0
\$\begingroup\$

Today I took an old computer power supply, stripped the yellow +12 V and the black ground, and twisted half of them together each into 4 separate pairs.

I then took a lead-acid battery, which seemed to have no fluid left, and tested it to have 0 V. I then filled it with diluted* battery acid and tested it immediately above 12 V.

I connected the red+ terminal of the battery to my unplugged +12 V PSU and the black- terminal to the ground of the same, still unplugged, PSU.

It began to spin the fans of the PSU and I could tell it was being powered. I reversed the direction to place the black- to the +12 V PSU and the red+ to the black ground of the PSU, still unplugged. I saw a small whiff of smoke come from the PSU and removed the connection as quickly as I could.

Neither of these results is what I expected, my anticipation was that I should be connecting red+ to the +12 V and the black- to the ground in order to charge the battery, but it did reach 12 V immediately after being filled.

My question is - what happened here? Was there an excess of latent electrical charge on the plates and the battery acid was enough to reinvigorate them without any further charging? Was there such an excess of electricity that it was flowing "against stream" in a way since there was no current flowing when I connected the terminals? Do I just not know anything about how electricity flows?

I appreciate any elaboration you can provide this novice.

Edit: I should note that I also twisted together the green and ground on the computer PSU in order to indicate an "on" status for it. It was unplugged from the electrical supply the entire experiment though.

*Diluted battery acid using distilled water and sulfuric acid. It arrived this way already. The packaging is uncertain but reviews indicate ~30-35% sulfuric acid by volume so mostly just distilled water.

\$\endgroup\$
8
  • 1
    \$\begingroup\$ why does your post title say recharging sealed lead-acid battery with recycled computer PSU? ... you never did any charging ... and the body of your post is asking about reviving a battery by adding distilled battery acid, whatever that is \$\endgroup\$
    – jsotola
    Sep 13, 2020 at 6:31
  • \$\begingroup\$ I meant diluted** battery acid = sulfuric acid and distilled water. My intention was to charge the battery by using a recycled computer PSU as a DC power source. I was about to plug it into the house's electrical and the intention was to use the +12V line from the DC source to charge the 12V battery, however it seemed to possess a charge as soon as I refilled it which boggled me. \$\endgroup\$ Sep 13, 2020 at 6:43
  • \$\begingroup\$ Mostly I am unsure why the power supply turned on at that point, it seemed to me that the current was flowing backwards and I was afraid to plug in the electrical source. Would I do damage to anything with this configuration? \$\endgroup\$ Sep 13, 2020 at 6:47
  • 2
    \$\begingroup\$ Battery clearly could back-feed current into the power supply. Good chance that the fan is just directly connected between yellow and black wires to get 12V. Please note that computer power supply is not a battery charger, it is not designed to be used for charging batteries, and thus it should not be used to charge batteries or connect to batteries. Then you finally killed the PSU by connecting the battery in reverse. \$\endgroup\$
    – Justme
    Sep 13, 2020 at 7:19
  • \$\begingroup\$ What exactly were you trying to accomplish here? \$\endgroup\$
    – copper.hat
    Sep 13, 2020 at 7:34

4 Answers 4

4
\$\begingroup\$

When you fill the battery, it will start showing a voltage across its poles; this voltage will push a current into your PSU and power parts of it. Most PSUs don't like this very much.

If you connect things wrong, a large current may flow into the PSU and fry it.

Also, the PSU will generate a voltage of 12 V, and will only be able to charge a (nominally) 12 V battery that has been discharged to below 12 V. When charging, the battery's voltage will (try to) rise very quickly to more than 12 V, and the charging will stop. At this point, very little charge will have gone into the battery and it will only be charged partially.

So, a 12 V PSU is not suited for charging (nominally) 12 V lead-acid batteries. These batteries should be charged with a current source until their voltage reached about 13 to 14 V, see battery specification.

