Can I use a charger meant for lithium ion batteries (eg a charger for a drill) to charge a lead acid car battery. It charges at 14.4V which is what I'm looking for (and will limit to 2Ah with resistor if needed).

I'm starting to lose hope in finding a transformer to build a charger and wondering if the above is an option.



closed as unclear what you're asking by Leon Heller, JIm Dearden, Scott Seidman, PeterJ, Keelan Jan 18 '15 at 12:48

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  • \$\begingroup\$ 2 atta-Henries is not a meaningful unit of anything related to a battery charger. Even Ah doesn't make sense here. We do engineering here, which includes being careful with units. \$\endgroup\$ – Olin Lathrop Jan 17 '15 at 18:00
  • \$\begingroup\$ My apologies. I meant ampere, corrected aH to Ah in my question. \$\endgroup\$ – Gil Jan 17 '15 at 18:05
  • \$\begingroup\$ As I said, even Ah doesn't make sense here unless the charger actually measures the total charge delivered to the battery and shuts off after 7.2 kC (2 Ah). However, that is unlikely to be set by a resistor. Do you mean Amps perhaps? \$\endgroup\$ – Olin Lathrop Jan 17 '15 at 18:41
  • \$\begingroup\$ IF it is a 4S LiIon charger the battery is nominal 4x 3.6 = 14.4V BUT the charger will charge to a peak of 4.2 x 4 = 16.8V. SO follow it with a Constant voltage unit and it will charge to whatever CV you set. 13.7V is safe for floating a car battery, giving 2.4V headroom. A higher spec LM317 (or 2 in parallel) or and LM350 and 2 resistors will do this. The LiIon charger will charge at CC until your battery reaches CV . The battery current will reduce and the charger will proably terminate early but this is a start and will be mostly charged. Discuss further if desired. (An input R to the CV... \$\endgroup\$ – Russell McMahon Jan 17 '15 at 20:33
  • \$\begingroup\$ ... stage will hold the charger on indefinitely and give a full float charge. \$\endgroup\$ – Russell McMahon Jan 17 '15 at 20:33

Are you sure it charges at 14.4, or perhaps it is set for 4S lithium of the 3.6V nominal type, in which case it would charge to about 17ish volts.

Usually your biggest issue using a lipo charger for lead acid, is that the consumer type for electric tools and the such, will hopefully have safety mechanisms that can make it problematic to use for your application.

Also, you are unlikely to get a perfect match for battery pack voltage, so you need to do some modifications. Don't make the mistake of confusing the nominal voltage of any battery technology for the charge voltage.

Yeah, it will be unlikely to work. There are insufficient details to really answer your question, though.

  • \$\begingroup\$ Thanks. I think it would have them as the charger is for a heavy duty drill battery. I measure 13V with a multimeter when plugging in the charger, but once the battery is attached it drops to the battery's voltage (10.8). I believe a safety mechanism did kick in and stop the charging (the led also didn't turn on). I'll put up a question about the circuit inside, it looks fairly simple and hopefuly I can make it work :). It's got a serious transformer inside so I think it should be able to handle it. \$\endgroup\$ – Gil Jan 17 '15 at 23:48

You can charge discharged car battery with 14.4V but you'd have to monitor the battery and disconnect it from the charger when current to the battery drops. Don't leave charged battery connected to this charger though since the voltage is too large for trickle charging for lead acid (14.1 max, 13.8V better).

  • \$\begingroup\$ We charge 12V car batteries for solar arrays to 14.8V all the time. \$\endgroup\$ – Craigo Jan 17 '15 at 21:05
  • \$\begingroup\$ From what I read up, I should disconnect the charger once the 12V battery's Voltage reaches near the charger's Voltage. I plan to monitor it closely for V and temp, will not leave it unattended for long periods. \$\endgroup\$ – Gil Jan 17 '15 at 23:50
  • \$\begingroup\$ @Craigo For deep discharge use 14.8V is fine enougth - and you will also ideallu use t oppingt cycle as well. For float use 13.7V is more usual. What is needed depends on what he is doing, which needs specifying. \$\endgroup\$ – Russell McMahon Jan 18 '15 at 0:32

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