I speak with reference to this (Electrical4u - blog tutorial) and this (All about electronics - video tutorial) on the Schmitt trigger.

From what I have understood, in the non-inverting schmitt Trigger,

  • When Vin > Vut the output is High (Vout = VH)
  • When Vin < Vlt the output is Low (Vout = VL)

And it is exactly the opposite in inverting Schmitt trigger,

  • When Vin > Vut the output is Low(Vout = VL)
  • When Vin < Vlt the output is High(Vout = VH)

where Vin --> Input Voltage, Vut --> Upper Threshold, Vlt -- Lower Threshold,Vout --> Output voltage.

Here in the picture shown, enter image description here

The first line gives the output equation. Here they assume that the output is in high state. Then they find the Vut as shown. Here is where my question is, for the inverting case the output should be VL for Upper threshold. But in the expression for the Upper Threshold it is VH (Voltage high or +Vsat). Why is it so?

  • 2
    \$\begingroup\$ Well, that's because of positive feedback and simply comes out from direct circuit analysis. Another good explanation with analysis and design steps is in ti.com/lit/an/snoa997a/snoa997a.pdf \$\endgroup\$ May 24, 2022 at 7:35
  • \$\begingroup\$ I am sorry that I couldn't able to follow the resource you provided. I understood that we make it Vh because after that the voltage turns towards -Vsat. So until that point it is in Vsat. Is my understanding right? \$\endgroup\$
    – Aghilan
    May 24, 2022 at 8:26
  • 1
    \$\begingroup\$ Yes, the idea is that when the comparator switches the output itself moves the reference point around. It's all basic circuit analysis after that, just do the math \$\endgroup\$ May 24, 2022 at 8:37
  • \$\begingroup\$ Oh ok Thank you for your time! \$\endgroup\$
    – Aghilan
    May 24, 2022 at 8:49

1 Answer 1


The circuit acts like a comparator. If you want to show VL at the output you have to push the input voltage over the threshold, because the "-" input shall be greater than the "+" input. The threshold to exceed is the voltage on the "+" input of the op-amp, that is the application of the Kirchhoff's voltage law. Suppose that the output is in the high state: $$V_{out} = V_H$$ then the voltage on the "-" input shall exceed the voltage on the "+" input, that is: $$V_{UT} = {R_1\over{R_1+R_2}}\bullet V_H$$ This is why the upper voltage is used


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