Hello,

I have built classic desulfator 12v circuit with external power source and have very old 12v car acid battery with easy each cell access (see picture) for testing and recovery.

Could somebody advise how to modify desulfator circuit to use it with external 12V DC power source and desulfating single 2v cells separately, to see how desulfating process is going for each cells.

It is obvious that 2v cell can not power 12v desulfator in classic circuit and vice versa it is also not good idea to connect constant 12v DC to the 2v cell in order to power on the desulfator.

attached is LTspice circuit I have built.

Please advise

 

You need to separate the back emf from the coil to feed it to the 2v cell,

 

L2 or L1?

more details please if possible how to apply this

 

I think I got it...

Please, could somebody check it

 

No, I think this one is the correct circuit

Now DC voltage applied to the cells will always be appropriate to the number of cells connected or total battery voltage i.e.

1 cell / 2V -> 12-10=2v
2 cell / 4V -> 12-8=4v
...
6 cell / 12v -> 12-12=0v

Thus desulfator powered from 12v DC external power source will act as "desulfator" only and not "charger" as well, still it will be possible to charge the particular cell(s) while desulfating them, by connecting additional external regulate DC voltage source directly to the cells in parallel to the desulfator.

Please, check the circuit and correct me if I am wrong.

 

Please, check the circuit and correct me if I am wrong.

 

looks ok , why do you need R3, in the negative line for the 555?

 

looks ok , why do you need R3, in the negative line for the 555?

to drive led (not shown) and 555 i.e. limit the current, comes from original circuit

thanks

 

Your circuit is wrong. You are dumping 4-5 amps continuously, into your battery, from your power supply, through L1 and L2. When the mosfet turns on it is suppose to charge L1 up from C1. Then when the mosfet opens up the inductive kickback from L1 is routed via D1 and C4 to the battery. The current through L1 has no significant change, so no pulse is generated.

Here is a link to the Lead acid desulfator forum that explains in detail how this circuit is suppose to work. (Written by your truly.)

http://leadacidbatterydesulfation.y...e-origional-kick-back-desulfator#.VWk55EZQDds

 

I have looked at this circuit for 20 minutes to see if there is a way to make it do what you want and I don't see it. The topology is wrong.

Also, even if you were to get it to work, it is too small of a desulfator to do anything to that huge battery.

Here is how it works,
small battery - small desulfator = slim to none
small battery - big desulfator = good chance of improvement
Big battery - small desulfator = you are wasting your time
Big battery - really really big desulfator = Now you are talking.

It takes a lot of energy to fix a battery, so, go big or go home.

People build that design and it is so small that it rarely works. So, now they think all desulfators are junk and snake oil. Of course, they didn't notice that the circuit fits into a sardine can and they have it hooked to a battery that weights 50 pounds. Anyone notice the disparity??

 

Your circuit is wrong. You are dumping 4-5 amps continuously, into your battery, from your power supply, through L1 and L2

Not true
Which circuit do you check from the list of attached to this thread? Check carefully 02.jpg and 03.jpg, L2 is no longer connected directly to the battery

Here is a link to the Lead acid desulfator forum that explains in detail how this circuit is suppose to work. (Written by your truly.)

http://leadacidbatterydesulfation.yuku.com/topic/1176/Optimizing-the-origional-kick-back-desulfator#.VWk55EZQDds

I read this article already any many others before build my desulfator


If you could help to modify classic circuit to perform separate cells desulfation do it please, all the rest how it works is well known already

 

I have looked at this circuit for 20 minutes to see if there is a way to make it do what you want and I don't see it. The topology is wrong.

Topology is based on classic circuit by Alastair Couper and see what he says in his article:

If you want to power the circuit from an auxiliary voltage source, so that the battery being
treated remains trickle charged, simply remove R3 and place 12 volts across C1.

This would also dump 4-5 amps continuously into your battery from the power supply, through L1 and L2

 

........ Check carefully 02.jpg and 03.jpg, L2 is no longer connected directly to the battery........

02.jpg and 03.jpg are two different circuits.

Lets look at 03.jpg.
The positives of the power supply and the battery are connected. The negative of the power supply and the battery are connected through L1, L2 and D1. The only current limiting is the resistance of the inductors, so L2 is going to get really hot really fast. Your own simulation shows a constant 4 to 5 amps of current being drawn through this path. Your own simulation shows the current decreasing, NOT INCREASING, during the pulse.

Topology is based on classic circuit by Alastair Couper and see what he says in his article........

The Cooper design is simple because the charging source and the "target" are the same device; the battery. What you are doing is different. You want to use a separate power supply to charge the circuit and discharge it through a different path to the battery.

.......I read this article already any many others before build my desulfator.....

You may have read it but you don't understand it. In order to get the energy out on the inductor, the field has to collapse. How can it collapse if it is being held in saturation by the 4 to 5 amps of charging current.

Again, look at your own simulations.

 

and what is the solution?

What about this circuit?

 

In the normal Cooper circuit the following is true:
L1, C4, D1 and the battery are in series. C4 is kept charged up to equal the battery voltage by L2. Because C4 is at the same voltage as the battery, no current flows, especially with D1 in the circuit. Now, when the mosfet turns off, The field of L1 collapses producing a current which adds to the voltage of C4. When that total voltage exceeds the battery voltage, plus the forward voltage of D1, current will flow into the battery. This is the desulfating pulse.

What you are trying to do is, keep C4 at 12 volts with a battery voltage of 2 volts. In this condition, D1 will always conduct and L1 will always have current flowing through it.

Simply put - a Cooper circuit will not permit you to use a supply voltage higher than the battery voltage.

There is a design on the lead acid desulfator forum called,"Direct Drive". This design, by its nature, uses a much higher supply voltage than the battery and will produce currents high enough to actually have a chance at revving your battery.

May I suggest the "Simple Direct Drive". There is a fellow on EBay that sells the PCB for this design.

http://www.ebay.com/itm/12-48V-Simple-Direct-Drive-Desulphator-Desulfator-PCB-/191170470658

 

How do you explain the following from Cooper:

If you want to power the circuit from an auxiliary voltage source, so that the battery being
treated remains trickle charged, simply remove R3 and place 12 volts across C1.

 

Where did that come from?

 

Where did that come from?

From Cooper

 

The operational word here is "trickle charge", as in low current, as in voltages very close to the battery voltage. The amount of energy you get out of an inductor is proportional to the change in flux. So, with a trickle charger, instead of going from a large flux to zero, you will go from a large flux to a lesser flux which results in a lower desulfating pulse. You can literally control the amplitude of the desulfating pulse by the amount of trickle charge current. More trickle current equals lower a desulfating pulse.
But In your case you want to go way beyond that. A 12 volt supply being applied across a 2 volt cell with only the resistance of the inductors to limit the current. Your own simulation shows about 4.5 amp DC across the inductors. First, that is more than enough to keep the inductors in saturation, resulting in no desulfating pulse, which your own simulation shows. And secondly, you are dissipating a total of about 45 watts, mainly in the 1mH inductor. It won't last long.

May I suggest a good experiment for you using LTspice. Simulate the original Couper desulfator circuit. Then connect a voltage source in series with a resistor in parallel with C4. Observe the desulfating pulse amplitude as you increase the voltage.

 

Thank you for your opinion, however I do not agree with your statements above, will be waiting for others...

There is LTspice circuit attached above, try yourself and check outputs from 04.jpg