Dr. Pavel Imris thesis about the "conservation of energy"

[conradelektro]

So the energy from the capacitor cannot cause the normal recharging of the battery due to the reversed polarity, in this respect the battery is a load which however does not show dissipated heat. 

Gyula, you are right, I did not understand the circuit presented in this patent application correctly.

So, my explanation is wrong. The mystery remains, or the author of the patent application is not telling the truth.

May be I try it. I have a rather small led acid battery (5 Ah) and some old big electrolytic capacitors (25 V, 10.000 µF). I suspect that the switches create huge sparks and eventually burn through.

I will and can not do the temperature measurement. I just want to see whether the cap, the switches or the wires become hot as Gyula suggests. If they become hot, the patent application is nonsense.

Greetings, Conrad

[gyulasun]

Hi Conrad,

Okay, and in case you tend to test the circuit and happen to have 4 high current MOSFETs (or bipolar power transistors) of the same type, with at least 25-30 Amper drain (or collector) current ratings, then you may wish to use a H-bridge to do the switching.
A H-bridge just does exactly what the double switch in the patent application does: changes the polarity of the input supply voltage across the load at its output. The load here would be the capacitor and the supply voltage to the H-bridge would be your small lead acid battery.
Notice that electrolytic capacitors are not suitable in this circuit because of the polarity change across the capacitor. An AC type or a non-polar electrolytic would need to be used.
Alternatively, you may use two electrolytic caps in series (with 'steering' diodes in parallel with each) to protect them from the opposite polarity, see a possible solution: http://i.stack.imgur.com/UxfjU.png from this link: http://electronics.stackexchange.com/questions/21928/can-you-make-a-non-polar-electrolytic-capacitor-out-of-two-regular-electrolytic ).
Of course the resulting capacitor would be only half the value of any of the two identical caps. The current rating for the diodes should also be in the 20-30 A range, basically the peak current would only be limited in the circuit shown by the contact resistances of the switches and by the inner resistance of the battery and by the ESR of the capacitor(s). Estimating the sum of all these resistances to be under say 1 Ohm, from a 12V battery the current would be a minimum of 12 A or higher.

If someone is not familiar with H-bridges, here are some circuits:
http://www.talkingelectronics.com/projects/H-Bridge/H-Bridge-1.html 

Of course you can use a mechanical relay with two independent on-off contact pairs (double Morse if I recall correctly). Sparking would surely occur at the contacts but may not be as severe as would be with coil switching.
The use of heavy wires to handle the high peak currents is a must.

Gyula

[conradelektro]

Gyula,

combining two electrolytic capacitors is a neat idea. I also like the H-bridge for switching polarity.

I did a simple test with a 10 µF AC capacitor (see the attached photo).

- I touched the two poles of the battery with the leads of the capacitor. There was a spark.

- Then I turned the capacitor around (switching polarity) and touched the two poles of the battery again. There was again a spark.

And of course, one can do that over an over again.

I did not observe anything strange, all was as expected.

Before building an H-bridge and doing longer tests I will try some calculations:

I seems to be common knowledge that halve of the charge in the capacitor is lost in the resistance of the lead wires and the inner resistance of the capacitor (when charging it). I want to calculate or estimate the heat generated by that. Then I want to calculate the heat generated if a 5 Ah 12 battery is discharged through a 1 Ohm resistor.

I have a question for the experts:

Let's say the capacitor is charged to 12 V. And then when it is turned around (changing polarity) and connected to the 12 V battery, will there be needed the double charge to recharge it from -12 V to 12 V in comparison to charging it from 0 V to 12 V?

Greetings, Conrad

[gyulasun]

Hi Conrad,

Thanks for doing the simple test.

I am going to attempt an answer to your question on whether a double charge is needed?

Well, I think that the moment the charged up capacitor is switched onto the battery with reversed polarity, the capacitor would be the 'generator' and would pump current into the inner resistance of the battery. So the direction of the current flow would be towards the battery till the moment all the charge from the capacitor is "consumed".
So during this discharge process of the capacitor I do not think the battery would need to lose charge, at least not significant charge like the capacitor has just lost.

And just after this discharge process will the capacitor be able to take up the normal charge with the polarity equal to that of the battery and this now obviously consumes energy from the battery till the voltage level across the capacitor reaches that of the battery, the usual capacitor charge-up process takes place. 
And then the process starts again by operating the double switch again.

Though I may be wrong, this is how I think and this would mean that a double charge would not be taken from the battery. I am not good in solving chemical formulas which may be needed to consider possible effects what the appearance of the opposite polarity voltage source would have on the materials of the electrodes and the acid liquid or gel in the battery. Of course the chemical formulas and the events taking place during the charge-up of a battery from a normal polarity source are known.

Regarding your wish to estimate the generated heat, is your LCR meter able to measure the ESR of capacitors? because that would be a good approach to know the loss in your capacitor. Of course a good estimation of the inner resistance of your 5 Ah 12 V battery would also be needed (by known resistor load tests) so that the max peak currents could be figured in advance before you consider the switching components of a H-bridge.

Maybe a series 1 Ohm or 0.1 Ohm current shunt resistor inserted in series with say the battery negative and monitor the voltage drop across it with your scope would be enough to evalute possible losses in the battery and the capacitor (and in the switches of the H-bridge) once the series loss resistances are known for them with a good approximation. 

Gyula

[conradelektro]

I did some research into applying a current with the wrong polarity to a lead acid battery:

Over-charging a lead acid battery is equivalent to applying the current with the wrong polarity and causes the formation of lead sulphate (as during charging), but this lead sulphate will be mainly hard lead sulphate which can not be reversed when discharging the battery.

So, when the capacitor in the patent is witched to the wrong polarity and connected to the battery the same chemical process happens as when charging the battery . This "charging" happens from -12 V till 0 Volt, then the capacitor will be charged to + 12 V.

It follows, that the energy stored in the capacitor during charging will be consumed "as charging energy" in the battery and will produce little heat.

What happens in the patent:

- Step 1: When the capacitor is charged from 0 V to +12 V, halve of the energy is lost as heat in the wires, the battery and the capacitor, the other halve is stored in the capacitor as charge.

- Step 2: When the capacitor "is turned around" and connected to the battery with the wrong polarity, this halve energy is consumed by the battery in the chemical process which generates lead-sulphate (but "hard lead sulphate" which is lost). This happens when the capacitor discharges into the battery from -12 V to 0V. Then from 0 V to 12 V step 1 happens again.

Note: step 2 does not charge the battery, but depletes it further because the created lead-sulphate is lost (it sinks to the bottom of the battery and can not be consumed as charge when drawing energy from the battery).

Conclusion: a lot of energy is lost in the chemical process which happens when "over-charging" the battery (creation of hard lead-sulphate) by connecting the "turned around capacitor" with the wrong polarity to the battery. This seems to be the mystery of the strange patent.

Greetings, Conrad