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Hi I am working on small project where I want to run ws2812b module via arduino and lipo battery. I have purchased 64 matrix for ws2812b from eBay and lipo battery with 500mah.

I am looking to find how I can calculate how much power does WS2818b modules will consume with all leds running and will the battery be sufficient for it. How long will it run on full charge.

Can someone please point me to the formula how I can calculate this. Also to note I am using a power up converter from 3.7 to 5v.

closed as off-topic by per1234, SDsolar, user31481, MatsK, jfpoilpret Nov 24 '17 at 20:28

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If all LEDs are in full bright white, assume 60mA per LED. 64 would mean a maximum current draw of 64*60mA=3840mA. This figured would not include the Arduino itself.

a 500mAh battery could power this for no more than 7.8 minutes (500mAh / 3840mA = 0.17 hours = 7.8 minutes)

If your usage will not be full bright 100% if the time, then you could run longer.

You didn’t say how long you want it to run. Perhaps start with the design goal before specifying parts.

For a more realistic view of how much power it takes, measure with a multimeter the current draw from the battery or bench power supply to your boost converter while displaying a typical usage on the matrix. Use calculations as above to determine the expected battery lifespan.

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    The battery capacity is rated at 3.7V and the LEDs run at 5V. You need to account for that. And can that small battery even provide almost 4A? – gre_gor Nov 21 '17 at 0:53
  • @gre_gor my arduino has 5v output pin can i not use it? As per answer above i need 3840mA to run all LED. Is it not possible will the battery get damaged? – Rahul Nov 21 '17 at 13:25
  • Whether your battery will run it for any amount of time depends on the battery specifications. What is the "discharge rate" of the battery? This is usually measured in "C", or multiples of the battery capacity. If you have a 5C discharge rate for a 500mAh battery, then you can discharge at 5*500mA, or 2500mA. What is the rated discharge rating of the battery? – jose can u c Nov 21 '17 at 13:43

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