Depletion region and space charge: Worked examples

This page is a worked-example gym for the parent depletion-region note. We will not re-derive the theory — instead we hit every case the topic can throw at you: symmetric, asymmetric, one-sided, forward bias, reverse bias, zero bias, degenerate limits, and a real-world twist.

Before we start, one promise: every symbol below is spelled out first, and I restate the two formulas we lean on so you never have to scroll away.

Plain-word decoder (read this before the formulas — every symbol on the page is here):

Symbol	Says out loud	Means
$N_{A}$	"n-a"	acceptor doping (holes' side), per cm³
$N_{D}$	"n-d"	donor doping (electrons' side), per cm³
$n_{i}$	"n-i"	intrinsic carrier density of pure silicon
$x_{p}$	"x-p"	how far the zone reaches into the p-side
$x_{n}$	"x-n"	how far the zone reaches into the n-side
$W$	"width"	$x_{p} + x_{n}$ , total thickness of the zone
$V_{bi}$	"v-b-i"	built-in potential (barrier with no battery)
$V_{R}$	"v-r"	reverse-bias voltage applied by an external battery
$V_{F}$	"v-f"	forward-bias voltage applied by an external battery
$V_{junction}$	"v-junction"	the total voltage across the junction (built-in ± applied)
$E_{ma x}$	"e-max"	peak electric field, right at the junction
$A$	"area"	cross-sectional area of the diode (cm²)
$ε_{s}$	"epsilon-s"	how easily silicon stores field (permittivity)
$k T / q$	"k-t over q"	thermal voltage $= 0.0259$ V at 300 K
$q$	"q"	electron charge $1.6 \times 1 0^{- 19}$ C

Constants for silicon at 300 K used throughout: $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ , $ε_{s} = 1.05 \times 1 0^{- 12} F/cm$ , $k T / q = 0.0259 V$ .

Intuition The bias sign convention — why

$V_{R}$ adds and $V_{F}$ subtracts The built-in field already points n→p, pushing carriers back. Think of $V_{bi}$ as a hill the carriers must climb.

Reverse bias connects the battery so it pulls electrons out of the n-side and holes out of the p-side — away from the junction. This exposes more ions and stacks on top of the built-in hill: $V_{junction} = V_{bi} + V_{R}$ → the barrier and width grow.
Forward bias connects the battery the other way, pushing carriers toward the junction to re-fill the ions — it fights the built-in hill: $V_{junction} = V_{bi} - V_{F}$ → the barrier and width shrink.
No battery: $V_{junction} = V_{bi}$ .

Mnemonic: the applied voltage adds when it agrees with $V_{bi}$ 's direction (reverse) and subtracts when it opposes it (forward).

The scenario matrix

Every depletion-region problem is one of these cells. The examples below are labelled with the cell they cover, and together they fill the whole grid.

Cell	Case class	What's special	Hit by
A	Symmetric junction $N_{A} = N_{D}$	$x_{p} = x_{n}$ , region centred	Ex 1
B	Asymmetric $N_{A} \neq = N_{D}$	region bulges into light side	Ex 2
C	One-sided $p^{+} n$ (huge ratio)	region "lives" in light side	Ex 3
D	Zero bias (equilibrium)	$V_{junction} = V_{bi}$ only	Ex 1, 2
E	Reverse bias $+ V_{R}$	barrier & width grow	Ex 4
F	Forward bias $- V_{F}$	barrier & width shrink	Ex 5
G	Degenerate limit: $V_{F} \to V_{bi}$	$W \to 0$ , region collapses	Ex 6
H	Real-world word problem (varactor)	capacitance from width	Ex 7
I	Exam twist: peak field / breakdown	find $E_{ma x}$ , not just $W$	Ex 8

The decision tree below is your GPS: answer the questions top-down and it drops you into the right cell.

Figure — Depletion region and space charge

Decision tree for picking a scenario cell: first ask "is a battery attached?" (bias branch), then "is the doping equal?" (symmetry branch), landing you on the correct worked example.

Read the "Forecast" line and guess before reading the steps — that guess is where the learning happens.

