
As the temperature increases, the value of the current gain $\beta$
A. increases
B. decreases
C. remains constant
D. cannot be determined
Answer
600.3k+ views
Hint: In this question, we need to understand the relationship between the current gain of which there are no direct equations, although there are various relations with which we can understand this and one way to do that is by using the Ebers-Moll model.
Complete step-by-step answer:
Here, we need to find out the effect of temperature on the current gain in a transistor. Let us use the Ebers-Moll model which determines the relationship between the emitter current ${I}_{E}$, collector current ${I}_{C}$ and the forward emitter current gain ${\beta}_{F}$ and is given by
${I}_{C}=\dfrac{{\beta}_{F}}{{\beta}_{F}+1}{I}_{E}$
which can also be written as ${\beta}_{F}=\dfrac{1}{\dfrac{{I}_{E}}{{I}_{C}}-1}$ ………. (i)
Now, we know the relation between the emitter current ${I}_{E}$, reverse saturation current ${I}_{ES}$, base emitter voltage ${V}_{BE}$ and thermal voltage ${V}_{T}$, which is given by
${I}_{E}={I}_{ES}(e^{\dfrac{{V}_{BE}}{{V}_{T}}}-1)$ ………. (ii)
Also, ${V}_{T}=\dfrac{{k}_{B}T}{q}$ ………. (iii)
where q is the charge and ${k}_{B}$ is a constant.
Now, from equation (iii), we can see that with increase in temperature, thermal voltage increases which in turn in equation (ii) decreases the value of emitter current. So, an increase in temperature decreases the value of emitter current. In equation (i), we can notice that with decrease in the value of emitter current, current gain increases.
Therefore, we can say from above deductions, that increase in temperature increases the current gain in a transistor.
Hence, option a is the correct answer.
Note: It should be noted that with increase in temperature, the collector current increases because it starts driving more positive charge towards base increasing the forward bias on the base-emitter diode.
Complete step-by-step answer:
Here, we need to find out the effect of temperature on the current gain in a transistor. Let us use the Ebers-Moll model which determines the relationship between the emitter current ${I}_{E}$, collector current ${I}_{C}$ and the forward emitter current gain ${\beta}_{F}$ and is given by
${I}_{C}=\dfrac{{\beta}_{F}}{{\beta}_{F}+1}{I}_{E}$
which can also be written as ${\beta}_{F}=\dfrac{1}{\dfrac{{I}_{E}}{{I}_{C}}-1}$ ………. (i)
Now, we know the relation between the emitter current ${I}_{E}$, reverse saturation current ${I}_{ES}$, base emitter voltage ${V}_{BE}$ and thermal voltage ${V}_{T}$, which is given by
${I}_{E}={I}_{ES}(e^{\dfrac{{V}_{BE}}{{V}_{T}}}-1)$ ………. (ii)
Also, ${V}_{T}=\dfrac{{k}_{B}T}{q}$ ………. (iii)
where q is the charge and ${k}_{B}$ is a constant.
Now, from equation (iii), we can see that with increase in temperature, thermal voltage increases which in turn in equation (ii) decreases the value of emitter current. So, an increase in temperature decreases the value of emitter current. In equation (i), we can notice that with decrease in the value of emitter current, current gain increases.
Therefore, we can say from above deductions, that increase in temperature increases the current gain in a transistor.
Hence, option a is the correct answer.
Note: It should be noted that with increase in temperature, the collector current increases because it starts driving more positive charge towards base increasing the forward bias on the base-emitter diode.
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