# Frequency of an Integrator Op-Amp Circuit

Discussion in 'Homework Help' started by Fraser_Integration, Jan 31, 2010.

1. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Hello again.

Please see the attached diagram which shows a square wave and triangle wave generator formed from a schmitt trigger and integrator circuit.

The bottom question asks to evaluate the period of the square wave. I know that this is probably the same period as the triangle wave, and that this is probably found by:

f = 1 / (2pi.R.C)

but my question is, if the oscillation of the schmitt trigger is started by noise, how does this end up being a periodic wave at all?

Tried looking this up before I asked, couldn't find anything concrete.

Cheers, Fraser.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
The Schmitt Trigger output will be bi-stable between values ±Vc where Vc is the possible output magnitude.

Switching of the Schmitt Trigger stage is subject to the integrator output voltage value - which is time varying as a triangle wave.

Suppose the Schmitt Trigger output has just gone from +Vc to -Vc. What integrator output voltage would have triggered the change?

The actual trigger values would be ±Vt. Can you derive the relationship between Vt and Vc for a given transition (pos-to-neg or neg-to-pos) case? You need to include the Schmitt Trigger resistor values in the analysis.

Once you know this you can deduce the period of the circuit output waveform.

I calculate the output frequency as 3.33MHz. Which would be exceptional for a bog standard op-amp implementation.

Last edited: Feb 1, 2010
3. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
I found Vt by applying KCL at the non-inverting input: Vi/22k = -Vo/66k therefore Vi = -Vo/3 so switching values of +5V and -5V, for a 15V supply.

Therefore the triangle waves from the integrator will have an amplitude of +-5V, but I'm still not 100% sure on how to work out the frequency of the triangle, and leading on from that, the square waves.

4. ### thyristor Active Member

Dec 27, 2009
94
0
I agree with TNK that the frequency is 3.33MHz. I recently used the same circuit as part of a 12v lamp dimmer for a boat where it performed superbly, operating at 500Hz. I have attached my hand-written notes in which I calculated the frequency from first principles.

Using YOUR component labelling, your frequency = R2 / (4 x R x R1 x C)

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5. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Thank you for that worked solution, I understand now. I've not done any problems before where they don't explicitly define values for dv and dt, but it seems so obvious now, as usual. I won't forget it in a hurry.

6. ### sgardner025 Well-Known Member

Nov 5, 2009
79
4
Not trying to hijack here, but could you explain how the frequency formula is derived?

7. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
It's in his attachment.

When you get a value of the trigger voltage for the Schmitt trigger, you can then start up the equation for the integrator circuit, as the trigger voltage for the schmitt is the output of the integrator, and the output of the schmitt is the input of the integrator.

Last edited: Feb 1, 2010
8. ### sgardner025 Well-Known Member

Nov 5, 2009
79
4
I've been studying on it and I understand now. Thanks for the quick help.