Recall The neutrality split — the one rule behind every "which side is wider"

The parent note proves it, but here is the whole idea in one breath: the crystal was neutral before the junction formed, and diffusion only rearranged charge — it never created any. So the total exposed positive charge (donor ions, $q N_{D}$ per volume, over depth $x_{n}$ ) must equal the total exposed negative charge (acceptor ions, $q N_{A}$ per volume, over depth $x_{p}$ ): $N_{A} x_{p} = N_{D} x_{n} \Rightarrow \frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}}$ Read it as: the side with fewer ions per volume must reach deeper to expose the same charge — so the zone always bulges into the lightly-doped side.

Cell A + D — Symmetric junction at equilibrium

Worked example Example 1 — the balanced case

Silicon: $N_{A} = N_{D} = 1 0^{16} cm^{- 3}$ , no battery attached. Find $V_{bi}$ , $W$ , and confirm $x_{p} = x_{n}$ .

Forecast: Equal doping → the zone should sit dead centre, and $V_{bi}$ should land near the famous $\approx 0.7$ V of a silicon diode. Does it?

Step 1 — Built-in potential. $V_{bi} = 0.0259 ln \frac{( 1 0 ^{16} ) ( 1 0 ^{16} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln (4.44 \times 1 0^{11}) \approx 0.69 V$ Why this step? No battery means the only voltage across the junction is the one the ions build themselves — that is exactly $V_{bi}$ .

Step 2 — Width. With $\frac{1}{N _{A}} + \frac{1}{N _{D}} = 2 \times 1 0^{- 16} cm^{3}$ : $W = \frac{2 ( 1.05 \times 1 0 ^{- 12} ) ( 0.69 )}{1.6 \times 1 0 ^{- 19}} (2 \times 1 0^{- 16}) \approx 4.2 \times 1 0^{- 5} cm = 0.42 μ m$ Why this step? We plug the equilibrium voltage $V_{bi}$ into the width formula — nothing external to add.

Step 3 — The two half-widths. Neutrality says $N_{A} x_{p} = N_{D} x_{n}$ . Equal doping ⇒ $x_{p} = x_{n} = W /2 = 0.21 μ m$ . Why this step? Same number of ions per volume on each side, so the same depth exposes the same charge — perfect symmetry.

Verify: Units of $W$ : $F/cm \cdot V \cdot cm^{3} / C = C/cm \cdot cm^{3} / C = cm^{2} = cm$ ✓. And $x_{p} + x_{n} = 0.21 + 0.21 = 0.42 μ m = W$ ✓.

Cell B + D — Asymmetric junction

Worked example Example 2 — bulge into the light side

$N_{A} = 1 0^{17} cm^{- 3}$ , $N_{D} = 1 0^{15} cm^{- 3}$ , equilibrium. Find $V_{bi}$ , $W$ , and the split $x_{p} : x_{n}$ .

Forecast: The n-side is 100× lighter. Which way does the zone bulge — into the heavy p-side or the light n-side? Guess before Step 3.

Step 1 — Built-in potential. $V_{bi} = 0.0259 ln \frac{( 1 0 ^{17} ) ( 1 0 ^{15} )}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln (4.44 \times 1 0^{11}) \approx 0.69 V$ Why this step? Notice: the product $N_{A} N_{D} = 1 0^{32}$ is the same as Example 1, so $V_{bi}$ is unchanged. Only the product matters for the barrier height.

Step 2 — Width. Now $\frac{1}{N _{A}} + \frac{1}{N _{D}} = 1 0^{- 17} + 1 0^{- 15} \approx 1.01 \times 1 0^{- 15} cm^{3}$ — dominated by the light side. $W = \frac{2 ( 1.05 \times 1 0 ^{- 12} ) ( 0.69 )}{1.6 \times 1 0 ^{- 19}} (1.01 \times 1 0^{- 15}) \approx 9.44 \times 1 0^{- 5} cm = 0.94 μ m$ Why this step? The $1/ N$ terms mean the lightly-doped side dominates the width — the same reason wider spacing appears where ions are scarce.

Step 3 — The split. Using the neutrality split $\frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}} = \frac{1 0 ^{17}}{1 0 ^{15}} = 100$ . The zone reaches 100× deeper into the light n-side — it bulges into the lightly-doped side (see figure). Why this step? To expose equal charge, the side with fewer ions per volume must reach further.

