Shortcut Methods and Tricks

1. Temperature Dependence of Resistivity

• Remember the approximate value of the temperature coefficient of resistivity for copper: 0.0039/°C.
• For small temperature changes, the change in resistivity can be approximated as:

$$\mathrm{\Delta }\rho ={\rho }_0\ast \alpha \ast \mathrm{\Delta }T$$ where:

• $\mathrm{\Delta }\rho$ is the change in resistivity
• ${\rho }_0$ is the resistivity at the reference temperature
• $\alpha$ is the temperature coefficient of resistivity
• $\mathrm{\Delta }T$ is the change in temperature.

2. Mobility and Resistivity

• The mobility of charge carriers is inversely proportional to their mass.
• The resistivity of a material is inversely related to the mobility of its charge carriers. This means that a material with high mobility will have a low resistivity, and vice versa.

Sample Problems

1. Temperature Dependence of Resistivity

A copper wire has a resistance of 1 Ω at 25°C. What will its resistance be at 50°C?

Solution:

Using the formula for the change in resistivity:

$$\mathrm{\Delta }R=R_0\alpha (T_2-T_1)$$

where $$R_0=1\mathrm{\Omega },$$ $$\alpha =0.0039/°C,$$ $$T_1=25°C,\text{ and }{T}_2=50°C$$

Substituting these values into the equation, we get:

$$\mathrm{\Delta }R=1\mathrm{\Omega }\ast 0.0039/°C\ast (50°C-25°C)=0.0975\mathrm{\Omega }$$

Therefore, the resistance of the copper wire at 50°C will be:

$$R=R_0+\mathrm{\Delta }R=1\mathrm{\Omega }+0.0975\mathrm{\Omega }=1.0975\mathrm{\Omega }$$

2. Mobility and Resistivity

A semiconductor material has a mobility of 0.1 m^2/V-s and a resistivity of 10^-3 ohm-m. What is the concentration of charge carriers in the material?

Solution:

The conductivity of the material can be calculated using the formula:

$$\sigma =ne\mu$$

where:

• $\sigma$ is the conductivity in S/m
• $n$ is the concentration of charge carriers in m^-3
• $e$ is the elementary charge in coulombs
• $\mu$ is the mobility in m^2/V-s

Substituting the given values into the equation, we get:

$${10}^{-3}S/m=n\times 1.602\times {10}^{-19}C\times 0.1m^2/V-s$$ $$n=\frac{{10}^{-3}S/m}{1.602\times {10}^{-19}C\times 0.1m^2/V-s}=6.24\times {10}^{15}{m}^{-3}$$

Therefore, the concentration of charge carriers in the semiconductor material is 6.24 x 10^15 m^-3.