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I have

  • 8, 5 watt LEDs that run from a 12VDC source.

  • 3, 18650 batteries at 3000mAh each.

I want to know the approx run time of the LEDs.

Thanks.

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Your load is 8*5=40W. Your source has 3A-h*12V = 36 W-h. In ideal case your LEDs will shine for 60*36/40 = 54 min = 0.9hr.

In reality, a 3,000 mAh battery has 3.7 * 3 *3 = 33W-h at best, since the average discharge voltage per Li-Ion cell is about 3.7V.

More, in best case the LEDs must be driven by switching type drivers, likely each drives two typical LEDS in series, and 4 chains. Switching-mode (the best!) will likely have 85-90% efficiency in 12:8 conversion mode, so overall your LEDs will run 60 * 33/40 *0.85 = 42 min. If you have passive ballast resistors to drive the LEDs, it will be quite less than 40 min.

The above assumes that the LEDs are driven to truly 5W each.

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  • \$\begingroup\$ Lights have run for 45 minutes. Battery voltage is now at 11.93 volts. Looks like it will run far more than 54 minutes. \$\endgroup\$ – fixit7 Apr 21 at 11:38
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    \$\begingroup\$ @fixit7, then it looks like you run your lights at far less than 5W each. How do you know that it is 5 W? \$\endgroup\$ – Ale..chenski Apr 21 at 14:00
  • \$\begingroup\$ ebay.com/itm/… \$\endgroup\$ – fixit7 Apr 22 at 11:27
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    \$\begingroup\$ @fixit7, so, what is the actual power draw by your lamps? \$\endgroup\$ – Ale..chenski Apr 23 at 19:15
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    \$\begingroup\$ @fixit7 "Current used is 87 milliamps for one led" - then 0.087 x 12V = 1W. Not 5W. Multiply my answer by 5 then. 33 W-h / 8W = ~ 4 hours. \$\endgroup\$ – Ale..chenski Apr 24 at 16:09
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P = IV relates power, voltage and current, rearranged:

Led current = P/V = 5W/12A 0.4A/LED

Battery capacity of 9AH,

9AH/0.4A gives the time for one LED Divided by 8 = 22.5H/8

Should be about 2 hours and 45 minutes.

Please correct me if I am wrong, I need to know!

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    \$\begingroup\$ I would guess the cells are in series, so the capacity would only be 3 Ah. \$\endgroup\$ – Oskar Skog Apr 21 at 5:36
  • \$\begingroup\$ @Callum I think Callum's calculation will be closer. \$\endgroup\$ – fixit7 Apr 21 at 11:39
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    \$\begingroup\$ The mistake in the above calculations is that a battery of 3 cells in series will have the amper-hour capacity as one cell, 3 Ah. Not 9 Ah. \$\endgroup\$ – Ale..chenski Apr 21 at 14:12

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