Hello, I found a board with the LM358 with a configuration that was a little strange for me. I would like to know if it is a non-inverter or inverter and calculate its gain.
Aschematic of the board

 

The op amp is being used as a non-inverting comparator.
Since there is no negative feedback, the gain equals the open-loop gain of the LM358.
R3 provides a small-amount of hysteresis to the trip point, but I'm not sure what D1 and D2 do since the D1 anode connection is not shown.

 

The power supply connections are also missing. Since the inverting (-) input is connected to a voltage divider between VDC1 and GND one might presume that the LM358 is powered by a single supply. Although that might be the case it would be good to know the actual situation.

 

LM358 won't work well with a 3V supply.

 

Additionally… Without an input impedance, usually a resistor, on the noninverting input, the hysteresis value cannot be determined.
This is an incomplete circuit, it can’t even be classified as a circuit snippet.

 

Inverting or non-inverting op-amp?

If the signal goes to the -ve (inverting) input, then it is inverting.
If the signal goes to the +ve (non-inverting) input, then it is non-inverting.

Amplifier gain is determined by the feedback network from the output to the inverting input.
Since there is no feedback in the circuit, the gain is the open loop gain of the opamp and that is usually in excess of 100,000.

 

I forgot to put a photo of the board itself D2 is D1, D1 is D2, R20 is R1, R21 is R2 and R13 is R3 in the schematic, I assumed an ideal opamp in the schematic

 

How do you know that the IC is an op-amp?
What is the marking on the IC?

 

How do you know that the IC is an op-amp?
What is the marking on the IC?

Yes it is a TI LM358 but my camera it so bad

 

So where is the D1 anode connected?

 

Hi dark,
Your circuit is missing VCC to pin #8
and Gnd on pin #4
E
Update:
Where on the PCB is your out label???

 

So where is the D1 anode connected?

I put the output in wrong place, I make the correct schematic, but the vcc value it unknown

 

hi dark,
Recheck the R3.D2 feedback path from pin #1 to where ???

E

 

hi dark
your circuit showed this

E

 

hi dark
your circuit showed this

E
View attachment 310500

I put the output where it wasn't, the one you made




and the one I published is the correct one

 

hi dark,
I believe your circuit is designed like this ??

E

 

hi dark,
I believe your circuit is designed like this ??

E
View attachment 310504

R3 it conected to non inverting pin

 

Eric - R3 is connected to the non-inverting input.

In a normal 2-resistor hysteresis circuit, the two transition levels are symmetrical about the reference voltage. For example, with a reference voltage of approx. 0,8 V as in this circuit, the two transition levels could be 0.7 V and 0.9 V. As above, the amount of hysteresis depends on the source impedance of the input signal.

Enter diode D2. Now there is no current through R3 in one of the high output state. Thus, the transition levels no longer are symmetrical. In the negative-going direction, the output will change when Vin = 0.8 V. In the positive-going direction, the output is low and D3 is conducting. This adds 0.6 V to the feedback equation, which again is dependent on the input resistor.

ak

 

Hi dark,
OK,
So this is an option.
E

 

With no impedance between the non-inverting input and V2, the V2 source impedance of 0 ohms swamps out the positive feedback, yielding identical transition levels.

ak