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Can I use 11.1v battery to power a 12v component?

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Deeg

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This may be a stupid question but here goes: I'm working on a project that uses a remote RF relay board. Specs on the board are here (warning: pdf; see sixth page). It requires a 12v battery but draws very little current--6ma.

For environmental reasons I'd like to use a Li-ion battery but it's $*@# hard to find a 12v Li-ion battery pack. There are, however, plenty of 11.1v battery packs, like here and here. My question is can I use the 11v pack? I assume the time between charges will be shorter but is there a way to calculate how much shorter?
 
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RCinFLA

Well-Known Member
The 11.1 v battery sounds like three Li-Ion cells in series (3.7v per cell).

Fully charged it will be 4.2v times 3 or 12.6 vdc.

Answer depends on device you are trying to power and what its current draw demain versus size of battery (mA-Hrs).

For example, GSM cellphone typically has a single cell Li-Ion 750 mA-Hr battery. The maximum pulse current on the battery is close to 2 amps worse case (when you are far from a cell tower). Typically the phone circuitry has to work down to 3.2 to 2.9 vdc to account for the heavy peak current through the Rs of battery (0.15 to 0.35 ohms) dropping the voltage on the circuitry.

The larger the A-Hr the battery the lower its Rs. (series resistance). Rough number you can use for Li-Ion polymer batteries is 0.12 ohms for a 1000 mA-hr battery. A 2000 mA-Hr will be about 0.06 ohms Rs, a 500 mA-hr will be about 0.24 ohms. This does not include any safety circuit that may be in battery pack. This adds about 0.05 ohms. The simplest protection circuitry is a series MOSFET and voltage detector that opens the MOSFET when battery voltage drops below about 2.5 vdc. Discharging below this level will permanently damage cell. Charging a Li-Ion that was below 1.5 vdc can result in battery explosion. Charging above 4.2 vdc will also damage battery.

Rs of Li-Ion gradually increases for number of charge-discharge cycles. After 200-400 cycles the Rs can double to triple the original new battery value.
 
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