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So assuming that the battery is fully dead, is it possible to charge abattery rated at 42V 1A using 12V 1A charger.

I have seen similar questions that are almost using same volts or slightly higher, with different amperage but none that asks about keeping the amperage the same and using lesser volts. The purpose of voltage is to have a potential drop across the battery, what if the potential provided is lesser? Will it never reach the capacity of the battery (due to internal battery resistance) or will the voltage of the battery will never exceed 12V?

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    \$\begingroup\$ The simple answer is, "No. This won't work." The more complex answers will soon arrive, though. Regardless of the current ratings, themselves something needing further clarification, the DC voltage ratings are incompatible. \$\endgroup\$ – jonk Feb 23 at 20:30
  • \$\begingroup\$ Discharged battery has some potential on terminal. Internal resistance going up during discharge and then it so high, battery can hot provide sufficient current. If you measuring "0" with high resistance voltmeter, the battery is completely dead. Charge will not be taken. \$\endgroup\$ – user263983 Feb 23 at 20:35
  • \$\begingroup\$ What do you mean by a battery rated at 1A? Do you mean charging current of 1A, or capacity of 1 Ah? If you have a 42V battery, wouldn't it be already damaged if it had less than 12V? And do you mean a lithium battery, lead-acid or some other type? And do you mean a 12V 1A charger, or a 12V 1A power supply, which is not a charger and must never be connected to batteries? And no, the battery would never exceed 12V because the charger charges it to 12V and can't charge any more. \$\endgroup\$ – Justme Feb 23 at 20:35
  • \$\begingroup\$ @Justme this is hoverboard battery 42 volts and charge supplying power at 1A \$\endgroup\$ – BiologyEnthusiast Feb 23 at 20:39
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    \$\begingroup\$ @BiologyEnthusiast that brings no new information that is relevant. Try answering my questions. Edit the clarifications into the original post. \$\endgroup\$ – Justme Feb 23 at 20:53
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This is a bit like asking if you can fill a 42 foot high tank through a spout 12 feet off the ground. No. Your source has to be higher than the destination or you need a pump (a voltage booster in your case).

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  • \$\begingroup\$ ok will boost converter decrease the amperage from 1A down to 1/4A right while stepping up the volts from 12 to 42V right? \$\endgroup\$ – BiologyEnthusiast Feb 23 at 20:40
  • \$\begingroup\$ You have learned well! Yes the power available is 12 x 1 = 12 W. The power out will be P = P_IN} x Eff so if you found an 80% efficiency booster you'd only have 9.6 W available and 2.4 W to keep you warm. Max current would be 9.6 / 42. In general though there is more to it than this. The charger needs to be designed for the specific battery chemistry and have voltage and current limiting control to prevent damage, fire or explosion - depending again on the chemistry. Buy the right charger. \$\endgroup\$ – Transistor Feb 23 at 20:46
  • \$\begingroup\$ A boost converter would step up the voltage and therefore less current will be available. However, a boost converter is not a battery charger, so it can't be used to charge batteries. For that, you need a charger. \$\endgroup\$ – Justme Feb 23 at 20:57
  • \$\begingroup\$ @Justme, it makes sense, so as Transistor mentioned charger takes other consideration into account such as battery chemistry type, so heating is avoided etc. so this is what you meant that boost converter in not a battery charger right...ideally it will still charge correct? \$\endgroup\$ – BiologyEnthusiast Feb 23 at 21:02
  • \$\begingroup\$ Ideally, the boost converter will protect itself and the battery from overcurrent and damage. Or burn up. No, it will not charge. A boost converter would try to keep the voltage at the pre-set voltage and so it will try to push all the current it possibly can into the load, but it can't as the battey will keep voltage below setpoint and can sink a lot of current. Any sane boost converter would hit overcurrent or undervoltage limit and shut down. \$\endgroup\$ – Justme Feb 23 at 21:15

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