How to determine the saturation level when an opamp schmitt trigger starts worknig?

Question

The Opamp inverting schmitt trigger is constructed as shown.

The output signal is:

So I ask why is the output at the start of operation saturated to positive?

Also, if the input voltage is moved from the negative to positive terminal the starting output will saturate to negative. Why?

Is there a rule to follow? If so, what causes this rule?

What will happen if R1 is replaced with a capacitor, as in a monostable multivibrator? What will be its starting saturation level that will charge the capacitor. What will be the UTP and LTP then?

Answer 1

The Schmitt trigger works similar to a comparator, so imagine you have a comparator with a Vref in the non inverting input. When the input voltage falls below that threshold, the output of the comparator saturates high. If the input voltage then goes above the threshold, the output goes low. If you were to put the Vref into the non inverting input, the same rule applies, but backwards, so when the input goes below the threshold, the output goes low, and if it then goes above, the output goes high.

The general rule is if the non inverting input is less than the inverting input, the output will be high.

The Schmitt trigger works in almost the same way, except it has upper and lower limits. As you can see from your diagram, the input voltage (sine wave) starts at 0, which means the inverting input is lower than the non inverting, hence the output starts high. Once it goes above the high threshold, the output goes low. With a Schmitt trigger, the output will stay low untill the lower threshold is met.

As with the comparator, if you were to swap the input voltage to the non inverting input, it would be the other way round.

I believe this is what you were asking, but please correct me if I am wrong as the English was rather poor, as Olin pointed out.

Answer 2

How to determine the saturation level when an opamp schmitt trigger starts workig?

With the "theoretical" circuit you have shown and the input at 0 volts, upon power being applied, there is no real way to determine what the output state will be. The output anomaly will resolve itself when the input signal passes one of the two thresholds. Once this has resolved, the output is unambiguous.

Many applications will want the output to be unambiguous at power-up (in the absense of a significant input signal) and circuit measures can be taken to force a high or low situation at the output but, on a simple theoretical circuit like the one shown, it is impossible to determine.

Answer 3

For the sake of theory, we assume an initial state of output.

Here in your circuit for inverting schmitt trigger, it is assumed to be \$+V_{sat}.R_2/(R_1+R_2)\$ at the non-inverting terminal, at t= 0, and the circuit waveforms are analysed.

Even if you assume it to be \$-V_{sat}.R_2/(R_1+R_2)\$ at t =0, you will eventually end up in the same waveform, as the outputs are resolved itself. The same happens in real circuits too.