Depletion region and space charge: Question bank — Hardware deep dive

Before the traps, we pin down the exact coordinate system, symbols, and sign conventions used on this page. Nothing below makes sense without them, so read this box first.

Definition Coordinate system and symbols (fixed for this whole page)

Look at the figure below.

The x-axis runs left → right, with $x = 0$ placed exactly at the metallurgical junction (where p-material meets n-material).
The p-side is on the left ( $x < 0$ ), the n-side is on the right ( $x > 0$ ). Positive $x$ therefore points from p toward n.
$x_{p}$ = how deep the depletion region reaches into the p-side; it lives at $x = - x_{p}$ (negative side). $x_{n}$ = how deep it reaches into the n-side, at $x = + x_{n}$ . Both $x_{p}$ and $x_{n}$ are written as positive lengths (cm); the minus sign in " $x = - x_{p}$ " is just the coordinate, not the length.
Total width $W = x_{p} + x_{n}$ .
$ρ (x)$ = volume charge density — charge per unit volume (units $C/cm^{3}$ ). It is $- q N_{A}$ on the p-side (negative acceptor ions) and $+ q N_{D}$ on the n-side (positive donor ions).
$ε_{s}$ = permittivity of the semiconductor — how much charge a unit of the material "absorbs" per unit field (units $F/cm$ ); for silicon $ε_{s} \approx 1.05 \times 1 0^{- 12} F/cm$ .
$E$ = electric field. Convention: $E > 0$ means the field arrow points in the +x direction (p→n); $E < 0$ means it points n→p.

Figure — Depletion region and space charge

Now three words we lean on everywhere below — each was earned in the parent note:

True or false — justify

TF1. The depletion region carries the diode's forward current.

False. At equilibrium net current is zero; the region is depleted of mobile carriers. Forward current flows only once bias shrinks the barrier and carriers are injected across it — see Diode I-V Characteristics.

TF2. The depletion region is symmetric about the metallurgical junction.

Only if $N_{A} = N_{D}$ . Neutrality forces

N_{A} x_{p} = N_{D} x_{n}

, so the region bulges deeper into the lightly-doped side.

TF3. Inside the depletion region the electric field is uniform, like a parallel-plate capacitor.

False. The charge is a volume density

ρ

(

C/cm^{3}

), so Poisson gives

d E / d x = ρ / ε_{s}

— the field is linear in

x

, peaking at the junction (a triangle), not flat.

TF4. The electric field points from the p-side toward the n-side (i.e. in the +x direction).

False. By our convention +x is p→n, but the field points the other way, n → p (so

E < 0

): it runs from the positive donor ions on the right toward the negative acceptor ions on the left, opposing further diffusion — that's what makes it self-limiting.

TF5. Reverse bias makes the depletion region wider.

True. Reverse bias adds to the barrier (

V_{bi} \to V_{bi} + V_{R}

, both positive), and since

W \propto V_{bi} + V_{R}

(the barrier sits inside the root),

W

grows.

TF6. Heavier doping on both sides makes the depletion region wider.

False. More dopant ions per unit volume means fewer are needed to build the same charge, so

W

shrinks: increasing

N_{A}, N_{D}

shrinks

\frac{1}{N _{A}} + \frac{1}{N _{D}}

inside the root.

TF7. The built-in potential

V_{bi}

can be measured by putting a voltmeter across the diode.

False.

V_{bi}

is internal; the contact potentials at the meter's probes exactly cancel it, so you read zero. It shows up only in device behaviour.

TF8. At equilibrium, both drift and diffusion currents are individually zero.

False. Each is large; they exactly cancel so the net is zero — a dynamic balance, not a dead calm (see Drift and Diffusion Currents).

TF9. The total exposed positive charge equals the total exposed negative charge.

True. Charge was only rearranged, not created, so

q N_{A} x_{p} = q N_{D} x_{n}

— the whole crystal stays neutral.

TF10. Forward bias widens the depletion region because you are "adding voltage."

False. Forward bias opposes

V_{bi}

(

V_{bi} \to V_{bi} - V_{F}

V_{F}

positive), so the term inside the root shrinks and

W

shrinks, letting carriers cross more easily.

Spot the error

SE1. "The field is zero at the junction and maximum at the edges."

Backwards.

E = 0

at the neutral edges (

x = - x_{p}

and

x = + x_{n}

) and reaches

∣ E_{ma x} ∣

at the junction

x = 0

, where the two linear ramps meet.

SE2. "

V_{bi} = \frac{1}{2} ∣ E_{ma x} ∣ W

where

W

is the width of the more heavily doped side."

Error:

W = x_{p} + x_{n}

is the total width (both sides). The triangle's base spans the whole depletion region.

SE3. "Since

N_{A} x_{p} = N_{D} x_{n}

, the heavily-doped side extends further."