Charge density $ρ (x)$ : the p-side ion block is tall and thin (heavy doping $N_{A}$ , small $x_{p}$ ); the n-side block is short and wide (light doping $N_{D}$ , large $x_{n}$ ). Equal shaded areas encode $N_{A} x_{p} = N_{D} x_{n}$ .

Verify (with explicit units): $x_{p} = W /101 = 0.0093 μ m$ , $x_{n} = 100 W /101 = 0.935 μ m$ ; sum $= 0.944 μ m = W$ ✓. Now the charge balance with units, converting $μ m \to cm$ ( $1 μ m = 1 0^{- 4} cm$ ): $N_{A} x_{p} = 1 0^{17} \frac{1}{cm ^{3}} \times 0.0093 \times 1 0^{- 4} cm = 9.3 \times 1 0^{11} \frac{1}{cm ^{2}}$ $N_{D} x_{n} = 1 0^{15} \frac{1}{cm ^{3}} \times 0.935 \times 1 0^{- 4} cm = 9.35 \times 1 0^{11} \frac{1}{cm ^{2}}$ Both are $\approx 9.3 \times 1 0^{11}$ ions/cm² — equal to rounding ✓. (Both sides carry units of ions per unit area, exactly what "exposed sheet charge" should be.)

Prerequisite reminder: this split comes straight from Doping (Donors and Acceptors) and the neutrality argument in Poisson's Equation and Gauss's Law.

Cell C — One-sided $p^{+} n$ junction

Worked example Example 3 — the "

$p^{+} n$ " limit $N_{A} = 1 0^{19} cm^{- 3}$ (very heavy), $N_{D} = 1 0^{15} cm^{- 3}$ . Show the depletion region effectively lives entirely in the n-side, and compute $W$ .

Forecast: With a 10,000× doping ratio, what fraction of $W$ is on the p-side — 50%? 1%? 0.01%?

Step 1 — Split. From the neutrality split, $\frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}} = 1 0^{4}$ . So $x_{p}$ is $1/1 0^{4}$ of $x_{n}$ — utterly negligible. Why this step? The heavily-doped $p^{+}$ side exposes all its charge in a razor-thin sliver.

Step 2 — One-sided approximation. Since $\frac{1}{N _{A}} = 1 0^{- 19} ≪ \frac{1}{N _{D}} = 1 0^{- 15}$ , drop the $1/ N_{A}$ term: $W \approx \frac{2 ε _{s} V _{bi}}{q N _{D}}$ Why this step? Whenever one side is $\geq 100 \times$ heavier, its $1/ N$ term vanishes — this is the standard engineering simplification.

Step 3 — Numbers. $V_{bi} = 0.0259 ln \frac{1 0 ^{19} \cdot 1 0 ^{15}}{( 1.5 \times 1 0 ^{10} ) ^{2}} = 0.0259 ln (4.44 \times 1 0^{13}) \approx 0.81 V$ . $W \approx \frac{2 ( 1.05 \times 1 0 ^{- 12} ) ( 0.81 )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{15} )} \approx 1.03 \times 1 0^{- 4} cm = 1.03 μ m$ Why this step? Heavier p-doping nudged $V_{bi}$ up (bigger log argument), so the width edges up slightly too.

Verify: Keeping the full formula, $\frac{1}{N _{A}} + \frac{1}{N _{D}} = 1.0001 \times 1 0^{- 15}$ — a 0.005% difference from $1 0^{- 15}$ . The one-sided answer matches the full one to 4 digits ✓.

Cell E — Reverse bias

Worked example Example 4 — pull the junction open

Take the Example 1 junction ( $V_{bi} = 0.69$ V) and apply $V_{R} = 8$ V reverse bias. Find the new width $W_{R}$ .

Forecast: Reverse bias adds to the barrier. Will $W$ grow by a factor of 8, or by $8$ -ish, or barely at all?

Step 1 — Update the junction voltage. Reverse bias reinforces the built-in field: $V_{junction} = V_{bi} + V_{R} = 0.69 + 8 = 8.69 V$ . Why this step? The external battery pulls carriers away from the junction, exposing more ions — same direction as $V_{bi}$ (see the bias sign-convention box above).

Step 2 — Scale the width. $W \propto V_{junction}$ , so $W_{R} = W_{0} \frac{V _{bi} + V _{R}}{V _{bi}} = 0.42 \frac{8.69}{0.69} = 0.42 12.6 \approx 1.49 μ m$ Why this step? Everything except $V$ stays fixed, so we just rescale by the square-root ratio — no need to recompute from scratch.

