Representation of intrinsic source follower by thevenin equivalent (Behzad Razavi)

[BartlebyScrivener]

On page 72 of the razavi book, figure 3.3. shows a MOSFET amplier in source follower configuration.

The input to the gate is Vin, the voltage between the gate and source is measured as V1 (+V1 at gate, -V1 at source) with a drain to source current gmv1.

This is turned into a thevenin equivalent at the source node with voltage Vth = Vin and resistance Rth =1/gm.

If I try to do this myself,

The open circuit voltage at the source node w.r.t ground is Vin-V1 so I get Vth = Vin-V1

The short circuit current is -gmVin

So Vth = Vin-V1 and Rth = Vin-V1/-gmV1

Where am I going wrong!? Thanks.

 

[sam_icdesign]

I think you're considering vgs = vin. As I understood from what you wrote vgs = V1, so now its not important what vin is , the transistor considers vgs and therefore produces a drain source current of ids=gm*vgs.
Therefore the vth will be vin (the main input ) and the rth will be 1/gm. I think you have considered the source voltage to be V1 and therefore vgs = vin-V1.

You could also say that in a source follower you output is the source so if you want to calculate the output resistance, the gate will be ground (vin = 0), the drain is also ac ground so vgs of transistor becomes -vs and rout = Vt/it = 1/gm || rout = 1/gm