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p-type minority carrier suppression and conductivity

Short answer

p-type: doped with acceptors, majority holes. From np = n_i² and p ≈ N_a, minority electron concentration n = n_i²/N_a ≪ n_i (suppression). Conductivity σ ≈ e N_a μ_h, because hole term dominates; electron contribution is negligible.

Define p-type semiconductor. Derive an expression for minority carrier suppression and hence prove that the conductivity in p-type semiconductor is mainly due to the hole.

A p-type semiconductor is an intrinsic semiconductor (Si, Ge) doped with a small amount of a trivalent (group III) impurity such as boron, aluminium, gallium or indium. Each impurity atom forms only three covalent bonds, so one bond is incomplete; it easily accepts an electron from the valence band, creating a hole. Such impurities are called acceptors (NaN_a). Holes are the majority carriers and electrons the minority carriers.

 Ec ------------------------
 EFi - - - - - - - - - - - -
 EFp .......................
 Ea  - o - o - o - (acceptors, ~0.05 eV)
 Ev ------------------------

Derivation of minority carrier suppression

1. Mass action law. In equilibrium, for any non-degenerate semiconductor,

np=ni2np = n_i^2

2. Charge neutrality. Positive charges = negative charges:

p=n+Na−p = n + N_a^-

At room temperature all acceptors are ionised, Na−=NaN_a^- = N_a.

3. Solve. Put n=ni2/pn = n_i^2/p:

p2−Nap−ni2=0  ⇒  p=Na2+Na24+ni2p^2 - N_ap - n_i^2 = 0 \;\Rightarrow\; p = \frac{N_a}{2} + \sqrt{\frac{N_a^2}{4}+n_i^2}

For normal doping Na≫niN_a \gg n_i, so

p≈Na,n=ni2Nap \approx N_a,\qquad n = \frac{n_i^2}{N_a}

4. Suppression. Compare with the intrinsic value:

nni=niNa≪1\frac{n}{n_i} = \frac{n_i}{N_a} \ll 1

Adding acceptors raises pp above nin_i by the factor Na/niN_a/n_i and lowers nn below nin_i by the same factor. The extra holes recombine with the thermally generated electrons. This reduction of the minority electron concentration is called minority carrier suppression.

Example: Si with Na=1016N_a = 10^{16}: p=1016p = 10^{16}, n=(1.45×1010)2/1016=2.1×104 cm−3n = (1.45\times10^{10})^2/10^{16} = 2.1\times10^{4}\ \text{cm}^{-3}.

Conductivity is due to holes

σ=e(nμe+pμh)=e(ni2Naμe+Naμh)\sigma = e(n\mu_e + p\mu_h) = e\left(\frac{n_i^2}{N_a}\mu_e + N_a\mu_h\right)

Ratio of the two terms:

nμepμh=ni2Na2⋅μeμh\frac{n\mu_e}{p\mu_h} = \frac{n_i^2}{N_a^2}\cdot\frac{\mu_e}{\mu_h}

For the example: (1.45×1010/1016)2×3≈6×10−12(1.45\times10^{10}/10^{16})^2\times3 \approx 6\times10^{-12}. Even though μe≈3μh\mu_e \approx 3\mu_h, the electron term is negligible, so

σ≈eNaμh\sigma \approx eN_a\mu_h

Hence the conductivity of a p-type semiconductor is almost entirely due to holes and is fixed by the acceptor concentration.

✓ Answer checked by Claude Opus 5.5 (high effort)

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