Question
Download Solution PDFDirections: It consists of two statements, one labelled as the ‘Statement (I)’ and the other as ‘Statement (II). Examine these two statements carefully and select the answer using the codes given below:
Statement (I): The ‘turn-on’ and ‘turn-off’ time of a MOSFET is very small.
Statement (II): The MOSFET is a majority-carrier device.Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFMOSFETs are majority carrier devices that mean flow of current inside the device is carried out either flow of electrons (N-Channel MOSFET) or flow of holes (P-Channel MOSFET). So, when the device turns off, the reverse recombination process will not happen. It leads to short turn ON/OFF times. As switching time is less, loss associated with it less and hence it gives the highest switching speed.
Therefore, both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I).
Important Points:
The turn-off times of different power electronic devices are given below.
- MOSFET has the lowest switching off time in the order of nanoseconds.
- BJT has the turn-off time in the order of nanoseconds to microseconds.
- IGBT has the turn-off time in the order of microseconds (about 1 μs).
- Thyristor has the turn-off time in the order of microseconds (about 5 μs).
Therefore, the increasing order of turn-off times is:
MOSFET > BJT > IGBT > Thyristor
Last updated on May 28, 2025
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.