Electric Conductivity

Electric conductivity is a fundamental property that measures a material's ability to conduct electric current. This ability is crucial for the efficient transmission of electrical energy and is typically quantified in Siemens per meter (S/m). Materials with high conductivity, such as copper and aluminium, are essential in various applications, from electrical wiring to electronic devices. In everyday life, electric conductivity influences the performance of everything from household appliances and computer systems to advanced technology like smartphones and electric vehicles. Understanding and optimizing conductivity allows for better design and functionality in these devices, contributing to advancements in technology and improvements in energy efficiency. This property not only affects the efficiency of electrical systems but also plays a key role in the development of new materials and technologies.

This Story also Contains

1. Electrical Conductivity (σ)
2. Applications of Electrical Conductivity
3. Solved Examples Based on Electrical Conductivity
4. Summary

Electrical Conductivity (σ)

Electrical conductivity is a measure of a material's ability to conduct an electric current. It depends on the presence of free charge carriers—such as electrons or ions—that can move through the material.

The semiconductor conducts electricity with the help of these two types of electricity or charge carriers (i.e. electrons and holes). These holes and electrons move in the opposite direction. The electrons always tend to move in opposite directions to the applied electric field.

Let the mobility of the hole in the crystal is μ_h and the mobility of the electron in the same crystal is μ_e

The current density due to the drift of holes is given by,

$J_h=e{n}_h{v}_h=e{n}_h{\mu }_{h}E$

The current density due to the drift of electrons is given by,

$J_e=e{\mathrm{n}}_e{\mathrm{v}}_e=e{\mathrm{n}}_e{\mu }_{e}E$

hence resultant current density would be

$\begin{array}{rl}& J=J_h+J_c=e{n}_h{v}_h+e{n}_e{v}_e=e{n}_h{\mu }_{h}E+e{n}_c{\mu }_{c}E=(n_h{\mu }_h+n_c{\mu }_c)eE\\ & \text{and}J=\sigma E\end{array}$

So, the general equation for conductivity is given as

$\sigma =e(n_c{\mu }_c+n_h{\mu }_h)$

where

$\begin{array}{rl}& n_c=\text{electron density}\\ & n_h=\text{hole density}\\ & {\mu }_e=\text{mobility of electron}\\ & {\mu }_h=\text{mobility of holes}\end{array}$

For intrinsic semiconductors (no impurities)

As the number of electrons will be equal to the number of holes.

$\begin{array}{rl}& \text{i.e}{n}_e=n_h=n_i\\ & \sigma =n_{i}e({\mu }_c+{\mu }_h)\end{array}$

Dependence of Electrical Conductivity

Material Types

• Conductors (e.g., metals like copper and silver) have high electrical conductivity because they have many free electrons.
• Insulators (e.g., rubber, glass) have low electrical conductivity as they do not have many free charge carriers.
• Semiconductors (e.g., silicon) have intermediate conductivity that can be modified by doping or other means.

Temperature Dependence

In metals, conductivity usually decreases with increasing temperature because the increased thermal vibrations of the lattice atoms impede the flow of electrons. In semiconductors, conductivity often increases with temperature as more charge carriers are excited into the conduction band.

Applications of Electrical Conductivity

Understanding electrical conductivity is crucial in designing electrical circuits, choosing materials for electrical components, and in various industrial processes.

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Solved Examples Based on Electrical Conductivity

Example 1: The ratio of the concentration of electrons that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4, then what is the ratio of their drift velocities?

1) 4/7

2) 5/8

3) 4/5

4) 5/4

Solution:

$\begin{array}{rl}& n_e=\text{electron density}\\ & n_h=\text{hole density}\\ & {\mathrm{I}}_{\mathrm{e}}=\text{electron current}\\ & {\mathrm{I}}_{\mathrm{h}}=\text{hole current}\\ & I=n{v}_{d}eA\\ & v_d\to \text{drift velocity}\\ & \frac{I_c}{I_h}=\frac{n_c}{n_h}\cdot \frac{v_{d,c}}{v_{d,h}}\\ & \Rightarrow \frac{7}{4}=\frac{7}{5}\cdot \frac{v_{d,e}}{v_{d,h}}\\ & \Rightarrow \frac{v_{d,e}}{v_{d,h}}=\frac{5}{4}\end{array}$

Hence, the answer is the option (4)

Example 2: The mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If , for an n-type semiconductor, the density of electrons is 10¹⁹ m ^-3 and their mobility is 1.6 m²/(V.s) then the resistivity ( in $\Omega m$ ) of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored ) is close to :

1) 0.4

2) 4

3) 2

4) 0.2

Solution:

Electrical Conductivity (σ)

$\sigma =e(n_c{\mu }_c+n_h{\mu }_h)$

For intrinsic semiconductors (no impurities), the number of electrons will be equal to the number of holes.

