Equation of Charge Neutrality

Equation of Charge Neutrality

• Consider a semiconductor material which is doped with both type of impurities, donor as well as acceptor. The concentration of donor impurity is N_D and that of acceptor impurity is N_A.

• Each donor atom donates one electron and as it loses electron it becomes positively charged ion, due to the process of ionization.

• Thus assuming ionization of all the impurity atoms, the concentration of positive ions generated due to donor impurity becomes N_D.

• There are positive charge carriers i.e. holes present in the material whose concentration is p.

• Thus the concentration of total positive charge in the material is given by,

Total positive charge concentration = N_D + P            ……………(1)

• The acceptor impurity atom when added to the material accepts an electron and becomes negatively charged ion.

• Assuming ionization of all the impurity atoms, the concentration of negative ions generated due to acceptor impurity is N_A.

• There are negative charge carriers i.e. electrons present in the material whose concentration is n.

• Thus the concentration of total negative charge in the material is given by,

Total negative charge concentration = N_A + n                   ……………(2)

• The material as a whole is always electrically neutral hence the total positive charge concentration in the material must be equal to the total negative charge concentration in the material. Hence equating (1) and (2) we can write,

 N_D + p = N_A + n                …………..(3)

The equation (3) is called the equation of charge neutrality.

• Observations:

1. The equation is important when a semiconductor is doped with both the types of impurities, donor as well as acceptor.

2. The equation is effective in obtaining the actual concentrations of electrons and holes in a material which is doped with both the types of impurities.

3. The equation helps to determine whether the material doped with both the types of impurities will behave as n‒type or p‒type.

Ex. 1.15.1 In a sample of germanium at 300 °K, it is found that donor concentration is 2×10¹⁴ atoms/cm³ and acceptor concentration is 3× 10¹⁴ atoms/cm³.

Determine the actual concentrations of free electrons and holes in the sample. Will it behave as p‒type or n‒type? Assume n_i at 300 °K = 2.5 × 10¹³ per cm³.

Solution:

In this sample,

 N_D = 2×10¹⁴ atoms/cm³ = 2×10¹⁴ / 10^‒6 = 2×10²⁰ m³

 N_A = 3×10¹⁴ atoms/cm³ = 3×10¹⁴ / 10^‒6 = 3×10²⁰ m³

Using equation of charge neutrality, p + N_D = n + N_A.

N_A ‒ N_D = p‒n

3×10²⁰  ‒ 2×10²⁰ = p‒n

i.e. p = n + 1×10²⁰

where p and n are actual concentrations of holes and electrons.

 n_i = 2.5×10¹³ per cm³

 = 2.5×10¹³ / 10^‒6

= 2.5×10¹⁹ per cm³

According to law of mass action, np = n_i²

 n (n + 1×10²⁰) = (2.5 × 10¹⁹)²      i.e. n² + 1×10²⁰n − 6.25×10³⁸ = 0

Solving

n = 5.901×10¹⁸ electrons/m³

p = 1.059×10²⁰ holes/m³

 As holes are much more than electrons, sample will behave as p‒type.

Ex. 1.15.2 A sample of Ge is doped to the extent of 10¹⁴ donor atoms/cm³ and 7 × 10¹³ acceptor atoms/cm³. At the temperature of the sample, the resistivity of pure Ge is 60 ohm‒cm. If the applied electric field is 2 V/cm, find the total conduction current density.

Solution:

 N_D = 10¹⁴/cm³ and N_A =7×10¹³/cm³

 ρ_i = 60 Ω‒cm

 J = (nμ_n + Pμ_p) qE

   …….. μ_n = 3800, μ_p = 1800

According to law of electrical neutrality,

 N_D + P = N_A + n

i.e. n‒p = 3 × 10¹³

According to law of mass action, np =n_i²

σ_i = n_i(μ_n+μ_p) q

i.e. 1/ρ_i = n_i (μ_n+ μ_p) q

1/60 = n_i (3800 + 1800) × 1.6 × 10^‒19

i.e. n=1.86 × 10¹³/cm³

Solve equations (1) and (2) to obtain n and p as,

n = 3.88×10¹³/cm³ and

p=0.88×10¹³/cm³

 J = (3.88 × 10¹³ × 3800 + 0.88 × 10¹³ × 1800) × 1.6 × 10^‒19 × 2 = 0.0522 A/cm

Review Question

1. Explain the equation of charge neutrality.