Law of Mass Action

Law of Mass Action

• If n is the concentration of free electrons and p is the concentration of holes then the law of mass action states that the product of concentrations of electrons and holes is always constant, at a fixed temperature.

Mathematically it is expressed as,

np = n_i²

where n_i is intrinsic concentration

• Important Observations

1. The law can be applied to both intrinsic and extrinsic semiconductors.

2. In case of extrinsic semiconductors, n_i is the intrinsic concentration of the basic semiconductor material used.

3. For n‒type material, n=n_n while p = P_n hence law can be stated as,

 n_nP_n = n_i²

4. For p‒type material, n=n_p while p = P_p hence law can be stated as,

 n_pP_p = n_i²

5. The law is applicable irrespective of amount of doping.

6. As n_i depends on temperature, the law is applicable at a fixed temperature.

7. The law can be used to find both majority and minority carrier concentrations in an extrinsic semiconductor.

Carrier Concentrations in Extrinsic Semiconductors

• Let us obtain the concentrations of minority and majority carriers in n‒type and p‒type materials using law of mass action.

n type material:

• For n type material it is seen that, n_n=ND:

• At any fixed temperature, according to law of mass action, n_n × P_n = n_i²

where n_n is electrons i.e. majority carrier concentration while P_n is hole i.e. minority carrier concentration. Using n_n = N_D, we can write minority carrier concentration as,

 N_D P_n = n_i²

i.e. P_n = n_i² / N_D

• Knowing n_i and N_D, the number of holes in n type material i.e. minority carrier concentration can be obtained.

p‒type material:

• For p‒type material it is seen that, n_p=ND:

• According to law of mass action, n_p × P_p = n_i²

where n_p is electrons i.e. minority carrier concentration while P_p is holes i.e. majority carrier concentration. Using P_p = N_A, we can write,

n_PN_A = n_i²

i.e. n_P = n_i² / N_A

Knowing n_i and N_A, the number of electrons in p‒type material i.e. minority carrier concentration can be obtained.

Ex. 1.14.1: Calculate the majority and minority carrier concentrations in silicon at room temperature of 27° C if

a) N_A =10¹⁷/cm³ and b) N_D = 5×10¹⁵/cm³

Solution:

a) As impurity is acceptor, the material is p‒type.

Pp = Majority carrier concentration = N_A =10¹⁷/cm³

For p‒type,

n_p×P_p = n_i²

                .... Law of mass action

n_i = 1.5 × 10¹⁰ /cm³

                …….From Table 1.13.1

n_p = n_i² / P_p = 2.25 × 10³ /cm³

                Minority carrier concentration

b) As impurity is donor, the material is n‒type

 n_n = Majority carrier concentration = N_D = 5 × 10¹⁵ /cm³

For n‒type,

 n_n × P_n = n_i²

               Law of mass action

 n_i = 1.5 × 10¹⁰ /cm³

               Basic material same

 P_n = n_i² / n_n = 45 × 10³ /cm³

               Minority carrier concentration

Ex. 1.14.2: A bar of silicon 0.1 cm long has a cross‒sectional area of 8 × 10^‒8 m², heavily doped with phosphorous. What will be the majority carrier density resulting from doping if bar is to have a resistance of 1.5 kΩ?

Given: For silicon at room temperature, μ_n = 0.14 m²/ V‒sec, μ_p = 0.05 m²/V‒sec, n_i = 1.5×10¹⁰ per cm³.

Solution: :

 R = ρl / A

where

 l = 0.1 cm = 0.1× 10⁻² m,

 A = 8×10^‒8 m², R = 1.5 kΩ

 ρ = RA/l = [ 1.5×10³ × 8×10^‒8  ]  / 0.1× 10^‒2

 = 0.12 Ω ‒ m

Conductivity.

σ = 1/ ρ  = 1/ 0.12 = 8.333 (Ω‒m)^‒1

But

σ = σ_n = (n_nμ_n+P_pμ_P)q

This is because phosphorous is donor impurity and will form n‒type material.

