I know the following equation.
Av=gm*Rd
Rs are my load resistors.

Is the equation really as simple as gm * Rs?
Example from (a): (3000uS)(4700ohms)


My textbook and lecture notes offer no help. I'm an hour in and am more confused.
I am only solving for Av or gain.

 

I know the following equation.
Av=gm*Rd
Rs are my load resistors.

Is the equation really as simple as gm * Rs?
Example from (a): (3000uS)(4700ohms)


My textbook and lecture notes offer no help. I'm an hour in and am more confused.
I am only solving for Av or gain.

Well . . . it would be, if you had a common-source amplifier, but what you have drawn is a common-drain amplifier, for which the gain will be slightly less than 1, and will vary by only a small amount if you change gm, Rs or both.

 

I know the following equation.
Av=gm*Rd
Rs are my load resistors.

Is the equation really as simple as gm * Rs?
Example from (a): (3000uS)(4700ohms)


My textbook and lecture notes offer no help. I'm an hour in and am more confused.
I am only solving for Av or gain.

You may "know" that equation, but it's apparent you don't know where it comes from, what it means, or when it is and is not applicable.

As a result, instead of "knowing" that it applies to a specific circuit topology in which there is a drain resistor (hence Rd), you choose to simply through whatever resistance happens to be lying around at whatever equation comes to mind.

Not the way to approach engineering.

So take a step back and LEARN how to analyze these circuits, not just regurgitating equations to throw at them.