LM358 Opamp configuration

Thread Starter


Joined Dec 20, 2020
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 board1703167384996.png


Joined Mar 14, 2008
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.


Joined Feb 24, 2006
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.

schmitt trigger

Joined Jul 12, 2010
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.


Joined Oct 2, 2009
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.


Joined Aug 1, 2013
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.



Joined Aug 1, 2013
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.