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advanced35 min studyLesson 513 in Physics

Derive a junction's space charge and built-in potential

Connect depletion charge, field direction, voltage, and capacitance in an abrupt homojunction.

Before you begin
On this page

The question for today

What you’ll understand

  • Relate equilibrium band bending to a flat Fermi level.
  • Derive depletion widths using neutrality and Poisson's equation.
  • Distinguish built-in potential from an externally available terminal voltage.

Joining p-type and n-type regions initially allows majority carriers to diffuse across their concentration gradients. The uncovered ionized dopants create a field that opposes further diffusion. At equilibrium the resulting barrier balances carrier transport rather than stopping all microscopic motion. The depletion approximation makes this coupled problem hand calculable by retaining fixed dopant charge while neglecting mobile charge within a finite layer, with neutral material on either side.

Fix directions before drawing a band diagram

Place p-type material at negative x and n-type material at positive x, with the metallurgical junction at zero. Depleted acceptors on the p side carry negative charge; depleted donors on the n side carry positive charge. The field therefore points from n toward p, along negative x. Since the field is minus the potential gradient, electrostatic potential rises from p to n. Electron band energies decrease in that direction because their electrostatic energy is minus q times potential.

At thermal equilibrium a single Fermi level is flat across the homojunction. Boltzmann statistics in the distant neutral regions then relate their carrier-density ratio to the built-in potential. Complete dopant ionization and nondegenerate statistics give a logarithm of the doping product divided by the intrinsic-density square. This potential is internal band bending. A voltmeter connected through equilibrium contacts does not obtain a battery voltage from it: other contact potentials complete the electrochemical balance around the circuit.

Vbi=VTln⁡NANDni2,ρ(x)={−qNA,−xp<x<0,qND,0<x<xn,0,outside the depletion layer,NAxp=NDxnV_{\rm bi}=V_T\ln\frac{N_AN_D}{n_i^2},\qquad \rho(x)=\begin{cases}-qN_A,&-x_p<x<0,\\qN_D,&0<x<x_n,\\0,&\text{outside the depletion layer},\end{cases}\qquad N_Ax_p=N_Dx_n
N_A and N_D are fully ionized uniform dopant densities of an abrupt homojunction; V_T = k_BT/q. The depletion approximation neglects mobile charge inside and net charge outside. Equal and opposite integrated space charge follows from a vanishing field at both outer depletion boundaries.↔ Scroll wide equations horizontally.

Integrate charge into field and voltage

Poisson's equation makes the field linear on each side of a uniformly doped depletion region. The field is continuous at the junction when there is no added sheet charge, and reaches its largest magnitude there. Integrating the triangular field profile gives the barrier voltage. Combining this area with integrated charge neutrality determines the total width. The more lightly doped side occupies more of that width because a larger distance is needed to supply the same opposite charge per area.

Let applied junction bias VaV_a be positive in forward bias, meaning the p side is raised relative to n. The remaining barrier drop is Vbi−VaV_{\rm bi}-V_a within the depletion model; reverse bias increases it. The square-root width formula assumes this drop remains positive and that depletion remains a useful approximation. It does not describe strong forward injection, tunneling, or avalanche breakdown. Fixed permittivity and abrupt, spatially uniform dopants are also part of the calculation.

W=xp+xn=2ε(Vbi−Va)q(1NA+1ND),∣Emax⁡∣=2(Vbi−Va)W,Cj=εAWW=x_p+x_n=\sqrt{\frac{2\varepsilon(V_{\rm bi}-V_a)}q\left(\frac1{N_A}+\frac1{N_D}\right)},\qquad |E_{\max}|=\frac{2(V_{\rm bi}-V_a)}W,\qquad C_j=\frac{\varepsilon A}{W}
W is a length, ε is permittivity in F/m, A is junction area, and C_j is the small-signal depletion capacitance. The field formula uses the triangular profile. Capacitance refers to depletion-charge change with bias and excludes forward minority-carrier storage capacitance.↔ Scroll wide equations horizontally.

Interpret a changing depletion layer

Increasing reverse bias widens the region of exposed dopants and decreases depletion capacitance. The derivative defining capacitance counts charge on one side, rather than summing the magnitudes on both sides and introducing an erroneous factor two. In a strongly asymmetric junction, the lightly doped side dominates the voltage drop and total width. This one-sided approximation simplifies capacitance measurements used to infer doping, but requires checking that the opposing side is indeed much more heavily doped.

