Mosfet amp calculations.

[demonstar]

I'm reading a book on understanding amps and I'm struggling a little bit to understand how the gain is set on a booster circuit such as the amz mosfet boost. Could anybody help me out a little please? I understand the biasing of it etc and pretty much understand the negative feedback. Just caught up on setting the gain.

I believe the voltage gain is equal to the transductance * 2k7 * Vgs (correct?)

Also is Vgs the DC bias voltage plus the AC signal or are they treated separate?

Any help would be much appreciated!

[soggybag]

That's a great question and something that I was thinking about lately.

A related question, how to calculate unity gain buffer with a transistor?

[Sir H C]

Transconductance = current out/voltage in IIRC.  Therefore to get voltage out, it is current out * output load,  so:

voltage gain = transconductance * output load.

[alanlan]

vgs is the small signal AC voltage across gate-source i.e. it ignores any DC biasing.

The output voltage vd (again small signal voltage) is gm x R4 x vgs (where gm is the transconductance).

So the ratio: vd / vgs = gm x Rd.

If the gain pot R6 is set to 0 ohms (the maximum gain setting), then the gain of the circuit is gm x Rd because as far as AC is concerned, vs = 0.  i.e. vgs = vin.

This does not work though if R6 is non-zero because vs is no longer 0 (AC wise).  i.e. negative feedback starts to occur and so gain is reduced.  If you do the maths you will find that without R6 and the bypass capacitor (C5), the gain approximates R4 / R5 i.e. unity gain.  If you get really stuck I can post the maths but if you're like me you never learn anything unless you go through the pain of finding out for yourself.

Bear in mind that because of non-linearities gm is never the same at all points on the transfer characteristic.

[soggybag]

Hey thanks Alanlan, that was just the information I was looking for.

[demonstar]

Thanks Alanlan thats great.

I'll go away and fiddle with some circuits and start using what I've just learnt and if  I feel I'm ready I'll come back and see if some one can show me the maths if they don't mind.

Thanks again everyone!

[demonstar]

Alanlan would it be possible that you could show me the maths please? I've been away and done lots more reading but I can't seem to find how to calculate the negative feedback and gain values for different values of the 5k pot.

I understand what you said about calculating the gain if there is just a cap bypassing the source resistor. How is the value for the source resistor calculated? It's for stability isn't it?

Thanks

[alanlan]

OK, I'll try...

first of all let me repeat that gm or transconductance or amount of small signal drain current for a given small signal voltage between the gate and source (vgs as we talked about before) varies a lot between devices so it's not easy to predict with any certainty.  The best hope you have is to measure it yourself or gleen it from datasheets by drawing lines on a tangent to the transfer characteristic (see the datasheet for a J201 for example).  Modelling using PSpice or equivalent can also be useful.

Also, you can generally create designs which are not too sensitive to variations between individual devices.

Now on to the theory:
if R6 is non-zero then as I said, negative feedback occurs in the small signal sense.  So let's generalise a little bit and call the parallel combination of R6 and R5 = Rs.  This ignores any effect of C5.

Also, we will call R4 Rd again just to generalise away from this particular circuit.

So,
vg the voltage at the gate = vin (our input signal)
vg = vin..................................................................1)

id = gm.vgs..................................................................2)

Our output voltage vo (or you could call it vd):
vo = -gm.vgs.Rd ..................................................................3)
(the minus sign indicates inversion because as vi rises, vo falls).

BUT vgs is the difference between the gate and source i.e.
vgs = vg - vs or from 1)
vgs = vin - vs..................................................................4)

substitute 4) into 3)
so vo = -gm.(vin - vs).Rd..................................................................5)

To solve we need to get rid of the vs term which we can do as follows:

vs = id.Rs..................................................................6)

from 2)
vs = gm.(vin - vs).Rs..................................................................7)

solve for vs:
vs = vin.gm.Rs / ( 1 + gm.Rs )..................................................................8 )

substitute 8 ) into 5)
vo = -gm.( vin - vin.gm.Rs / ( 1 + gm.Rs ) ).Rd..................................................................9)

simplify a little bit...
vo = -vin.gm.( ( 1 + gm.Rs - gm.Rs ) / ( 1 + gm.Rs ) ).Rd..................................................................10)

and finally...
vo = -vin.gm.Rd / ( 1 + gm.Rs )..................................................................11)

so 11) is the final gain in terms of Rd, Rs and gm of the device.

To test this, let Rs = 0:

vo = -vin.gm.Rd which is what we originally said was the gain for R6 set to 0.

If Rs is non-zero and gm.Rs is significantly greater than 1 then:

vo = approx -vin.Rd / Rs which is a well known result and if Rd = Rs then we have a unity gain buffer.

Sorry it is a bit of a slog but I have tried to include all the maths without taking any short cuts.

[demonstar]

Thankyou very much for giving some of your time to show me that. That's great!

Thankyou!

[alanlan]

No problem.

One other thing worth pointing out:
The maths assumes that gm is a constant which of course it isn't because the curve of the so called transfer characteristic (id vs vgs) is a curve not a straight line.  But, what you will notice is that as you increase Rs so that gm.Rs in the equation becomes significantly greater than 1, you can start to ignore the 1 i.e.

if gm.Rs >> 1 then the equation becomes:
vo approximates to -vin.gm.Rd / ( gm.Rs ) which = -vin.Rd/Rs.

So the point is that this shows how negative feedback (increasing Rs) linearises the circuit by reducing the effect of the non-linear gm of the device (albeit it also reduces the gain).  i.e. it reduces the significant amount of 2nd harmonic distortion you get without negative feedback.  However you can cascade multiple sections together to increase gain.

Would it sound any better?  That's a completely different question.

Just thought I'd point that out.