advanced33 min studyLesson 512 in Physics

Balance drift, diffusion, and electrochemical drive

Track electron and hole current signs and derive the thermal connection between mobility and diffusion.

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What you’ll understand

  • Distinguish carrier flux from conventional current.
  • Apply electron and hole drift-diffusion signs consistently.
  • Derive and delimit the classical Einstein relation.

A semiconductor can have a large electric field and a steep carrier-density gradient while carrying no net current. Drift and diffusion can cancel exactly at thermal equilibrium. Understanding this cancellation requires distinguishing the direction electrons move from the direction conventional electrical current points. It also reveals why mobility and diffusion are related: independent transport coefficients cannot be chosen arbitrarily if the material is to relax toward the correct equilibrium distribution.

Start with signed particle fluxes

Take the positive elementary charge as q>0q>0 and electric field E\mathbf E. An electron has charge minus q and low-field drift velocity −μnE-\mu_n\mathbf E, with positive mobility μn\mu_n. Multiplying electron particle flux by its negative charge makes its drift current point along the field. Holes drift along the field and carry positive charge, so their drift current also points along the field. Opposite particle velocities therefore do not imply opposite drift-current directions.

Diffusion transports particles down their own density gradient. The electron flux is −Dn∇n-D_n\nabla n; its conventional current consequently points up the electron gradient. The hole diffusion current instead points down the hole gradient. These signs follow directly from charge times particle flux and do not depend on which carrier happens to be the minority. Mobilities have units square metres per volt-second, diffusion coefficients square metres per second, and current densities amperes per square metre when densities use SI units.

Jn=qμnnE+qDn∇n,Jp=qμppE−qDp∇p\mathbf J_n=q\mu_n n\mathbf E+qD_n\nabla n,\qquad \mathbf J_p=q\mu_p p\mathbf E-qD_p\nabla p
n and p are local number densities; μ_n and μ_p are positive mobilities. This local, low-field, isothermal drift-diffusion description assumes carriers relax rapidly enough to admit local transport coefficients. It excludes ballistic flight across the entire device and strong velocity saturation.↔ Scroll wide equations horizontally.

Require zero current at thermal equilibrium

For a uniform material at constant temperature, an equilibrium Fermi level is spatially constant. The conduction edge varies as Ec=Ec0−qϕE_c=E_{c0}-q\phi, while E=−∇ϕ\mathbf E=-\nabla\phi. Differentiating the Boltzmann electron density then gives ∇n=−qnE/(kBT)\nabla n=-qn\mathbf E/(k_BT). Substitution into the electron current makes it vanish only when diffusion divided by mobility equals the thermal voltage. The hole derivation reaches the same ratio despite its opposite diffusion-current sign.

This Einstein relation is a consequence of dilute equilibrium statistics and linear response, rather than a universal conversion independent of regime. A useful generalized equilibrium expression replaces thermal voltage by carrier density divided by its thermodynamic density response. Here the chemical potential is an energy, and must not be confused with mobility despite the shared Greek letter often used for both. Degenerate carriers have a different density response; hot-carrier distributions or strong nonequilibrium can invalidate even a local equilibrium generalization.

Dnμn=Dpμp=kBTq≡VT,Dnμn=nq(∂n/∂μchem)Tin local equilibrium\frac{D_n}{\mu_n}=\frac{D_p}{\mu_p}=\frac{k_BT}{q}\equiv V_T,\qquad \frac{D_n}{\mu_n}=\frac{n}{q(\partial n/\partial\mu_{\rm chem})_T}\quad\text{in local equilibrium}
V_T is in volts, and μ_chem in the generalized expression is a chemical energy in joules. The first equality requires nondegenerate isothermal statistics. The generalized form assumes a consistent local equilibrium density response and linear transport, not arbitrary driven distributions.↔ Scroll wide equations horizontally.

Separate equilibrium from absence of gradients

A concentration gradient can coexist with zero electrical current because the equilibrium electrostatic potential adjusts to oppose diffusion. Conversely, a uniform density under an applied field carries drift current unless another driving term intervenes. Under injection, quasi-Fermi-level gradients organize the combined drive. They should not be interpreted as a license to set electron and hole densities independently while ignoring continuity equations, recombination, and Poisson electrostatics. A device solution requires transport and charge conservation together.

