advanced34 min studyLesson 511 in Physics

Count intrinsic and doped semiconductor carriers

Combine band-state counting, Fermi occupation, and charge neutrality while checking ionization and degeneracy.

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What you’ll understand

  • Separate density of states from occupation probability.
  • Solve compensated nondegenerate carrier statistics.
  • Identify freeze-out and degeneracy limits of Boltzmann formulas.

Doping does not supply a freely adjustable number of conducting electrons independently of temperature and band structure. The carrier density follows from available states, their occupation, and the electrostatic requirement that bulk charge nearly cancel. These ingredients explain why an intrinsic semiconductor conducts, why compensation reduces the effect of donors, and why formulas that work at moderate temperature fail during freeze-out or when the Fermi level approaches a band edge.

Count states before filling them

Let EcE_c and EvE_v denote conduction and valence edges, with gap Eg=Ec−EvE_g=E_c-E_v. Electrons occupy a state of energy E with Fermi probability f(E)=[1+e(E−EF)/(kBT)]−1f(E)=[1+e^{(E-E_F)/(k_BT)}]^{-1}. The electron density integrates the conduction density of states against f; the hole density integrates the valence density against one minus f. A hole counts an unoccupied valence state, rather than a second electron species with an independently chosen equilibrium chemical potential.

Near an isotropic parabolic minimum, the three-dimensional density of states grows as the square root of energy above the edge. Its thermal integral produces the effective density NcN_c. For identical valleys, a factor gvg_v counts equivalent minima and a separate factor two counts unresolved electron spin. If a tabulated density-of-states mass already incorporates valley multiplicity, adding that factor again is wrong. A conductivity effective mass describes acceleration and need not equal this state-counting mass.

n=∫Ec∞Dc(E)f(E) dE≃Nce−(Ec−EF)/(kBT),Nc=2gv(2πmDOSkBTh2)3/2n=\int_{E_c}^{\infty}D_c(E)f(E)\,dE\simeq N_c e^{-(E_c-E_F)/(k_BT)},\qquad N_c=2g_v\left(\frac{2\pi m_{\rm DOS}k_BT}{h^2}\right)^{3/2}
D_c is states per volume per energy, n and N_c are number densities, and m_DOS here is the mass of one identical isotropic valley. The approximation requires dilute conduction occupation; the explicit two is spin multiplicity. Energies and k_BT must use the same units.↔ Scroll wide equations horizontally.

Impose neutrality and thermal equilibrium together

In the nondegenerate equilibrium regime, holes obey the corresponding Boltzmann expression with effective valence density NvN_v. Multiplying the electron and hole formulas eliminates the common Fermi level and gives the mass-action law. Intrinsic material has equal electron and hole populations, although its Fermi level is not exactly at midgap unless the two effective densities match. Differences in band curvature move the intrinsic level without producing net charge in an otherwise ideal intrinsic crystal.

For uniformly doped neutral bulk, the positive charge of holes and ionized donors balances electrons and ionized acceptors. With complete ionization, the difference between donors and acceptors controls majority carriers, not their sum. Solving neutrality together with mass action gives a quadratic equation with one positive physical electron density. When the net donor density greatly exceeds the intrinsic density, that exact solution reduces to majority electrons approximately equal to the net donor count, with minority holes fixed by mass action.

np=ni2=NcNve−Eg/(kBT),n−p=ND+−NA−,Ei−Ec+Ev2=kBT2ln⁡NvNcnp=n_i^2=N_cN_v e^{-E_g/(k_BT)},\qquad n-p=N_D^+-N_A^-,\qquad E_i-\frac{E_c+E_v}{2}=\frac{k_BT}{2}\ln\frac{N_v}{N_c}
These equilibrium mass-action and intrinsic-level expressions assume nondegenerate parabolic-band statistics. N_D⁺ and N_A⁻ are ionized dopant densities, not automatically total chemical dopant concentrations. E_i is the intrinsic Fermi level, and all densities must share a volume unit.↔ Scroll wide equations horizontally.

Check the regime after obtaining a number

At low temperature, donor electrons can remain bound to donor levels. Their ionized fraction depends on the Fermi level and impurity occupation statistics, so replacing ND+N_D^+ by total NDN_D fails. At strong doping, conduction occupation is no longer dilute and the full Fermi integral replaces the Boltzmann exponential. These are different failures: freeze-out concerns dopant ionization, whereas degeneracy concerns occupation of band states. Heavy doping may also change the bands themselves beyond this simple model.

A practical calculation should therefore check the inferred Fermi-level separation from both relevant edges and the assumed ionization range. Under illumination or injection, electrons and holes may require separate quasi-Fermi levels; their product need not equal the equilibrium intrinsic square. Charge neutrality remains useful in broad bulk regions, but cannot be imposed locally inside a depletion layer where space charge generates an electric field. Statistics and electrostatics complement each other rather than providing interchangeable constraints.

