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Why Vgs = Vtn in NMOS here?

Discussion in 'Mathematics and Physics' started by anhnha, Feb 19, 2014.

  1. anhnha

    anhnha Member

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    Hi,
    Please help me with the question in the picture. It is from a lecture in VLSI course. Thank you.
    ?temp_hash=69a4562539f36675aed5db9204fb4d05
     

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  2. Miles Prower

    Miles Prower Member

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    Look at the schemo again: you have the gate and drain at the same voltage, therefore turning it full on. That causes Vds to drop to its lowest voltage: Vth.
     
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  3. steveB

    steveB Well-Known Member Most Helpful Member

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    Another way to look at it is to realize that this is the diode configuration of a transistor. Think back to your bipolar transistor theory. A diode can be made from a transistor by tying the base to the collector, and the voltage drop is Vbe, which makes the emitter voltage Vc-Vbe.

    This circuit is a FET diode basically. The same principle applies and the voltage drop is Vth : the threshold voltage for Vgs.

    You can do a simple analysis to show the real formula and see why what they show is an approximation.

    Take the current equation for a FET, Id=k(Vgs-Vth)^2.

    Then realize that once the gate is connected to the drain, Vgs=Vds, hence Id=k(Vds-Vth)^2.

    Then solve for Vds and you get Vds=Vth + sqrt(Id/k)

    So, really the actual voltage depends on the drain current, but the variation is small for reasonable currents. This is similar to a bipolar diode where we say the voltage drop is 0.7 V (assuming it's silicon). Really the actual voltage drop on a diode (whether bipolar or FET based) depends on current.
     
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  4. dave

    Dave New Member

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  5. anhnha

    anhnha Member

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    Thank you.

    Miles Prower:

    Could you explain more about the bold part? How can we know that the transistor is full on as VDG is maximum?

    SteveB:

    Id here is still unknown, how can we know that sqrt(Id/k) is very small and can be ignored?
     
  6. steveB

    steveB Well-Known Member Most Helpful Member

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    The only way to know for sure is to use typical numbers and compare. What is k? What is Vth? What current are you operating at? You actually are making a good point because the dependence on Id is a square root function, not a logarithm function, as in the case of a bipolar transistor. There will be some variation in a real operating circuit. But, for small currents, the voltage will be near the threshold voltage.
     
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  7. Miles Prower

    Miles Prower Member

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    The transistor is saturated since the gate and drain are at the same potential. Actually, Vdg is at a minimum here: 0V. Once it's saturated, Vds drops to a minimum, very close to Vth.
     
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