Depletion region and space charge: Exercises — Hardware deep dive

This page is a self-testing ladder. Each rung is harder than the last. Try each problem with the solution hidden, then open the [!recall]- callout to check every step. All the physics comes from the parent note — if a symbol here feels unfamiliar, that note builds it from zero.

Constants used throughout (silicon, 300 K unless stated): $q = 1.6 \times 1 0^{- 19} C$ , $k T / q = 0.0259 V$ , $n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ , $ε_{s} = 1.05 \times 1 0^{- 12} F/cm$ .

Prerequisite threads if you get stuck: Doping (Donors and Acceptors), Poisson's Equation and Gauss's Law, Built-in Potential and Fermi Level Alignment, Diode I-V Characteristics, Junction Capacitance and Varactors, Drift and Diffusion Currents.

L1 — Recognition

Can you name the pieces and read the signs?

Exercise 1.1

A silicon junction sits at equilibrium. Inside the depletion region, list what charge is present and what charge is absent, on each side.

Recall Solution

WHAT we're doing: just recalling the definition — no maths.

n-side of the region: the free electrons have diffused away and recombined. Left behind are the donor cores that gave up an electron → fixed positive ions, $N_{D}^{+}$ .
p-side of the region: the holes have diffused away. Left behind are the acceptor cores that captured an electron → fixed negative ions, $N_{A}^{-}$ .
Absent everywhere in the region: mobile carriers (free electrons and free holes). That is the whole point of the word depletion.

Answer: n-side = fixed $+$ donor ions; p-side = fixed $-$ acceptor ions; no mobile carriers anywhere inside.

Exercise 1.2

The built-in electric field inside the region points from which side toward which side? What does it do to a stray electron that wanders in from the p-side?

Recall Solution

WHAT we're doing: reading the sign of $E$ . The n-side ions are positive, the p-side ions are negative. Field lines start on positive charge and end on negative charge, so $E$ points from the n-side to the p-side (n → p).

A stray electron carries negative charge, so the force on it is $F = - q E$ , i.e. opposite to the field — it gets pushed back toward the n-side. That is exactly the drift force that cancels diffusion at equilibrium.

L2 — Application

Plug into one formula, get a number.

Exercise 2.1

A one-sided junction has $N_{A} = 5 \times 1 0^{17} cm^{- 3}$ and $N_{D} = 5 \times 1 0^{15} cm^{- 3}$ . Find the ratio $x_{n} / x_{p}$ and state which side the region bulges into.

Recall Solution

WHY neutrality: no net charge was created, only rearranged, so exposed $+$ and $-$ charge must match: $N_{A} x_{p} = N_{D} x_{n}$ . $\frac{x _{n}}{x _{p}} = \frac{N _{A}}{N _{D}} = \frac{5 \times 1 0 ^{17}}{5 \times 1 0 ^{15}} = 100.$ Answer: $x_{n} / x_{p} = 100$ . The region reaches 100× deeper into the lightly-doped n-side. Fewer ions per cm³ there means you must expose a thicker slab to store the same total charge.

Exercise 2.2

Silicon, $N_{A} = N_{D} = 1 0^{17} cm^{- 3}$ . Compute $V_{bi}$ at 300 K.

Recall Solution

WHY the log formula: the Fermi level must be flat at equilibrium; the band-bending needed equals $q V_{bi}$ , and Boltzmann statistics turn a concentration ratio into a logarithm. $V_{bi} = 0.0259 ln (\frac{( 1 0 ^{17} ) ( 1 0 ^{17} )}{( 1.5 \times 1 0 ^{10} ) ^{2}}) = 0.0259 ln (4.44 \times 1 0^{13}) .$ $ln (4.44 \times 1 0^{13}) \approx 31.43$ , so $V_{bi} \approx 0.0259 \times 31.43 \approx 0.814 V .$ Answer: $V_{bi} \approx 0.81 V$ . Higher doping than the parent's $1 0^{16}$ example nudged $V_{bi}$ up from $0.69$ V toward $0.8$ V — the log grows slowly, so even $10 \times$ more doping adds only ~0.12 V.

