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Question

Match List I with List II

LIST I LIST II
A. When negative VG is applied to gate of MOSI. Depletion
B. When zero-bias is applied to gate of MOSII. Surface Inversion
C. When a small positive gate bias VG is applied to gate of MOSIII. Accumulation
D. When a higher positive gate VG is applied to gate of MOSIV. Neutral MOS

 

Choose the correct answer from the options given below:

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

A-I, B-IV, C-III, D-II

The task is to match the conditions (List I) applied to the gate of a Metal-Oxide-Semiconductor (MOS) device with the associated effects (List II). Let's analyze and understand each condition and its effect on the MOS structure:

  1. When negative \(V_G\) is applied to gate of MOS:
    • When a negative gate voltage is applied, it causes the holes (majority carriers in p-type substrate) to gather at the oxide interface, depleting the region of free electrons.
    • This condition is known as Depletion.
    • Therefore, A-I is the correct match.
  2. When zero-bias is applied to gate of MOS:
    • In this scenario, there is no influence from the gate voltage on the MOS capacitor, so the charge carriers remain in their neutral state.
    • This condition reflects a Neutral MOS.
    • Thus, B-IV is the correct match.
  3. When a small positive gate bias \(V_G\) is applied to gate of MOS:
    • A small positive voltage attracts electrons toward the oxide interface, causing an accumulation of negative charge at the surface.
    • This is known as Accumulation.
    • Hence, C-III is the correct match.
  4. When a higher positive gate \(V_G\) is applied to gate of MOS:
    • A high positive gate voltage significantly attracts electrons to form a conductive channel at the surface, resulting in the strong inversion of the surface.
    • This condition is termed as Surface Inversion.
    • Thus, D-II is the correct match.

After considering the explanations above, the correct matching is:

  • A-I
  • B-IV
  • C-III
  • D-II

Hence, the correct answer is: A-I, B-IV, C-III, D-II.

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