Semiconductors Question 4
Question 4 - 25 January - Shift 2
Statement I : When a Si sample is doped with Boron, it becomes $P$ type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and $N$-type has excess electrons.
When such P-type and N-type semiconductors are fused to make a junction, a current will not automatically flow, but a potential difference will be created which can be detected with an externally connected voltmeter.
In the light of above statements, choose the most appropriate answer from the options given below.
Options:
(1) Both Statement I and statement II are incorrect
(2) Statement I is incorrect but statement II is correct
(3) Both Statement I and statement II are correct
(4) Statement I is correct but statement II is incorrect
Show Answer
Answer: (4)
Solution:
Statement $-I$ is correct
When $P-N$ junction is formed an electric field is generated from $N$-side to $P$-side due to which barrier potential arises \& majority charge carriers cannot flow through the junction due to barrier potential so current is zero unless we apply forward bias voltage.
BETA