Exercise 2.3

Using $V_{bi} = 0.81 V$ from Ex 2.2 and $N_{A} = N_{D} = 1 0^{17} cm^{- 3}$ , find the total width $W$ .

Recall Solution

WHY this formula: it is Poisson's equation integrated twice under the depletion approximation, then solved for width. $\frac{1}{N _{A}} + \frac{1}{N _{D}} = 2 \times 1 0^{- 17} cm^{3} .$ $W = \frac{2 ( 1.05 \times 1 0 ^{- 12} ) ( 0.81 )}{1.6 \times 1 0 ^{- 19}} (2 \times 1 0^{- 17}) .$ Numerator inside root: $2 \times 1.05 \times 1 0^{- 12} \times 0.81 = 1.70 \times 1 0^{- 12}$ . Divide by $1.6 \times 1 0^{- 19}$ : $1.063 \times 1 0^{7}$ . Times $2 \times 1 0^{- 17}$ : $2.13 \times 1 0^{- 10} cm^{2}$ . $W = 2.13 \times 1 0^{- 10} \approx 1.46 \times 1 0^{- 5} cm = 0.146 μ m .$ Answer: $W \approx 0.15 μ m$ . Heavier doping than the parent's $0.42 μ m$ example gave a thinner region — more ions per cm³ means less thickness needed.

L3 — Analysis

Chain two ideas, or reason about a limit.

Exercise 3.1

For the junction of Ex 2.2–2.3 ( $W = 0.146 μ m$ , $V_{bi} = 0.81 V$ ), find the peak electric field $∣ E_{ma x} ∣$ at the metallurgical junction.

Recall Solution

WHY a triangle: by Poisson, $E$ grows linearly through each charged slab and peaks at $x = 0$ . The potential $V_{bi}$ is the area under the $∣ E ∣$ -vs- $x$ triangle, base $W$ and height $∣ E_{ma x} ∣$ : $V_{bi} = \frac{1}{2} ∣ E_{ma x} ∣ W \Rightarrow ∣ E_{ma x} ∣ = \frac{2 V _{bi}}{W} .$ Look at the triangular field in the figure below — base $W$ , apex at the junction.

Figure — Depletion region and space charge

Convert $W$ to cm: $0.146 μ m = 1.46 \times 1 0^{- 5} cm$ . $∣ E_{ma x} ∣ = \frac{2 ( 0.81 )}{1.46 \times 1 0 ^{- 5}} \approx 1.11 \times 1 0^{5} V/cm .$ Answer: $∣ E_{ma x} ∣ \approx 1.1 \times 1 0^{5} V/cm$ — comfortably below silicon's ~ $3 \times 1 0^{5} V/cm$ breakdown field, so this junction is safe at equilibrium.

Exercise 3.2

Apply $V_{R} = 10 V$ reverse bias to the junction of Ex 2.3 ( $V_{bi} = 0.81 V$ , $W_{0} = 0.146 μ m$ ). Find the new width $W$ and the new peak field.

Recall Solution

WHY replace $V_{bi} \to V_{bi} + V_{R}$ : reverse bias adds to the barrier, pulling more carriers away and exposing more ions. Since $W \propto V_{bi} + V_{R}$ : $W = W_{0} \frac{V _{bi} + V _{R}}{V _{bi}} = 0.146 \frac{10.81}{0.81} = 0.146 13.35 \approx 0.146 \times 3.65 \approx 0.534 μ m .$ New peak field: $∣ E_{ma x} ∣ = \frac{2 ( V _{bi} + V _{R} )}{W}$ with $W = 5.34 \times 1 0^{- 5} cm$ : $∣ E_{ma x} ∣ = \frac{2 ( 10.81 )}{5.34 \times 1 0 ^{- 5}} \approx 4.05 \times 1 0^{5} V/cm .$ Answer: $W \approx 0.53 μ m$ , $∣ E_{ma x} ∣ \approx 4.0 \times 1 0^{5} V/cm$ . Analysis: width grew $3.65 \times$ but the field grew even faster ( $1.1 \to 4.0$ , a $3.65 \times$ jump too) because the voltage in the numerator scaled up 13×. This field now exceeds the ~ $3 \times 1 0^{5} V/cm$ breakdown threshold — reverse-biasing this hard would likely avalanche.

