In the schematic below:



The circuit is behaving as a comparator. When V+>V- the output is at 12V. When V+<V- the output is 0. (the positive pin power supply of the op-amp is connected to 12V and the nagative pin power supply is connected to ground).

The problem is when I try to do the math derivation to justify the behavior of the op-amp, I don't get what that comparator effect.
Here is the math derivation:
At V- (pin6), we have 6V because of the voltage divider R33/(R31+R33). Which means that we should get 6V at the V+(pin 5).
To make it simple, no currents flow in pins 5 and 6 (in reality there are some pAmps). the current flowing in R35 is (7V-6V)/R35=0.14mA.
then this current is flowing also in R34, since V5 is at 6V, the voltage at pin 7 (the output) is 2.93V (20kR*0.14mA). In LTspice I got at pin 7 12V.
I would like to know how I can justify the behavior of such circuit.

 

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At V- (pin6), we have 6V because of the voltage divider R33/(R31+R33). Which means that we should get 6V at the V+(pin 5).

Why do you think that there are also 6V at pin 5 ?

 

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

 

It works so:

 

It works so:
View attachment 234273

I know that it works, but how about the math derivation

 

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

Yes - that is correct, but for NEGATIVE feedback only. Positive feedback causes saturation and the mentioned rule does NOT apply anymore.

 

Yes - that is correct, but for NEGATIVE feedback only. Positive feedback causes saturation and the mentioned rule does NOT apply anymore.

Ah okay , but can we prove it mathemtically ? I mean how can we reach the saturation in this case.

 

Ah okay , but can we prove it mathemtically ? I mean how can we reach the saturation in this case.

Saturation means that the opamp output reaches its limits (power supply rail). The amplifier has left his linear range. Because of the very large dc gain and due to missing negative feedback), some tenth of µVolts between the inputs are enough to reach this state .
Question: Are you using dual or single supply for the opamp?

 

Saturation means that the opamp output reaches its limits (power supply rail). The amplifier has left his linear range. Because of the very large dc gain and due to missing negative feedback), some tenth of µVolts between the inputs are enough to reach this state .
Question: Are you using dual or single supply for the opamp?

I am using a single supply, I just want to know what is the voltahe at the pin 5 (+). (I know that it's greater than V-), but what is the exact value, so that we can have that voltage difference at the inputs.

 

I am using a single supply, I just want to know what is the voltahe at the pin 5 (+). (I know that it's greater than V-), but what is the exact value, so that we can have that voltage difference at the inputs.

Voltage at pin 5 (+) under which condition? 7 Volts DC at R35 ?

 

The voltage on pin 5 depends on the input voltage and the output voltage only. Write an expression for it in terms of those two voltages.

Bob

 

The voltage on pin 5 depends on the input voltage and the output voltage only. Write an expression for it in terms of those two voltages.
Bob

Yes - that`s why I have asked for the input voltage .
Its nothing else than a case for superposition.

 

Yes - that`s why I have asked for the input voltage .
Its nothing else than a case for superposition.

The DC input voltage is 7V at R35, I still don't see the voltage difference at the inputs

 

Capacitors value is changed:

ADDED:
See how works ideal opamp.
Pay attention to R2 value:
R2=(R1/Vref)*(Vth-Vref) this formula is accurate for ideal opamp only.
Vth is high going threshold voltage.

 

Yes - that`s why I have asked for the input voltage .
Its nothing else than a case for superposition.

My post was directed to the TS, I know you can do it

Bob

 

My post was directed to the TS, I know you can do it

Bob

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

 

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

You have two separate voltage divider problems involving two resistors for each divider.
Are you saying you don't know how to do that?

 

You have two separate voltage divider problems involving two resistors for each divider.
Are you saying you don't know how to do that?

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 unkown voltages. If it's simple to you explain it to me

 

The op amp acting as a comparator with positive feedback, so the output has two states, either positive saturation (plus input higher than the negative input) or negative saturation (plus input lower than the negative input).

So you calculate the input voltage difference for both conditions of the output, and see which one leads to the input voltage difference that is correct one for the two possible output voltages.

Make sense?

 

So when there is always a positive feedback, the op amp operates as a comparator ?