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I have a DC power supply from either a WIFI router or printer (not sure) rated for 13.5V @ 1.2A output.

This is a replacement charger for an electric ride on car as it was originally 6V and I've upgraded to a 12Vbattery and system.

Without load, the charger puts out about 14.8V and roughly 14.6V when plugged into the original DC plug to charge the SLA battery.

I'd like to utilise this charger but require a voltage reduction to charge the battery. I'd also prefer 13.5V charging voltage so I don't need to watch the battery temperature all the time, but am concerned about heat dissipation for that kind of voltage drop.

So:

  1. What method is best to reduce the voltage, and/or the most cost effective way? I'd like to stay away from a buck converter if possible. I've also looked into using resistors or a potentiometer as a voltage divider.
  2. What specifically will a resistor affect my charging current, and how much will it drop?
  3. In terms of heat, is it even viable to use a resistor to drop to 13.5V?

Further info can be provided if needed.

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  • \$\begingroup\$ You can't use a resistor to drop voltage reliably in a battery charging application. You need something 'better'. Is it a standard lead-acid battery like most ride-on toys??? \$\endgroup\$
    – Kyle B
    Aug 17 '20 at 5:29
  • \$\begingroup\$ You know you can buy a purpose-made automatic battery charger for like $10-15 if you shop around a bit. \$\endgroup\$
    – Kyle B
    Aug 17 '20 at 5:31
  • \$\begingroup\$ Cheers for the info. It is a standard lead acid and the best price I can find for an automatic charger is $22 AUD on eBay. That will be my last resort. \$\endgroup\$
    – BlueBell
    Aug 17 '20 at 8:29
  • \$\begingroup\$ One thing to consider is storage. If you get one of those automatic chargers, you can leave the battery connected "all the time". That will prevent over-discharge from damaging the cells (sure you know, if you leave a lead-acid battery on a shelf for a few months, when you come back it'll be ruined). The automatic chargers can be left on for a 'maintenance' charge. Wreck one battery and you'll spend alot more than $22 to replace it \$\endgroup\$
    – Kyle B
    Aug 17 '20 at 13:37
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You want to lose about 1.1 to 1.3V from the charger voltage. A resistor will only work if the current is constant, as you need to use the current to calculate the resistance you need. R = V/I where V is the voltage you want to drop and I is the current.

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A simpler alternative would be two silicon rectifier diodes in series. Each diode drops about 0.6V, giving 1.2V drop in total. Make sure you get ones rated for the current you need, perhaps 3A to allow a bit of safety margin.

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    \$\begingroup\$ That sounds like a great solution! I'll steer away from resistors. What you said makes a lot of sense as i have wondered what would happen when the battery starts to top off. I also don't know much about diodes and the different types, but will just make sure i look at datasheets before i buy so i get the right ones. I'm assuming these won't affect current though? \$\endgroup\$
    – BlueBell
    Aug 17 '20 at 8:44
  • \$\begingroup\$ Diodes won't affect current. \$\endgroup\$
    – Kyle B
    Aug 17 '20 at 13:34
  • \$\begingroup\$ @BlueBell the diodes don't limit the current. But you need to make sure the ones you get can handle the maximum current you will put through them. The common small rectifier diodes (1N4007, etc.) are only rated for 1A. \$\endgroup\$
    – Simon B
    Aug 17 '20 at 14:05
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    \$\begingroup\$ Do note though: even diodes have 0V voltage drop at 0A current. So you might still manage to overcharge your battery with 2 diodes in series, if nothing is using enough current to maintain the voltage drop. \$\endgroup\$
    – user253751
    Aug 17 '20 at 14:27
  • \$\begingroup\$ @Simon B, I have just gotten around to testing the voltage drop of 2 different diodes I have on hand (1N5401 And 1N5404, both rated for 3A). The first one drops 0.26v per diode on my 12v battery and 0.3v drop for the 2nd one. However, I conducted a no load test with these on the charger and the voltage rose! My '13.5v' rated power supply put out About 98v with the 1N5401 and about 20v with the other one. Why did it rise so much? Due to no load on the power supply? \$\endgroup\$
    – BlueBell
    Aug 18 '20 at 10:17

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