Foundations — Depletion region and space charge

This page assumes nothing. Before you read the parent note, every letter, arrow, and idea it uses is built here from the ground up. Read top to bottom — each item leans on the one above it. (Words like p-type, n-type, and space charge appear in the parent note's title; they are defined below in Sections 1 and 2 before we ever lean on them.)

0. What is a semiconductor, really?

Why the topic needs it: the whole depletion story is a tale of blue marbles (electrons) and red marbles (holes) moving, meeting, and vanishing. If you don't know what they are, nothing else lands.

The figure below draws both players in the silicon lattice. Look at the two markers: the solid blue disc is a free electron (label e-), and the dashed red circle is a hole — a seat with the electron missing, which is why it reads as a +. The point to extract: an electron is a thing that is there; a hole is a thing that is absent, yet both move and carry charge.

1. Doping — donors and acceptors

The key, easy-to-miss point: the dopant atom itself cannot move. It is bolted into the crystal lattice. Only the electron/hole it releases is free.

Why the topic needs it: the depletion region is built entirely out of these stuck, charged dopant atoms. Understanding that "the carrier leaves, the ion stays" is the heart of the whole chapter.

2. The symbols $N_A$, $N_D$, $q$, $n_i$, and "space charge"

Why the "per volume"? Because to know how much total charge a slab of depletion region holds, we need charge-per-atom ($q$) times atoms-per-volume ($N_A$ or $N_D$) times the volume. That product will appear everywhere.

The tiny superscripts you'll meet:

• $N_D^{+}$ = a donor that has become a positive ion (gave up its electron).
• $N_A^{-}$ = an acceptor that has become a negative ion (captured an electron).

3. Diffusion and drift — the two motions

Why the topic needs it: the depletion region is the outcome of a tug of war — diffusion pushes carriers across, drift (from the field that builds up) pushes them back. Equilibrium is when these two exactly cancel.

The figure shows the moment just after contact, before any field has built up. Watch the two arrows crossing the yellow junction line: the blue arrow is electrons streaming from the crowded n-side (right) into the empty p-side (left); the red arrow is holes streaming the other way. The lesson to extract: nothing is pushing them — they move purely because one side is crowded and the other empty.

4. Electric field $E$ — what the arrows mean

In the junction, positive donor ions sit on the n-side and negative acceptor ions on the p-side, so the field points from n toward p. That direction is exactly what pushes electrons back into the n-side and holes back into the p-side — the "wind that stops the marbles".

Before the boxed law: recall two quantities it will use. $\rho$ (Greek "rho") = charge density = charge per unit volume (how much net charge is packed into a tiny box at position $x$). ${\epsilon }_s$ (Greek "epsilon-s") = permittivity of the semiconductor, a material constant measuring how much the silicon "soaks up" a field. With those in hand:

Why this tool and not another? We want to go from "here is a slab of fixed charge" to "here is the field it makes". Gauss's law is precisely the law that turns charge → field. No other rule does that job.

5. Potential $\varphi$ and the derivative $\frac{d}{dx}$

Why we need integration. Gauss's law gives us the slope of the field. To get the field itself we must undo the slope — that's integration (the reverse of the derivative). Integrating once more turns field into potential. So the parent note's chain is: $\rho \stackrel{\text{integrate}}{\to }E\stackrel{\text{integrate}}{\to }\varphi .$

Two symbols the parent note will lean on the moment it builds this chain — meet them now:

6. Built-in potential $V_{bi}$ and the width $W$

The figure has two stacked panels sharing the same $x$-axis (origin at the junction). Top panel: the two slabs of fixed charge — red on the p-side ($-x_p$ up to $0$), green on the n-side ($0$ up to $x_n$); the double-headed arrow labels the full width $W$. Bottom panel: the resulting field magnitude $|E|$ — note it is a triangle, zero at both outer edges and peaking at $E_{max}$ right at the junction, and the shaded area equals $V_{bi}$. Extract this: charge slabs (top) produce the triangular field (bottom), and the field's area is the built-in voltage.

Why the topic needs these: the entire "HOW to derive" section chases three numbers — the peak field $E_{max}$, the built-in voltage $V_{bi}$, and the width $W$. Knowing what each symbol pictures (a distance into a side, a voltage hill, a triangle peak) keeps the algebra meaningful instead of a symbol soup.

7. The natural-log $\ln$ and $kT/q$

Why the log appears. The number of carriers follows an exponential law with energy (Boltzmann statistics). To recover an energy/voltage from a carrier ratio, we must undo the exponential — and the log is exactly the un-doer of the exponential. That gives $V_{bi}=\frac{kT}{q}\ln \left(\frac{N_A{N}_D}{n_i^2}\right)$.

How the foundations feed the topic

Equipment checklist

Cover the right side and test yourself. If any answer is fuzzy, re-read that section before the parent note.