Conductivity of Extrinsic Semiconductor

Conductivity of Extrinsic Semiconductor

• In an extrinsic semiconductors, there are two types of materials n‒type and p‒type.

1. Conductivity of n‒Type Material

• It is known that in n‒type material, the free electrons are majority carriers while the holes are minority carriers.

Let n_n = Concentration of free electrons in n type

P_n = Concentration of holes in n type

N_D = Concentration of donor atoms

Key Point: In the symbol, main letter n or p indicates concentration of type of charge carrier electron or hole while the suffix indicates the type of material i.e. n‒type or p‒type. Thus n_n indicates electron concentration in n‒type material while n_p indicates electron concentration in p‒type material and so on.

• From the basic equation of conductivity, the conductivity of n‒type material can be expressed as,

 σ_n = (n_nμ_n + P_nμ_p)q

• But P_n<<n_n as holes are in minority hence,

 σ_n ≈ n_nμ_nq

• The number of free electrons is dominantly controlled by donor atoms added than the thermal at generation room temperature. Hence concentration of donor atoms N_D added can be approximately assumed to be equal to the concentration of free electrons n_n in n type materials.

• Thus as N_D >> n_i we can write,

 n_n = N_D and σ_n ≈ N_Dμ_nq

2. Conductivity of p‒Type Material

• For a p type material, holes are in majority and electrons are in minority.

Let

n_P = Concentration of free electrons in p type

p_P = Concentration of holes in p type

N_A = Concentration of acceptor atoms

• Thus the conductivity of p type material can be expressed as,

 σ_p = (n_pμ_n + P_Pμ_P) q

• But n_p<< P_P as free electrons are in minority hence,

 σ_p ≈ P_p μ_p q

The number of holes is dominantly controlled by added acceptor impurity than the thermal generation. Each added impurity atom creats a hole hence N_A >> n_i. Thus all the holes generated can be approximately assumed to be equal to the concentration of acceptor N_A. Thus,

P_p = N_A

And

σ_p = N_A μ_p q

Properties of Silicon and Germanium

Table 1.13.1 Properties of germanium and silicon

Ex. 1.13.1 If a donor impurity is added to the extent of one atom per 10⁸ germanium atoms, calculate its resistivity at 300 °K. If its resistivity without addition of impurity at 300 °K is 44.64 Ω‒cm, comparing two values, comment on the result.

Assume: μ_n = 3800 cm²/V‒sec.

Solution:

Referring to Table 1.13.1 of properties of germanium, germanium has

 4.4× 10²² atoms/cm³.

For 10⁸ germanium atom there is 1 atom impurity added, as given.

Thus, for 4.4× 10²² germanium atoms, we have,

= 4.4×10²² / 10⁸⁰

= 4.4× 10¹⁴ atoms of impurity/cm³

This is nothing but concentration of donor atoms i.e. N^D

 N_D = 4.4 × 10¹⁴ per cm³

 = 4.4 × 10¹⁴ / 10^‒6

= 4.4× 10²⁰ per m³

Now as donor impurity is added, n‒type material will form,

 σ_n = n_n μ_n q = N_D μ_n q

where

n_n = N_D

and

μ_n = 3800 cm²/V‒sec

= 3800 × 10^‒4 m²/V‒sec

σ_n = 4.4× 10²⁰ × 3800 × 10^‒4 × 1.6 × 10^‒19

= 26.752 (Ω‒m)^‒1

Resistivity = ρ_n = 1/σ_n

= 1 / 26.752

= 0.0373 Ω‒m = 3.73 Ω ‒ cm

Comment: Comparing this with resistivity of intrinsic germanium it can be observed that resistivity reduces considerably due to addition of impurity. Hence conductivity of n‒type material is much higher and hence it can carry more current as compared to the intrinsic semiconductor. By controlling amount of doping we can control the conductivity.

Review Questions

1. Obtain the expressions for conductivity of n‒type and p‒type materials.

2. Consider silicon at room temperature of 27 °C. Assuming μ_n = 1350 cm³/V‒s and μ_P=0.048m²/V‒s, find conductivity if silicon is doped with

a) N_D = 5×10¹⁶/cm³,

b) N_A=5×10¹⁶/cm³

[Ans.: 10.8 (Ω‒cm)^‒1, 3.84 (Ω‒cm)^‒1]