Verify: $12.6 \approx 3.55$ ; $0.42 \times 3.55 = 1.49 μ m$ . A 13× voltage increase gave only a 3.5× width increase — the square-root softens it, exactly as should ✓. This widening under reverse bias is the engine of the varactor.

Cell F — Forward bias

Worked example Example 5 — squeeze the junction shut

Example 1 junction, apply $V_{F} = 0.5$ V forward bias. Find the new width $W_{F}$ .

Forecast: Forward bias fights $V_{bi}$ . The remaining barrier is $0.69 - 0.5 = 0.19$ V. Does $W$ shrink to about a quarter, a half, or stay nearly the same?

Step 1 — Update junction voltage. Forward bias opposes the built-in field: $V_{junction} = V_{bi} - V_{F} = 0.69 - 0.5 = 0.19 V$ . Why this step? The battery pushes carriers toward the junction, letting them re-fill and re-neutralise ions — the opposite of Example 4 (again, the sign-convention box explains why we subtract here).

Step 2 — Scale the width. $W_{F} = W_{0} \frac{V _{bi} - V _{F}}{V _{bi}} = 0.42 \frac{0.19}{0.69} = 0.42 0.275 \approx 0.22 μ m$ Why this step? Same rescaling trick; here the ratio is below 1, so the width shrinks.

Verify: $0.275 \approx 0.525$ ; $0.42 \times 0.525 = 0.22 μ m$ — roughly half. A thinner barrier lets carriers cross more easily, which is why forward bias switches a diode on (see Diode I-V Characteristics and Drift and Diffusion Currents) ✓.

Cell G — Degenerate limit $V_{F} \to V_{bi}$

Worked example Example 6 — how far can forward bias go?

Same junction. What happens to $W$ as $V_{F}$ approaches $V_{bi} = 0.69$ V? Try $V_{F} = 0.68$ V.

Forecast: As the barrier drops toward zero, does $W$ approach some minimum thickness, or collapse to zero?

Step 1 — Take the limit. $V_{junction} = V_{bi} - V_{F} \to 0$ , so $W_{F} = W_{0} \frac{V _{bi} - V _{F}}{V _{bi}} ⟶ 0.$ Numerically at $V_{F} = 0.68$ : $W_{F} = 0.42 \frac{0.01}{0.69} = 0.42 0.0145 \approx 0.051 μ m$ . Why this step? Mathematically the depletion width vanishes as the barrier disappears.

Step 2 — Physics reality check. The depletion approximation breaks down here: with no barrier, huge diffusion current flows, and our "empty box of ions" picture no longer holds. The formula is only valid while $V_{F} < V_{bi}$ . Why this step? Always test whether a limiting formula still respects its own assumptions — here it does not once carriers flood back in.

Verify: $0.0145 \approx 0.120$ ; $0.42 \times 0.120 = 0.050 μ m$ ✓. The zone has shrunk to ~12% of its equilibrium width — trending correctly toward zero.

Cell H — Real-world varactor word problem

Worked example Example 7 — tuning a radio

A $p^{+} n$ diode ( $N_{D} = 1 0^{15} cm^{- 3}$ , area $A = 1 0^{- 4} cm^{2}$ , $V_{bi} = 0.81$ V) is used as a voltage-tuned capacitor. Find its capacitance $C$ at $V_{R} = 8$ V reverse bias.

Forecast: The depletion region acts like a parallel-plate capacitor $C = ε_{s} A / W$ . More reverse bias → wider $W$ → smaller $C$ . Roughly how many picofarads?

Step 1 — Width at 8 V (one-sided). $W = \frac{2 ε _{s} ( V _{bi} + V _{R} )}{q N _{D}} = \frac{2 ( 1.05 \times 1 0 ^{- 12} ) ( 8.81 )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{15} )} \approx 3.40 \times 1 0^{- 4} cm$ Why this step? $p^{+} n$ ⇒ one-sided approximation from Ex 3, with $V_{junction} = V_{bi} + V_{R} = 8.81$ V.

