Depletion region and space charge: Visual walkthrough

We will lean on three prerequisite ideas as we go: Doping (Donors and Acceptors) (where the fixed ions come from), Poisson's Equation and Gauss's Law (the master tool), and Built-in Potential and Fermi Level Alignment (what sets the voltage). Nothing else is assumed.

Step 1 — Draw the two crystals before they touch

WHAT. We start with two blocks of silicon. The left block is p-type: it was doped with acceptor atoms, so it is full of free holes (draw them as pink circles) and, locked into the lattice, an equal number of negative acceptor ions (we write these $N_{A}^{-}$ ). The right block is n-type: full of free electrons (blue circles) plus positive donor ions $N_{D}^{+}$ .

WHY. Before anything moves, each block is electrically neutral: the free carriers exactly cancel the fixed ions. We must see this starting state clearly, because the whole story is about what happens when the free carriers leave and the fixed ions get uncovered.

PICTURE. In the figure, notice that inside each block the mobile carriers (small filled circles) sit right on top of the fixed ions (small squares) — so every region is neutral. The dashed line in the middle is the metallurgical junction: the plane where p meets n.

Figure — Depletion region and space charge

Step 2 — Let them touch: diffusion carves out the empty zone

WHAT. Now we join the blocks. Right at the junction, blue electrons see empty space to their left (few electrons there) and pink holes see empty space to their right. So they spill across and recombine — an electron meets a hole and both vanish.

WHY. This spilling is diffusion: particles drift from where they are crowded to where they are sparse (see Drift and Diffusion Currents). It is the same reason a drop of ink spreads in water. It happens because of the concentration difference, and it stops nothing on its own — something else will have to.

PICTURE. Look at the middle strip in the figure: the mobile circles are gone, but the fixed squares remain. On the left of the junction we are left with bare negative acceptor ions; on the right, bare positive donor ions. This stripped strip is the depletion region. Outside it, the blocks are still neutral (carriers still cover their ions).

Step 3 — Model the charge as two clean boxes (the depletion approximation)

WHAT. Set up a horizontal axis $x$ with $x = 0$ at the junction. We call the depletion edge on the p-side $- x_{p}$ and on the n-side $+ x_{n}$ (both are positive distances, measured outward from the junction). We now claim the charge density $ρ (x)$ — charge per unit volume — looks like two flat rectangles:

$ρ (x) = ⎩ ⎨ ⎧ - q N_{A} + q N_{D} 0 - x_{p} < x < 0 (p-side box) 0 < x < x_{n} (n-side box) everywhere else$

Reading the equation left to right: $- q N_{A}$ says "on the p-side, every unit of volume holds $N_{A}$ acceptor ions, each carrying charge $- q$ — so a negative block." $+ q N_{D}$ says "on the n-side, $N_{D}$ donor ions each carrying $+ q$ — a positive block." Outside $[- x_{p}, x_{n}]$ the carriers still cover the ions, so $ρ = 0$ .

WHY. In reality the carrier count fades out smoothly over a short distance (a Debye length). But smooth curves make the integrals ugly. Chopping them into sharp boxes — fully depleted inside, fully neutral outside — turns the hard integrals into rectangle areas, and it barely changes the answer because the fade is so short. This shortcut is the depletion approximation.

PICTURE. The figure shows $ρ (x)$ as a downward pink rectangle on the left and an upward yellow rectangle on the right. Their heights are $q N_{A}$ and $q N_{D}$ ; their widths are $x_{p}$ and $x_{n}$ .

Step 4 — Balance the two boxes (charge neutrality)

WHAT. The two rectangles cannot have any shape they like. The total negative charge (area of the left box) must equal the total positive charge (area of the right box):

$height of left box q N_{A} \cdot its width x_{p} = height of right box q N_{D} \cdot its width x_{n} \Rightarrow N_{A} x_{p} = N_{D} x_{n}$

WHY. Joining the crystals created no new charge — we only rearranged existing charge. So the plus and minus totals must still cancel. This single line has a big consequence: solving for the ratio, $\frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}},$ which says the depletion region reaches further into the lightly-doped side. Fewer ions per volume there means you must expose a thicker slab to match the same total charge.