\$\endgroup\$
3
  • \$\begingroup\$ I also have a +3.3V Rail that I'm not using - Could I introduce a resistor and bridge these 2 rails somehow without shorting out the PSU? When I tried that on my bench it detects the short and just shuts off right away. I've got hundreds of old power supplies laying around but not much money for a real battery charger. \$\endgroup\$ Sep 13, 2020 at 16:12
  • 2
    \$\begingroup\$ You can't bridge two rails in the same supply, because their grounds are the same. \$\endgroup\$
    – mguima
    Sep 14, 2020 at 0:32
  • 2
    \$\begingroup\$ I've deleted 3 comments here from 2 people (neither being ocrdu). You've both read what the other said and the comments now are better absent (imho). \$\endgroup\$
    – Russell McMahon
    Sep 14, 2020 at 10:47
3
\$\begingroup\$
  1. the fan started working because it's connected to the 12V rail. so either the power supply provides the supply for the fan, or [in your case] you provide the 12V suppy for fan. the power supply itself didn't turn on, it was just an external fan and it has nothing to do with the power supply's main circuitry.
  2. there are many parts that can be damaged because of the reverse voltage. (e.g. switching ICs on the low side, mosfets ...)
\$\endgroup\$
4
  • \$\begingroup\$ So what you're saying is that I should have ignored the fan and continued with supplying an electrical source to the PSU because everything was just fine? \$\endgroup\$ Sep 13, 2020 at 7:16
  • \$\begingroup\$ @TechnoNewbie no. you should power up the PS and wait for a few seconds, then connect the battery to it. and that aside, a battery charging unit in between should be used to maintain currents to a resonable value for prolonging battery life or preventing unexpected damages to PW. \$\endgroup\$ Sep 13, 2020 at 7:46
  • 2
    \$\begingroup\$ @TechnoNewbie No, you should have stopped and investigated, researched and contemplated what was occurring. Science isn't doing a bunch of experiments, it's doing 1 experiment and then finding out why the outcome occurred. \$\endgroup\$ Sep 13, 2020 at 17:31
  • \$\begingroup\$ Harper, that is, infact, what I'm doing here. As soon as the result wasn't in my realm of expectation I stopped. \$\endgroup\$ Sep 13, 2020 at 21:05
1
\$\begingroup\$

Before attempting to charge a lead-acid battery, we have to understand its construction and working. A typical cell of a lead-acid battery consists of plates, separators, Electrolyte (dilute Sulphuric acid), and an outer case of hard Rubber or Plastic. The positive plates are made of lead dioxide (PbO2) and the negative Plates are of spongy lead (Pb). The separators are of porous insulating material which allows the electrolyte to flow freely among the plates. When there is no electrolyte ( ie when the battery is dry) it will only show 0 volts across the positive and negative terminals as was observed by you.

A typical cell has an open-circuit voltage of around 2.1 Volt. Its capacity (in Ah) depends on the number of plates and their size. A 12-volt battery consists of 6 such cells and its open-circuit voltage will be around 12.6 volts. During discharge due to chemical reaction (PbO2+Pb+ 2H2SO4= 2PbSO4 + 2H2O), sulphate gets deposited on both the plates, and water is released and therefore during discharge specific gravity of electrolyte and voltage goes on falling. During charging exactly opposite reaction take place and therefore specific gravity (SG) of the electrolyte and voltage goes on rising. During charging when the SG remains constant for 2 to 3 hours battery is said to be fully charged. Typically when a cell is fully charged its voltage is around 2.5 volts with the charger in on position (for some batteries it goes to even 2.75 Volts per cell). Therefore, a charger should be able to supply at least 15 volts in order to drive the charging current for fully charging the 12-volt battery. I hope it is clear as to why the 12-volt power supply of your old computer will not charge the battery.

\$\endgroup\$
0
\$\begingroup\$

You should have used a 1n4007 diode had you put a diode on the positive side power would have only been allowed to flow in one direction. I've never put fluid into an empty battery before so can't help with that. There's another thing here as well to charge a sealed lead acid you need a voltage of around I think it's 13.5 to 14 volts because a fully charged battery is 13.5 volts. I don't think I would ever try charging a battery with a desktop PSU never mind a DIY PSU. I just got into sealed lead acid a few weeks ago.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ The diode would have helped to not push current in reverse from battery into computer supply and prevent damage. However, if the computer power supply is then turned on, it would try to output 12V minus diode drop into battery, but an empty battery would have limited the voltage to around 11.8V, and the power supply would try to push a lot of amps to battery via 1A diode to try to raise the voltage to 12V, which the battery prevents. Thus the power supply would just likely go into overcurrent or undervoltage shutdown to protect itself and any load and wiring, as it is not a battery charger. \$\endgroup\$
    – Justme
    May 16 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.