I need them to power a small handheld soldering iron, which had an old battery that died.
The original battery was NiCD 1200 mAh 2.4v
I am replacing them with 2 NiMH-520rs in series
First question: Will putting the two batteries in series, allow them to be charged with 2.4v or do i have to charge them with 1.2v?
Secound: Does it make any difference that the battery in the device was a NiCD and the ones im replacing it with are not?
Secound problem is that the wall charger is broken as well. When ever i try to draw power from it, it dies and outputs 0v no matter what.
It says on the back, that it supply 3v at 600 mAh
But when i hook it up to a multimeter it says 6v…
So my first thought was that, if it could convert it to be charged via USB 5v, i would solve the problem with the wall charger.
So my next question is how to transform or drop the voltage from 5v to something i can charge the batteries with…
Best Answer
Yes, you can charge the two cells as a series 2.4V pack. During charging the voltage will rise to ~3V.
The charger should be current limited, to avoid charging the battery too fast and overheating it. The circuit could simply be a resistor in series which limits current to a safe 'trickle' level (10 hour rate = 100mA for a 1000mAh battery). The resistor might be inside the charger (perhaps explaining why the voltage drops when you try to draw power from it) or built into the iron - then the 'charger' is just a DC power supply with high enough voltage to make up for loss in the resistor.
While you have the iron disassembled you could look for a current limiting circuit. If it doesn't have one (ie. the battery is connected directly to the charge socket) then you must use a charger which is current-limited. Don't try to use a DC power supply, even if its ratings appear to be the same.
Most USB ports will deliver 100mA without any negotiation, so to charge from USB you just need a resistor in series which limits current to <=100mA. Assuming the cells are 1.1V when flat, the resistor must drop 5V-2.2V = 2.8V. Calculating the value using Ohm's Law gives 2.8V/0.1A = 28Ω (the nearest preferred value of 27Ω should be close enough). The resistor will dissipate up to 2.8V*0.1A = 0.28 Watts, so it should be rated at 0.5W or higher. As the battery charges up the voltage difference will reduce causing charge current to drop, so to get a full charge you may have to leave it on for 12~14 hours.
The original Nicad battery probably had lower internal resistance and could hold a slightly higher voltage under load, so you may find the iron takes a bit longer to heat up. However NiMH has higher capacity than Nicad per volume, so you might considering using similar size NiMH cells that have higher capacity - rather than smaller cells which have higher internal resistance and may not last as long. Avoid ultra-high capacity AA cells, as these are optimized for capacity rather than power.