Arduino "Phantom Powering"

Question

I am using an Arduino Uno in my project. I recently found that if I disconnect the power from my Arduino (I am powering it by USB port, from a 5V regulator - 7805), then it remains under power from the input pins. I also found that if I disconnect the 3 digit 7 segment common anode display (which is also powered from same 7805) and if I remove the LM358 (which is again powered from the same 7805) from the socket, then my Arduino is not powered anymore. My questions are:

1. Is my setup correct - if I am powering it from the 5V regulator (7805) through the USB port ?
2. My Arduino and the other components from the schematic - the display, the LM358, the mosfet - will be powered on at the same time, when the 230V switch that is placed on the mains input of the 24V SMPS is closed. Could this be a problem in terms of "phantom powering" ? Is there any better way to power the Arduino when I have 5V regulated supply on the schematic ? I will always keep connected the Arduino to the 5V output of the 7805 to the USB port.

Please find attached 2 schematics, one is the main PCB (the one with Arduino) and the other schematic is the one with the display, which is placed on the front panel. The J2 connector from first schematic is connected pin to pin to the J2 from second schematic.

https://ibb.co/QnLHsM0

https://ibb.co/KxLFKsQ

Answer 1

Is my setup correct - if I am powering it from the 5V regulator (7805) through the USB port ?

Yes.

My Arduino and the other components from the schematic - the display, the LM358, the mosfet - will be powered on at the same time, when the 230V switch that is placed on the mains input of the 24V SMPS is closed. Could this be a problem in terms of "phantom powering" ? Is there any better way to power the Arduino when I have 5V regulated supply on the schematic ? I will always keep connected the Arduino to the 5V output of the 7805 to the USB port.

Not really.

Your problem only occurs when you "break" the system by separating the 5V from the Arduino - at which point current flows from the output of the op-amp and through the LED displays, through the IO pins, up through the clamping diodes inside the IO pins, and to the Arduino's 5V.

Since it only happens if you disconnect the 5V and you don't intend to disconnect the 5V then it's not a problem. However if you wanted to improve it then you could take the output of the 5V reguator and feed that direct into the Arduino's USB port. You could then power all the other devices that run at 5V from the Arduino's 5V pin. That way if you remove the 5V from the Arduino then you also remove it from all the other 5V devices that potentially feed power back in.

Answer 2

When you disconnect the arduino from 5V you must also remove either

• disconnect any other device that is connected to the arduino and has its own power supply
• or disconnect every device that is connected to the arduino from its power supply

When you disconnect 5V from the arduino while devices connected to a gpio are still powered, this gpio might see a voltage larger than zero. But the voltage at any pin must not exceed the arduino's actual supply voltage by more than 0.3V. Otherwise the atmega's supply rail will be pulled high through esd protection diodes and the atmega will try to operate. It will then probably draw more current than what the esd diodes are rated for and they will eventually fail (destroying that gpio).