Schmitt Triger with dynamic hysteresis [help]

I'm making an assignment on this topic but so far I've encountered a few problems that I can't figure out.
For the OA I've used LM324 whose output voltage is Vcc-1.5V.So for my case I've made these calculations with input voltage of 10 V.
I have τ=100uS.UTHL= ±1mV UTHH=±1V
I can't reach the required working frequency range of 1kHz-100kHz
These are my calculations so far I've probably messed some formula to achieve so low frequency.

What exactly is the circuit supposed to do?

An LM324 will not work well at a frequency of 100kHz.

Well in the assignment it's said I should only calculate the values of the resistors which determinate my low and high threshold. Ive read at a few places that lm324 it's not the fastest op amp but I only need the calculations and not make any pspice or breadboard testing

What is the power supply voltage for the LM324?

I've chosen it to be 10V.In this case the output based on the datasheet is 8.5V

I believe your math may be a bit off (using your formulae from above):



therefore:

(I think this where the error came into play)

therefore:



therefore:


thus:



Does that frequency more closely meet your expectations?

Yeah but R3 has a value of 810 kΩ

What is the purpose of C1 in your circuit?

ak

This circuit is more like a non-retriggerable one-shot than a Schmit-Trigger. Once it triggers, it effectively locks-out a small signal from re-triggering it again until the C1 discharges via R1 and R2 (it has not much to do with R3). The actual triggering point is very close to 0V for either transition, unlike a real Schmit trigger with hysteresis, where there are two separate triggering levels. It is a very weird circuit, because its sensistivity to being triggered increases with time...

Look at V(ni); triggering (both ways) as V(ni) crosses zero V. Notice all the small cycles of the input waveform that do not cause a trigger. Note that it would trigger much sooner if the amplitude of V(in) was higher.

The output of an unloaded LM324 will pull a bit higher than 8.5 as shown by V(out). Note that V(ni) is just V(d) reduced by the voltage divider R1/(R1+R2).

isNomad said:

Yeah but R3 has a value of 810 kΩ

Click to expand...

You're right. My mistake.

In that case, Fo works out to be 3.538570417551309e-6Hz.