N-channel and p-channel MOSFET

Discussion in 'Homework Help' started by dortmundBVB, Apr 28, 2013.

  1. dortmundBVB

    Thread Starter New Member

    Apr 23, 2013
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    http://imgur.com/5hunyGU

    The circuit in the url provided contains both a n-channel and a p-channel MOSFET. The n-channel device has parameters K2 = 2mA/V^2 and VTR2 = 2 V; the p-channel device has parameters K1 = -1.5 mA/V^2 and VTR1 = -2.5 V.

    a) in what region of operation does Q1 operate?
    b) if VDD = 5 V and v1 = 2.5 V, find the values of i2 and v2

    I realize that is in Constant current region.

    Am I right in assuming that id1 = id2?
    also Vin = VGS2; therefore
    id2 = K2(VGS2 - VTR2)^2
    and id2=id1

    Since the gate is connected with the drain for Q1, VGS1 = VDS1 which means that id2 = K1(VDS1 - VTR1)^2

    Is it right to assume that V2 = VDS1?

    Thanks for reading all of this
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Why do you say that? I'm not saying you're wrong, but you should get in the habit if justifying your answers and claims.

    Again, why are you making that assumption? If you can state why you are making that conclusion, you will likely find that you don't need anyone to tell you whether it is correct or not because it will become obvious to you.

    Are you making any assumptions about what region Q2 is operating in?
    No problems here.

    Does knowing the voltage difference between the drain and the source of a transistor tell me the voltage of the drain relative to ground?
     
    screen1988 likes this.
  3. dortmundBVB

    Thread Starter New Member

    Apr 23, 2013
    3
    0
  4. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    If that is the attitude you are going to take toward someone that is trying to help you learn how to sanity check your own work so that you can have confidence in the parts of your solution that are correct, can spot when you are making good versus bad assumptions, can focus on the parts that are more likely to have problems, and who asks you a simple question so that you can spot your own error, then so be it.
     
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