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im kind new to the arduino and im experiencing the following problem. Me and my friend build a circuit with a distance sensor which gets the distance and visualize it. The sensor requires a 5V power and i've connected it directly to the arduino. Now i want to connect a 10K potensiometer to get the analog value but i dont know if i can connect it at the same line.

enter image description here

As you can see i've connected the VCC to the 5V the ground and the echo cable. I've placed the potentiometer and i want to connect the 5V wire next to the sensor's 5V wire. Can i do it or im risking damaging the arduino? Thanks

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I can't see it, sorry. Consider posting a schematic instead. \$\endgroup\$ Commented Jul 7, 2014 at 1:23
  • \$\begingroup\$ Are you connecting both the potmeter and the distance sensor to the same Arduino input? \$\endgroup\$
    – RJR
    Commented Jul 7, 2014 at 1:34
  • \$\begingroup\$ To the same 5V input. \$\endgroup\$ Commented Jul 7, 2014 at 10:38

2 Answers 2

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Sure. You should be fine as long as you don't connect the 5v power rail of the arduino to ground or any external supply voltages.

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It isn't generally an issue to connect multiple devices to the Arduino's 5V connection. Mind however, the voltage regulator on an Uno can't source infinite current though.
I'd say make sure you draw less than 500mA or so. To be sure, look at the sensors specs, and use Ohm's law to calculate the current the potmeter draws (shouldn't be much, assuming you use a 10k or higher pot).
Note that maximum current draw depends on how you power the device. See the Uno documentation here, especially the 'Power' paragraph.

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  • \$\begingroup\$ the sensor is using 15mAh (from this datasheet: micropik.com/PDF/HCSR04.pdf) \$\endgroup\$ Commented Jul 7, 2014 at 12:12
  • \$\begingroup\$ 15 mAh? You're trying to charge a battery? \$\endgroup\$ Commented Jul 7, 2014 at 18:00
  • \$\begingroup\$ 15mA I assume - should not be an issue at all then. \$\endgroup\$
    – RJR
    Commented Jul 8, 2014 at 5:19

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