MOSFET Guide for Idiots? Part 2

[R.G.]

since gain with and Unbypassed source resistor is -rd/rs, then is it safe if assume that if we have a bypassed potentiometer as a gain control the formula would be
-Rd/(Rs||RsByp)

Yes, except that the gain is asymptotic to the transconductance times the drain resistor - that is, the expression you have there is an approximation, leaving several minor terms out, and as the effective value of the source resistances gets towards zero, the gain comes ever nearer Yfs*Rd, not ever nearer infinity. Which is good, because infinite gain is an oscillator - output signal with zero input signal. 

If this is the case is it better to have the Rd and Rs at equal values to allow for a larger swing so the minimum swing would be 2.7/(2.7||0) and close to unity gain when the pot is wide open?

This gets into the differences between AC and DC conditions. You want your DC conditions to be as solid and predictable as possible. So gain changers in this kind of circuit tend to be AC coupled so that the DC conditions are not affected. The biggest swing you can ever get is the power supply, and if you have Rd and Rs equal, it's half the power supply, as the MOSFET can only change from fully off to fully on. When it's off, the output at the drain is the power supply, and when it's on the output at the drain is half the power supply. As you make the voltage across the source resistor smaller, you get a bigger swing, but you also get a bigger minimum gain. They're interrelated.

Lastly if I put a put a resistor in series with a pot can i then control the minimum resistance (or set the maximum it could go to)?

Yes.

[Dylfish]

Just as a test I'll is what I think i've learnt has got me anywhere

Assumptions
Vcc = 9v
Max Guitar Input Signal  = 800mv Peak to Peak (I know, Large, but a test none the less)
Idle / Bias Current Desired = 4ma
Vth = 1.65v (Made this up as a test value)

Therefore

Vs = 800mv/2
Vs = 400mv

Rs = 0.4v/0.004a
Rs = 100 ohms

Vg = Vs + Vth
Vg = 0.4v + 1.65v
Vg = 2.05v

Vgs = Vg - Vs
Vgs = 2.05v - 0.4
Vgs = 1.65v (Vth...Suprise!)

Therefore the remaining "swing room" is:
9v - 0.4v = 8.6v
and to bias it midway for maximum swing
8.6/2 = 4.3v

Therefor again, using the previous assumptions...
Rd = Desired voltage drop / Id
Rd = 4.3v/0.004
Rd = 1075 ohms

since Rd is about 10x Rs, we should get about 10x gain on the input signal. Therefore a 800mv signal should be sitting at around 8v, when we have an available 8.6 clean swing room.

I take it is we bypass the Rs resistor then we have a large resistance which can reduce this factor of 10 further? Eg.  Rd = 10k and Rs(b) = 5k then we get a gain of 2?

Although i hope this is closer to the answer im still a little stuck on how the available gm is having an effect on gain? unless I've just explained it and misinterpreted it's workings?

Cheers

I would struggle using figures like the above to have any clear cut control over gain wouldn't I? having a large swing if fine but trying to get a descent parallel resistance with a 100 ohms rs would be an issue (being able to get the gain from the ~7x down to near unity) What other considerations could be taken?  I was thinking if I had a 1K bypassed pot i could get it down towards unity, but obviously since 100 ohms is so low parallel resistance will be >100, but if I increase Rs then I lose headroom and have a lower tollernance of having a "clean" boost.

[PRR]

> gain with and Unbypassed source resistor is -rd/rs

Rs is in-series with 1/Gm.

While electronics can be an abstract puzzle (like crosswords chess or 4-color maps), most interest is in what you can Build and Do with electronics.

I really think you want a breadboard, clean signal generator, and meter/monitor. Predict the gain, measure it, explain the difference.

The pot+resistor question can be settled with a pot, a resistor, and an ohm-meter.

[Dylfish]

Hey Guys,

I've got a bit of a stupid question inreagrds to biasing but here it goes anyway.

Lets say I design the below circut with the below details, with the goal of the FET being in the saturation region.

Vcc = 9v
Rd = 3k
Rs =  1.5k
Ids = ~1ma

Therefore I should have the below voltages:

Vd = 9v - (0.001x2500) = 6.5v
Vs = (0.001x1500) = 1.5v

To make sure this is baised correctly is it best to work out the Vt of the fet and potential input swing for the gates voltage, Or is it just as easy to put a multimeter over Rs (or Rd) and fine tune until the correct voltage is met at certain nodes? I think i'm overcomplicating this a little too much usually and this might be all i really need to do.

Cheers =)