Question 18 of 209
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.
Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ammeter.
In the light of above statements, choose the most appropriate answer from the options given below.
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Answer: (D) Statement I is correct but statement II is incorrect
Doping silicon with boron leads to holes without any additional electron. This results in P-type semiconductors. Arsenic is pentavalent therefore when added with silicon it leaves one electron as a free electron. This results in N-type semiconductors.
When such P-type and N-type semiconductors are fused to make a junction, an electric field is produced from the N-side to the P-side, causing a barrier potential to develop. Because of the barrier potential, the majority of charge carriers are unable to flow through the junction, hence current is zero, unless we apply forward bias voltage. We cannot detect the current through ammeter as battery is not connected.
Hence, statement I is correct but statement II is incorrect.
When such P-type and N-type semiconductors are fused to make a junction, an electric field is produced from the N-side to the P-side, causing a barrier potential to develop. Because of the barrier potential, the majority of charge carriers are unable to flow through the junction, hence current is zero, unless we apply forward bias voltage. We cannot detect the current through ammeter as battery is not connected.
Hence, statement I is correct but statement II is incorrect.