Intuitive explanation for gmb

idsat square law prove

I am having trouble understanding how, in the presesnce of body effect, the resistance looking into the source of a mosfet is reduced.
Why is body effect modeled by a current source between the source and the drain? I thought it just shifts Vth?


Thanks in advance,

Sharas

Let us start from current source . In order to understand this we need two equations. First one is the saturation current equation(square law) and threshold equation considering Vsb & Vtho. Now substitute Vth equation in Idsat and simlify. considering a fractional change in Id current apply Taylor series expansion. Neglecting higher order we have three terms . on studing these three terms we will have a circuit. From the equation it will be clear that a current source has to be used to model boby current.

Thanks, but outside the math, what's the physical explanation for this current source?

Wow, this last explanation with the Taylor expansion and all would be envied by any good physicist. An engineer, probably would not pay much attention to it - too complicated. Well, here are my 2 cents. you may want to take a piece of paper and a pen as you read further.
You know already that if the source is different and higher than the bulk voltage there will be an increase in the threshold voltage of the transistor because in this case the reverse voltage across the source-bulk junction increases and the depletion region becomes bigger, so you'll need more gate-source voltage to push the same amount of electrons (for NMOST) into the channel - so effectively increasing the threshold voltage. Now, assume that the gate is at ac ground, the bulk is at ground and there is a voltage source in the MOST source. If we increase the voltage of the source it will become more positive than the bulk by some delta V and because the threshold will increase or in other words the depletion region will become wider, the drain current will decrease by some delta I. For the NMOS transistor we usually assume the positive direction of the current flowing from drain to source. In this case if it is decreasing that means there is that delta I going in the opposite direction i.e. from source to drain. In the small signal equivalent model this can be modeled by a controlled current source between the drain and source and pointing to the drain IF the controlling voltage is with its positive terminal connected to the source and negative terminal to ground (remember, bulk we assumed is at gnd). In the same small signal model we usually have the gm*Vgs current source also between drain and source but pointing to the source IF the gate source voltage is positive. But in our set-up here we have gate at ground and source positive, so the gate -source voltage is negative and then the gm*Vgs current source reverses direction i.e. points to the drain and can be described as gm*Vsource (Vsource=-Vgs), i.e. it is controlled by the same source voltage. Now, both controlled current sources are controlled by the same voltage and going in the same direction which means that effectively the source voltage sees an increase in the current going out of it and into the transistor's source, meaning it sees smaller resistance compared to the case when the source and bulk were tied together i.e. all this results in decrease of the resistance looking in the source.

Thanks sutapanaki!

This was a great explenation.