gamezawy
Newly Enlightened
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- Mar 27, 2013
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can a 3x ( 2500ma 18650 battery ) connected in series (12.6V) give me 5000ma if i used them with a 5000ma buck driver ? or the maximum i can get is the 2500ma that every cell gives ?
confused about battery in series and current draw please help
- Thread starter gamezawy
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mattheww50
Flashlight Enthusiast
You are confusing two different terms. The driver can deliver 5000ma (whether it really can from 3 x 18650's will depend upon the driver design). That is current. The 2500ma of the cell is really 2500milli-amp-hours. It is a measure of the capacity of the cell, and says nothing about the current that the cell can deliver. The 2500mAh cell capacity means the cell can delivery 2500mA for 1 hour. In theory it could delivery 5000ma for half an hour, but unless you using an IMR 18650, it is doubtful you could safely draw 5000ma from most 18650's.
Having said that, the inpput of the driver is probably not 12.6 volts (for that matter at realistic current, it is probably closer to 11 volts under load). Volts x amps= power, so the output
from the driver will be the driver output voltage x the 5000ma. If that is 3.5 volts, than the driver needs 17.5 watts. If the driver is reasonably well designed, it should not take much more than
17.5 watts input, which at 11 volts is about 1.6 amps, and well within the capability of just about any 18650 battery.
Having said that, the inpput of the driver is probably not 12.6 volts (for that matter at realistic current, it is probably closer to 11 volts under load). Volts x amps= power, so the output
from the driver will be the driver output voltage x the 5000ma. If that is 3.5 volts, than the driver needs 17.5 watts. If the driver is reasonably well designed, it should not take much more than
17.5 watts input, which at 11 volts is about 1.6 amps, and well within the capability of just about any 18650 battery.
gamezawy
Newly Enlightened
- Joined
- Mar 27, 2013
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how much ? sorry i am kinda very confused
Not accounting for losses in the driver, but for the most part, watts in == watts out.
If you have an LED running at 1000mA with a forward voltage of about 3.2v, you have 3.2W (watts = volts * amps). 3.2W draw at the LED creates 3.2W draw at the batteries. If you have a pair of lipo cells at say 3.8v under load, you take the watt draw and divide by the pack voltage. In this case, you divide 3.2W by 7.6v. That gives you 421mA current draw from the cells.
If you have an LED running at 1000mA with a forward voltage of about 3.2v, you have 3.2W (watts = volts * amps). 3.2W draw at the LED creates 3.2W draw at the batteries. If you have a pair of lipo cells at say 3.8v under load, you take the watt draw and divide by the pack voltage. In this case, you divide 3.2W by 7.6v. That gives you 421mA current draw from the cells.
