# Origin of the signal in an oscillator composed of a Schmitt trigger and an integrator

Following instructions, we built this oscillator as part of our electronics course:

I was surprised at first that we could get those signals without any initial signal connected to the op-amps input. Is it correct that the op-amp (via the positive feedback on the Schmitt trigger) is simply amplifying random noise (always present in the circuit, for various reasons such as ambient magnetic radiation or temperature gradient which results in movement of 'electrons) until it first flips the switch?

EDIT: I'm only talking about the very first moment the signal is created and the very first flip of the switch. I agree that the regular operation of the Schmitt trigger and the integrator take over without any link with this phenomenon afterward.

Is it correct that the op-amp (via the positive feedback on the Schmitt trigger) is simply amplifying random noise (always present in the circuit, for various reasons such as ambient magnetic radiation or temperature gradient which results in movement of 'electrons) until it first flips the switch?

No, that is not entirely correct.

Once the circuit has overcome the (theoretical only) initial start-up problem of an op-amp/comparator with positive feedback, the circuit is implicitly unstable without any of the influences you mention It's a type of relaxation oscillator and this link explains about the theoretical unstable equilibrium problem: -

The system is in unstable equilibrium if both the inputs and outputs of the comparator are at zero volts. The moment any sort of noise, be it thermal or electromagnetic noise brings the output of the comparator above zero (the case of the comparator output going below zero is also possible, and a similar argument to what follows applies), the positive feedback in the comparator results in the output of the comparator saturating at the positive rail.

• I was only talking about the very first moment the signal is created (which I think your link explains exactly the same thing as I wrote ?), sorry if my message was unclear. I agree that after that the regular operation of the Schmitt trigger and the integrator take over without any connection to these phenomena. Commented Mar 17, 2023 at 13:36
• I still mark your answer as accepted since your link completely confirm what I thought of it. Commented Mar 17, 2023 at 13:40
• @c.leblanc OK, I understand and thank you Commented Mar 17, 2023 at 13:40

The random noise together with the unavoidable asymmetries of the internal circuitry of any real opamp play a role only during power-up.

After the power-up transient has ended, the output of the Schmitt trigger will end up (randomly) in either of two states, defined by the two saturation level of its output (Vee or Vcc). From that moment on, the integrator will simply integrate that initial Vout level, producing either a negative or a positive ramp, which is then fed-back to the Schmitt trigger, whose function is simply that of inverting the input of the integrator periodically, so that to produce a periodic signal (triangle wave).

Of course this in turn also produces a square wave as the trigger output (Vout).

So, in the end, random noise has no crucial role in this circuit besides during the startup transient. It's not even essential for that. Even if noise were absent (impossible in real circuit) it would be enough that the asymmetries of the opamp internal circuitry kept the output in saturation during power-up, i.e during that brief period of time when the supplies go from 0 to their nominal level.

But this is only of theoretical relevance, since the two factors (noise + asymmetries) are always present together, so it is not possible in general to ascribe the startup process to either one alone.

• I was only talking about the start-up, I totally agree that what's coming after is completely different. Thank you for the explanation about asymmetries, I had no idea that this could be enough (for a theoretical case) to initiate the signal. I have a clearer vision of the phenomena with your explanation! Commented Mar 17, 2023 at 13:34
• @c.leblanc For real opamps, see offset voltage, which is one parameter found in datasheets that is direct consequence of those asymmetries (depending on application you should consider also offset current). Commented Mar 17, 2023 at 13:53
• I see, what I still don't get is how it could oroduce the start of the signal? I mean, there may be a mechanical difference between the transistors in the op-amp that produce an offset between the 2, but when there is no signal at all in the circuit there is nothing to amplify right ? Commented Mar 17, 2023 at 16:51
• @c.leblanc Research opamp offset voltage. The presence of offset voltage is like a spurious voltage generator in series with one of the inputs of the opamp. For a jellybean opamp could be some tens of millivolts, much less for speciallty ones. That voltage end-up being amplified by the opamp. For a positive feedback circuit like the Schmitt trigger this means saturation even with no input signal. Commented Mar 17, 2023 at 17:29

Is it correct that the op-amp (via the positive feedback on the Schmitt trigger) is simply amplifying random noise ...

It is not amplifying "random noise", but really "integrating" (through C1) a signal (which is a two-level voltage) at input Vout ...

• I'm really talking about the very first time any signal is emitted. My understanding was that it is amplifying the noise through the positive feedback of the trigger Schmitt first, only then when it triggers it the integrator will integrate whats coming out of the schmitt trigger. Actually, if you try connecting only the Schmitt trigger on a loop it will saturate. And if you try to connect only the integrator nothing happens. Commented Mar 17, 2023 at 13:25