
Diffusion current in a p-n junction is greater than the diffusion current in magnitude:
A. If the junction is forward biased.
B. If the junction is reversed biased.
C. If the junction is unbiased.
D. In no case.
Answer
578.4k+ views
Hint: p-n junction is an active element of the circuit, which is used where we want the flow of current in one direction and not in both. It is called a p-n junction because it is built using a p-type and a n-type semiconductor. When this diode is forward biased, only in that case it allows the current to flow through it.
Complete step by step answer:
In forward biased condition, the positive terminal of the battery is connected with the p part of the p-n junction diode, as shown in figure.
Now, in case of forward bias, the corresponding charge density on the two sides will increase as a result of which diffusion current will increase since it depends on the charge density. Drift current will be less as the junction charge density will be less and hence a low electric field is present in the opposite direction. The reason for this is the positive external voltage.
So, the correct answer is “Option A”.
Note:
Depletion region is the region of meeting of p and n type material. In forward biased conditions, its width is minimal. Hence it allows the passage of electrons very easily. One must also note that all other options are very true, but for reverse biased conditions. In short we can say the battery pushes more electrons (donors) into the n-type material side. We can also say by using the relation V = Ed, where d is the width of depletion region. If the width of the depletion region increases, the electric field will decrease.
Complete step by step answer:
In forward biased condition, the positive terminal of the battery is connected with the p part of the p-n junction diode, as shown in figure.
Now, in case of forward bias, the corresponding charge density on the two sides will increase as a result of which diffusion current will increase since it depends on the charge density. Drift current will be less as the junction charge density will be less and hence a low electric field is present in the opposite direction. The reason for this is the positive external voltage.
So, the correct answer is “Option A”.
Note:
Depletion region is the region of meeting of p and n type material. In forward biased conditions, its width is minimal. Hence it allows the passage of electrons very easily. One must also note that all other options are very true, but for reverse biased conditions. In short we can say the battery pushes more electrons (donors) into the n-type material side. We can also say by using the relation V = Ed, where d is the width of depletion region. If the width of the depletion region increases, the electric field will decrease.
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