The local model is most reliable when carrier momentum relaxes over distances short compared with the scale of density and potential variation. Contacts, abrupt interfaces, large fields, and nanometre channels can require boundary or kinetic treatments. Even within the model, a negative net electron current does not mean negative mobility: it can simply mean that diffusion outweighs positive drift current. Compute signed contributions first, then interpret their sum, rather than assigning a direction from the majority-carrier label alone.

An electron current that cancels

In one dimension take n=1021 m−3n=10^{21}\,\mathrm{m^{-3}}, μn=0.10 m2/(V s)\mu_n=0.10\,\mathrm{m^2/(V\,s)}, VT=0.025 VV_T=0.025\,\mathrm V, dn/dx=−2×1023 m−4dn/dx=-2\times10^{23}\,\mathrm{m^{-4}}, and Ex=5 V/mE_x=5\,\mathrm{V/m}. Use q=1.6×10−19 Cq=1.6\times10^{-19}\,\mathrm C. Find the diffusion coefficient and both current contributions.

  1. The Einstein relation gives Dn=μnVT=0.0025 m2/sD_n=\mu_n V_T=0.0025\,\mathrm{m^2/s}. This uses the stipulated nondegenerate, low-field thermal regime; mobility alone would not fix D outside those assumptions.

  2. The drift contribution is Jn,drift=qnμnEx=80 A/m2J_{n,\rm drift}=qn\mu_nE_x=80\,\mathrm{A/m^2}. Electron drift velocity is −0.50 m/s-0.50\,\mathrm{m/s}, illustrating that electron velocity and conventional current have opposite signs.

  3. The diffusion contribution is Jn,diff=qDn(dn/dx)=−80 A/m2J_{n,\rm diff}=qD_n(dn/dx)=-80\,\mathrm{A/m^2}. Thus the total electron current is zero. Its underlying particle diffusion points toward positive x and cancels the negative electron drift flux.

The result

The diffusion coefficient is 0.0025 m²/s and the signed currents are +80 and −80 A/m². Zero total current is compatible with both a nonzero field and a nonzero concentration gradient.

Explore the model

Interactive lab

Interactive lab · vectors

Keep the two current signs visible

Reset and set both controls to angle 0° and magnitude 1; predict x projection +1. Reset and set both controls to angle 180° and magnitude 1; predict x projection −1. Interpret these separate vectors as normalized opposite current contributions whose sum would cancel. The instrument displays one vector per run and does not calculate carrier densities, mobility, or a self-consistent electric field.

Keep the two current signs visiblePlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forCancellation concerns signed contributions; the lab does not turn a current arrow into a carrier trajectory.

Check your understanding

A positive electron-density gradient produces which conventional diffusion-current direction in one dimension?

Try it yourself

For holes take p=5×1020 m−3p=5\times10^{20}\,\mathrm{m^{-3}}, μp=0.040 m2/(V s)\mu_p=0.040\,\mathrm{m^2/(V\,s)}, VT=0.025 VV_T=0.025\,\mathrm V, dp/dx=1023 m−4dp/dx=10^{23}\,\mathrm{m^{-4}}, and Ex=2 V/mE_x=2\,\mathrm{V/m}. Find the net hole current and the field that would make it zero. Use the same q as the example.

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

Give me a hint

Hole diffusion current carries a minus sign. Set drift equal to its magnitude to find the balancing field.

Compare with the solution

Here Dp=0.0010 m2/sD_p=0.0010\,\mathrm{m^2/s}. Drift is qpμpEx=6.4 A/m2qp\mu_pE_x=6.4\,\mathrm{A/m^2}, while diffusion is −qDp(dp/dx)=−16 A/m2-qD_p(dp/dx)=-16\,\mathrm{A/m^2}. The net is therefore −9.6 A/m2-9.6\,\mathrm{A/m^2}. Cancellation needs Ex=Dp(dp/dx)/(μpp)=5 V/mE_x=D_p(dp/dx)/(\mu_pp)=5\,\mathrm{V/m} in the positive direction. The negative net current at the original field reflects dominant diffusion, not negative hole mobility.

Take it with you

  • Electron particle motion and conventional current have opposite signs.
  • The classical Einstein relation requires dilute isothermal linear transport.
  • A zero-current equilibrium can contain large opposing drift and diffusion terms.

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