Depletion region and space charge

WHAT is happening?

WHY does it form?

Free electrons are dense on the n-side, holes dense on the p-side.
Concentration gradient → diffusion. Electrons diffuse into the p-side, holes into the n-side.
They recombine, wiping out mobile carriers near the boundary.
The now-uncovered fixed dopant ions can't move — they stay put as space charge.
This charge makes an E-field pointing n→p which pushes carriers back (drift).
Equilibrium = drift exactly cancels diffusion. Net current = 0.

Figure — Depletion region and space charge

HOW to derive the field, potential, and width

We use Poisson's equation in 1D. This is the master tool — everything else falls out.

WHY this equation? Gauss's law says charge creates diverging field ( $\nabla\cdot E = \rho/\varepsilon$ ). In 1D that's $dE/dx = \rho/\varepsilon_s$ . And $E = -d\phi/dx$ by definition of potential.

Step 1 — The depletion approximation

Assume the space charge is a clean box: fully depleted inside, neutral outside. $\rho(x) = \begin{cases} -qN_A & -x_p < x < 0 \ (\text{p-side}) \\ +qN_D & 0 < x < x_n \ (\text{n-side}) \end{cases}$

Why this step? Real carrier profiles are smooth, but treating the edges as sharp turns messy integrals into simple ones — and it's accurate because carriers vanish fast (over a Debye length).

Step 2 — Charge neutrality

The whole crystal is neutral, so total positive = total negative charge: $qN_A x_p = qN_D x_n \quad\Rightarrow\quad ==N_A x_p = N_D x_n==$

Why this step? No net charge was created — we only rearranged it. So the exposed positive and negative charge must balance. Consequence: the depletion region extends further into the lightly-doped side.

Step 3 — Integrate for the E-field

Integrate $dE/dx = \rho/\varepsilon_s$ . On the p-side ( $\rho=-qN_A$ ), with $E=0$ at $x=-x_p$ : $E(x) = -\frac{qN_A}{\varepsilon_s}(x + x_p), \qquad -x_p < x < 0$ On the n-side ( $\rho=+qN_D$ ), with $E=0$ at $x=x_n$ : $E(x) = -\frac{qN_D}{\varepsilon_s}(x_n - x), \qquad 0 < x < x_n$

The field is maximum at the junction ( $x=0$ ), triangular in shape: $E_{max} = -\frac{qN_A x_p}{\varepsilon_s} = -\frac{qN_D x_n}{\varepsilon_s}$

Why this step? The field starts at zero at the edges (neutral region), grows linearly through each charged slab, and peaks at the metallurgical junction. The two expressions match at $x=0$ because of Step 2's neutrality.

Step 4 — Integrate for the potential (built-in voltage)

$V_{bi}$ is the area under the $|E|$ vs $x$ curve (a triangle): $V_{bi} = -\int_{-x_p}^{x_n} E\,dx = \frac{1}{2}|E_{max}|\,W$ where $W = x_p + x_n$ is the total width.

Step 5 — Solve for the total width

Substitute $x_p, x_n$ (from neutrality) into $V_{bi}=\tfrac12|E_{max}|W$ and solve: $\boxed{W = \sqrt{\frac{2\varepsilon_s V_{bi}}{q}\left(\frac{1}{N_A}+\frac{1}{N_D}\right)}}$

Why this form? Higher $V_{bi}$ → wider region (more charge needed to build the field). Higher doping → narrower region (fewer ions needed to expose the same charge).

Worked Examples

Worked example Example 2 — Compute

$V_{bi}$ and $W$ Silicon at 300 K: $N_A = N_D = 10^{16}\,\text{cm}^{-3}$ , $n_i = 1.5\times10^{10}\,\text{cm}^{-3}$ , $\varepsilon_s = 1.05\times10^{-12}\,\text{F/cm}$ , $kT/q=0.0259$ V.

Step 1 — Built-in potential: $V_{bi}=0.0259\ln\!\frac{(10^{16})(10^{16})}{(1.5\times10^{10})^2} = 0.0259\ln(4.44\times10^{11}) \approx 0.69\ \text{V}$ Why: plug into the log formula; the huge ratio of doping to $n_i^2$ gives ~0.7 V, the classic Si value. Step 2 — Width: with $\tfrac1{N_A}+\tfrac1{N_D}=2\times10^{-16}\,\text{cm}^3$ : $W=\sqrt{\frac{2(1.05\times10^{-12})(0.69)}{1.6\times10^{-19}}(2\times10^{-16})}\approx 4.2\times10^{-5}\ \text{cm}=0.42\ \mu\text{m}$ Why: just substitution; sub-micron width is typical for moderate doping.

