saturation when disconnecting the serial monitor

Hello, I use an arduino nano with an ethernet shield as a black box, on which I feed the signal from the Omron zx1 sensor, the arduino measures the 4-20mA signal with the help of an additional ADC (16bit), and then puts the value of the AD conversion into the arduino where it is converted to mm and sent to the website. When I measure the transient characteristics with the laptop connected to the arduino, the saturation does not appear in any way, when I disconnect the arduino, the saturation appears, see pictures.

The arduino (5V) and the sensor (24) are powered by a single source (24V), the arduino is powered via step-down DC-DC.

my question is: Why does saturation appear when disconnecting the laptop?



note: the pinheader is for connecting the sensor and the LED for signaling the functionality of the ADC

I doubt anyone can answer your question based on the information given above.

Please read the forum guide in the sticky post. That will tell you important things to include, for example code, schematic and links to the specs of major components, and how to include those things correctly in your post.

My guess at this point is that you have a power supply problem. When connected to serial monitor, the Nano receives power from the PC/laptop. May be there is a problem with your 24V power supply or the DC converter.

sorry, I didn't notice forum guide, I'll edit the post.

Please do not edit previous posts. Just add a new message.

wiring diagram (the ethernet shield is connected to the arduino via pin headers) I hope you don't mind.

data sheets:
LM2596:datascheet
ADS1115:datascheet

What is the output of your LM2596? If it is 5V it should be connected to the 5V pin on the Nano. The Vin pin on the Nano goes to a 5V regulator and needs a minimum of about 7 volts.

output of LM2596 is 5V

There you go. Connect the output to V+ instead of Vin and see if it works better.

But 5V pin on arduino is for output, not for input. Or i got it wrong

See it now, I got it wrong. Thanks for all your help

The 5V is the output when the onboard regulator is being used. When the onboard regulator is not used it can be an input (which may be confusing). The 5V from the USB is connected to it through a diode to provide power when using the USB input. Schematic is here

...if you don't read the documentation or look at the schematic.