How does Q point affect gain?

[EricKnabe]

Let’s say I have a transistor class A amplifier circuit using a depletion mode FET. I have the load and the source resistor values already picked out to provide the right amount of gain and the source resistor is cold biased to provide asymmetrical clipping. Now let’s say suddenly I must substitute my depletion FET for a similar enhancement type FET. When I set my gate bias up so that the Q point is in the middle of the active region, will that change the way the current drain and source resistor values affect the gain of my circuit? Thanks.

 

[Mark Tillotson]

The gain is set by the transconductance, the curve for which will be in the datasheet normally.

 

[sgrossklass]

Usually the ratio of R_S to (1/gm + R_D) should provide a decent approximation for the gain. If you have enough current available that drain output resistance r_D = 1/gm is << R_D (i.e. a lot of degeneration), then this in turn can be approximated by R_S / R_D.

With an input bias adjustment and enough current, you can make a circuit with a low R_S to R_D ratio work with almost anything - JFET, MOSFET, BJT, tube. Inverters use R_S = R_D no matter what kind of amplifying device (or three-legged critters as I like to call them).

 

[EricKnabe]

Usually the ratio of R_S to (1/gm + R_D) should provide a decent approximation for the gain. If you have enough current available that drain output resistance r_D = 1/gm is << R_D (i.e. a lot of degeneration), then this in turn can be approximated by R_S / R_D.

With an input bias adjustment and enough current, you can make a circuit with a low R_S to R_D ratio work with almost anything - JFET, MOSFET, BJT, tube. Inverters use R_S = R_D no matter what kind of amplifying device (or three-legged critters as I like to call them).

So basically you’re saying that the way I bias transistor, as long as it’s in the active region, won’t affect the gain if I keep the resistor values I’m using now?