PICTURE. The figure re-draws the two boxes with equal shaded areas but different heights and widths — the tall skinny box (heavily doped) and the short wide box (lightly doped). The equal-area shading is the visual meaning of neutrality.

Step 5 — Turn charge into field (integrate Poisson once)

WHAT. Now we bring in the master tool. Gauss's law in one dimension says $\frac{d E}{d x} = \frac{ρ ( x )}{ε _{s}},$ where $E$ is the electric field (force per unit charge) and $ε_{s}$ is the permittivity of silicon — a number measuring how much field a given charge produces in this material.

WHY this tool and not another? We have a charge layout $ρ (x)$ and we want the field it creates. The one equation that connects "charge here" to "field's rate of change" is Gauss's law. Reading it: $d E / d x$ means "how fast the field grows as you step in $x$ ." The equation says that growth rate is set by the local charge density. Where there is charge, the field ramps; where $ρ = 0$ , the field is flat.

Doing the integral, p-side first ( $ρ = - q N_{A}$ , and the field is zero at the outer edge $x = - x_{p}$ because outside is neutral): $E (x) = - \frac{q N _{A}}{ε _{s}} (x + x_{p}), - x_{p} < x < 0$ Term by term: the slope is $- q N_{A} / ε_{s}$ (negative, so the field grows more negative as $x$ increases across the p-box); the factor $(x + x_{p})$ is zero at $x = - x_{p}$ , confirming $E = 0$ at the edge.

n-side ( $ρ = + q N_{D}$ , field zero at the outer edge $x = + x_{n}$ ): $E (x) = - \frac{q N _{D}}{ε _{s}} (x_{n} - x), 0 < x < x_{n}$ Here $(x_{n} - x)$ is zero at $x = x_{n}$ (edge) and largest at $x = 0$ .

Both lines meet at the junction with the most negative value: $E_{m a x} = - \frac{q N _{A} x _{p}}{ε _{s}} = - \frac{q N _{D} x _{n}}{ε _{s}}$ (the two forms are equal because of Step 4's neutrality, $N_{A} x_{p} = N_{D} x_{n}$ ).

PICTURE. The field is two straight lines forming a triangle: zero at both edges, dipping to a sharp point $E_{m a x}$ at the junction. The field points from n toward p (it is negative on our axis), which is exactly the "wind" that pushes carriers back and stops diffusion.

Step 6 — Turn field into potential (integrate again)

WHAT. The potential $ϕ$ relates to the field by $E = - d ϕ / d x$ . So the potential drop across the region — the total climb — is the (negative) area swept out by $E$ : $V_{bi} = - \int_{- x_{p}}^{x_{n}} E d x = \frac{1}{2} ∣ E_{m a x} ∣ W, W = x_{p} + x_{n}$

Reading it: $V_{bi}$ (the built-in potential) is the area of the field triangle. A triangle's area is $\frac{1}{2} (base) (height)$ ; here the base is the total width $W = x_{p} + x_{n}$ and the height is $∣ E_{m a x} ∣$ .

WHY. Potential is "how much uphill work per charge." Since $E$ tells the slope of that hill at every point, adding up (integrating) the slope over the region gives the total height climbed — and geometrically, adding up a triangular slope profile is just the triangle's area. No calculus mystery: it is a triangle.

PICTURE. The figure stacks two panels: on top the triangular $E (x)$ with its area shaded; below, the resulting $ϕ (x)$ — a smooth S-curve (parabola-joined-to-parabola) rising by exactly $V_{bi}$ from the p-side up to the n-side. The steepest part of the S sits right at the junction, where $∣ E ∣$ is biggest.

Step 7 — Solve for the total width $W$

WHAT. We now have two facts about the region:

Neutrality: $x_{p} = \frac{N _{D}}{N _{A} + N _{D}} W$ and $x_{n} = \frac{N _{A}}{N _{A} + N _{D}} W$ (splitting $W$ by the doping ratio).
Triangle area: $V_{bi} = \frac{1}{2} ∣ E_{m a x} ∣ W$ , with $∣ E_{m a x} ∣ = \frac{q N _{A} x _{p}}{ε _{s}}$ .

