Why Vgs = Vtn in NMOS here?

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

1. anhnhaMember

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

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2. Miles ProwerMember

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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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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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5. anhnhaMember

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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?

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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 ProwerMember

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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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