Step 2 — Capacitance. The two neutral regions are the "plates", the depletion zone is the "gap": $C = \frac{ε _{s} A}{W} = \frac{( 1.05 \times 1 0 ^{- 12} ) ( 1 0 ^{- 4} )}{3.40 \times 1 0 ^{- 4}} \approx 3.09 \times 1 0^{- 13} F = 0.31 pF$ Why this step? A block of fixed charge with an empty gap is a capacitor — see Junction Capacitance and Varactors.

Verify: Units: $\frac{( F/cm ) ( cm ^{2} )}{cm} = F$ ✓. At $0$ V bias the same diode has $W_{0} = 8.81 \to 0.81$ smaller, i.e. $W_{0} = 3.40 \times 1 0^{- 4} 0.81/8.81 = 1.03 \times 1 0^{- 4}$ cm, giving $C_{0} = 1.02 pF$ — tuning from 1.02 pF down to 0.31 pF as you sweep bias. That 3.3× swing is exactly how you retune a station ✓.

Cell I — Exam twist: peak field & breakdown

Worked example Example 8 — will it break down?

For the Example 4 reverse-biased junction ( $N_{A} = N_{D} = 1 0^{16} cm^{- 3}$ , $W_{R} = 1.49 μ m$ at $V_{R} = 8$ V), find the peak field $E_{ma x}$ at the junction. Silicon breaks down near $∣ E ∣ \approx 3 \times 1 0^{5} V/cm$ — is this junction safe?

Forecast: The field is triangular, peaking at the junction. Is the peak field a few thousand V/cm, or already near the breakdown ceiling?

Step 1 — Peak field from the triangle. The area under the triangular $∣ E ∣$ -vs- $x$ curve equals the total voltage: $V_{junction} = \frac{1}{2} ∣ E_{ma x} ∣ W \Rightarrow ∣ E_{ma x} ∣ = \frac{2 V _{junction}}{W}$ Why this step? From the parent note, $V = area of the triangle$ ; invert it for the peak.

Triangular field profile: $∣ E ∣$ starts at zero at each edge of the depletion zone, grows linearly through each charged slab, and peaks at the junction. The shaded area equals $V_{junction}$ , giving $∣ E_{ma x} ∣ = 2 V_{junction} / W$ .

Step 2 — Numbers. $V_{junction} = V_{bi} + V_{R} = 0.69 + 8 = 8.69$ V, and $W = W_{R} = 1.49 μ m = 1.49 \times 1 0^{- 4}$ cm: $∣ E_{ma x} ∣ = \frac{2 ( 8.69 )}{1.49 \times 1 0 ^{- 4}} = \frac{17.38}{1.49 \times 1 0 ^{- 4}} \approx 1.17 \times 1 0^{5} V/cm$ Why this step? Direct substitution — no new physics needed. (Watch the units: volts divided by centimetres gives V/cm.)

Step 3 — Verdict. $1.17 \times 1 0^{5} V/cm < 3 \times 1 0^{5} V/cm$ , so the junction is safe — it sits at about $\frac{1.17}{3.0} \approx 39%$ of the breakdown field. Why this step? Compare against the material limit; exceeding it triggers avalanche breakdown.

Verify: Cross-check via $E_{ma x} = q N_{D} x_{n} / ε_{s}$ with $x_{n} = W /2 = 0.745 \times 1 0^{- 4}$ cm: $E_{ma x} = \frac{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} ) ( 0.745 \times 1 0 ^{- 4} )}{1.05 \times 1 0 ^{- 12}} = 1.13 \times 1 0^{5} V/cm$ — matches the triangle method to within rounding ✓.

Active Recall

Recall Which cell is each phrase?

"Region bulges into the light side" — which cell? ::: Cell B (asymmetric) " $W \to 0$ as barrier vanishes" — which cell? ::: Cell G (degenerate forward limit) " $W$ grows like $V_{bi} + V_{R}$ " — which cell? ::: Cell E (reverse bias) "Drop the $1/ N_{A}$ term" — which cell? ::: Cell C (one-sided $p^{+} n$ )

Recall The one bias rule everyone forgets

Under reverse bias, do you add or subtract $V_{R}$ from $V_{bi}$ ? ::: Add — reverse bias reinforces the barrier ( $V_{bi} + V_{R}$ ), widening $W$ . Forward bias subtracts ( $V_{bi} - V_{F}$ ), shrinking it.