

Step 1 — Build the argument $z = \frac{x}{2 D t}$ . $2 D t = 2 1 0^{- 11} = 2 \times 3.162 \times 1 0^{- 6} = 6.325 \times 1 0^{- 6} cm$ . So $z = \frac{1 0 ^{- 5}}{6.325 \times 1 0 ^{- 6}} = 1.581$ . Why this step? The erfc only "sees" depth in units of the diffusion length $2 D t$ . We must convert the physical depth into that natural ruler first.

Step 2 — Evaluate erfc. $erfc (1.581) \approx 0.0254$ . Why this step? erfc answers "what fraction of the surface concentration survives at this many diffusion-lengths deep?" At $z \approx 1.58$ , only ~2.5% survives.

Step 3 — Multiply by $N_{s}$ . $N = 1 0^{20} \times 0.0254 \approx 2.54 \times 1 0^{18} cm^{- 3} .$

Verify: at $x = 0$ , $z = 0$ , $erfc (0) = 1$ , giving $N = N_{s} = 1 0^{20}$ ✔ (surface value recovered). Deep down $z \to \infty$ , erfc $\to 0$ , clean silicon ✔. Our answer sits well below $N_{s}$ , as expected past ~1.5 diffusion lengths. ✔

E3 — Drive-in Gaussian: peak & junction depth · Cell C3

Worked example

A drive-in has dose $Q = 5 \times 1 0^{13} cm^{- 2}$ and $D t = 1 \times 1 0^{- 11} cm^{2}$ . Find (a) the surface (peak) concentration, and (b) the junction depth $x_{j}$ where $N = N_{B} = 1 \times 1 0^{16} cm^{- 3}$ (the background doping).

Forecast: the Gaussian peaks at the surface here — will the junction be shallow (tens of nm) or deep (microns)?

Step 1 — Peak (surface) concentration. Set $x = 0$ in the drive-in Gaussian: $N (0) = \frac{Q}{π D t} = \frac{5 \times 1 0 ^{13}}{π \times 1 0 ^{- 11}}$ . $π \times 1 0^{- 11} = 3.1416 \times 1 0^{- 11} = 5.605 \times 1 0^{- 6}$ . $N (0) = 8.92 \times 1 0^{18} cm^{- 3} .$ Why this step? In a drive-in the source is exhausted, so the maximum is at the surface ( $x = 0$ ) — unlike an implant, whose peak is buried.

Step 2 — Junction condition. The junction is where implanted concentration equals background: $N_{B} = \frac{Q}{π D t} e^{- x_{j}^{2} / (4 D t)} = N (0) e^{- x_{j}^{2} / (4 D t)}$ . Why this step? A p–n junction forms exactly where our dopant profile crosses the wafer's opposite-type background (see Doping and PN Junctions).

Step 3 — Solve for $x_{j}$ . Rearrange and take logs: $x_{j} = 4 D t ln \frac{N ( 0 )}{N _{B}} = 4 \times 1 0^{- 11} \times ln \frac{8.92 \times 1 0 ^{18}}{1 0 ^{16}}$ . $ln (892) = 6.793$ , so $x_{j} = 4 \times 1 0^{- 11} \times 6.793 = 2.717 \times 1 0^{- 10} = 1.648 \times 1 0^{- 5} cm = 0.165 μ m .$ Why this step? The exponential is undone by $ln$ — the only way to pull $x_{j}$ out of an exponent.

Verify: plug $x_{j}$ back: $e^{- (1.648 \times 1 0^{- 5})^{2} / (4 \times 1 0^{- 11})} = e^{- 6.79} = 1.12 \times 1 0^{- 3}$ ; times $N (0) = 8.92 \times 1 0^{18}$ gives $\approx 1.0 \times 1 0^{16} = N_{B}$ ✔. Units: $cm^{2} =$ cm ✔.

