Electrical – Lead acid battery cutoff voltage circuit

Battery heating can be roughly defined using its internal resistance. We can get a rough estimate of a battery's internal resistance using it's cold cranking amp rating. This is rated at -18C, which is close to -20C.

For a nominal 12V battery: $$ R = \frac{12V - 7.2V}{CCA} $$

Let's assume this is constant vs. current draw.

Let's assume at time t=0, the battery is completely insulated from the world and has a heat capacity similar to pure water.

The amount of energy required to heat 17 kg of water from say even -20C to -18C is: $$ E = 4.2 \frac{J}{g K} \cdot 17 kg \cdot (20^oC - 18^oC) = 142.8 kJ $$

For a 9A draw, this would take:

$$ t = \frac{E}{(9A)^2 \cdot R} $$

Suppose you have a 200 CCA battery. Then we can find t = 73 seconds.

Problems with this analysis: Obviously the full weight of the battery is not just the liquid inside. Thus the time required to heat the battery 2 degrees C would be less. However, a larger problem is that you probably don't have a perfectly insulated battery. The 9A draw only produces ~2W, which could easily be lost from the battery (the cables are extremely good conductors away from the battery).

At temperatures much colder you're going to have to start worrying about fluids freezing, thus many things could just stop working. You're much better off following proper car winterization guide.

You don't say what you use the batteries for, and if you are running your motor at full speed all the time. It's often better to have more voltage than you need and then modulate or regulate the current to run over a longer period of time. Then as the battery voltage decays you can go longer without recharging. A d.c. motor will rotate at a speed creating a counter voltage that equals the source voltage. Often motors are under some form of feedback, like a servo amplifier. I have controlled high inductance motors that put power back into the control amplifier. That extra energy needs to be put somewhere when slowing down the motor - it acts like a generator. Extra energy needs to be released as heat, or put back into the powering battery. You don't need any extra voltage as long as you can run at the speed you desire. Don't forget that voltage times current equals force, so can you usefully use extra torque? Do you mind accelerating faster than you have before? The insulation on a d.c. motor is at least twice the rated voltage, usually much more. The constraints have more to do with force, top speed and control. Lighting a lamp with overvoltage will burn it out sooner, but that is not usually the main constraint in motors.