I have a PIR motion sensor and looking at its specifications it lists the following:

  1. Voltage: 12V DC

  2. Standby current: < 3uA

  3. Output current: 5A

I am unsure of the term 'output current.'

If I'm not mistaken the outputted current will depend on the resistance of the load, so how can the current be stated?



1 Answer 1


The output current rating would be the amount of current the output of the sensor can safely source or sink. So you need to avoid attaching loads that would cause currents greater than 5A to flow in the output of the sensor.

Since you did not link a data sheet it is not possible to explain more.

  • \$\begingroup\$ Thank you! That makes sense. I wish I had more specifications but I bought it online and details on the webpage were scarce and no data sheet came with it. Thanks again. \$\endgroup\$
    – user122374
    Sep 10, 2016 at 12:06
  • 1
    \$\begingroup\$ Ok then. Without data sheet you are going to be hindered by not knowing if the sensor output is a current sourcing type, current sink, isolated output, contact closure between two pins, voltage rating of output plus others. \$\endgroup\$ Sep 10, 2016 at 12:11
  • \$\begingroup\$ I believe it outputs the voltage provided by the battery as the reading I got from my voltmeter read almost 9V (I was using a 9V battery). When it detects motion it clicks quite loudly so would that mean it uses a contact closure mechanism? And if so, would that give any additional clues to its functionality? \$\endgroup\$
    – user122374
    Sep 10, 2016 at 12:19
  • \$\begingroup\$ It may give you a clue. Not to me because I do not have it in hand to test and check. BTW....if it says the part operates on 12V why are you trying to use it on a 9V source??? \$\endgroup\$ Sep 10, 2016 at 12:22
  • \$\begingroup\$ Because I had a 9V battery on my table...I assumed it wouldn't be a problem, that I would just get a lower output voltage. Did I overlook something? It wasn't hooked up to anything special, just an LED to see if the thing worked. \$\endgroup\$
    – user122374
    Sep 10, 2016 at 12:25

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