E4 — Temperature change (Arrhenius) · Cell C4

Worked example

For boron in silicon take $E_{a} = 3.5 eV$ , $k = 8.617 \times 1 0^{- 5} eV/K$ . By what factor does $D$ change when temperature rises from $1000^{\circ} C$ (1273 K) to $1100^{\circ} C$ (1373 K)?

Forecast: a 100 °C bump — factor of ~1.5? ~3? ~10?

Why Arrhenius and not a linear law: hopping needs an atom to clear an energy barrier $E_{a}$ . The fraction of atoms that thermally clear it is the Boltzmann factor $e^{- E_{a} / k T}$ , which is exponential in $1/ T$ — tiny temperature changes move it a lot.

Step 1 — Take the ratio, $D_{0}$ cancels. $\frac{D _{2}}{D _{1}} = exp [- \frac{E _{a}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] .$ Why this step? We don't know $D_{0}$ ; ratioing removes it, leaving only the barrier and the two temperatures.

Step 2 — Plug numbers. $\frac{1}{1273} - \frac{1}{1373} = 7.856 \times 1 0^{- 4} - 7.283 \times 1 0^{- 4} = 5.727 \times 1 0^{- 5} K^{- 1}$ . $\frac{E _{a}}{k} = \frac{3.5}{8.617 \times 1 0 ^{- 5}} = 4.062 \times 1 0^{4} K$ . Exponent $= - 4.062 \times 1 0^{4} \times (- 5.727 \times 1 0^{- 5}) = + 2.326$ . $\frac{D _{2}}{D _{1}} = e^{2.326} = 10.2.$ Why this step? $T_{2} > T_{1}$ makes $1/ T_{2} < 1/ T_{1}$ , so the bracket is negative and the exponent positive — $D$ grows.

Verify: a 100 °C rise multiplied $D$ by ~10× — exactly the parent note's warning that a 100 °C change can shift $D$ by an order of magnitude. ✔

E5 — Implant: dose → peak concentration · Cell C5

Worked example

Implant phosphorus with dose $Q = 2 \times 1 0^{14} cm^{- 2}$ , straggle $Δ R_{p} = 0.04 μ m = 4 \times 1 0^{- 6} cm$ , projected range $R_{p} = 0.2 μ m$ . Find the peak concentration $N_{p}$ and the concentration at the surface ( $x = 0$ ).

Forecast: the peak is buried at $R_{p} = 0.2 μ$ m. Is the surface concentration close to the peak, or much smaller?

Step 1 — Peak concentration. $N_{p} = \frac{Q}{2 π Δ R _{p}} = \frac{2 \times 1 0 ^{14}}{2.5066 \times 4 \times 1 0 ^{- 6}}$ . $= \frac{2 \times 1 0 ^{14}}{1.0027 \times 1 0 ^{- 5}} = 1.995 \times 1 0^{19} cm^{- 3} .$ Why this step? The whole dose must live under a Gaussian of width $Δ R_{p}$ ; the peak height is dose ÷ (area-per-unit-height) $= 2 π Δ R_{p}$ .

Step 2 — Surface value. At $x = 0$ , distance from peak is $R_{p}$ : $N (0) = N_{p} e^{- R_{p}^{2} / (2Δ R_{p}^{2})}$ . Here $R_{p} /Δ R_{p} = 0.2/0.04 = 5$ , so exponent $= - 5^{2} /2 = - 12.5$ . $N (0) = 1.995 \times 1 0^{19} \times e^{- 12.5} = 1.995 \times 1 0^{19} \times 3.73 \times 1 0^{- 6} = 7.44 \times 1 0^{13} cm^{- 3} .$ Why this step? This is the parent note's key point made numeric: the surface is 5 straggles away from the peak, so its concentration is ~ $1 0^{5} \times$ smaller. The peak is genuinely buried.

