It doesn't actually matter, as Vgs is fixed, and above Vgs(th), and the input signal is on the drain terminal.

Isn't this being used as a current controller to test as a load or something?

 

Isn't this being used as a current controller to test as a load or something?

A part of the test's challenge is not telling us what's the circuit for, I have to make an initial hypothesis (saturation) and if the voltages don't add up then it's in triode and then I proceed

 

A part of the test's challenge is not telling us what's the circuit for, I have to make an initial hypothesis (saturation) and if the voltages don't add up then it's in triode and then I proceed

Oh sorry about that, I got this thread mixed up with another thread. Not sure how that happened. Ignore that post

 

You have the formula for the gain of the MOSFET. As the gate-source voltage remains constant throughout, the MOSFET is just a fixed resistor.

Thank you though, I definitely needed to confront myself with others, I understand much more now. Of course there are a lot more exercises I could do (like coupled differential amplificators etc), I'd post more, with better details, ok? Thank you again

 

It doesn't actually matter, as Vgs is fixed, and above Vgs(th), and the input signal is on the drain terminal.

Hey I just wanted to say that the teacher's assistant corrected it, confirmed that it was pretty difficult/confusing as a circuit and most have to give it again. Our steps were correct though, so thank you for your insight, the NMOS is considered as a triode! The F=Vo/Vin was correct and so were my Bode's graph, needed to calculate the right starter numbers hehe.

PS to find the Vg he didn't do the KVL on the entire branch 10V - I*R1 -(-4,3V) -0,7V -1V =0;
The guy simply did Vg = -(-4,3V) + 0,7V -1V. I guess I have some basic voltage/current exercises I have to do to hone into perfection but I'm glad I'm on the right path.

 

The guy simply did Vg = -(-4,3V) + 0,7V -1V.

Where does he get the -1 from?
The op-amp will have both its inputs at the same voltage if the feedback is operating.
With 5V on the gate, it will no longer be in the triode region.

 

Where does he get the -1 from?
The op-amp will have both its inputs at the same voltage if the feedback is operating.
With 5V on the gate, it will no longer be in the triode region.

the -1V is where the ground is supposed to be, mean trick of the exam.
In the gate there's 4V, it is definitely not in saturation and it's in triode. With the current going from the Vin till the Vo means Vo < Vin so it'll be negative in DC

 

the -1V is where the ground is supposed to be, mean trick of the exam.
In the gate there's 4V, it is definitely not in saturation and it's in triode. With the current going from the Vin till the Vo means Vo < Vin so it'll be negative in DC

The non-inverting input is at -1V.
That means that the op-amp feedback will operate to put the inverting input at -1V.
The source therefore at -1V.
The gate is at 4V with respect to earth, but 5V with respect to the source.
No MOSFET is still in its triode region at 5V, it's fully enhanced.