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Assume we have a field effect transistor.We connect the source terminal to the anode of a battery and the drain terminal to the cathode of a battery.We connect the gate terminal to cathode of a battery.Now the electrons will start flowing from the gate terminal towards the p type region.

Due to the attraction, the electrons will start flowing towards the insulating material,though creating a bridge for electrons from source to go to drain.

Question:

Will the depletion region of the p type region and the drain be destroyed?

Help appreciated.

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    \$\begingroup\$ Welcome to EE.SE! A schematic is worth more than a thousand words here. \$\endgroup\$ – winny Apr 20 at 7:55
  • \$\begingroup\$ What you describe would usually be a fine way of heating and/or destroying a MOSFET and would not be something one would usually do. Why are you asking this question? \$\endgroup\$ – Russell McMahon Apr 20 at 9:00
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What you describe would usually be a fine way of heating and/or destroying a MOSFET and would not be something one would usually do.
Why are you asking this question?

You need to say what sort of MOSFET is involved (P or N type).

You need to define what you mean by battery anode. That is less clear than you may think See this article

Based on hints in your text I'll assume that the battery anode is negative relative to it's cathode and that you have an NChannel FET.
So - the FET Drain is connected to battery +ve and FET source is connected to battery negative. The FET gate is connected to battery positive.

If the battery voltage is below the FET "threshold voltage (Vgs_th) the FET will not turn on, no current will flow and no magic smoke will be generated.

If the battery voltage is significantly greater than Vgs_th the FET will be fully enhanced (turned on). The FET will draw a large amount of current. How large depends on the battery short circuit current capacity, the MOSFET on resistance (Rds_on). If the battery is say 1 12V car battery (lead acid or LiIon or LiFePO4 or ...) then in most cases the MOSFET will be destroyed. It does not make much sense to ask if the p type depletion region will be destroyed. In cases such as this the FET will become a molten smoking blob with no well defined regions retained and all regions essentially destroyed.

What you describe would usually be a fine way of heating and/or destroying a MOSFET and would not be something one would usually do.
Why are you asking this question?

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If the FET was a P Channel FET or if the battery was of opposite polarity the FET would not be enhanced BUT the D-S body diode would conduct. If the battery had substantial current capability the MOSFET would be destroyed.
Why are you asking this question?

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  • \$\begingroup\$ If the depletion region isnt destroyed then how does current flow? \$\endgroup\$ – Altair Apr 20 at 11:03
  • \$\begingroup\$ 2 things must occur : A bridge must be made for the electrons to pass from the source to the drain and the depletion region to be destroyed. \$\endgroup\$ – Altair Apr 20 at 11:04
  • \$\begingroup\$ @Altair 1. What you describe would usually be a fine way of heating and/or destroying a MOSFET and would not be something one would usually do. Why are you asking this question? 2. The word destroyed is usually used to mean damaged permanently. You may be using it to mean "temporarily electrically removed". 3. What you describe would usually be a fine way of heating and/or destroying a MOSFET and would not be something one would usually do. Why are you asking this question? \$\endgroup\$ – Russell McMahon Apr 20 at 12:57
  • \$\begingroup\$ Is the depletion region temporarily electrically removed during the on mode of a MOSFET? \$\endgroup\$ – Altair Apr 20 at 14:12
  • \$\begingroup\$ @Altair You should add a drawing to your question. It is quite unclear what you are asking. Which depletion region are you refering to? To turn an enhancement mode mosfet on you must invert the substrate, which then has a depleted region below it. \$\endgroup\$ – Matt Apr 20 at 22:25

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