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I am an mid experienced Programmer but I have no clue about electronic circuits. I want my Arduino to do sth if he gets a signal from an external source. The external source has 5V and is 0V or 5V and when it has 5V I want my Arduino to send me a Message, the Code for the Message is no problem, but please help me with the circuit: The external source is independent and runs on battery and the device has a 2.5mm jack, normally there is an LED attached that lights up, but I want my Arduino to receive the signal. Are there any opportunities for realising this? Thank you for your help

asked May 3, 2019 at 20:00
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  • its ground to the uno's ground, it's 5v to one of the uno's GPIO pin. Commented May 3, 2019 at 20:01
  • You may need an optocoupler if for whatever reason you cannot safely share grounds. Commented May 3, 2019 at 20:08
  • majenko.co.uk/blog/importance-sharing-grounds Commented May 3, 2019 at 20:08
  • what is "the external source"? Commented May 4, 2019 at 4:01
  • the external source is an indicator that sends a 5V signal via a 2.5mm mono jack, I think this just has +&- so no ground. But I only need to measure if it is on or of, not the intensity Commented May 4, 2019 at 7:20

2 Answers 2

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Unless there are exceptional circumstances you can just connect the two ground connections together. Then the 5V of your other circuit will be the same voltage as the 5V of the Arduino and you can just connect your signal direct to an input.

However, if there are exceptional circumstances (ground isn't really ground, for example) then you can use an opto-coupler to act like an optically-controlled button:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 can be replaced by the internal pullup resistor of the Arduino by setting the pinMode to INPUT_PULLUP.

Note that this inverts the logic of the input: a HIGH input signal is detected on the Arduino as a LOW logic level, and vice versa.

answered May 3, 2019 at 21:24
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the external source is an indicator that sends a 5V signal via a 2.5mm mono jack, I think this just has +&- so no ground. But I only need to measure if it is on or off, not the intensity

answered May 4, 2019 at 11:03

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