According to law of mass action for n‒type material,

n_n P_n = n_i²

p_n = n_i² / n_n

σ_n = (n_nμ_n+P_pμ_P)q 

σ_n = (n_nμ_n+( n_i² / n_n)μ_P)q 

σ_n = (n_n²μ_n+ n_i²μ_P)q 

 n_n × 8.333 = [ n_n² × 0.14 + (1.5 × 10¹⁰ / 10^‒6)² × 0.05 ] 1.602 × 10^‒19

  0.14n_n² ‒5.201× 10¹⁹ n_n + 1.125× 10³¹ = 0

Solving,

 n_n = 3.715 × 10²⁰ per m³ (neglecting other value as comparable to n_i)

n_n = N_D        

This is majority carrier density.

Ex. 1.14.3 Find the concentration of holes and electrons in a p type silicon at 300 °K assuming its resistivity in a p type silicon as 300 °K assuming its resistivity as 0.02 Ω‒cm, μ_p = 475 cm² / V ‒ sec, n_i = 1.45 ×10¹⁰ per cm³.

Solution:

 ρ = Resistivity = 0.02 Ω‒cm = 0.02 × 10^‒2 Ω‒m

The material is p type and its conductivity is,

 σ_P = 1/ρ = 1 / 0.02×10^‒2 = 5×10³ (Ω‒m) ^‒1

But

σ_P = N_Aμ_p q

where

 q = 1.6 x 10^‒19 C,

μ_p = 475 cm²/V‒sec = 475 × 10^‒4 m² / V‒sec

5×10³ = N_A × 475 × 1.6 × 10⁻¹⁹ ×10^‒4

 N_A = 6.5789 × 10²³ per m³

But

 P_p = N_A = 6.5789 × 10²³ per m³

Concentration of holes

Using law of mass action, Pp × n_p = n_i²

And

n_i =  1.45 × 10¹⁰ / 10^‒6 per m³

n_p = n_i² / Pp = (1.45 × 10¹⁶)²  / 6.5789×10²³

= 3.1958 × 10⁸ per m³

Concentration of electrons

Ex. 1.14.4 Find the concentration of holes and electrons in a p‒type Germanium at 300 °K, if the conductivity is 100 per ohm‒cm. Also find these values for n‒type silicon, if the conductivity is 0.1 per ohm‒cm. Given that

For Germanium n_i = 2.5×10¹³/cm, μ_n = 3800 cm²/v‒s, μ_p = 1800 cm²/v ‒ s

For silicon, n_i = 1.5×10¹⁰ per cm³, μ_n = 1300 cm²/v ‒ s and μ_p = 500 cm²/v ‒ s

Solution:

Case 1: p‒type Germanium

n_i = 2.5×10¹³/cm³ = 2.5×10¹³ / 10^-6 /m³

= 2.5×10¹⁹/m³

μ_n = 3800 cm²/V‒s = 3800×10^-4 m²/V‒s

μ_р= 1800 cm²/V‒s = 1800×10^‒4 m²/V‒s

σ_р = N_Aμ_pe  where σ = 100 (Ω‒cm)^‒1

100 / 10^‒2 = N_A×1800×10^‒4×1.6×10^‒19     i.e. N_A = 3.47×10²³/m³

P_p = N_A = 3.47 × 10²³/m³

       ... Concentration of holes

n_p = n_i² / P_P = n_i² / N_A

=  (2.5×10¹⁹)² /  3.47×10²³

= 1.8×10¹⁵/m³

... Concentration of electrons

case 2: n‒type Silicon

n_i = 1.5×10¹⁰/cm³ = 1.5×10¹⁰ / 10^-6 /m³

= 1.5×10¹⁶/m³

μ_n = 1300 cm²/V‒s = 1300×10^-4 m²/V‒s

μ_р= 500 cm²/V‒s = 500×10^‒4 m²/V‒s

σ_n = N_Dμ_nq  where σ_n = 0.1 (Ω‒cm)^‒1

0.1 / 10^‒2 = N_D×1300×10^‒4×1.6×10^‒19     i.e. N_D = 4.807×10²⁰/m³

n_n = N_D = 4.807 × 10²⁰/m³

       ... Concentration of electrons

p_n = n_i² / n_n = n_i² / N_D

=  (1.5×10¹⁶)² /  4.807×10²⁰

= 4.68×10¹¹/m³

... Concentration of holes

Review Questions

1. State and explain the law of mass action.

2. Explain the carrier concentrations in extrinsic semiconductors.