Razavi Micro feedback example

[shredder929]

This example in Razavi's Microelectronics book seems to break KCL to me. If I_in goes up, by KCL it must increase the Id of M2 right? So Vx will go down, and Vout goes up, and Id of M1 goes down as well, which makes sense because Vx went down.

But what the text states is I_in goes up and Id of M2 goes down. How is that possible? M1 has an infinite input impedance, where is that extra current going? When I try and simulate in SPICE it makes even less sense, neither of the transistors turns on.

 

[LvW]

Why do you think that Vx will "go down" when Id of M2 increases? In contrary, a larger Id will produce a slightly increased voltage drop across the Rds of M2. Because Vgs1 is identical to Vds2, the lower device (M2) will work in its quasi-linear resistive region.
Both transistors work in common source configurations - hence we have two phase inversions within the closed-loop. That means: Positive feedback.

 

[d123]

Hi,

Make a fool of myself (ignorance is bliss, until you publicly display how ignorant you are on a forum, which I will now do, then bliss turns into being trolled or just ridiculed)... Maybe M2 VDS rises as Iin rises, increasing due to its RDS x Iin, which in turn increases the voltage at the gate of M1, thereby turning M1 on more, which then draws/sinks more gate voltage away from M2, so M2 turns off more and M1 turns on more, and the end result is positive feedback. (I am not certain I understand the meaning of positive feedback here - positive feedback is usually not a desirable effect, surely?)
Or, maybe Iin increases M1 gate voltage by charging the gate capacitance more, with the same result of sinking more gate voltage away from M2 and so on.

Separately, why do you say M1 has infinite input impedance? - I do not think that is a realistic starting point, only a theoretical one of little practical use for real circuit analysis with this example - please correct me on that.

 

[FvM]

Same exercise problem was discussed recently: https://www.edaboard.com/threads/determining-polarity.397725

 

[shredder929]

Oh gosh that makes so much sense. My thoughts went to the CS stage with a resistor between the drain and Vdd, where an increase in Id means a decrease in Vd. Thanks everyone!