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Minority suppression and donor-defined conduction

Short answer

Minority carrier suppression is the reduction of minority carriers below n_i due to doping, from np = n_i². In n-type, charge neutrality n = p+N_d and N_d >> n_i yields n ≈ N_d, p = n_i²/N_d. Conductivity σ ≈ e N_d μ_e, controlled by donors.

What is minority carrier suppression? Prove electron concentration and conduction in n-type semiconductor is defined by impurity donor.

Minority carrier suppression is the reduction of the minority carrier concentration below the intrinsic value nin_i when a semiconductor is doped. In n-type material the holes are suppressed; in p-type the electrons are suppressed. It follows from the mass action law np=ni2np = n_i^2: if doping raises one carrier type, the other must fall in the same ratio.

Proof for n-type semiconductor

Consider Si doped with NdN_d donors per cm³ (P, As, Sb). Each donor level lies about 0.05 eV below EcE_c, so at room temperature practically all donors are ionised: Nd+=NdN_d^+ = N_d.

 Ec -------------------------
 Ed  - * - * - * - (donors, ~0.05 eV)
 EFn ........................
 EFi - - - - - - - - - - - - -
 Ev -------------------------

1. Charge neutrality:

n=p+Ndn = p + N_d

2. Mass action law (thermal equilibrium):

np=ni2,ni2=NcNve−Eg/kTnp = n_i^2,\quad n_i^2 = N_cN_ve^{-E_g/kT}

3. Solve: substitute p=ni2/np = n_i^2/n:

n2−Ndn−ni2=0  ⇒  n=12[Nd+Nd2+4ni2]n^2 - N_dn - n_i^2 = 0 \;\Rightarrow\; n = \frac{1}{2}\left[N_d + \sqrt{N_d^2+4n_i^2}\right]

4. Normal doping, Nd≫niN_d \gg n_i:

n≈Nd,p=ni2Nd≪nin \approx N_d,\qquad p = \frac{n_i^2}{N_d} \ll n_i

So the electron concentration is fixed by the donor concentration only, not by temperature or band gap, and the holes are suppressed by the factor ni/Ndn_i/N_d.

Conduction is defined by the donors

σ=e(nμe+pμh)=e(Ndμe+ni2Ndμh)\sigma = e(n\mu_e + p\mu_h) = e\left(N_d\mu_e + \frac{n_i^2}{N_d}\mu_h\right)

The ratio of the hole term to the electron term is ni2μhNd2μe\dfrac{n_i^2\mu_h}{N_d^2\mu_e}, which is tiny. Hence

σ≈eNdμe\sigma \approx eN_d\mu_e

Example: Si with Nd=1016N_d = 10^{16}: n=1016n = 10^{16}, p=2.1×104p = 2.1\times10^{4} cm⁻³; hole/electron current ratio ≈7×10−13\approx 7\times10^{-13}. σ=1.6×10−19×1016×1350=2.16\sigma = 1.6\times10^{-19}\times10^{16}\times1350 = 2.16 S cm⁻¹.

Conclusions

  • Electron concentration ≈Nd\approx N_d (donor defined).
  • Conductivity ≈eNdμe\approx eN_d\mu_e (donor defined), and it can be set accurately by controlling the doping.
  • This holds over the extrinsic range of temperature; at very high T, nin_i becomes comparable to NdN_d and the material turns intrinsic.

✓ Answer checked by Claude Opus 5.5 (high effort)

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