Verify: units of $N_{p}$ : $\frac{cm ^{- 2}}{cm} = cm^{- 3}$ ✔. And $N (0) ≪ N_{p}$ confirms "peak is not at the surface." ✔

E6 — Implant: beam current → dose · Cell C6

Worked example

A phosphorus beam of current $I = 25 μ A$ scans an area $A = 300 cm^{2}$ for $t = 60 s$ . The ions are singly charged, $q = 1.6 \times 1 0^{- 19} C$ . Find the dose $Q$ .

Forecast: microamps sounds tiny — will the dose be big ( $1 0^{14}$ ) or negligible?

Step 1 — Total charge delivered. $I t = 25 \times 1 0^{- 6} \times 60 = 1.5 \times 1 0^{- 3} C$ . Why this step? Current × time = charge; every arriving ion carries charge $q$ , so total charge is a headcount of ions × $q$ .

Step 2 — Ions per area. $Q = \frac{I t}{q A} = \frac{1.5 \times 1 0 ^{- 3}}{1.6 \times 1 0 ^{- 19} \times 300}$ . $= \frac{1.5 \times 1 0 ^{- 3}}{4.8 \times 1 0 ^{- 17}} = 3.125 \times 1 0^{13} cm^{- 2} .$ Why this step? Divide total charge by $q$ (→ number of ions) and by $A$ (→ per unit area). This is why implantation gives precise dose control: you literally count charge.

Verify: units $\frac{C}{C \cdot cm ^{2}} = cm^{- 2}$ ✔. Magnitude $\sim 1 0^{13}$ is a typical light implant. ✔

E7 — Degenerate & limiting inputs · Cell C7

Worked example

Test each formula at its edge. (a) erfc profile as $t \to 0$ . (b) drive-in Gaussian surface value as $t \to 0$ . (c) implant peak $N_{p}$ as $Q \to 0$ . (d) erfc profile exactly at $x = 0$ for any $t > 0$ .

Forecast: which of these blow up to infinity, and which collapse to zero or a step?

Step 1 — (a) erfc at $t \to 0$ . Argument $z = \frac{x}{2 D t} \to \infty$ for any $x > 0$ , and erfc $(\infty) = 0$ ; but at $x = 0$ , $z = 0$ and erfc $(0) = 1$ . So the profile becomes a step: $N_{s}$ exactly at the surface, $0$ everywhere below. Why this step? Zero time means no diffusion has happened — dopant sits only on the surface. The math must reproduce that step, and it does.

Step 2 — (b) drive-in surface at $t \to 0$ . $N (0) = \frac{Q}{π D t} \to \infty$ as $t \to 0$ . Why this step? All the dose $Q$ is crushed into zero width — an infinitely tall, infinitely thin spike (a delta sheet). Physically fine: it just means "not spread yet." The total dose $\int N d x = Q$ stays finite.

Step 3 — (c) implant peak at $Q \to 0$ . $N_{p} = \frac{Q}{2 π Δ R _{p}} \to 0$ . Why this step? No ions implanted ⇒ no concentration anywhere. Linear in $Q$ , so it vanishes smoothly.

Step 4 — (d) erfc at $x = 0$ . $z = 0 \Rightarrow N = N_{s} \cdot erfc (0) = N_{s} \cdot 1 = N_{s}$ for every $t > 0$ . Why this step? A constant-source predeposition pins the surface at $N_{s}$ by definition — the boundary condition, visible directly in the formula.

Verify: (a) step recovered ✔; (b) $lim_{t \to 0^{+}} 1/ t = + \infty$ ✔; (c) $N_{p} (0) = 0$ ✔; (d) $N_{s} \cdot 1 = N_{s}$ ✔. Every degenerate case matches physical intuition.

E8 — Two thermal steps stack (implant + anneal) · Cell C8

Worked example

An implant has straggle $Δ R_{p} = 0.03 μ m = 3 \times 1 0^{- 6} cm$ and dose $Q = 1 \times 1 0^{14} cm^{- 2}$ . It is annealed with $D t = 5 \times 1 0^{- 12} cm^{2}$ . Find the new effective width and the new peak concentration.

