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I am trying to run a sensor from my switch using POE. I have created a temp/humidity sensor using the MySensors directions https://www.mysensors.org/about/arduino. I decided to use a nano due to the USB connector. I have connected this to my switch via a TP-Link POE splitter set to 5 volt output.

When connected, the switch provides power for a second or two then turns off for 10 seconds or so before coming on again. I believe the Switch is testing the circuit not detecting enough power being drawn and shuts down. It then goes through the cycle again. Note: both the power LED on the switch and the nano LEDS light up while power is supplied. I believe the nano itself only uses about 20mA

I have tested with another device that requires 9volts (hand held phone) via the splitter and it works fine with power provided consistently.

I was wondering if there is some circuitry I can add before the Arduino that will draw a little more current but still provide a constant 5V to the nano.

Additional information. The switch is a Linksys LGS124P. I have tried to find more detail regarding the required power but can only find this datasheet that does not provide that level of detail http://downloads.linksys.com/downloads/datasheet/1224699456700/25685_LINKSYS_unmanaged_POE_switches_PROD_Revised.pdf

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    First you need to work out exactly how much current needs to be drawn in order to make it stable. For that you really want an adjustable dummy load with current display. When you know how much current you can calculate the right size of resistor to use across the supply to draw the amount of current without wasting current or overheating and burning.
    – Majenko ♦
    Aug 6, 2017 at 9:37
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    You may alternatively find the specification in the manual for the switch, but since you don't tell us what the switch is we can't help you there. Also this is not an Arduino question but a general electronics question. It should be on EE.SE.
    – Majenko ♦
    Aug 6, 2017 at 9:39

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A resistor or a light bulb in parallel with the Arduino will increase the power consumption.

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  • Combination of this and Majenko's comments give me the answer I need. thanks
    – C Ellis
    Aug 8, 2017 at 10:33

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