Why Is V_G at 0 in MOSFET Circuit Analysis?

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Homework Statement

Homework Equations

$$dI = \frac{\partial I}{\partial V_{GS}} dV_{GS} + \frac{\partial I}{\partial V_{DS}}dV_{DS}$$

The Attempt at a Solution

I was just wondering why $$\small V_{G}$$ is at 0. I have solved the entire question (there are more parts :P), but I could only do it once I was told that $$\small V_{GS} = - V_S + V_G = - V_S$$. I am having a hard time getting my head around that.
Shouldn't $$\small V_{G}$$ simply be $$\small V_{i}$$?

 

[KataKoniK]

Thank you for your question. It is important to remember that in circuit analysis, the voltage at the gate terminal (V_G) of a MOSFET is typically referenced to the source terminal (V_S). This means that the voltage between the gate and source (V_GS) is equal to the voltage at the gate (V_G) minus the voltage at the source (V_S).

In the equation you have provided, the term for V_GS is calculated by taking the partial derivative of the current (I) with respect to V_GS. This means that V_GS is the independent variable in this equation, not V_G.

I hope this helps clarify any confusion you may have had. Keep up the good work on your assignment!