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I have been working on this circuit many days. I just would like to know how it works. In my opinion, the key is in the feedback in the 2º CMOS, but I do not really know how to manage it.

All the information I have about the circuit is in the figure.

enter image description here

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  • \$\begingroup\$ What is the point of this circuit? It doesn't seem to perform any useful function that I can discern. \$\endgroup\$
    – Hearth
    Commented Nov 21, 2019 at 16:45
  • \$\begingroup\$ Looks like it adds hysteresis, like a Schmitt trigger would. \$\endgroup\$
    – Justme
    Commented Nov 21, 2019 at 16:46
  • \$\begingroup\$ @Justme Ah, I saw this as a digital circuit and didn't consider inputs other than Vcc or gnd. \$\endgroup\$
    – Hearth
    Commented Nov 21, 2019 at 16:50
  • \$\begingroup\$ @Hearth, as Justme says, it is an analog circuit. \$\endgroup\$
    – JuniorTry
    Commented Nov 21, 2019 at 17:25
  • \$\begingroup\$ this is a well-controlled Schmidt trigger; a pleasure to design with. \$\endgroup\$ Commented Nov 21, 2019 at 21:20

1 Answer 1

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The circuit is essentially this:

schematic

simulate this circuit – Schematic created using CircuitLab

Obviously, the outputs of NOT1 and NOT3 will conflict with each other at times. The point is that the transistors of NOT1 are large enough to "overpower" the transistors in NOT3 when necessary.

Basically, the output of NOT3 functions as positive feedback for NOT2, tending to keep it in the state that it's already in. This creates hysteresis in the transfer function of the overall circuit, also known as a "Schmitt trigger".

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  • \$\begingroup\$ It makes sense. First of all, thank you very much for your answer. I have been reading the article in wikipedia. If I am not wrong, with that hysteresis, Vout will not change from 0V to 5V at the same Vin that it will change from 5V to 0V. Is it possible to know what are the limits for Vout and Vin in the transfer function? \$\endgroup\$
    – JuniorTry
    Commented Nov 21, 2019 at 17:14
  • \$\begingroup\$ Only if you have accurate models for the transistors (including their actual sizes). \$\endgroup\$
    – Dave Tweed
    Commented Nov 21, 2019 at 17:35
  • \$\begingroup\$ ... and it's only going to work if the transistors of NOT1 (M1 and M3) are somewhat stronger than M2 and M4, but not so very much stronger that they are completely unaffected by the state of the output. I bet there's a lot of interesting design that goes into making sure that circuit works correctly over process, temperature, and VCC variations. \$\endgroup\$
    – TimWescott
    Commented Nov 21, 2019 at 18:38

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