General Expression for Conductivity

General Expression for Conductivity (σ)

• Consider a tube of metal with number of free electons as shown in Fig. 1.5.1.

Let

A = Area of cross‒section

L = Length in m

V = Voltage applied in volts

T = Time required by an electron to travel distance L

V = Drift velocity = L/T and E = V/L

• Consider any cross‒section as shown in Fig. 1.5.2.

• Let N be the number of electrons passing through area A in time T. So number of electrons crossing the area A in unit time is N/T.

 q= Charge on one electron = 1.6×10^‒19 C

• The total charge crossing the area A in unit time is,

 dq = Nq/T C

• I = Charge in unit time / Unit time

I = [Nq/T] / (1 second) = Nq/T

• The current density J is current per unit cross‒sectional area hence,

 J = I/A = Nq / AT  but T= L/v

J = Nqv / LA

but LA = Volume of the tube

• Hence N/LA is number of electrons per m³ which is called concentration of free electrons denoted as 'n'.

J = nqv

where n = N/LA m³

 J = nqμB A/m²

• current density is related to an electric field E by Ohm's law as,

 J = σE

 σ = Conductivity in (Ω‒m)⁻¹

• Comparing the above equations, the conductivity is given by,

 σ = nqμ (Ω‒m)^‒1     ……….Conductivity

 ρ = 1/σ = 1 / nqμ    Ω‒m      ………. Resistivity

• The conductivity indicates the ease with which current can flow through the given material.

Ex. 1.5.1: A bar of n type silicon has length of 4 cm and circular cross‒section of 10 mm². When it is subjected to a voltage of 1 V applied across its length, the current flowing through it is 5 mA. Calculate i) Concentration of free electrons. ii) Drift velocity of electrons.

Assume ‒ Charge on one electron as 1.6 × 10^‒19 C.

Mobility of free electron as 1300 cm²/V‒s.

Solution:

The bar is shown in Fig. 1.5.3.

 Current density J = I/A = 5×10^‒3 / 10×10^‒6 = 500 A/m²

The current density is also given by,

J = nqμE         ………..(1)

where

E = Field intensity = V/L

= 1 / 4×10^‒2

 = 25 V/m

q = Charge on electron = 1.6×10^‒19 C

 μ = Mobility = 1300 cm²/V‒s = 0.13 m²/V‒s

Substituting all the values in equation (1),

500 = n×1.6×10^‒19 × 0.13×25

i.e.

 n = 0.9615×10²¹ per m³

Drift velocity v = μ E = 0.13×25 = 3.25 m/s        ... Drift velocity

Review Questions

1. Derive the general expression for drift current density and conductivity of a material.

2. The resistance per unit length of a piece of copper wire with circular cross section with diameter of 1.03 mm is 2.5×10^‒4Ω/cm. Concentration of free electrons is 8.4×10²⁸ per m³. The current density is 2.1× 10⁶ A/m². Calculate current flowing, conductivity, velocity of free electrons and mobility.

[Ans. : 1.749 mA, 4.8×10⁷(Ω‒m)^‒1, 1.56×10^-4 m/s, 3.567 × 10^‒3 m² N‒sec]