Compensation changes the majority density

At a temperature where dopants are fully ionized, take ni=1010 cm−3n_i=10^{10}\,\mathrm{cm^{-3}}, ND=2×1016 cm−3N_D=2\times10^{16}\,\mathrm{cm^{-3}}, NA=5×1015 cm−3N_A=5\times10^{15}\,\mathrm{cm^{-3}}, Nc=3×1019 cm−3N_c=3\times10^{19}\,\mathrm{cm^{-3}}, and kBT=0.025 eVk_BT=0.025\,\mathrm{eV}. Estimate n, p, and the conduction-edge separation from the Fermi level.

  1. The net donor density is ND−NA=1.5×1016 cm−3N_D-N_A=1.5\times10^{16}\,\mathrm{cm^{-3}}, far above the intrinsic density. Neutrality therefore gives n≃1.5×1016 cm−3n\simeq1.5\times10^{16}\,\mathrm{cm^{-3}}. The correction from minority holes is negligible at the precision of the supplied data.

  2. Use mass action to obtain p=ni2/n=1020/(1.5×1016)=6.67×103 cm−3p=n_i^2/n=10^{20}/(1.5\times10^{16})=6.67\times10^3\,\mathrm{cm^{-3}}. Adding donors and acceptors would have overestimated n and underestimated this minority density.

  3. Invert the electron Boltzmann formula: Ec−EF=kBTln⁡(Nc/n)=0.025ln⁡(2000) eV=0.190 eVE_c-E_F=k_BT\ln(N_c/n)=0.025\ln(2000)\,\mathrm{eV}=0.190\,\mathrm{eV}. This is 7.60 thermal energies below the edge, consistent with nondegenerate conduction occupation. Complete ionization was separately stipulated.

The result

The material is n-type, with about 1.5 × 10¹⁶ electrons and 6.67 × 10³ holes per cm³. Its inferred Fermi level lies 0.190 eV below the conduction edge under the stated model.

Explore the model

Interactive lab

Interactive lab · network

Inspect the occupation sigmoid

Reset and set both controls to input x = 0.5 and weight w = 1; predict output 0.5. Reset and set both controls to x = 1.5 and w = 1; predict output about 0.731. The exact model is σ(wx − 0.5). Identifying its logit with (E_F − E)/(k_BT) illustrates a single occupation probability; it neither integrates band states nor solves dopant neutrality.

Inspect the occupation sigmoidOne inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Current readings: Product · wx: 1.2; Logit · wx − 0.5: 0.7; Sigmoid output: 0.6682. All parameters are adjustable with the labeled controls below.One input. One weight. One sigmoid.x1.20× w1.20+ b0.70σ0.668output0.668w = 1.00b = −0.50σ(z) = 1 / (1 + exp(−z))
Computation graphSignal trace
Product · wx
1.2
Logit · wx − 0.5
0.7
Sigmoid output
0.6682
t = 0 s

One inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. 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 forA Fermi occupation probability must be combined with a density of available states to obtain a carrier density.

Check your understanding

In fully ionized, compensated n-type material with net doping much larger than the intrinsic density, which quantity approximately sets the electron density?

Try it yourself

An intrinsic nondegenerate semiconductor has Nc=4×1018 cm−3N_c=4\times10^{18}\,\mathrm{cm^{-3}}, Nv=1018 cm−3N_v=10^{18}\,\mathrm{cm^{-3}}, Eg=1.0 eVE_g=1.0\,\mathrm{eV}, and kBT=0.025 eVk_BT=0.025\,\mathrm{eV}. Find its intrinsic density and the displacement of its Fermi level from midgap. The given effective densities already include all degeneracies.

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

Give me a hint

Take the square root of mass action, then use the logarithmic valence-to-conduction density ratio for the intrinsic level.

Compare with the solution

The intrinsic density is ni=NcNve−Eg/(2kBT)=2×1018e−20=4.12×109 cm−3n_i=\sqrt{N_cN_v}e^{-E_g/(2k_BT)}=2\times10^{18}e^{-20}=4.12\times10^9\,\mathrm{cm^{-3}}. The level shift is (0.025/2)ln⁡(1/4) eV=−0.0173 eV(0.025/2)\ln(1/4)\,\mathrm{eV}=-0.0173\,\mathrm{eV}, so the intrinsic Fermi level is 17.3 meV below midgap. Electrons and holes still have equal densities. No additional spin or valley factor should multiply the supplied effective densities.

Take it with you

  • Density of states and occupation probability are separate ingredients.
  • Compensation, ionization, and degeneracy must be checked independently.
  • Equilibrium nondegenerate mass action does not apply unchanged to injected or degenerate carriers.

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