Regarding your response to 2, is there anything I could read further regarding the subject? I get the math equation in it's simplest form - being I can use it to calculate something but I don't completely get "cut-off point at frequency" and what that means.

EDIT: In looking that equation up - I understand how to use it to calculate a corner frequency for a R-C filter - how does that work in conjunction with the Jfet and what Rs and Cs to use?

It works the same way as in

BJT Emitter degenerated amplifier..

(you can search it using something like above underlined description..)

In breef:

We sometimes place a resistor between Emitter(Source) and GND to more accurately set bias needs (Collector/Drain current, Gate-Source voltage, etc.) and also have an amount of negative feedback but that results in lowering amp gain (which is R

_{C}/R

_{E} or R

_{D}/R

_{S}) so we want the lowest possible resistor value for the higher affordable bias variation..

Instead of messing with not always convenient compromises, we by-pass Emitter(Source) resistor with a capacitor (whose reactance is ideally(*) zero in AC) so we have establish our DC requirements and also have obtain the maximun amp gain..

(which gain should be considered as R

_{C}/

*r*_{e} in case of BJT or

*g*_{m}*R

_{D} in case of FET -

*r*_{e} can be calculated from 0.025/I

_{C} and

*g*_{m} can be found on datasheets..)

(*) In real world, capacitors exhibit frequency (

*f*) variable resistance (capacive reactance) of Z

_{C} = 1/(6.28*C*

*f*)..

Given that, we don't completely by-pass Emitter(Source) resistor if we don't do it with an infinite value capacitor..

Such a capacitor should require infinite room and infinite money so we shake hands with our wallet for an impedance value of, 10% say, of Emitter(Source) resistor value, 91% effectively by-passing it..

If you face that RC combination as a voltage divider, corner frequency is the point at which R & C values are resistively equal so we have a 50% domination..

The higher the capacitor value (provided a fixed resistor value) the lower its reactance hence the more effective resistor by-passing resulting in higher gain..

With the formula of cut-off point you can calculate actual capacitor AC resistance for a given frequency and place it in parallel with Emitter(Source) resistor to find the total AC effective resistance and deside if the gain satisfies you..

P.S.

I don't further refer to partially by-passed Emitter(Source) resistor and the reason for that 'cause I've already deviate enough from "in breef" mean..