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SOURCE FOLLOWER

kbcheong

Mar 24, 2012
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May I enquire the formula for the circuit as attached for iDS in relation to K, vIN, vOUT and VT?

Is iDS = K/2 (vIN - vOUT - VT) ^2 ?
 

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  • circuit.doc
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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In this case Vgs will be Vin - Vrs, with Vrs = Id*Rs

You'll find that you have 2 equations, one giving Id in terms of Vgs and Vrs, and hence Irs, the other giving Id in terms of Vgs and Vth. Since Id = Irs, you can equate these two equations and solve a quadratic equation.

There will be two solutions, but only one will satisfy another important relationship...

The short answer is No, that's not correct. You use the regular formula.
 

Harald Kapp

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Sorry steve,
but in my view this time kbcheong is right.
The equation is
Id = K/2*(Vgs-Vt)^2.
As you rightfully observed, Vgs=Vi-Vrs and Vrs=Id*RS. But also: Vrs=Vout. Therefore
Vgs-Vt = Vi-Vrs-Vt = Vi-Vout-Vt.

Harald
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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That's true, but I'm not sure it's helpful as it leaves you with 1 equation and 2 unknowns.

I am presuming that Vin is known and Vout is not. Even if Vin is not known, you can get the answer as a function of Vin.

What I suggested is

Id = (k/2)(Vgs-Vt)^2 [1] and Vin=Vgs+Id*Rs [2]

Therefore (by [2]) Id = (Vin-Vgs)/Rs

So now with 2 equations for Id, knowing that they must be equal we can solve for Vgs. (which turns out to be done by solving a quadratic equation).

Having established Vgs, everything else falls out using (1) and (2).

I would be curious how you would solve it using your approach. This isn't my area of expertise (I only learned it yesterday :p) so maybe there's something very obvious I'm missing.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Harald and I have been discussing this privately. The practical upshot is that the equation you wrote was a correct relationship.

Harald is currently walking me through a demonstration of its usefulness :)
 

Laplace

Apr 4, 2010
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Three equations govern this source follower:

Vin=VGS+Vo

Vo=Ids x Rs

Ids=(K/2)(VGS-VT)^2

Solving for Vo yields a quadratic solution, see attachment. It would seem that only one of the solutions should correspond to a condition that VGS>VT.

Also the source follower output has an offset relative to the input, and the linearity is better as Rs increases.
 

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  • Mathcad - MathCad_EP-16.pdf
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kbcheong

Mar 24, 2012
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What about iDS, how to define the solution for iDS from the above formula, if iDS = 0 when vIN = VT?
 

Harald Kapp

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If Ids=0 then Vout=0 becaus Vout=Ids*Rs.

Harald
 

Raven Luni

Oct 15, 2011
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Theres a nice tool for solving complicated equations and stuff. I wouldnt call it cheating (no more than using a calculator to do arithmetic anyway) you still need to know what to feed the thing. Google for wxmaxima.
 

Raven Luni

Oct 15, 2011
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I think the maths for this course is going to end up being too much for me though. Its the one subject my brain refuses to deal with nomatter how much I read. I remember at school going from top grades to absolute 0 as soon as it hit calculus. I still have that problem to this day. There are a small range of subjects that I cant seem to learn unless they are presented in a very specific way - gave me some trouble and cost me a year of college. The old learning difficulty strikes again :(
 
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