A Battery Rejuvenator

[rvcrawford]

J.J.

That math is correct but only if the 100W light bulb is energized with it's rated voltage. The tungsten filament has low resistance at room temperature and increases as it heats.
I just measured a 65W bulb at 16.5 ohms. If it stayed at that resistance it would draw 7.3 amps and would burn 873 watts not 65 watts.

I have no idea how much resistance a 100 watt light bulb would have when connected in series with a half wave rectifier and the impedance of the battery but my guess is that it's less than 144 ohms.

This has been a very interesting thread and I'm going build me one and try it on an old boat battery that I have.

Thanks
Roger

 

[J_J]

Probably, some of you test a battery charger by touching the clips together to see a spark, and if you do, you assume that it is working.

Have any of you came across a battery charger that seem to not put out any voltage? Well I did and thought the charger that I purchased was bad/broke.

A friend of mine showed me something interesting that I did not know. Apparently some battery chargers need some voltage to start the charging process.

The charger reads no voltage not connected to battery, and when connected to battery, it is putting out about 14 V.

Anyway, didn't know if anybody had came across one of these chargers, and thought it was no good, but try connecting the clips on the battery and see if there is any charging voltage.

 

[dcyrilc]

A 100 w bulb, using 120 v AV, will have a current of .83333 amps, and a resistive load 144 ohms.

Three light bulbs will develop 300 W, and use 2.5 amps.

In this circuit, a half wave rectifier is used to rectify the AS voltage to 60 v DC.

So 60 v divided by 144 ohms, is .41 amps through each leg, three legs will draw 1.23 amps, and three bulbs will develop 221 W, and the bulbs will not be as bright.

I believe this correct. If you find mistakes, please correct. It's been a while since messing with electronics circuits

If you short the diode, then you are putting 120 V AC at 2.5 amps on the battery.

Now that I am actually awake and have thought about it. It should be 120V pulsating DC. On a 120V AC circuit the sign wave goes from 120V+ to 120V- referenced to 0V (neutral). The diode will cut off either the + or - half of the sign wave depending on orientation. I believe the purpose of the diode is to prevent the reversing of current through the battery during the second half of the AC cycle.

 

[J_J]

Rectifier circuits : DIODES AND RECTIFIERS

 

[dcyrilc]

Rectifier circuits : DIODES AND RECTIFIERS

Full-wave rectifier, center-tapped design.

This circuit's operation is easily understood one half-cycle at a time. Consider the first half-cycle, when the source voltage polarity is positive (+) on top and negative (-) on bottom. At this time, only the top diode is conducting; the bottom diode is blocking current, and the load sees the first half of the sine wave, positive on top and negative on bottom. Only the top half of the transformer's secondary winding carries current during this half-cycle.


This is effectively what we are doing and only using the first half as 120V AC in the home is one side of a 240V center tapped transformer with neutral being the center tap. Our 1/2 voltage is 120V because of the center tap.

 

[ModMech]

Have any of you came across a battery charger that seem to not put out any voltage? Well I did and thought the charger that I purchased was bad/broke.

A friend of mine showed me something interesting that I did not know. Apparently some battery chargers need some voltage to start the charging process.

The charger reads no voltage not connected to battery, and when connected to battery, it is putting out about 14 V.

My Schumacher is that way, unless the battery has about 1/2 charge voltage, it won't even try

 

[ETpilot]

Well, on the other forum the discussion has turned to this is a potentially deadly thing. They deleted the post on the circuit. I told them to just delete the entire thread.

So this is the only thread on the subject. I have found 2 maybe 3 batteries. Two are old and one has a little life. I am going to test them to see if it is true that they would explode. Sure don't need that. No where except that thread has a possible explosion been mentioned.

I just have to test this theory and see if it is true. So everyone needs to use caution with this.

A battery does have the potential to explode. Mostly it is poor handling and having something sparking nearby. They should be charged in a well ventilated area away from any sparks.

 

[mmurphy]

(removed), you are making me hungry,might get me something to eat before Spot comes around!!!!!:laughing:

 

[greasemonkeyok]

Couldn't you put two diodes in series in the circuit to double your safety factor?

 

[patrick_g]

You should state that the diode is what rectifies the AC voltage, into DC voltage. The purpose of the light bulbs is to regulate the current. If you short the diode, you are applying 120 volts AC to the DC battery, and it could cause some damage.

The diode is only using 1/2 of the AC cycle, and if the voltage is 120 AC, then the diode will rectify that to 60 v DC

Therefore, 60 volts is going through the single 100 w bulb, and will pass or limit the current to .506 ma, and is used as a current limiter.

So essentially, what you are doing is applying a higher charging voltage to the battery.

What happens to all that sulfate that falls to the bottom, and shorts out the cell?

Right on regarding the sulfate. Your math skills are OK but...

Tungsten filament light bulbs demonstrate a very non linear resistance under varying currents/heating so your application of Ohm's law, however well intentioned is severely flawed.

There is no effective filtering (except the battery.) The effective voltage of the applied sine wave (half wave rectified is not 60 volts. The 60 volts you talk about is off a bit. The battery dc voltage is "BUCKING" at what ever its current state of charge happens to be (say 10 volts) so the bulb will light as if connected to something more like 50 volts.

After rejuvenation, fully charge the battery, pour out the electrolyte, flush all cells with distilled water, filter the electrolyte through acid proof filter material and put it back into the battery. The battery will probably last considerably longer than just rejuvenation and ignoring the sulphate build up.

Long life cells are built with extra room at the bottom so sulphate build up takes longer to be a problem.

Battery minders that keep your battery fully charged but will not charge a depleted battery nor hurt a fully charged one over time are often "on sale" for less than $5 at Harbor Freight. I have at least 4-5 and no problems in several years. If you don't like HF you can get these for $20-$30 at parts stores, Wally World, etc (still made in China!) I personally know of none that have failed and they are popular around here. There use extends battery life and gives convenience of full charged when needed. They do not "boil off" electrolyte.

Pat