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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...

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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.

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  • \$\begingroup\$ The iron does not have any resistors, the battery is directly connected to a simple button and then the solder head. The charger it self. The Charger is broken, the circuit board is fried and the capacitor is blown. The charger is dead beyond repair. But i hooked up my arduino nano, with 5v to see if i could drop the voltage from 5v to 3.~ And now the problem seems to get the right resistors :) \$\endgroup\$ – Thue Iversen Jul 24 '15 at 18:42
  • \$\begingroup\$ I then put some different resistors together, and got 18 ohms and gets 2.95v, they are drawing 0.160A making the output current 0.5-0.16 = 0.340A So i think i solved it! \$\endgroup\$ – Thue Iversen Jul 24 '15 at 18:57

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