I have a DC power supply which outputs 21V 2.2A.

I want to use it to charge a car battery so I need to get the voltage down to 14V.

I have a 3.9Ohm 100W resistor but am am I correct that that would draw around 5A.

What resistor should I use to drop the voltage to 14V?

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    \$\begingroup\$ You cannot and should not charge a car battery this way. The voltage is too high (more than 13.8 V) so you'll overcharge the car battery and damage it. The resistor will not limit the voltage to 14 V when the battery is full. Also explosive gasses could escape. Car battery chargers are cheap, just get one of those. \$\endgroup\$ Commented Aug 12, 2020 at 12:39
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    \$\begingroup\$ Resistors don't work that way. At the very least you need a voltage regulator, which could be a single component or a circuit of multiple components (some of which may be resistors). But a single resistor will not do what you want. \$\endgroup\$
    – brhans
    Commented Aug 12, 2020 at 12:47
  • \$\begingroup\$ (21-12)/2A=4.5 Ohm. (21-11.5V)/3.9 = 2.43 A . overcurrent on low battery, manual cuttoff needs monitor in real-time. \$\endgroup\$ Commented Aug 12, 2020 at 12:57
  • \$\begingroup\$ Try to limit the charge current by connecting a car lamp or a combination of lamps between the power supply and the battery. For example, try with two 21 W lamps in parallel or two 60 W in series... or something else (as a bonus you will also have free lighting:) You can make simple calculations but keep in mind that the lamp resistance is nonlinear (it increases with temperature). I have done it many times in the past when I was dealing with cars... and not once did a battery explode... You can enlarge this idea with some else useful ballast load (e.g. a fan:) \$\endgroup\$ Commented Aug 12, 2020 at 13:26
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    \$\begingroup\$ I'm with @Bimpelrekkie. This is NOT a suitable application for a resistor, don't even try it. Or if you do, wear a full hazmat suit and eye protection so the battery acid doesn't permanently disfigure you. Dude, you can literally buy a "real" battery charger for like $10. Sometimes you just gotta suck it up and take the pre-engineered solution. \$\endgroup\$
    – Kyle B
    Commented Aug 12, 2020 at 19:14

1 Answer 1


Your understanding of the setup is wrong,

A single resistor will not help, all that can achieve is current limiting which would be I = V/R = 21V / 3.9R = 5.4A @ 21V. This would not change the voltage in the slightest.

You would need to create a voltage divider to drop the voltage, to do this you would need something like a 3.9R (R1) & ~6R (R2) to get around ~12.7V (Vout) with Vin @ 21V. This is not recommended as the battery will act as a load and will affect the division based on its internal resistance. See the image below, enter image description here

To do this correctly you would need a regulator to step down the voltage from 21V to 14V, by the time you get ones of these you would probably be better off buying a Car Battery Charger. In the UK it looks like something dirt cheap costs around £20 and looks like this. enter image description here

  • \$\begingroup\$ This would not change the voltage in the slightest - don't be too general about this. I know what you mean but you need to spell it out when connected to the battery. \$\endgroup\$
    – Andy aka
    Commented Aug 12, 2020 at 13:58
  • \$\begingroup\$ Well maybe slightly FOXSUR 12V 5A Pulse Repair Charger \$\endgroup\$ Commented Aug 12, 2020 at 14:02
  • \$\begingroup\$ Good point Andy, that was a bit to generalised. Obviously, I was describing the scenario before the described setup is connected to a battery. 21V through the 3.9R resistor not connected would measure as 21V. The moment you connect the other end of the 3.9R Resistor you are introducing a new resistance to the circuit (Quick google suggests that car batteries are around 1R) so now you would have R1 3.9R & R2 1R giving you in theory 4.285V over R2. Regardless of my abbreviated description this is not the solution. \$\endgroup\$ Commented Aug 12, 2020 at 14:04
  • \$\begingroup\$ car battery 1R? (not) if CCA = 500A at 7.5V then ESR=<0.01R = (12.5-7.5)/500, the FOXsur is a cheap solution to desulphation if done shortly after detected or as preventive in parallel with alternator. 5A is a narrow pulse at a rate which can breakdown lead sulphate crystals that raise ESR.. \$\endgroup\$ Commented Aug 12, 2020 at 19:05

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