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I have recently come across the op amp Schmitt trigger presented in 3rd edition art of electronics. I decided to build the circuit myself in circuit lab, and I have some questions regarding its operation.

Schmitt trigger with one threshold

schematic

simulate this circuit – Schematic created using CircuitLab

First, when the voltage at the non inverting input is greater than the voltage at the inverting input, the op amp finds itself in positive saturation, and has an output equal to +5V. If the voltage at the inverting input is equal or greater than the voltage at the non inverting input (4V), then the op amp enters negative saturation, with an output close to ground.

Schmitt trigger with two thresholds-positive feedback

schematic

simulate this circuit

enter image description here

When the voltage at the inverting input is lower, and the op amp is in positive saturation, the threshold, or the voltage at the non inverting input is approximately equal to +4V. When the input is high, the output is low, near ground and the new threshold is equal with 9090/(10k+9090) x8V=3.809V , where 9090k is the value of 10kII100k. The AoE author says that this circuit works with positive feedback. Why does this circuit work with positive feedback only? I am not very aquainted to positive feedback, so can someone explain to me why this circuit wouldn't work with negative feedback?

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  • \$\begingroup\$ You can't have a Schmitt trigger with one threshold. Schmitt triggers always have two thresholds. \$\endgroup\$
    – Andy aka
    Oct 30, 2023 at 9:13
  • \$\begingroup\$ @Andyaka-From what I see this only applies if you add feedback to the circuit. If you don't, like in the first circuit (which I have copied from AoE), then there will be only one threshold at the non inverting input, equal to the voltage divider equation, or 4V in this case. \$\endgroup\$
    – David Ab
    Oct 30, 2023 at 9:46
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    \$\begingroup\$ @DavidAb I would call the first circuit a comparator circuit (not a Schmitt Trigger). A Schmitt Trigger is formed from a comparator circuit by adding some positive feedback to it. \$\endgroup\$
    – user350400
    Oct 30, 2023 at 10:42

1 Answer 1

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With negative feedback the output will move in a direction which reduces any difference voltage between the inputs to keep the two inputs at a very similar voltage to each other.

When positive feedback is employed, any difference voltage between the op amp's two inputs will drive the output in a direction which further increases the difference voltage between the inputs.

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  • \$\begingroup\$ Ooh, the reason why with positive feedback the gain of the op amp is at its maximum? \$\endgroup\$
    – David Ab
    Oct 30, 2023 at 9:42
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    \$\begingroup\$ @DavidAb The gain of the op amp (Vout/Vdiff) at DC is the same, whether used with positive or negative feedback, during the output transition phase before saturation occurs. It's just that, with positive feedback, a changing output voltage increases the difference voltage between the inputs which drives the output harder in the same direction. \$\endgroup\$
    – user350400
    Oct 30, 2023 at 9:56

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