Schmitt trigger equations

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If you start off with the input LOW. then the voltage on the + input sets the switching point. So doing a series equivalent of Vref, you get Vref X R2/R2+R1 call this Veq, with an internal impedance of R1//R2 call this Req. Now you have the output high, so circuit is +Vcc feeding via R3 into Req to Veq. The voltage on the + input is the switching point. If you increase the input voltage until its Vcc, then reduce it The upper switching point will be -Vcc feeding via R3 into Req and Veq.
Frank
 


Frank, trying to get my head around what you have said here.

It sounds as though you are saying that the equations should be:

Vut = -Vcc x (R2 / R2 + R3)
Vlt = Vcc x (R2 / R1 + R2)

where Vut = upper threhold voltage and Vlt = lower threshold voltage.
And by the looks of that calculator I found, they seem to choose the value for R2 of 10k.

So two equations with two unknown values.

The calculator pages seem to refer to Vref and Vcc but, as far as I can see, Vcc = Vref in my case.

And since I am not using a positive/negative power supply, -Vcc = 0V and Vcc = 15V.
 

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