Temperature coefficient of resistance of a semiconductor is
A. Zero
B. Constant
C. Positive
D. Negative
Answer
647.7k+ views
Hint: We must know that temperature coefficient of a material is proportional to the difference in resistance of the material at temperature (t) and absolute zero temperature ($0K$) and inversely proportional to the product of temperature and resistance at absolute zero. The temperature coefficient for metals and alloys are positive. In the case of insulators it is negative.
Formula used:
$\alpha =\dfrac{{{R}_{t}}-{{R}_{0}}}{{{R}_{0}}t}$
Complete answer:
We know that the temperature coefficient of resistance is equal to the change in resistance of the wire of resistance one ohm at $0{}^\circ C$ when the temperature changes by $1{}^\circ C$ .
It is denoted by $\alpha $and given as,
$\alpha =\dfrac{{{R}_{t}}-{{R}_{0}}}{{{R}_{0}}t}$
Where, ${{R}_{t}}$ is the resistance of the material at temperature (t).
${{R}_{0}}$ is the resistance of the material at absolute zero temperature.
$t$ is the temperature.
Let’s consider a case,
If ${{R}_{0}}=1$ and $t=1{}^\circ C$. Then, $\alpha ={{R}_{t}}-1$.
In general, for two different temperature and resistance, we can write the equation as,
$\alpha =\dfrac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}t-{{R}_{2}}t}$
From this equation, we can say that as the temperature of a semiconductor increases, resistance decreases. Then the value of ${{R}_{2}}-{{R}_{1}}$becomes a negative value. Therefore, the entire value becomes negative in the case of the semiconductor. Thus, the temperature coefficient of resistance ($\alpha $) of a semiconductor is always negative.
So, the correct answer is “Option D”.
Note:
We should know that resistance is the opposition offered by the substance to the flow of electric current. The resistance of a conductor depends on the nature of material and temperature. But in the case of a semiconductor, the effect of increase in the number of free electrons per unit volume is much higher than the effect of decrease in relaxation time. So, the resistance of a semiconductor increases with increase in temperature and vice versa.
Formula used:
$\alpha =\dfrac{{{R}_{t}}-{{R}_{0}}}{{{R}_{0}}t}$
Complete answer:
We know that the temperature coefficient of resistance is equal to the change in resistance of the wire of resistance one ohm at $0{}^\circ C$ when the temperature changes by $1{}^\circ C$ .
It is denoted by $\alpha $and given as,
$\alpha =\dfrac{{{R}_{t}}-{{R}_{0}}}{{{R}_{0}}t}$
Where, ${{R}_{t}}$ is the resistance of the material at temperature (t).
${{R}_{0}}$ is the resistance of the material at absolute zero temperature.
$t$ is the temperature.
Let’s consider a case,
If ${{R}_{0}}=1$ and $t=1{}^\circ C$. Then, $\alpha ={{R}_{t}}-1$.
In general, for two different temperature and resistance, we can write the equation as,
$\alpha =\dfrac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}t-{{R}_{2}}t}$
From this equation, we can say that as the temperature of a semiconductor increases, resistance decreases. Then the value of ${{R}_{2}}-{{R}_{1}}$becomes a negative value. Therefore, the entire value becomes negative in the case of the semiconductor. Thus, the temperature coefficient of resistance ($\alpha $) of a semiconductor is always negative.
So, the correct answer is “Option D”.
Note:
We should know that resistance is the opposition offered by the substance to the flow of electric current. The resistance of a conductor depends on the nature of material and temperature. But in the case of a semiconductor, the effect of increase in the number of free electrons per unit volume is much higher than the effect of decrease in relaxation time. So, the resistance of a semiconductor increases with increase in temperature and vice versa.
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