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I have a question about the MOS source-follower / common-drain stage. In particular, it's behaviour as a level shifter:

enter image description here

The book mentions the following: enter image description here

I'm having some trouble figuring out how the drain current may increase by a factor of 2? The source-follower has a built-in source degeneration, right? I recall from the common-source stage with source degeneration that one advantage of it was that it 'softens' or 'linearises' the drain current variation with Vin:

enter image description here

So surely, if Vin changes from 0.7 V to 1 V, Id should change by some factor times Vin since we essentially have negative feedback?

How come the benefit of the CS stage with source-degeneration (being robust against input DC level changes) does not appear here, even though we have a source resistor? The statement after the highlighted one is also not clear to me. It is like they are almost ignoring the fact that higher Vin -> higher Id > will bring the Vs(=Vout) up and so Vgs should be stable!

I fully understand the small-signal analysis of this circuit but I find that most other textbooks don't talk about the large-signal analysis of these circuits, Razavi is one of the only ones I found.

I would be very grateful if someone could explain the large-signal behaviour of the source-follower. I'm just getting very lost now.

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  • \$\begingroup\$ In source follower, we want to achieve Vout/Vin = 1. But in a single transistor stage, this is not possible. In this case Vout = Vin - Vgs.And as you know the Vgs voltage is not constant but it will vary together with the source current. And to overcome this we can bias the source by using a constant current source. Thus, the Is current and the Vgs voltage will be constant. Thus, we will have a constant dc-offset between Vin and Vout. \$\endgroup\$
    – G36
    Commented Mar 17, 2021 at 15:51
  • \$\begingroup\$ Why do you think that the CS stage with source-degeneration is robust against input DC level changes? This is not true. And changes in the DC level at the source will be amplified by a factor of RD/RS and "showed" at the output. \$\endgroup\$
    – G36
    Commented Mar 17, 2021 at 16:01
  • \$\begingroup\$ It might be helpful to just do the math for the large-signal analysis to get a better understanding. It was basically already done for you in the book, look at equation 3.82. There you could replace Vout with Rs*Id if that helps. \$\endgroup\$
    – Lars Hankeln
    Commented Mar 17, 2021 at 16:07
  • \$\begingroup\$ @LarsHankeln That's part of where my confusion is. The effect of the drain current increasing due to the input voltage increasing will be combated by the current through source resistor increasing, so Vout (=Vs) goes up, hence Vgs returning back down to some extent, lowering drain current. So why does the author say "drain current of M1 heavily depends on input voltage"? \$\endgroup\$
    – AlfroJang80
    Commented Mar 17, 2021 at 16:12
  • \$\begingroup\$ @G36 Regarding your first comment, why is the Vgs not constant? Let's say for the source-follower, you increase Vin by some amount V0 (large-signal), Vgs increases, higher drain current flows through the source resistor, higher voltage drop across the source resistor, Vs (=Vout) increases. Now, Vin = Vg has increased and so has Vs = Vout, so Vgs should return to some extent back down, lowering the drain current back down. So why does the author say "drain current of M1 heavily depends on input voltage" if the negative feedback of the source resistor will regulate it? \$\endgroup\$
    – AlfroJang80
    Commented Mar 17, 2021 at 16:18

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