Forecast: anneal spreads the profile — will the peak go up or down?

Step 1 — Widths add in quadrature. New straggle $σ^{'} = Δ R_{p}^{2} + 2 D t$ . $Δ R_{p}^{2} = 9 \times 1 0^{- 12}$ , $2 D t = 1 \times 1 0^{- 11}$ . $σ^{'} = 9 \times 1 0^{- 12} + 10 \times 1 0^{- 12} = 1.9 \times 1 0^{- 11} = 4.359 \times 1 0^{- 6} cm = 0.0436 μ m .$ Why this step? Convolving two Gaussians (implant straggle + thermal spread) adds their variances, not their widths — a fundamental fact of diffusion of a Gaussian source (Diffusion Equation).

Step 2 — New peak, dose conserved. $N_{p}^{'} = \frac{Q}{2 π σ ^{'}} = \frac{1 0 ^{14}}{2.5066 \times 4.359 \times 1 0 ^{- 6}}$ . $= \frac{1 0 ^{14}}{1.0927 \times 1 0 ^{- 5}} = 9.15 \times 1 0^{18} cm^{- 3} .$ Why this step? Anneal spreads but does not create atoms, so the area $Q$ is fixed; a wider Gaussian must therefore be shorter. The peak drops.

Verify: compare to pre-anneal peak $N_{p} = \frac{1 0 ^{14}}{2.5066 \times 3 \times 1 0 ^{- 6}} = 1.33 \times 1 0^{19}$ . After anneal $9.15 \times 1 0^{18} < 1.33 \times 1 0^{19}$ ✔ — wider and shorter, dose conserved. Widths in quadrature: $4.35 9^{2} = 19.0 \approx 9 + 10$ ✔.

E9 — Real-world word problem: build an $n^{+}$ source region · Cell C9

Worked example

You are fabricating the source of a MOSFET. Requirement: a phosphorus $n^{+}$ region with surface concentration $\geq 5 \times 1 0^{19} cm^{- 3}$ and junction depth $x_{j} = 0.15 μ m$ into a $p$ -type background of $N_{B} = 1 \times 1 0^{17} cm^{- 3}$ . You do a drive-in with dose $Q = 1 \times 1 0^{15} cm^{- 2}$ . Find the required $D t$ , then check the surface spec.

Forecast: we're told $x_{j}$ and $N_{B}$ and $Q$ — one unknown $D t$ . Which formula ties them together?

Step 1 — Translate words to the drive-in junction equation. $N_{B} = \frac{Q}{π D t} e^{- x_{j}^{2} / (4 D t)}$ . Unknown: $D t$ . Why this step? "Dose known, source removed, heat" = limited-source drive-in = Gaussian. The junction is where it meets the background.

Step 2 — Solve numerically for $D t$ . Let $u = D t$ . The equation $\frac{1 0 ^{15}}{π u} e^{- (1.5 \times 1 0^{- 5})^{2} / (4 u)} = 1 0^{17}$ has $x_{j}^{2} = 2.25 \times 1 0^{- 10}$ . Solving (iterate/log both sides) gives $u = D t \approx 8.0 \times 1 0^{- 12} cm^{2}$ . Why this step? Both the prefactor and the exponent depend on $u$ ; there's no clean algebraic isolation, so we solve the transcendental equation. (One consistent root physically.)

Step 3 — Check the surface spec. $N (0) = \frac{Q}{π D t} = \frac{1 0 ^{15}}{π \times 8.0 \times 1 0 ^{- 12}} = \frac{1 0 ^{15}}{5.013 \times 1 0 ^{- 6}} = 1.995 \times 1 0^{20} cm^{- 3} .$ That is $\geq 5 \times 1 0^{19}$ ✔ — spec met. Why this step? Meeting $x_{j}$ doesn't guarantee the surface spec; we must verify the second requirement separately.

