Yes but, the positive feedback resistor determines the hysteresis around the transition voltage. That is the goal of the question if this is homework. Smaller resistors give more hysteresis until an infinite resistor gives no hysteresis.

 

I am asking about how to calculate the voltage at pin 5, if we consider that we don't know the output voltage we are having two unknown voltages. If it's simple to you explain it to me

You do know the output voltage; it is on the datasheet.

Because there is no negative feedback, the output voltage is the input voltage difference (Vin+ - Vin-) times the open-loop gain of the device, which is in the datasheet. Because the output cannot swing to hundreds of thousands of volts (for a 1 V differential voltage, the opamp would try to make an output of 3 million volts), the real output will be either the high or low output saturation voltage. This means that there is not one single, simple equation; there are two, one for each output case. Again, this is not a simple negative-feedback situation, so all of that rule-of-thumb stuff is out the window.

In both output cases, R34 and R35 form a voltage divider between the output voltage and the input voltage. Using Ohm's Law you can determine the input voltage necessary for the non-inverting input voltage to be equal to the inverting input voltage (which in your case is 6 V). This is called the transition voltage, and there are two of them. The fact that there are two of them is why this is a Schmitt Trigger circuit rather than a simple comparator.

Two equations, one for Vout-high and one for Vout-low.

Two transition voltages, one for a positive-going input, and one for a negative-going input.

ak

 

I could not do it that's why I asked for help

There are two resistors in series between two voltage sources. I think, it is not too difficult to find the current through the resistor chain. Having the current, it should be possible to calculate the voltage across one of the resistors.....

 

@LvW , that's how an amplifier operates when there is a feedback, it tries to maintain the V+ and V- at the same potential

An amplifier uses negative feedback to its - input. But your circuit uses positive feedback to its + input.

 

Hello there

And you can get them in two flavors.