L4 — Synthesis

Combine width, capacitance, and bias into one story.

Exercise 4.1

The depletion region acts like a parallel-plate capacitor with the ions as the plates: $C_{j} = ε_{s} A / W$ per the Junction Capacitance and Varactors model. For the junction of Ex 2.3 with area $A = 1 0^{- 4} cm^{2}$ : (a) find $C_{j}$ at zero bias, and (b) find $C_{j}$ at $V_{R} = 10 V$ . Comment on the ratio.

Recall Solution

WHY $C_{j} = ε_{s} A / W$ : the two edges of the depletion region hold equal-and-opposite fixed charge separated by distance $W$ — geometrically identical to a parallel-plate capacitor.

(a) Zero bias ( $W_{0} = 1.46 \times 1 0^{- 5} cm$ ): $C_{j} = \frac{( 1.05 \times 1 0 ^{- 12} ) ( 1 0 ^{- 4} )}{1.46 \times 1 0 ^{- 5}} \approx 7.19 \times 1 0^{- 12} F = 7.2 pF .$

(b) At $V_{R} = 10 V$ ( $W = 5.34 \times 1 0^{- 5} cm$ from Ex 3.2): $C_{j} = \frac{( 1.05 \times 1 0 ^{- 12} ) ( 1 0 ^{- 4} )}{5.34 \times 1 0 ^{- 5}} \approx 1.97 \times 1 0^{- 12} F = 2.0 pF .$

Ratio: $C_{j} (0) / C_{j} (10) = 7.2/2.0 \approx 3.65$ — exactly the width ratio, since $C \propto 1/ W$ and $W \propto V_{bi} + V_{R}$ . Synthesis: the capacitance tunes with voltage as $C_{j} \propto (V_{bi} + V_{R})^{- 1/2}$ . This voltage-controlled capacitor is a varactor — sweep the reverse bias and you sweep the resonant frequency of an LC circuit. One junction, three concepts (width, field, capacitance) chained.

Exercise 4.2

The same diode is now put under forward bias $V_{F} = 0.5 V$ . Predict qualitatively whether $W$ , $∣ E_{ma x} ∣$ , and $C_{j}$ go up or down, then compute the new $W$ .

Recall Solution

WHY $V_{bi} \to V_{bi} - V_{F}$ : forward bias opposes the built-in barrier, letting carriers flood back in and re-neutralise the edges.

$W$ shrinks (barrier voltage falls).
$∣ E_{ma x} ∣ = 2 (V_{bi} - V_{F}) / W$ : numerator falls, so field shrinks.
$C_{j} = ε_{s} A / W$ : $W$ shrinks, so $C_{j}$ rises.

Compute: $V_{bi} - V_{F} = 0.81 - 0.5 = 0.31 V$ . $W = W_{0} \frac{V _{bi} - V _{F}}{V _{bi}} = 0.146 \frac{0.31}{0.81} = 0.146 0.383 \approx 0.146 \times 0.619 \approx 0.090 μ m .$ Answer: $W \approx 0.090 μ m$ (down from $0.146$ ). A thinner, weaker barrier is exactly what lets forward current surge — the Diode I-V Characteristics exponential turn-on.

L5 — Mastery

Design backwards, or push to a limit.

Exercise 5.1 (Design)

You need a varactor whose zero-bias depletion width is exactly $W_{0} = 0.30 μ m$ on a symmetric silicon junction ( $N_{A} = N_{D} = N$ ). Find the required doping $N$ . (Note: $V_{bi}$ itself depends on $N$ , so this needs iteration.)

Recall Solution

WHY iterate: $W$ depends on both $N$ (explicitly) and $V_{bi}$ (which itself depends on $N$ ). We can't isolate $N$ in closed form, so we guess, compute, refine.

For a symmetric junction $\frac{1}{N _{A}} + \frac{1}{N _{D}} = \frac{2}{N}$ , so $W_{0} = \frac{2 ε _{s} V _{bi}}{q} \cdot \frac{2}{N} = \frac{4 ε _{s} V _{bi}}{q N} .$ Target $W_{0} = 3 \times 1 0^{- 5} cm$ , so $W_{0}^{2} = 9 \times 1 0^{- 10} cm^{2}$ .

