# Getting a square wave as a output of op-amp

I have the following op-amp circuit and i am trying to find out what is the open loop gain.

In order to make this simulation properly, we need to make sure that the DC operation of the amplifier is the correct one. Applying negative feedback is a possibility, since it is capable of automatically set the right operation point, but it needs to operate only at DC, otherwise the open-loop gain could not be measured. To acomplish that i used the configuration shown below.

The problem is that the ouput I am getting is a square wave and I can't understand why. Isn't it supposed to be a sinusoidal wave like the input? I already applied several amplitudes do the input but the output is always the same.

Input wave: green

Ouptut wave: blue

• 1. Why are you using an LC divider for the feedback? 2. Why are you using any feedback if you are trying to measure open-loop? 3. If you do get a correct open-loop configuration at what input signal level would you expect the output to saturate? Maybe edit your thoughts into your question rather than post them in the comments. Commented Nov 25, 2023 at 15:38
• Man, you are using ~3V input signal to find an open-loop gain? For poor 80db (==*10^4) amp non-clipped output must be a ~30kV! Try to set input to uV range. Commented Nov 25, 2023 at 20:12

The output amplitude is too high and the sinewave is clipping. Basically the op-amp cannot supply an output voltage bigger than its power rails minus 1 to 4 volts (op-amp dependent). Your closed loop gain is rather large at 5 kHz ($$\1+\omega^2 LC\$$). Do the math.