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SOURCE FOLLOWER

Discussion in 'Electronics Homework Help' started by kbcheong, Mar 31, 2012.

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  1. kbcheong

    kbcheong

    25
    0
    Mar 24, 2012
    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 ?
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
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    Jan 21, 2010
    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.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    11,662
    2,697
    Nov 17, 2011
    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
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    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.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    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 :)
     
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    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.
     

    Attached Files:

  7. kbcheong

    kbcheong

    25
    0
    Mar 24, 2012
    What about iDS, how to define the solution for iDS from the above formula, if iDS = 0 when vIN = VT?
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

    11,662
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    Nov 17, 2011
    If Ids=0 then Vout=0 becaus Vout=Ids*Rs.

    Harald
     
  9. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    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.
     
  10. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    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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