confused about CMOS analysis

Discussion in 'Homework Help' started by screen1988, Apr 14, 2013.

  1. screen1988

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    Mar 7, 2013
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    I get confused with the circuit bellow.
    [​IMG]
    Here is the link to the lecture analyzing the circuit:http://www.youtube.com/watch?v=ZMEpc2o9_gU&feature=youtu.be
    When Vin = 0V => Vgs (NMOS) = 0 < Vt => NMOS is in CUT OFF.
    And therefore Ids = 0 A.
    For PMOS, Vsg = Vs - Vg = 5V > |Vt| => PMOS is NOT in CUT OFF.
    PMOS is in TRIODE or SATURATION.
    In the lecture:
    PMOS has Vsg > |Vt| and null current (Ids = 0) => He concluded that PMOS is in TRIODE.
    He doesn't explain the reason why it is in triode. Therefore here is what I guess:
    With PMOS in triode region:
    Ids = -kn(Vsg -|Vt| - Vds/2)Vds (1)
    With PMOS in saturation:
    Ids = -1/2kn (Vsg - |Vt|)^2 (2)
    As for the case in the video, Vsg =5V > Vt => PMOS is in triode or cut off.
    Now because Vsg - Vt≠ 0 => Ids ≠ 0 => This case don't happen.
    => PMOS is in triode

    But here is what is confusing me.
    With Vin = 0V and NMOS is in CUT OFF therefore Ids = 0.
    Can I now consider that D and S of the MOS is not connected and means that it acts as a open switch?
    If it can be considered as an open switch then drain of PMOS D is not connected to ground and it is also in CUT OFF.
    I think in real life these transistors is not ideal and they has resistance and capacitance but now let consider that they are all ideal.
     
    Last edited: Apr 14, 2013
  2. Jony130

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    Feb 17, 2009
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    For Vin = 0V NMOS is OFF so you can remove it from the circuit.
    PMOS in "ON" because Vgs > Vt. But no current is flow through PMOS so there is no voltage drop across Vsd. And since Vsd < (Vsg - Vt) the PMOS is in triode mode.
    Only if Vsd > (Vsg - Vt) the PMOS is in saturation.
     
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  3. screen1988

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    Mar 7, 2013
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    NMOS is OFF and you said that I can remove it? Do you mean that I can remove NMOS and left D of PMOS open?
    Can you tell me how to calculate Vsd in PMOS?
     
  4. Jony130

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    Yes, for NMOS OFF and without any external load the PMOS drain is left open.
    In witch situation?
    Without any external load Vds = 0V and Vout = 5V.
     
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  5. screen1988

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    Mar 7, 2013
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    I meant that Vds when NMOS is left open.
    I think you computed Vds by Ohm's law? Vds = Id* Rds = 0 because Id = 0?
     
  6. Jony130

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    Yes, you are right.
     
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  7. screen1988

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    Mar 7, 2013
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    I think I need your help again. Is it correct to apply Ohm's law with MOSFET because it is not a linear device?
    And with the assumption that PMOS is in triode then I can apply the equation:
    Id = kn(Vsg - |Vt| - 1/2Vsd)Vsd
    With Id = 0 then Vsd = 2(Vsg - |Vt|)= 2*(5 - |Vt|)≠ 0???
     
    Last edited: Apr 14, 2013
  8. Jony130

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    But MOS in triode mode act just like a resistor.
    Rsd(on)≈ 1/kp*(Vgs - Vt)
     
    Last edited: Apr 14, 2013
  9. screen1988

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    Mar 7, 2013
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    Yes, but before that we don't know if the transistor is in triode or saturation?
    How you come to know that PMOS is in triode?
     
  10. screen1988

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    Mar 7, 2013
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    Hello Jony,
    Can you reprove what you did in post #2?
    Then we still don't know that PMOS is in triode region and therefore how can you compute Vds?
    I think Ohm's law is only applied when we know that it is in triode.
     
  11. Jony130

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    For me it is obvious that PMOS is in the triode region. How can PMOS be in saturation region if no current is flow through MOS ?
     
  12. screen1988

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    Mar 7, 2013
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    Yes. I have just figured it out. My calculation is OK, I know that with Vsd = 2*(Vsg - Vth) then Id = 0 but when I look at PMOS characteristic bellow I see no point that appropriate with above.
    [​IMG]
    Now I have draw Id - Vsd for PMOS with Id is the current flow from S to D.
    The characteristic is similar to the one bellow and it is appropriate with the equation.
    [​IMG]
     
  13. Jony130

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    How can this be true?
     
  14. screen1988

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    Mar 7, 2013
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    Oh, it is my mistake!
    With the original circuit we have Id = 0. Therefore, I wrote equation Id, Vsd, Vsg for PMOS both in triode and saturation.
    Then from Id = 0, I solve these equation and see if it is happen. Then I see that there is only PMOS is in triode is appropriate.
    When Id= 0 => Vsd = 2(Vsg -Vt) ≠ 0
    With the first characteristics that I posted in #12, I can't determine the point in the picture.
    However with the second chacteristics, it is obvious that there is a point in the triode that has Id = 0 and Vds = 2(Vsg -Vt) ≠ 0. Therefore it make me more confident to say that PMOS is in triode.
     
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