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Another NMOS-transistor problem

  1. Nov 17, 2007 #1
    [SOLVED] Another NMOS-transistor problem

    I'm sorry for asking again so soon, but these transistors really give my head a spin.

    1. The problem statement, all variables and given/known data
    The problem is pretty much summed up by the following photograph:
    [​IMG]

    3. The attempt at a solution
    To be quite honest I'm really stumped here. I can go over what I know (Or at least what I think I know):

    Since everything displayed here is in series, I'm presuming the current over each element is the same. The voltage over R1 should be 10 - v1, basically meaning if I could find v1 I should be able to find the current R1 and thus the current in every other element connected aswell, which would make the problem easy.

    Vgs for the first transistor is 5 volts, and vgs=vds for the second transistor. R1 and R2 should have the same voltage drop across them.
    All in all, 15 volts dissipate over this circuits as vdd = 10 volts and vss = -5.

    After this, it completely stops. I've got the correct answer, which is
    v1 = 6 v
    and
    v2 = 2 v

    So by my logic, the current over the first resistance is (10-6)/1000 = 4 mA.
    This is where I get confused, because if v2 = 2v, doesn't that mean the voltage drop across the first transistor is 4v aswell?
    But if it is, this indicates that 16 mA runs through it, which isn't really possible if only 4 mA runs through the first resistance.
    I tried checking what the voltage vds over the first transistor had to have been in order to allow for 4mA to pass through, and as far as I can remember the answer I got was sqrt(2)+1. Which is a fairly ugly number so I'm presuming this is wrong.

    Anyone got a hint that can push me in the correct direction? It would be greatly appreciated. I'm dying to understand these blasted transistors.
     
  2. jcsd
  3. Nov 18, 2007 #2

    dlgoff

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    Gold Member

    "Vgs for the first transistor is 5 volts..." Vgg=5 volts. Wouldn't Vgs=Vgg-V2?
     
  4. Nov 19, 2007 #3
    Aha, I see now, of course. Thanks!
    I managed to solve it now by using four different equations, was quite a hustle but I've never figured it out if not for this!

    Thanks again!
     
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