Reversed. Solving for the depths, the lightly-doped side extends further:

x_{n} / x_{p} = N_{A} / N_{D}

, so a large

N_{A}

makes

x_{n}

large.

SE4. "The depletion region is where recombination is happening right now, continuously."

Error: recombination carved it out during formation. Once formed and depleted, there are no mobile carriers left there to recombine at equilibrium.

SE5. "Because carriers left, the depletion region is electrically neutral (nothing's there)."

Error: removing the mobile carriers is exactly what uncovers the fixed ion charge. It is highly charged — that charge is the whole point.

SE6. "

E = - q N_{A} x / ε_{s}

on the p-side, with

E = 0

at the junction."

Boundary chosen wrong.

E = 0

at the outer edge

x = - x_{p}

(a neutral edge), not at the junction, giving

E = - \frac{q N _{A}}{ε _{s}} (x + x_{p})

SE7. "For a

p^{+} n

junction we ignore the depletion region entirely."

Error: we ignore only its penetration into the $p^{+}$ side (negligibly thin); nearly all of

W

lives in the lightly-doped n-side, and we very much keep that.

Why questions

WY1. Why does the depletion region extend further into the lightly-doped side?

To expose equal total charge on both sides (

N_{A} x_{p} = N_{D} x_{n}

), the side with fewer ions per volume must span a greater width.

WY2. Why does the field "self-limit" instead of growing forever?

More diffusion exposes more ions → stronger opposing field → less diffusion. Negative feedback settles at a fixed width where drift cancels diffusion.

WY3. Why is Poisson's equation the natural tool here, rather than just Coulomb's law?

We have a distributed charge

ρ (x)

and want the field/potential profile across it. Poisson relates

ρ

directly to

d^{2} ϕ / d x^{2}

in 1D — see Poisson's Equation and Gauss's Law.

WY4. Why does the depletion approximation (sharp box of charge) give accurate answers?

Carriers vanish over a very short Debye length, so the true smooth edge is thin compared to

W

; the sharp-box error is small but the integrals become trivial.

WY5. Why does

V_{bi}

appear in the log formula

\frac{k T}{q} ln (N_{A} N_{D} / n_{i}^{2})

The Fermi level must be flat at equilibrium; the band-bending needed to align it is

q V_{bi}

, and Boltzmann statistics turn the carrier ratio into a logarithm — see Built-in Potential and Fermi Level Alignment.

WY6. Why is

V_{bi}

larger for more heavily doped junctions?

The log rises with

N_{A} N_{D}

: more carriers means a steeper concentration gradient, so a bigger potential step is needed to hold them back.

WY7. Why does reverse bias make the junction a useful voltage-variable capacitor?

Reverse bias widens

W

, and the depletion charge behaves like a capacitor whose plate separation

W

grows with voltage — the basis of the varactor, see Junction Capacitance and Varactors.

WY8. Why is the potential

ϕ (x)

a smooth parabola-like curve while

E (x)

is straight lines?

Integrating a linear field once gives a quadratic potential; each doped slab contributes a parabolic arc, joined smoothly at the junction.

Edge cases

EC1. What happens to

W

as doping approaches zero (nearly intrinsic material)?

1/ N_{A} + 1/ N_{D}

blows up, so

W \to \infty

— with almost no ions to expose, the depletion region must spread very wide to build any field.

EC2. What happens to

V_{bi}

if the junction is heated (T rises)?

n_{i}

grows rapidly with temperature, shrinking the ratio

N_{A} N_{D} / n_{i}^{2}

, so

V_{bi}

falls — hot diodes have a lower barrier.

EC3. What is the field at the exact centre of a symmetric junction (

N_{A} = N_{D}

It is at its maximum

∣ E_{ma x} ∣

there, and the centre coincides with the junction

x = 0

; the field is symmetric but not zero.

EC4. For an extreme

p^{+} n

junction (

N_{A} ≫ N_{D}

), how does

W

compare to a single-side value?

1/ N_{A}

is negligible, so

W \approx 2 ε_{s} V_{bi} / (q N_{D})

— width is set almost entirely by the lightly-doped n-side.

EC5. Under very large reverse bias, what physical limit eventually stops

W

from growing?

The peak field

∣ E_{ma x} ∣

rises with bias until it triggers breakdown (avalanche or Zener); beyond that the junction conducts heavily and the depletion picture breaks down.

EC6. If

V_{bi} = V_{F}

exactly (forward bias equals the barrier), what does the model predict for

W

W \propto V_{bi} - V_{F} \to 0

: the barrier is nominally flattened. Real diodes never reach this because injected carriers dominate first, but it shows why

W

collapses under forward bias.

Recall One-line self-test

Cover every answer above, run down the list, and flag any where your reason (not just the yes/no) was shaky.