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Why does a MOSFET have 2 semiconductor leads?

Electrical Engineering Asked on October 29, 2021

I have a question on the mosfet construction. Considering a n-channel enhancement type mosfet, it has two n-type regions connected to source and drain terminals. Also there is a p-type substrate. If source is connected -ve and drain is connected +ve voltage, there is no current flow unless there is a positive voltage in the gate which attracts some electrons to form the n-channel and hence conduction occurs.

It seems the same effect could be achieved using only one n type region connected to drain and no n-type region connected to source in the mosfet. alternate mosfet implementation

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There would be an depletion region with p-n reverse biased (drain +ve and source -ve). Gate being in a positive voltage can create a n-channel and let current flow. The device could operate in pinch-off or linear or saturation region.

I was getting the feeling the source side n-region is redundant. Would like to know if it serves a purpose?

One Answer

What you are proposing is called a Schottky-Barrier Source and Drain MOSFET:

Compare this n/p well MOSFET

enter image description here

with this metal source/drain MOSFET

enter image description here

Images from https://www.semanticscholar.org/paper/Electrical-Characterisation-and-Modelling-of-metal-Pearman/f3d7a94a98b10c7c36e527c697eb3d158d69d4c0

See also https://people.eecs.berkeley.edu/~tking/theses/rvega.pdf

The metal S/D lowers the Rds and improves the ft, but the Schottky barrier doubles the threshold voltage.

Answered by P2000 on October 29, 2021

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