I understand that if I use a FET as a source follower, the signal will be slightly attenuated due to the fact that there is some small inherent resistance in the FET, creating a voltage divider effect between the FET's internal resistance and the source resistor Rs.
In the student manual for 'The Art of Electronics' (Horowitz), he shows a diagram that implies using a current source in the place of Rs can create a circuit with no attenuation. As is the case with many diagrams in this book, his explanations kind of glaze over what is going on, without going into great detail.
In this particular situation, he's aiming to have 2 mA Id current flow. He picked the 1.4k resistor from the curve of ∆Id vs ∆Vgs given this current requirement - making Vgs = -2.8V.
In the diagram, he traces his reasoning: (1) Here drops Vgs, (2) so same current also drops Vgs here ... so Vout = Vin.
I get why Vgs drops across these resistors, but I don't understand how he arrives at the conclusion that Vout = Vin. In particular, I don't see why the voltage at the source of the top FET should be equal to Vin+Vgs.
Can anyone shed some light on this confusion?
Here's the diagram: