Depletion Region: Width, Electric Field, Space Charge

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Meta description: Calculate PN junction depletion width, electric field, and space charge using consistent units, clear assumptions, and a worked silicon example.

The depletion region is the part of a PN junction where mobile carrier concentrations are greatly reduced, leaving fixed ionized donors and acceptors. These charges create an electric field and a potential barrier across the junction.

For a uniformly doped, abrupt PN junction under the depletion approximation, total depletion width is:

W = √[(2εs/q)(1/NA + 1/ND)(Vbi + VR)]

Here, VR is defined as positive for reverse bias. The depletion region becomes wider as reverse bias increases and extends farther into the more lightly doped side.

Assumptions and Symbols

The equations below assume a one-dimensional abrupt junction with uniform doping on each side, fully ionized dopants, and negligible mobile charge inside the depletion region.

The built-in potential expression also assumes nondegenerate carrier statistics.

Use consistent units throughout. If permittivity is in F/m, convert doping concentrations to m⁻³.

1 cm⁻³ = 10⁶ m⁻³

Why a Depletion Region Forms

When p-type and n-type semiconductor regions meet, carrier concentration gradients drive electrons and holes across the junction.

Electrons leaving the n side expose positively charged donor ions. Holes leaving the p side expose negatively charged acceptor ions.

The ions are fixed in the crystal lattice. Their charge creates an electric field directed from the n side toward the p side.

At thermal equilibrium, drift and diffusion currents balance. The net current is zero, but the internal electric field and built-in potential remain.

The physical depletion width is not a universal constant. It depends on doping, semiconductor material, temperature, and applied bias.

Space Charge Density and Coordinate Convention

Place the metallurgical junction at x = 0.

• The p side lies at negative x.
• The n side lies at positive x.
• The depletion region extends from −xp to xn.

Under the depletion approximation:

ρ(x) = −qNA for −xp < x < 0

ρ(x) = +qND for 0 < x < xn

Outside this region, approximate the semiconductor as charge neutral.

Charge neutrality across the depletion region requires:

qNAxp = qNDxn

Therefore:

NAxp = NDxn

This gives:

xn/xp = NA/ND

If the p side is much more heavily doped than the n side, most of the depletion width lies on the n side.

Electric Field Inside the Junction

Poisson’s equation relates field gradient and charge density:

dE/dx = ρ/εs

Using zero field at the depletion boundaries:

On the p side:

E(x) = −(qNA/εs)(x + xp)

On the n side:

E(x) = (qND/εs)(x − xn)

With this coordinate convention, the electric field is negative inside the depletion region because it points from n to p.

The field is continuous at the junction. Its maximum magnitude is:

|Emax| = qNAxp/εs = qNDxn/εs

The signed electric-field plot is triangular below the x-axis. The field-magnitude plot is triangular above it. Distinguishing these two plots prevents sign confusion.

Built-In Potential

For the ideal junction model:

Vbi = (kT/q) ln(NAND/ni²)

At T = 300 K:

kT/q ≈ 0.02585 V

Use NA, ND, and ni in the same concentration units.

Built-in potential depends on doping and temperature. It should not be replaced automatically by a fixed “silicon diode voltage” such as 0.7 V.

Deriving Total Depletion Width

Define the potential drop across the depletion region as:

Vdep = Vbi + VR

The area under the electric-field magnitude is:

Vdep = ½|Emax|W

where:

W = xp + xn

Combining this relationship with charge neutrality gives:

W = √[(2εs/q)(1/NA + 1/ND)Vdep]

The individual widths are:

xn = [NA/(NA + ND)]W

xp = [ND/(NA + ND)]W

An alternative expression for peak field magnitude is:

|Emax| = 2Vdep/W

This provides a useful calculation check.

Worked Example: Silicon PN Junction at Zero Bias

Assume:

These values illustrate the calculation. More precise material models may use different temperature-dependent values for ni.

Step 1: Calculate Built-In Potential

Vbi = 0.02585 ln[(10¹⁷ × 10¹⁶)/(10¹⁰)²]

Vbi = 0.02585 ln(10¹³)

Vbi ≈ 0.774 V

Step 2: Convert Doping to SI Units

NA = 10²³ m⁻³

ND = 10²² m⁻³

Thus:

1/NA + 1/ND = 1.1 × 10⁻²² m³

Step 3: Calculate Total Width

W = √[(2 × 1.04 × 10⁻¹⁰ / 1.602 × 10⁻¹⁹) × 1.1 × 10⁻²² × 0.774]

W ≈ 3.32 × 10⁻⁷ m

W ≈ 0.332 µm

Step 4: Split the Width Between the Two Sides

xn = (10/11)W ≈ 0.302 µm

xp = (1/11)W ≈ 0.0302 µm

The n side is ten times less heavily doped, so its depletion width is ten times larger.

Step 5: Calculate Peak Electric Field

|Emax| = 2Vbi/W

|Emax| ≈ 2 × 0.774 / (3.32 × 10⁻⁷)

|Emax| ≈ 4.66 × 10⁶ V/m

|Emax| ≈ 4.66 × 10⁴ V/cm

Check the potential:

½|Emax|W ≈ 0.774 V

The result agrees with the built-in potential.

What Happens Under Reverse Bias?

For the same junction at VR = 5 V:

W(5 V)/W(0) = √[(Vbi + 5)/Vbi]

W(5 V)/W(0) ≈ √(5.774/0.774) ≈ 2.73

Therefore:

W(5 V) ≈ 0.907 µm

The depletion width increases with the square root of the junction potential, not linearly with voltage.

This result assumes the junction remains below breakdown and the depletion region does not reach another device boundary.

Connection to Junction Capacitance

For junction area A, the depletion capacitance is:

Cj = εsA/W

For an abrupt junction:

Cj(VR)/Cj(0) = √[Vbi/(Vbi + VR)]

Increasing reverse bias widens the depletion region and reduces capacitance.

This is depletion capacitance. Under significant forward bias, stored minority carriers introduce diffusion capacitance, which requires additional analysis.

Common Mistakes

• Combining doping in cm⁻³ with permittivity in F/m.
• Treating W, xn, and xp as interchangeable.
• Using a fixed built-in potential without checking doping and temperature.
• Reporting field magnitude while describing it as a signed field.
• Applying the abrupt-junction formula to a graded junction.
• Extending the simple model through breakdown, strong injection, or punch-through.

Frequently Asked Questions

Why does the depletion region extend into the lightly doped side?

Charge neutrality requires equal and opposite integrated charge. A lower dopant concentration requires a larger depleted width to provide the same charge per unit area.

Does forward bias shrink the depletion region?

Within the depletion model, replace Vbi + VR with Vbi − VF, where VF is positive forward bias. The width decreases, but strong injection limits the usefulness of this simple electrostatic model.

Can depletion width predict breakdown voltage?

It helps estimate the electric field, but accurate breakdown prediction requires the appropriate breakdown mechanism, doping profile, material properties, and device geometry. A single universal critical-field value is insufficient.

Next Steps

Use the calculated depletion width to estimate junction capacitance, then examine how reverse bias changes electric field and device behavior.

Continue with reverse-bias leakage and breakdown mechanisms, or connect junction electrostatics to the BJT Early effect.

References and Further Reading

• MIT OpenCourseWare — PN Junctions: Electrostatics
• NPTEL — PN Junction