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Can someone post a formula / help me figure out how to figure out the battery run time for the following scenario?

I have a single 12v led light bulb with a draw of 510mA @ 12v and 1400mA @ 8v. If I start with a AGM 12v 80Ah battery and let it run for say 10 hours how do I figure out the run time given that the amp draw will rise as the battery is discharged? I know battery discharge is non-linear so I am not sure how to calculate.

Also it is worth noting that I plan on using 4 of the these light bulbs so ultimately I need to also calculate that in as well.

Battery Specs:

Product ID: 24M-XHD Cranking Amps: 1000 Cold Cranking Amps: 800 Voltage: 12 Termination: Common Code M Weight (lbs): 44.6 Width (in): 6.88 Length (in): 11.00 Height (in): 9.50 ReserveCapacity-25: 135.00 WET/DRY: W

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  • \$\begingroup\$ Battery datasheet says? \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 17 '16 at 22:07
  • \$\begingroup\$ There is no datasheet for the battery. I added the specs that were published. \$\endgroup\$ – Mr.SCW Nov 17 '16 at 22:11
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What you need is the battery capacity, usually expressed in Amp-hours. If you want to run something that takes 510 mA for 10 hours, then it will drain 5.1 Ah from the battery.

Any 12 V car battery will handily exceed that minimum rating. Car batteries don't like to be deep discharged, so plan to use only half the rated capacity. At that level, the voltage will be fairly close to 12 V the whole time, so you shouldn't have to worry about the 10 V draw.

In any case, are you really sure about this light taking 1.4 A at 10 V, but only 510 mA at 12 V? That sounds really odd and implausible, like maybe you read something wrong. This implies it uses 6.1 W at 12 V but 14 W at 10 V. Where do the extra 7.9 W go? If it has a switching power supply I could believe a little less efficiency at 10 V compared to 12 V, but not such a whopping change with just a little less input voltage. Something isn't right in what you're telling us.

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  • \$\begingroup\$ I ran the bulb off my DC power supply. And adjusted the voltage to see the current draw at different voltages. I don't want to destroy the bulb to find out what is inside but my gut says there might be some circuity to boost the voltage back to 12v. \$\endgroup\$ – Mr.SCW Nov 17 '16 at 23:27
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As noted the current drawn by your LED bulbs seems to indicate something active inside.

But your battery (as far as I see) is rated at 75 Ah. Fully charged it will have a terminal voltage around 13.4 V and fully discharged you should not go below about 10V. Assuming the 0.5 A discharge rate is correct then to consume 75 Ah would take 150 hours. If you run 4 bulbs you'd expect 37.5 hours. The discharge curve for the battery is non-linear, but if you plot the bulb current from 10 V to 13.4 and take an middle point, you should not be far off.

I'd measure your lamps on your power supply again, you should measure a sharp knee below which the current (and light output) drops dramatically. If from the knee the current goes up linearly with applied voltage then you don't have active components in the bulb. If the current goes down as you increase the voltage then you may have some type of buck/boost regulator circuit in the bulb. ...though it seems a rare beast indeed.

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  • \$\begingroup\$ Amp-Henries is not a unit of battery capacity. \$\endgroup\$ – Olin Lathrop Nov 18 '16 at 11:59
  • \$\begingroup\$ You are absolutely right Amp-Henries does not apply here....the unit is Ampere-Hours and IS a unit of battery capacity. en.wikipedia.org/wiki/Ampere_hour \$\endgroup\$ – Jack Creasey Nov 18 '16 at 15:21
  • \$\begingroup\$ If you are actually saying I should have typed Ah instead of AH.....then say so (you are correct)....don't distort a comment just to look smart. \$\endgroup\$ – Jack Creasey Nov 18 '16 at 15:23
  • \$\begingroup\$ You wrote "AH" twice, so it's not just a typo. You also consistently left out the space between numbers and their units. We do engineering here, where sloppiness with units is not tolerated. If you don't like being called on it, then next time be more careful and lose the "eh, screw you all, the standards don't apply to me" attitude with important technical content. Despite your complaining and acknowledging the error, I see you still haven't fixed it. I didn't downvote, and gave you a chance to fix it. My patience is running out. \$\endgroup\$ – Olin Lathrop Nov 18 '16 at 16:28
  • \$\begingroup\$ You are absolutely right, I fixed the error. Can you now get down off your high horse? Anyone with any grace at all would have simply pointed out the error as ...it should be Ah and not AH....since in context it could not have meant Ampere-Henries. \$\endgroup\$ – Jack Creasey Nov 18 '16 at 17:25

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