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Why Vgs = Vtn in NMOS here?

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hanhan

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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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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.
 
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.
 
Thank you.

Miles Prower:

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

SteveB:

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.

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

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.
 
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:

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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