your professor only teach common source circuit?

@@myerwerl lol

i have my test tmr too lol

Please contact me I need some explanation

Thank you:)

Thanks! What was the piece of understanding you were missing that this filled in?

@@JordanEdmundsEECS It's embarrassing, but it was the part that used KVL. Thank you once again.

Glad to hear :) Anything specific you think helped you understand?

You are welcome! What specifically did you enjoy about it?

Glad you liked the explanation! Anything in particular you found helpful about it?

Thanks! We don't necessarily need Vgs to be zero, we just need the *input* voltage vin to equal zero. This is because output resistance measures how a change in the *output* affects the system, while completely ignoring the input (setting it to be zero).

@@JordanEdmundsEECS Got it, thanks!!!

Yes, it's in saturation. Expected for Common Source but could have been explicit. Bias resistors aren't necessary for MOSFETs if we assume saturation, meaning we are certain that Vds > (Vgs - threshold) and Vgs > Vthreshold. We aren't calculating the Q-point either where a bias voltage may help. Vdd is needed for Id, which is needed for gm and ro but he doesn't calculate gm and ignores ro. Current source to ground is because there is no Rsource resistor. It isn't as necessary as in the BJT equivalent. In Thevenin/Norton + small signal analysis, we short circuit constant voltage sources, including Vdd, and open circuit constant current sources. Since the current source is voltage-dependent and Vgs = 0V from finding Vthevenin aka Vtest, it must have no current and be an open circuit. He's deliberately choosing the most simple example. That's what's causing confusion.

Yup!

just thought I'd let you know, it may depend on a lot on the school, but my degree started active components (op amps) late second year and moved in BJTS and MOSFETS in third

For this you need to know where all the capacitors are - then you can figure out what the -3dB frequency is where the transfer function falls to 1/sqrt(2).

Yup, I have a few videos on the body effect (none on channel length modulation I don't think). In general, I find it's good practice to simplify things like my life depends on it, and then introduce complexity as needed. One thing I could do better and I always liked is sprinkle in little hints here and there about that complexity without talking about it too much.

Mostly. MOSFETs are characterized by a thin oxide layer that galvanically isolates the gate from the source and drain - thus the infinite input impedance. JFETs physically connect the bits of semiconductor without an oxide dielectric, so the input impedance is less than infinite.

@@jacobfaseler5311 Thanks for the reply. Finally, I found a website where the whole process of biasing a jfet comes explained, and that's what I did, and it works fine.

please very important

what i understood is to get rid of the influence of input voltage

Because that's the definition of Rout. To find Rout, we always need to short Vin.

It's because we are using the Thevenin equivalent circuit to get Rout. Voltage sources are considered short circuit and current sources are considered open circuit.

because that's how two-port analysis is performed

LOL

It’s used in essentially *all* analog circuits xD. It’s also the simplest case to analyze and so a good place to start.

Then someone asks you to calculate the cascaded noise figure for multiple MOSFET devices--this is huge!