1

I was experimenting with external power supplies and noticed that the only way the Arduino can communicate with an externally powered sensor (through SPI) is when the GND pin under power AND the GND pin under AREF on the UNO is connected to the external ground?

From what I understand, both these pins are connected to the boards mutual ground (i.e. they are already internally connected)?

In the attached image both the orange and yellow wires were required and would not work individually.

I feel that either this shouldn't be the case or I am simply missing something and would appreciate an explanation.

Thanks!

Physical setup

2
  • 2
    Looks like orange is plugged into VIN to me... – Majenko Sep 13 '20 at 19:42
  • you may be misunderstanding why a common ground is required when communicating with external devices – jsotola Sep 13 '20 at 19:55
1

The circuit in the image shows the plan of the Arduino Uno, as you can see the GND pin under AREF and in the GND under the Power pin VCC are connected.enter image description here

you can double-check this with a tester and figure it out. so yes, they are internally connected.

now, in your breadboard you have connected GND to VIN, that's the problem, Vin is an input voltage that is regulated to supply 3.3V and 5V. enter image description here

So remember to share grounds with external voltage supplies (DC-DC), that set the circuit to the same reference.

4
  • I would think connecting VIN to ground would cause the power supply to fail. – Duncan C Sep 15 '20 at 16:49
  • yes, sir, it would. – joshua andres blanco jerez Sep 16 '20 at 17:36
  • ...And it might even let out the "magic smoke" that makes electronics work. – Duncan C Sep 16 '20 at 18:09
  • haha almost! but the MOSFET down there doesn't let the "magic smoke" come out. – joshua andres blanco jerez Sep 16 '20 at 23:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.