
In n-p-n transistor, in CE configuration, identify true statements from the following:
(a) The emitter is heavily doped than the collector
(b) Emitter and collector can be interchanged
(c) The base region is very thin but is heavily doped
(d)The conventional current flows from base to emitter
(A) (a) and (b) are correct
(B) (a) and (c) are correct
(C) (a) and (d) are correct
(D) (b) and (c) are correct
Answer
558.9k+ views
Hint: To answer this question, we need to use the properties of the three regions of a transistor, namely, their doping levels and their thickness. The conventional current direction can be answered using the type of the transistor given in the question.
Complete step-by-step solution
We know that a transistor has three different regions; the emitter, the base, and the collector. Out of these, the collector is the widest, the base is the thinnest, and the emitter is of moderate thickness. Talking about the doping level, the emitter is most heavily doped, the base is least doped, and the collector is moderately doped.
So clearly the emitter is heavier than the collector.
Therefore, the statement (a) is correct.
Now, since the thickness and the doping levels of the emitter and the collector are different, they cannot be exchanged.
Therefore, the statement (b) is incorrect.
The base is thinnest out of the three regions, but it is only slightly doped.
Therefore, the statement (c) is also incorrect.
Now, the given transistor is of n-p-n type. We know that this indicates the type of majority charge carriers respectively for the emitter, the base, and the collector. So for an n-p-n transistor, the emitter is of n-type, the base is of p-type, and the collector is of n-type semiconductor. We know that the conventional current is shown in the transistor symbol between the emitter and the base. Since the current flows from the positive to the negative region, the conventional current for the given transistor should flow from the base to the emitter.
Therefore, the statement (d) is correct.
Thus, (a) and (d) are the correct statements.
Hence, the correct answer is option C.
Note
The configuration of the transistor given in the question is just the extra information. It does not have any relation with the solution of this question.
Complete step-by-step solution
We know that a transistor has three different regions; the emitter, the base, and the collector. Out of these, the collector is the widest, the base is the thinnest, and the emitter is of moderate thickness. Talking about the doping level, the emitter is most heavily doped, the base is least doped, and the collector is moderately doped.
So clearly the emitter is heavier than the collector.
Therefore, the statement (a) is correct.
Now, since the thickness and the doping levels of the emitter and the collector are different, they cannot be exchanged.
Therefore, the statement (b) is incorrect.
The base is thinnest out of the three regions, but it is only slightly doped.
Therefore, the statement (c) is also incorrect.
Now, the given transistor is of n-p-n type. We know that this indicates the type of majority charge carriers respectively for the emitter, the base, and the collector. So for an n-p-n transistor, the emitter is of n-type, the base is of p-type, and the collector is of n-type semiconductor. We know that the conventional current is shown in the transistor symbol between the emitter and the base. Since the current flows from the positive to the negative region, the conventional current for the given transistor should flow from the base to the emitter.
Therefore, the statement (d) is correct.
Thus, (a) and (d) are the correct statements.
Hence, the correct answer is option C.
Note
The configuration of the transistor given in the question is just the extra information. It does not have any relation with the solution of this question.
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