So $n_e=n_h=n_i$
$\sigma =n_{i}e({\mu }_c+{\mu }_h)$

For N -type semiconductors electrons are the majority carriers.
Conductivity $\sigma \approx ne\mu$

Resistivity
$\rho =\frac{1}{\sigma }=\frac{1}{n_{e}e\mu }$
$\begin{array}{rl}& =\frac{1}{{10}^{19}\times 1.6\times \times {10}^{-19}\times 1.6}\\ & =0.4\mathrm{\Omega }\mathrm{m}\end{array}$

Hence. the answer is the option (1).

Example 3: In the ratio of the concentration of electrons that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4 then what is the ratio of their drift velocities?

1) 4/7

2) 5/8

3) 4/5

4) 5/4

Solution:

Electrical Conductivity (σ)

$\sigma =e(n_e{\mu }_e+n_h{\mu }_h)$
wherein
$\begin{array}{rl}& n_e=\text{electron density}\\ & n_h=\text{hole density}\\ & {\mu }_c=\text{mobility of electron}\\ & {\mu }_h=\text{mobility of holes}\end{array}$

So,
$\begin{array}{rl}& \frac{n_e}{n_h}=\frac{7}{5},\frac{I_e}{I_h}=\frac{7}{4}\\ & I=n{v}_{\alpha }eA\\ & \frac{I_e}{I_h}=\frac{n_e{v}_{d,e}}{n_h{v}_{d,h}}\Rightarrow \frac{v_{d,e}}{v_{d,h}}=\frac{I_e}{I_h}\cdot \frac{n_h}{n_e}\\ & \frac{v_{d,e}}{v_{d,h}}=\frac{7}{4}\times \frac{5}{7}=\frac{5}{4}\end{array}$

Hence, the answer is the option (4).

Example 4: The effect of the increase in temperature on the number of electrons in the conduction band $\mathrm{(n_{e})}$ and resistance of a semiconductor will be as:

1) Both $n_e$ and resistance increase
2) Both ${\mathrm{n}}_e$ and resistance decrease
3) $n_e$ decreases, resistance increases
4) $n_e$ increases, resistance decreases

Solution:

In semi-conductors,
$\mathrm{T}\uparrow ,{\mathrm{n}}_{\mathrm{e}}$ in Conduction Band increases
$\mathrm{T}\uparrow ,\mathrm{R}\downarrow$Hence, the answer is the option (4).

Example 5: Copper has face-centered cubic (fcc) lattice with interatomic spacing equal to 2.54 Å. The value of lattice constant for this lattice is:

1) 2.54 Å
2) 3.59 Å
3) 1.27 Å
4) 5.08 Å

Solution:

Given interatomic spacing $=2\mathrm{r}=2.54$ Å
$4\mathrm{r}=\sqrt{2}\mathrm{a}$, where a is lattice constant $2\mathrm{r}=\frac{\sqrt{2}\mathrm{a}}{2}=\frac{\mathrm{a}}{\sqrt{2}}$

$\mathrm{a}=2\mathrm{r}\sqrt{2}=(2.54\hat{\mathrm{A}})(1.414)=3.59\hat{\mathrm{A}}$

Hence, the answer is the option (2).

Summary

Electric conductivity quantifies a material's ability to conduct electric current and is expressed in Siemens per meter (S/m). It is inversely related to resistivity, indicating how strongly a material opposes current flow. In semiconductors, conductivity is influenced by the mobility and concentration of charge carriers (electrons and holes). Practical applications of conductivity include efficient wiring in electrical circuits and precise control in electronic devices. The given examples illustrate the calculation of drift velocities, resistivity, and effects of temperature on semiconductor conductivity.