Depletion does not mean exactly zero carrier density or zero current under every condition. It is an electrostatic approximation that neglects mobile charge relative to ionized dopants. Under bias, small minority populations can dominate current even while contributing negligibly to the electric field. Likewise, neutral regions may carry current with a small nonzero field. Keeping electrostatic leading terms separate from transport leading terms avoids the mistaken conclusion that a negligible charge density makes a region electrically irrelevant.

Work it through

A symmetric junction's triangular field

An abrupt junction has NA=ND=1022 m−3N_A=N_D=10^{22}\,\mathrm{m^{-3}}, ε=10−10 F/m\varepsilon=10^{-10}\,\mathrm{F/m}, and an equilibrium built-in drop of 0.64 V0.64\,\mathrm V. Use q=1.6×10−19 Cq=1.6\times10^{-19}\,\mathrm C. Find the total depletion width, its division, peak field magnitude, and charge magnitude per area on either side.

  1. The width square is [2(10−10)(0.64)/(1.6×10−19)](2×10−22)=1.6×10−13 m2[2(10^{-10})(0.64)/(1.6\times10^{-19})](2\times10^{-22})=1.6\times10^{-13}\,\mathrm{m^2}. Thus W=4.0×10−7 m=0.40 μmW=4.0\times10^{-7}\,\mathrm m=0.40\,\mu\mathrm m.

  2. Equal doping and charge neutrality give xp=xn=W/2=0.20 μmx_p=x_n=W/2=0.20\,\mu\mathrm m. The charge magnitude per area is qNDxn=3.2×10−4 C/m2qN_Dx_n=3.2\times10^{-4}\,\mathrm{C/m^2}, with positive sign on n and negative sign on p.

  3. The triangular-profile area requires ∣Emax⁡∣=2(0.64)/W=3.2×106 V/m|E_{\max}|=2(0.64)/W=3.2\times10^6\,\mathrm{V/m}. Independently, charge divided by permittivity gives the same peak magnitude; its signed x component is negative in the chosen geometry.

The result

The width is 0.40 μm, split equally between the sides, and the peak field is 3.2 MV/m toward p. Opposite charge sheets integrated through each depletion half have magnitude 3.2 × 10⁻⁴ C/m².

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Relate potential curvature to field slope

Reset and set both controls to coefficient a = 1 and position x = 0.5; predict tangent slope 1. Reset and set both controls to a = 1 and x = 1; predict slope 2. The model y = ax² illustrates how a quadratic potential segment has a linear derivative. A physical field would be minus that derivative. This single parabola has no junction matching, depletion edges, dopants, or voltage units.

Relate potential curvature to field slopeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forUniform space charge gives a quadratic potential segment; assembling the two junction halves requires separate matching conditions.

Pause & reason

Check your understanding

With p on the left and n on the right, what is the equilibrium depletion-field direction?

Your turn

Try it yourself

Apply reverse bias of 1.92 V1.92\,\mathrm V to the symmetric junction in the example, staying below breakdown. By what factors do total width, peak field, and depletion capacitance change? Also find the division of a 1.0 μm1.0\,\mu\mathrm m depletion layer in a separate junction with NA=4NDN_A=4N_D.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Reverse bias adds to the built-in drop. Width scales as its square root; the asymmetric division follows from equal integrated charge.

Compare with the solution

The reverse-biased drop is 0.64+1.92=2.56 V0.64+1.92=2.56\,\mathrm V, four times its equilibrium value. Width doubles to 0.80 μm0.80\,\mu\mathrm m, peak field doubles to 6.4 MV/m6.4\,\mathrm{MV/m}, and capacitance halves because it varies inversely with width. For the separate junction, 4xp=xn4x_p=x_n and xp+xn=1.0 μmx_p+x_n=1.0\,\mu\mathrm m, giving xp=0.20 μmx_p=0.20\,\mu\mathrm m and xn=0.80 μmx_n=0.80\,\mu\mathrm m. The lightly doped n side occupies most of the depletion layer.

Keep the idea

Take it with you

  • Equilibrium field and carrier diffusion balance across a flat Fermi level.
  • Depletion charge neutrality fixes the width division.
  • Reverse bias widens depletion and lowers its capacitance within the model's validity range.

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