How does gate-source voltage forward bias the source-substrate junction of MOSFET?

Question

In forward bias, the p-type is connected with the positive terminal and the n-type is connected with the negative terminal.

The junction that I'm observing is the source-substrate junction. The source is the n-type semiconductor and the substrate is the p-type semiconductor. In this case, \$V_{GS}\$ has the positive terminal to the gate and the negative terminal to the source.

So my doubt is about the negative terminal of \$V_{GS}\$ because it is connected at the same time with the source and the substrate, so they shouldn't have the same voltage?

So how does \$V_{GS}\$ forward bias the source-substrate junction?

EDIT Whit source-gate junction I was meaning the n-p junction (source-substrate) under the Oxide.

I do not understand how the p-type substrate is connected with the positive terminal of \$V_{GS}\$. From the image, the substrate is connected with the negative terminal I mean.

Answer 1

How does gate-source voltage forward bias the source-gate junction of MOSFET?

It doesn't; it never does (unless the MOSFET is damaged). Reason, the gate is insulated from the channel hence, the gate is insulated from the source; the gate-source region acts as a capacitor and not a conductor. A MOSFET is a type of IGFET (similar term used with bipolar transistors; IGBT). An IGFET is an insulated gate field effect transistor. IGFET was a name commonly used to describe MOSFETs in the 1960s and 70s but, these days, MOSFET is the preferred term.

Answer 2

The junction between the source and the substrate in your attached image is forward biased due the drawn polarity of Vds. The Drain-Substrate junction in the right is reversely biased; there's even a depletion region drawn (symbolically) as grey.

Positive voltage Vgs pulls (see NOTE1) electrons from the substrate to the top surface of the substrate. There are plenty of them, no matter it's P-type. The concentration of the conduction electrons grows so high just below the insulation glass that the current stopping functionality of the reversely biased junction between drain and substrate vanishes and there's a contiguous conductive channel between the drain and source.The P-type semiconductor has in a thin layer effectively changed to N-type due the +Vgs and there's no more a current stopping junction.

NOTE1: Essentially you expect an analysis of how mosfets work. An exact explanation should be based on solid state physics - I mean on the theory of material structures and properties based on quantum statistics. You find easily numerous different precision level presentations of it from semiconductor physics textbooks for different student levels.