I have a NiCad battery that I got from an old rc car.

I have some trouble charging it. the battery is rated 6v and the charger's voltage is rated 7.25 (why is it bigger)?

however, I measured the charger's real voltage and it is 12v! after a night of charging, the battery mesured voltage is 7. 2v.

Is everything normal here?

  • \$\begingroup\$ You charge batteries at a higher voltage than nominal so that it charges at faster than glacial speeds. 6V is the nominal voltage, and it is normal for the fully charged voltage to be higher than that. NiCd batteries are 1.2V nominal cell voltage and 1.4V full charge iirc. That would be 6V nominal for a 5 cell battery and 7V fully charged. this explains the 12V pretty well: robotshop.com/media/files/PDF/… \$\endgroup\$ – I. Wolfe Jan 11 '16 at 23:03
  • \$\begingroup\$ but 12v is far too big, right? (it's the official charger, but it is quite old, 10years, so maybe broken...) will the battery explode or release gas if the charger is broken? \$\endgroup\$ – n0tis Jan 11 '16 at 23:04
  • \$\begingroup\$ it is probably an unregulated power supply, and those will show a higher than rated voltage if you measure it open circuit. I would bet it reads 7.25V when the battery is connected (also if it really charged at 12V your battery would be quite dead after charging overnight). tl;dr - looks normal to me \$\endgroup\$ – I. Wolfe Jan 11 '16 at 23:06

NiCd batteries are charged at constant current. (Lead-acid batteries are charged at constant voltage.)

An easy way to make a constant current charger is to use a relatively high voltage and add in a series resistor to limit the current to a safe value.


simulate this circuit – Schematic created using CircuitLab

It's generally safe to charge NiCd batteries indefinitely at 1/16 of their Ah rating. For example, if we had a 1.6 Ah battery pack we could charge at 1.6/16 = 0.1 A. Putting a 60Ω resistor in series with the 12 V source would limit the current to 0.1 A when the battery is at 6 V. It will decrease a little as the voltage rises (good) but it would be 0.2 A if the battery were completely flat.

You can work out the power wasted and the required power rating in R1 using \$P = I^2 \cdot R\$.


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