Guess 1: $N = 1 0^{16}$ . Then $V_{bi} = 0.0259 ln \frac{1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 0.0259 ln (4.44 \times 1 0^{11}) \approx 0.694 V$ . $W_{0}^{2} = \frac{4 ( 1.05 \times 1 0 ^{- 12} ) ( 0.694 )}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} )} = \frac{2.915 \times 1 0 ^{- 12}}{1.6 \times 1 0 ^{- 3}} = 1.82 \times 1 0^{- 9}$ . Too big (want $9 \times 1 0^{- 10}$ ) → need higher $N$ .

Guess 2: $N = 2 \times 1 0^{16}$ . Then $V_{bi} = 0.0259 ln \frac{4 \times 1 0 ^{32}}{2.25 \times 1 0 ^{20}} = 0.0259 ln (1.78 \times 1 0^{12}) \approx 0.730 V$ . $W_{0}^{2} = \frac{4 ( 1.05 \times 1 0 ^{- 12} ) ( 0.730 )}{( 1.6 \times 1 0 ^{- 19} ) ( 2 \times 1 0 ^{16} )} = \frac{3.066 \times 1 0 ^{- 12}}{3.2 \times 1 0 ^{- 3}} = 9.58 \times 1 0^{- 10}$ . Just above target.

Guess 3: $N = 2.15 \times 1 0^{16}$ . $V_{bi} \approx 0.0259 ln (2.05 \times 1 0^{12}) \approx 0.734 V$ . $W_{0}^{2} = \frac{4 ( 1.05 \times 1 0 ^{- 12} ) ( 0.734 )}{( 1.6 \times 1 0 ^{- 19} ) ( 2.15 \times 1 0 ^{16} )} = \frac{3.083 \times 1 0 ^{- 12}}{3.44 \times 1 0 ^{- 3}} = 8.96 \times 1 0^{- 10} \approx 9 \times 1 0^{- 10}$ . ✔

Answer: $N \approx 2.1 \times 1 0^{16} cm^{- 3}$ (each side). Notice how weakly $V_{bi}$ moved during iteration (0.69 → 0.73 V) — the doping does almost all the work through the explicit $1/ N$ .

Exercise 5.2 (Limit)

In a $p^{+} n$ junction, $N_{A} ≫ N_{D}$ (say $N_{A} = 1 0^{19}$ , $N_{D} = 1 0^{15} cm^{- 3}$ ). Show that $W$ collapses to a one-sided formula depending almost entirely on the lightly-doped side, and give that formula.

Recall Solution

WHAT the limit does: start from $\frac{1}{N _{A}} + \frac{1}{N _{D}}$ . With $N_{A} = 1 0^{19}$ and $N_{D} = 1 0^{15}$ : $\frac{1}{N _{A}} = 1 0^{- 19}, \frac{1}{N _{D}} = 1 0^{- 15} .$ The first term is 10,000× smaller — negligible. So $\frac{1}{N _{A}} + \frac{1}{N _{D}} \approx \frac{1}{N _{D}} .$ Therefore $W \approx \frac{2 ε _{s} ( V _{bi} + V _{R} )}{q N _{D}} .$ Look at the geometry — almost the entire region lives in the n-side:

By neutrality, $x_{p} / x_{n} = N_{D} / N_{A} = 1 0^{- 4}$ , so $x_{p}$ is negligible and $W \approx x_{n}$ . Answer: the width is set by the lightly-doped side alone. This is why datasheets for $p^{+} n$ diodes quote only $N_{D}$ — the heavy $p^{+}$ side contributes an atomically-thin sliver of depletion.

Recall One-line summary of the whole ladder

Everything reduces to: neutrality sets the shape, $V_{bi}$ from the log sets the barrier, $W = \frac{2 ε _{s} ( V _{bi} \pm V )}{q} (\frac{1}{N _{A}} + \frac{1}{N _{D}})$ sets the size, and the triangle $V = \frac{1}{2} E_{ma x} W$ ties field, potential, and width together. Reverse widens & strengthens; forward narrows & weakens.