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Paul is posting his results to discuss them with others and exchange ideas, which is what this place is all about.Do you have something to say about Paul's test Bill and Wilby? If yes please post and we can discuss it.MileHigh
The proof of excess can only be proven beyond doubt to be useful by using one charged cap and transfer charge to others ..then back to the original It sounds simple but we have to remove thermal loss.The measurements look fantastic but they must be of practical use.We know so little about the limits of our theoryWhat ever you do, start with one charged capNice to see you have opened up Paul Great work
Yo Paul,I donÂ´t see any solid data to support OU claims in this area yet, do you?Maybe the whole boost cap thing is just an advertisement to sell caps.I have vouched $500 toward the OU prize money, at the moment I think my money is safe lol. Of course nothing would please me more than to see Stefan validate GMs bodacious claim.With regard to your RC timings indicating varying capacitance, I agree you need to use higher constant charging currents and also over a wider voltage range from say 0.5V to 2V. Also use a stout 10 ohm resistor to time the discharge over a data logger or slowtimebase DSO to audit the caps energy.If, (and thats a big if), any magic is happening then I suspect short sharp pulse charging might encourage it.
Paul:I looked at the ADC0809. Is it set up to convert from 0 to 5 volts or did you change the Vref+ and possibly the Vref-? For the sake of argument let's assume assume that the A/D converter sweeps from 0 to 5 volts. 5/256 = 19.5 millivolts per digital step. In addition, the A/D error is +/- one least-significant bit.The numbers above don't add up so did you change the A/D sweep range?Here is my main point: The accuracy of the A/D conversion depends on the programmable A/D sweep range and the fact that the built-in ADC error is +/- one least-significant bit. Then this has to be related back to your very small delta-V which is typically 10 millivolts.It is possible that your ability to measure a 10 milivolt change with the ADC0809 is +/-3% or +/-40%, it all depends on the ADC sweep range and the size of the delta-V you are trying to measure. Can you clarify this issue?Just the inherent inaccuracy in your ability to measure a 10 millivolt change in voltage ADC0809 can explain your fluctuations in capacitor value calculations.What about your current source? Are you using a bench power supply in current source mode and running that through a multimeter on current measurement? What is your error margin here?I am just trying to understand your measurement setup.MileHigh
Paul:The above example is a hypothetical example and not from your data.You are wrong here. If the capacitance of the capacitor is dynamic and can somehow change with respect to time (to be determined what the speculated change mechanism is) then this will not be indicative of an increase in stored energy.
Yucca made the following statement wherein lies the answer:There is no mechanism in the experiment for there to be any excess charge. You have complete control over the amount of charge stored in the capacitor with your timed current source.If the capacitance of the setup is indeed dynamic with respect to time and increases, then this increasing capacitance will cause a decrease in the voltage across the capacitor. Your example above is incorrect, the voltage will NOT remain at 1 volts if somehow the capacitance increases tenfold. The voltage will simply decrease so that the amount of energy in the capacitor remains the same. This test looks like a dead-end to me.I will repeat my challenge to everyone again: Try to suggest an alternative method for the apparent increase in measured capacitance in Paul's data.MileHigh
Paul:Correct me if I am wrong, but your line of investigation is that you charge the cap slowly and measure the capacitance and voltage you know approximately how much energy is in the cap. Then if you discharge the cap quickly or in some other fashion and can deduce that the capacitance is larger, the you have shown OU.
What I am saying is that if the capacitance changes and the amount of charge on the capacitor remains the same, then the voltage on the cap will go down all by itself without any discharging. You will measure a larger capacitance with delta-Q/delta-V, but the voltage on the cap at that instant in time will have dropped. You will not get more out of the cap than you put into it by changing the discharge rates.
With respect to the apparent increased measurement of the capacitance, I don't think that is related to the temperature of the cap going up. The effective series resistance of the cap is very low, so my gut feel is that the temperature of the cap will rise marginally, but this will not affect the measured capacitance.
I can offer up a theory. I am not an expert in supercaps, I only read up on them a tiny bit and I was also skimming. Since the separation between the "plates" is so small (something like 20-40 microns) and the dielectric layer is some sort of flexible layering of molecules, I am going to assume that it is "spongy" based on your results. Whatever the dielectric layer is made up of, for sure it is some sort of flexible ultra-thin membrane.As the capacitor charges up to higher and higher voltages, the attraction between the "plates" of the capacitor gets higher and higher because of the opposite charges attracting. So as the capacitor voltage gets higher and higher there is an ever increasing attraction force and this squeezes the dielectric layer and makes it a tiny bit thinner. Suppose for the sake of argument that the dielectric layer separation goes from 40 microns thick to 35 microns thick when the capacitor is charged to its maximum voltage.This decrease in the thickness of the dielectric layer results in the supercap having a higher measured capacitance. That's my theory for your consideration. I would not be surprised with enough Google searching that you would find the true explanation and I may not be right, but at least on track.Note also that this is what your data is showing. The higher the voltage the higher the measured capacitance. My theory is based on your data and my limited knowledge about supercaps and good overall knowledge of electronics.For those that don't understand why decreasing the thickness of the dielectric layer increases the capacitance, Wikipedia awaits you.MileHigh
Hey Paul,A few more thoughts for your consideration. I will assume that your printer port/ADC chip data logging setup can plot your acquired data. I will also assume that your current source is a true constant current source, especially considering the low voltages that you are working with.If the capacitance does not change as you do a slow charge then the voltage vs. time plot should look like a straight line (you can set it to 45 degrees for example).If the capacitance does increase then the voltage vs. time plot should look like some sort of curved line with the slope decreasing.If you get this plot than you can test your thermal theory to see if it explains the phenomenon.It's a bit of a pain but if you put the setup near your sink and somehow arranged for a continuous slow flow of tap water with the cap sitting in the water, then you will remove any excess heat from the cap as the test runs. Let's assume that your tap water will be a constant temperature.So with the cap in a flowing "water jacket" you rerun the test. You know that the cap temp will pretty much remain constant and you can check out what the voltage vs. time plot looks like to see if the capacitance is changing or not.Finally I want to mention that Poynt gave a link about the issues involved for measuring the values of capacitors. The effective series resistance does come into play because the higher your current the more energy lost in charging or discharging. I think that the paper stated that the "safe" way to measure a supercap using the 37% method was to do a very slow discharge over hours using a relatively high resistor value. It was a very informative paper and you might want to find the link if you haven't read it.MileHigh
Paul, Will you be posting any new results today.cat