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Sunday, January 17, 2016

Battery Desulfator - Capacitive Voltage Divider

A few years ago, I posted this video showing my first crude battery desulfator.


This works like a champ, but I wanted to point out there is a few ways to make this safer. The one in the video had no protection and it put out up to 170 volts pulsed DC at 1.1 amps. This one MUST be connected to a battery when you turned the power on.

I wanted to make it safer and limit the voltage to about 55 volts. This is done without a transformer and uses just run-type capacitors. You could also use a transformer, but, hey...this is just interesting. Besides, you can swap out capacitors to get different currents. Also, a transformer isn't constant-current like this circuit.

Update - the fuse should be on the hot side.

The top capacitor basically controls the amperage. This one would be 1.1 amps at 120 VAC at 60hz. The bottom controls the voltage. If the bottom capacitor is rated the same as the top, then the output will be cut in half. In this case, there is 120 VAC coming in and the output will be one-third of the input, or 40 volts AC. Actual DC is 1.4 times the AC value, so the full-wave bridge's output will be 56 VDC pulsed at 120 times per second.

Here is a real-world example. Let's say you have lots of 25 MFD run-type capacitors that you picked up from the local HVAC shop. You could use 5 of them as shown below. This gives you one fifth the input voltage that goes to the rectifier.  If you start with 120 VAC, this gives 24 VAC to the rectifier. Multiply that times 1.4 and you get 33.6 VDC. This example could charge any lead-acid battery up to and including 24 volts with no problem. It would do it at 1.1 amps.
Update - the fuse should be on the hot side.

IMPORTANT: This is not a smart charger. It will keep trying to charge a battery and it is like the terminator, it will not give up. You have to monitor the battery voltage and fluid levels. You don't want to overcharge or dry out the cells. OR, you could use an Arduino to monitor the voltage for you and switch this circuit off with a relay.



8 comments:

  1. hi Richard
    am now using 45mf capacitor , its give now a 11>>12 volt on battery ?why >without battery its give 275 dc v . the battery volt without charging is 11 v

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    1. Yossef, you are using just a single capacitor so you are limiting the current. So, under load, the voltage will pick a point that allows the set current to flow. For instance, if I had capacitor that allowed 1 amp to flow from 120 VAC source, and my load was a 10 ohm resistor, then the voltage across the resistor would measure at 10 volts (Ohm's law). If you have no load or a very high resistance, such as the air, then the voltage tries to max out to achieve that set current of 1 amp. But, if you only have 120VAC, that is what it will show (because there is no load applied). In actuality, 120 VAC is actually peaks of about 170 volts. Just take your AC value and multiply by 1.4. In your case, it is 200 VAC or so, just divide by 1.4 to get the peaks in DC..275 VDC in your case. I don't recommend that setup for safety reasons. Please build my new version that limits the voltage when no load is applied.

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  2. Richard,
    As per your diagram, I wired 5 -50 uf capacitors. I thought I was onto something as opposed to my previous configuration. A jumper wire that I had connected to my multimeter probe had a meltdown so I figured there must be greater current going thru. But on 2/6v batteries in series it started out as a 9.14v charge and progressively dropped down within a few hours to 8.24v. Will heavily sulfated batteries waver on the charge rate? Go up, drop down? What is the amp output on 50 uf? 2 amps? I am guessing even with 5/ 50uf capacitors, 2--450 ah batteries, if badly sulfated will take days if not weeks to come around.

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    1. Jiffer, if you are on 120 VAC running at 60 hz, then 50 MFD is just over 2 amps. Sulfation will increase internal resistance. As you blow off some of the sulfation, you are lowering resistance. To get the 2 amps to flow with lower resistance, voltage has to lower, initially. After some time the battery voltage should start rising again as can't continue to desulfate, but starts to actually charge.

      So, the charge rate isn't changing, it is still 2 amps. Just the voltage is changing to keep the 2 amps constant. I say 2 amps, but it is closer to 2.5 amps.

      At 2.5 amps, assuming no sulfation and just charging dead batteries, it would take 7.5 days. After about a week, you will have to watch this closely. The 2.5 amps might be too at that point and you might have to keep adding water or just switch to a set of lower value capacitors. I like to run it hot for a few days though. It really mixes the electrolyte. Just watch it closely.

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  3. Richard

    So, in essence, what is the basic effect of having multiple capacitors in parallel?

    If one 50uf capacitor with 120v /60hz input can output 170v/2a
    and with my configuration of 5/50uf capacitors having the current controlled by the first capacitor and the voltage reduced by 1/5th

    120v/5= 24v
    24vx 1.4= 33.6v

    33.6v  charge rate as you stated above, 2a in my case,
    ( if my 50uf setup figures the same as your example of 25uf capacitors),

    does that mean that the one capacitor is charging at a higher rate than the multiple setup? Increasing the mfd input doesn't change the amperage output and decreases the volts, as per your shown configuration? Does a multiple setup increase the pulsating factor?

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    1. Jiffer, the 50 MFD will allow almost 2.5 amps, while the 25 MFD will be half that. The multiple setup is just for safety to limit the voltage. And the pulsating factor doesn't change at all between the two configs.
      If you need more amps and don't want to limit the voltage, you can go back to my original design and just put all 5 in parallel. That will give you about 12 amps. But, it is very unsafe because it puts out 170 VDC if it isn't loaded by the battery.

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  4. Hello..
    I have a doubt.. How much power this charger will consume to charge a 12V battery with 10A? As per my calculation it will consume approximately 2 units per hour.

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    1. Nikhil, I can't detect any heat loss, so extremely efficient. At 14 volts and 10 amps, that would be 140 watts of power. I usually charge batteries at 1 to 2 amps, so closer to 10 to 30 watts.

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