Common Mistakes

Active Recall

Recall Feynman: explain to a 12-year-old

Imagine a room split in half. The left half is packed with red balls (holes) and the right half with blue balls (electrons). Open the door in the middle: the balls rush to the emptier side and pair up, vanishing. But each ball that leaves uncovers a fixed magnet stuck to the floor (the dopant ion) — red balls leave behind negative magnets, blue balls leave behind positive ones. These stuck magnets create an invisible "wind" that eventually blows so hard it stops any more balls from crossing. That calm empty strip in the middle, full of stuck magnets but no free balls, is the depletion region.

Flashcards

What is the depletion region?

A junction zone depleted of mobile carriers, containing only fixed ionized dopant ions (space charge).

What charge remains on the n-side of the depletion region?

Positive donor ions (

N_D^+

What charge remains on the p-side?

Negative acceptor ions (

N_A^-

What is the charge-neutrality condition?

N_A x_p = N_D x_n

Which side does the depletion region extend further into?

The lightly-doped side.

What starting equation derives E-field and potential in the junction?

Poisson's equation

d^2\phi/dx^2 = -\rho/\varepsilon_s

What is the shape of the E-field vs position?

Triangular — linear in x, maximum at the metallurgical junction.

Formula for built-in potential?

V_{bi} = (kT/q)\ln(N_A N_D / n_i^2)

Formula for total depletion width?

W = \sqrt{(2\varepsilon_s V_{bi}/q)(1/N_A + 1/N_D)}

How does reverse bias affect width?

Widens it; replace

V_{bi}

with

V_{bi}+V_R

, so

W\propto\sqrt{V_{bi}+V_R}

How does forward bias affect width?

Narrows it; use

V_{bi}-V_F

Net current through the junction at equilibrium?

Zero — diffusion current exactly cancels drift current.

Why does the depletion width shrink with heavier doping?

Fewer ions are needed to expose the same charge, so the region is thinner (

W\propto 1/\sqrt N

effect).

Connections

Doping (Donors and Acceptors) — supplies the $N_A$ , $N_D$ ions that become the space charge.
Built-in Potential and Fermi Level Alignment — the thermodynamic origin of $V_{bi}$ .
Poisson's Equation and Gauss's Law — the mathematical engine of this derivation.
Diode I-V Characteristics — depletion behavior under forward/reverse bias.
Junction Capacitance and Varactors — $C_j = \varepsilon_s A / W$ , directly from the width.
Drift and Diffusion Currents — the two balancing currents at equilibrium.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab p-type aur n-type semiconductor ko jodte ho, toh junction ke paas electrons (n-side se) aur holes (p-side se) ek dusre ki taraf diffuse karte hain kyunki concentration gradient hota hai. Yeh mil ke recombine ho jaate hain, aur peechhe reh jaate hain fixed ionized dopant atoms — n-side pe positive donor ions, p-side pe negative acceptor ions. Yeh carriers-se-khaali strip hi hai depletion region, aur inn fixed ions ka charge hi space charge kehlata hai.

Ab yeh exposed charge ek electric field banata hai jo n se p ki taraf point karta hai. Yeh field carriers ko wapas dhakelta hai (drift). Jab drift aur diffusion barabar ho jaate hain, tab equilibrium aa jaata hai aur net current zero ho jaata hai. Field ka shape triangular hota hai — junction pe maximum, edges pe zero — kyunki charge distributed hai, isliye Poisson's equation se $E$ linearly badhta hai.

Important baat: neutrality condition $N_A x_p = N_D x_n$ . Iska matlab depletion region us side mein zyada ghusti hai jahan doping kam hai. Yaad rakho: "light doping = long reach". Aur width ka formula $W=\sqrt{(2\varepsilon_s V_{bi}/q)(1/N_A+1/N_D)}$ — heavy doping se width chhoti, high voltage se badi. Reverse bias lagao toh $V_{bi}+V_R$ ho jaata hai aur region choda ho jaata hai; forward bias lagao toh $V_{bi}-V_F$ hoke region patli ho jaati hai aur current flow karne lagta hai. Yahi diode ka core funda hai.

WHAT is happening?

HOW to derive the field, potential, and width

Step 1 — The depletion approximation

Step 2 — Charge neutrality

Step 3 — Integrate for the E-field

Step 4 — Integrate for the potential (built-in voltage)

Step 5 — Solve for the total width

Worked Examples

Common Mistakes

Active Recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Doping & PN Junctions

Connections