Substitute $x_{p}$ into $∣ E_{m a x} ∣$ , then into the area, and solve for $W$ : $W = \frac{2 ε _{s} V _{bi}}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}})$

Reading the formula:

$V_{bi}$ on top → a bigger barrier needs a wider charge region to build it. $W$ grows like $V_{bi}$ .
$(\frac{1}{N _{A}} + \frac{1}{N _{D}})$ → higher doping shrinks $W$ : more ions per volume means you expose fewer of them to reach the same total charge. The lighter side dominates the sum (the bigger of the two $1/ N$ terms wins).
$ε_{s}$ on top → a more permissive material spreads charge over a wider region.

WHY the square root? Because we integrated twice (charge → field → potential), each integration adding one power of distance, so the potential grows like $W^{2}$ ; inverting that relationship gives $W \propto V_{bi}$ .

PICTURE. The figure shows $W$ swelling under reverse bias and shrinking under forward bias — a slider from narrow (forward) to wide (reverse). This $V$ behaviour is exactly what makes a varactor work.

Step 8 — The edge cases (never leave the reader stranded)

WHAT & WHY & PICTURE, one degenerate case at a time:

Symmetric junction, $N_{A} = N_{D}$ . Heights equal → widths equal → the region is a mirror image about the junction. The field triangle is isosceles.
One-sided junction, $N_{A} ≫ N_{D}$ (a $p^{+} n$ diode). Then $x_{n} ≫ x_{p}$ : almost the entire region lives in the lightly-doped n-side. We approximate $W \approx x_{n} = 2 ε_{s} V_{bi} / (q N_{D})$ — the heavy side contributes a negligible sliver.
Forward bias $V_{F}$ . The applied voltage opposes the built-in barrier: replace $V_{bi} \to V_{bi} - V_{F}$ . So $W$ shrinks, the field weakens, and carriers start crossing — this is the "on" state in the diode I-V curve.
Reverse bias $V_{R}$ . Replace $V_{bi} \to V_{bi} + V_{R}$ . $W$ grows, field strengthens, region empties further.

The one-picture summary

Everything on one axis, stacked so each layer causes the next:

$ρ (x)$ — two boxes of fixed ion charge (heights $q N_{A}$ , $q N_{D}$ , equal areas).
$E (x)$ — integrate the boxes → a triangle, peaking at the junction.
$ϕ (x)$ — integrate the triangle → a smooth S-curve rising by $V_{bi}$ .
$W$ — the base of the triangle, set by the boxed formula.

Recall Feynman: retell the whole walkthrough in plain words

Two crystals meet. Near the seam, the little movers (electrons and holes) rush across and pair off, disappearing — and each one that leaves uncovers a fixed charged atom stuck in the lattice. So the seam becomes an empty strip with a slab of frozen minus-charge on one side and plus-charge on the other. That charge slab, by the same rule that says "charge bends field lines," makes an electric field that ramps up straight to a peak right at the seam — a triangle. If you now ask "how much uphill voltage is that?", the answer is just the area of the triangle. We already know that voltage from the doping (Fermi levels must line up), so we run the whole thing backwards: known voltage → triangle area → triangle base → the width $W$ . Widen the barrier with reverse voltage and the triangle grows; oppose it with forward voltage and the triangle shrinks. And because we integrated twice, the width follows a square-root law — which is the secret behind the voltage-tunable capacitor.

Recall Quick self-test

Why is the field triangular and not flat? ::: The charge lives in a volume ( $ρ$ constant in each box), so Gauss's law $d E / d x = ρ / ε_{s}$ makes $E$ ramp linearly, peaking at the junction. Which side of the junction is wider, and why? ::: The lightly-doped side. Neutrality $N_{A} x_{p} = N_{D} x_{n}$ forces the region with fewer ions per volume to be thicker to hold equal total charge. Why does $W$ scale as $V_{bi}$ ? ::: Two integrations (charge→field→potential) make potential grow like $W^{2}$ ; inverting gives $W \propto V$ .