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Find K and VTR for MOSFET

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A MOSFET has the set of v-I characteristics shown in Fig. P5.5.

    What are the values of K and VTR?

    2. Relevant equations
    iD=K(VGS-VTR)2


    3. The attempt at a solution
    I have looked in the book and all of the example problems give VTR, but it doesn't show how to actually get the K and VTR values based on the graph. I really have no idea what to do.

    The answer in the book says K=0.25 mA/V2 and VTR=2V. Is VTR 2V because VGS increases by 2V?
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2013 #2

    rude man

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    First job for you is to figure out where on the Vds axis (the x axis) you're operating at.
    Hint: notice that Vds does not appear in your equation. What does that suggest?
     
  4. Feb 21, 2013 #3
    The transition from triode to constant current uses the equation VDS=VGS-VTR. So, substituting into the equation I gave in my first post:

    iD=KVDS2

    I also noticed that I can find VTR using the point where the graph goes horizontal. For VGS=4V, the point would be at 2V. Then 4-2=2V is what I want (verified this by looking at an example in the book where I was given VTR). This seems to work fine, but not sure if it is the "correct" way.
     
  5. Feb 21, 2013 #4

    rude man

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    OK, there are 2 ways to go:
    1. Use VT = Vgs - Vds. So look where the curves bend flat. As you point out, using the Vgs = 4V curve it looks like VT ~ 2V.

    But if you used the Vgs = 8V curve you'd get more like VT = 3V (the curve bends at about Vds = 5V so 8 - 5 = 3). The Vgs = 6V and 11V curves would also compute to about VT = 3.

    A better way to go is to use your original equation of Ids = k(Vgs - VT)^2. Use it twice, on two different curves, to solve simultaneously for k and VT.

    But as I said, first you have to decide where on the Vds axis you're sitting. That equation of yours holds only if Vds > (Vgs - VT).
     
  6. Feb 21, 2013 #5
    Do I just pick a region (i.e. constant current, saturation, triode) and use the corresponding equation?
     
  7. Feb 22, 2013 #6

    rude man

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    Yes, but remember what I said about the region where that equation is valid.

    There are separate equations fdepending on the region of the i-V characteristics you're looking at.

    For example, in the triode region your equation would not be appropriate.
     
    Last edited: Feb 22, 2013
  8. Feb 22, 2013 #7
    Ah, gotcha. I figured it out after using two different curves, as you said.

    Thank you for the help.
     
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