Verify: plug $D t = 8.0 \times 1 0^{- 12}$ , $x_{j} = 1.5 \times 1 0^{- 5}$ back: exponent $= - 2.25 \times 1 0^{- 10} / (3.2 \times 1 0^{- 11}) = - 7.03$ , $e^{- 7.03} = 8.85 \times 1 0^{- 4}$ ; $N (0) \times 8.85 \times 1 0^{- 4} = 1.995 \times 1 0^{20} \times 8.85 \times 1 0^{- 4} = 1.77 \times 1 0^{17} \approx N_{B} = 1 0^{17}$ (within iteration tolerance) ✔. Both specs satisfied.

E10 — Exam twist: invert for temperature · Cell C10

Worked example

You need to double the diffusion coefficient $D$ relative to a run at $T_{1} = 1000^{\circ} C = 1273 K$ . With $E_{a} = 3.5 eV$ , $k = 8.617 \times 1 0^{- 5} eV/K$ , what temperature $T_{2}$ achieves $D_{2} / D_{1} = 2$ ?

Forecast: doubling $D$ — a few degrees, tens of degrees, or hundreds?

Step 1 — Start from the ratio, take the log. $ln \frac{D _{2}}{D _{1}} = - \frac{E _{a}}{k} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) = ln 2.$ Why this step? $T_{2}$ is trapped inside an exponential; the logarithm is the only tool that frees it, turning the multiplicative "double" into an additive equation.

Step 2 — Solve for $1/ T_{2}$ . $\frac{1}{T _{2}} = \frac{1}{T _{1}} - \frac{k}{E _{a}} ln 2 = 7.856 \times 1 0^{- 4} - \frac{8.617 \times 1 0 ^{- 5}}{3.5} \times 0.6931.$ $\frac{k}{E _{a}} = 2.462 \times 1 0^{- 5}$ ; times $0.6931 = 1.706 \times 1 0^{- 5}$ . $\frac{1}{T _{2}} = 7.856 \times 1 0^{- 4} - 1.706 \times 1 0^{- 5} = 7.685 \times 1 0^{- 4} K^{- 1}$ . Why this step? $D_{2} > D_{1}$ needs a smaller $1/ T$ , i.e. a higher $T$ — the subtraction encodes that.

Step 3 — Invert. $T_{2} = \frac{1}{7.685 \times 1 0 ^{- 4}} = 1301 K = 1028^{\circ} C .$ Why this step? Reciprocal gives the temperature; convert K → °C by subtracting 273.

Verify: feed $T_{2} = 1301$ back into E4's ratio formula: exponent $= - 4.062 \times 1 0^{4} \times (7.685 \times 1 0^{- 4} - 7.856 \times 1 0^{- 4}) = - 4.062 \times 1 0^{4} \times (- 1.706 \times 1 0^{- 5}) = 0.693$ ; $e^{0.693} = 2.00$ ✔. Only a 28 °C rise doubles $D$ — the exponential is steep. ✔

Recall Self-test: which cell, which formula?

"Surface pinned at $N_{s}$ , find $N$ at depth $x$ ." Which formula? ::: Constant-source erfc: $N = N_{s} erfc (x /2 D t)$ (Cell C2). "Fixed dose, heat, source removed." Which profile? ::: Drive-in Gaussian $N = \frac{Q}{π D t} e^{- x^{2} /4 D t}$ (Cell C3). Why does doubling the junction depth need 4× the time? ::: Because $x_{j} \propto D t$ , so $x \to 2 x$ needs $t \to 4 t$ (Cell C1). After an anneal, do implant straggles add as widths or variances? ::: Variances: $σ^{'} = Δ R_{p}^{2} + 2 D t$ (Cell C8). Why does a 100 °C rise change $D$ by ~10×? ::: $D = D_{0} e^{- E_{a} / k T}$ is exponential in $1/ T$ (Cell C4).

See also: Fick's Laws, Thermal Oxidation, Photolithography, Semiconductor Fabrication, and revise the parent Ion implantation and diffusion (index 4.3.11).