I'm using a Luminus SST-90-W57S-F11-GN200 with this > http://www.ebay.com/itm/Constant-Cu...Parts_Accessories&hash=item4ac1f58536&vxp=mtr < driver. Looking for the longest run time possible (2-3 hours would be perfect) for a can light, but am having some trouble figuring out what the power draw is going to be (and therefore a battery pack). I'm looking at NiMH and Li-ion packs right now, but again I don't know whats required. Any advice?
Battery Help
- Thread starter MrDelsic
- Start date
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DIWdiver
Flashlight Enthusiast
If it actually puts out 9A (it may be somewhat under that), the output power would be about 9A * 3.6V = 32.4W. That's the voltage times current in the LED.
I would expect that driver to have an efficiency of 80-85% (or 0.8 - 0.85), so the input power could be as high as 32.4/0.8 = 40.5W.
If you want it to run for 3 hours, you need a battery with a capacity of 40.5W * 3H = 121.5WH. This is not to be confused with the AH (or mAH) rating of the battery, which is properly termed ampacity. To keep it simple however, the term capacity is used almost universally for both the WH and AH ratings ;-)
The WH capacity of a battery is the AH rating times the voltage. Since watts=volts*amps, WH = V * AH.
In all cell chemistries, the voltage drops as the cell is discharged. The effect is more dramatic in some chemistries than others. So the "voltage" to use in the conversion of AH to WH is somewhat arbitrary. A healthy LiIon cell under moderate load will average around 3.7V, so this is a number that's pretty commonly used. For NiMH, 1.1V is pretty good estimate, though the specified voltage of NiMH battery packs almost always use 1.2V for the cell voltage. That's pretty optimistic unless loads are very light. Likewise some LiIon cells and packs are specified as 4.2V per cell. That's the MAX charging voltage, and for all practical purposes, the voltage will always be less than that under load.
Hope that helps!
Oh, by the way, welcome to the forum!
I would expect that driver to have an efficiency of 80-85% (or 0.8 - 0.85), so the input power could be as high as 32.4/0.8 = 40.5W.
If you want it to run for 3 hours, you need a battery with a capacity of 40.5W * 3H = 121.5WH. This is not to be confused with the AH (or mAH) rating of the battery, which is properly termed ampacity. To keep it simple however, the term capacity is used almost universally for both the WH and AH ratings ;-)
The WH capacity of a battery is the AH rating times the voltage. Since watts=volts*amps, WH = V * AH.
In all cell chemistries, the voltage drops as the cell is discharged. The effect is more dramatic in some chemistries than others. So the "voltage" to use in the conversion of AH to WH is somewhat arbitrary. A healthy LiIon cell under moderate load will average around 3.7V, so this is a number that's pretty commonly used. For NiMH, 1.1V is pretty good estimate, though the specified voltage of NiMH battery packs almost always use 1.2V for the cell voltage. That's pretty optimistic unless loads are very light. Likewise some LiIon cells and packs are specified as 4.2V per cell. That's the MAX charging voltage, and for all practical purposes, the voltage will always be less than that under load.
Hope that helps!
Oh, by the way, welcome to the forum!
Packhorse
Flashlight Enthusiast
What DIW said...except the SST90's i used had a higher Vf at 9 amp. Maybe 4 volt?
Thanks for the explanation!
I've found a battery I like but it's good for a max current draw of 7A. Will this be fine, or do I need to find something that can supply 9A?
I've found a battery I like but it's good for a max current draw of 7A. Will this be fine, or do I need to find something that can supply 9A?
DIWdiver
Flashlight Enthusiast
Depends on the voltage. That driver is a buck type, so the input current will be lower than the output current. Iin = Vout/Vin * Iout / Efficiency. Efficiency is probably between 80 and 90% for that one (more likely at the low end of that range).
As you can see from the equation, if you raise the input voltage, the input current goes down.
The battery pack has a low end cutoff of 11V, so you'd be able to get 77W from that pack.
Recalculating using 4V from Packhorse, you'd get 45W max input. Should be fine.
As you can see from the equation, if you raise the input voltage, the input current goes down.
The battery pack has a low end cutoff of 11V, so you'd be able to get 77W from that pack.
Recalculating using 4V from Packhorse, you'd get 45W max input. Should be fine.
