Directions: 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.

This question was previously asked in
ESE Electrical 2014 Paper 2: Official Paper
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  1. Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)
  2. Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)
  3. Statement (I) is true but Statement (II) is false
  4. Statement (I) is false but Statement (II) is true

Answer (Detailed Solution Below)

Option 1 : Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)
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Detailed Solution

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MOSFETs 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

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