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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.

Null
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AlfroJang80
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  • 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. – G36 Mar 17 '21 at 15:51
  • 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. – G36 Mar 17 '21 at 16:01
  • 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. – Lars Hankeln Mar 17 '21 at 16:07
  • @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"? – AlfroJang80 Mar 17 '21 at 16:12
  • @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? – AlfroJang80 Mar 17 '21 at 16:18
  • Because this is true "drain current of M1 heavily depends on input voltage". If we increase the Vin voltage the new equilibrium state will now be present. And this "new state" must satisfy the KVL law Vin = Vgs + Is*RS and this Is = k(Vgs - Vth)^2 at the same time. And because the Vin increases the Vgs and Is also must increases (Vgs = Vth +√(Id/k) ) to satisfy the KVL law. Look here https://tinyurl.com/yk6qbb73 – G36 Mar 17 '21 at 16:44
  • I'm still not fully understanding this. Increasing Vin, which increases Vgs, which increases Id, which increases Vs, which reduces Vgs but not to the same extent. It is a linear dependence rather than square law just like in the CS stage with source degen. So Vgs will increase, but by a square law, in a linear fashion instead due the degeneration. – AlfroJang80 Mar 17 '21 at 23:07
  • The feedback loop is what is confusing me here. – AlfroJang80 Mar 17 '21 at 23:10
  • It is very strange that you think that increase in Vin voltage should somehow "set" (bring back) the via negative feedback the previous state. This is not true, because Vin is a "reference voltage" for a feedback loop. And this is why changing Vin will set the new equilibrium state. And now, when Vin = const, and some "external force" (e.g. Vgs changes due to change in temperature ) wants to "knock out" our circuit from equilibrium, the negative feedback will now "kicks in" and it will help maintain a state of equilibrium despite the fact that an "external force" want to change the state. – G36 Mar 18 '21 at 16:54
  • I understand. What I was trying to say in my last message was: Vin increases by V0, drain current increases, more drain current through Rs, Vs will increase by not exactly V0 but some fraction of V0. So yes, Vgs will increase compared to the previous state but it is not a square law dependency thanks to the source degeneration. There will be a new equilibrium higher current flowing in the circuit, a higher Vgs too but the effect is dampened due Rs. – AlfroJang80 Mar 18 '21 at 17:19

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