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I am making a hand held logger with a 12v lead acid battery inside. I want the user to be able to recharge by plugging unit into the car cigarette lighter. What is a simple way to limit the amount of charge the hand held battery receives from the car? Do I need to limit the charge? The hand held battery is similar to a camcorder battery (12V 2.0 Ah) Thanks.

You need to limit the current rush only. The battery voltages and charge voltages are the same, but a large resistor of 100 watts or so, 5 or ten ohms in value, will stop the large battery and wires from attempting to rush in and charge the small battery too quickly. Wires tend to melt when over loaded. <p>Other than that, they are a matched pair as far as their both lead acid, and 12 volts, and require the same conditions.<p>You could also place a large light bulb between the two batteries charging the battery through the bulb, as a regulator, of around the size of the amp hour rate of charge of the small battery.<p>You will need at least a twenty or more watt bulb to do the job unless you want to trickle charge the battery.<p>[ February 13, 2003: Message edited by: Chris Smith ]</p>

A battery charging regulator can also shut off when it compares voltage of the battery being charged to a reference and also the temperature of the battery in the event the battery no longer is cooperating. I do not recommend using only a resistor, the light bulb idea is inexpensive 40 watt 110VAC type measure 26 ohms the temp curve 3/2 for tungsten.<p>[ February 13, 2003: Message edited by: dim bulb ]</p>

I agree with Chris. If the power outlet on the car is used for charging only, then you could use a smaller (less wattage) higher resistance (like the 10 Chris suggests, albeit at a lesser wattage). The data logger draw in this state is the unknown which must be accounted for. The following is based on a 0 mA draw at the data logger circuitry in the charge (quiescent) mode. Please advise if this is not the case.<p>The rule of thumb is to never charge a battery continuously at over 10% of its' amp hour rating (trickle charge principle). Working at 50% voltage as a design point (the battery voltage when EXTREMELY exhausted) would yield a potential of 7 volts accross the "R" isolator. Using 200 mA (trickle current limit) that works out to be 14 Ohms, multiplying the short circuit (i.e. 0 volts on the output or shorted battery) current of 1 Amp times 14 volts would yield 14 Watts power disipation on the resistor (worst case). A 25 Watt 10 Ohm or 15 Ohm resistor should work pretty well.

Rick, you boofed the resistor calculation. It comes out to 35 ohms, 5.6 watts. You could use 33 ohms, 10 watts.<p>Ron

Thanks for all the advice. It sounds fairly simple.