Howdy all
I have just been doodling with some figures and I think I may have stumbled over something.
No doubt it has been seen before but if someone has been down this road and come unstuck then just let me know thanks.
I am using unrealistic values of capacitance to make the calcs easier.
Here goes
I take three capacitors of 3.333 farads each and charge them to 6 volts in paralell.
I effectively have one capacitor of 10 farads charged to 6 volts.
The joules of energy stored is 180 joules acording to E=(C*(V^2))/2.
If I now put the capacitors in series I effectively have one capacitor of 3.333 farads charged to 18 volts.
Acording to the formula for energy this is 539 joules of energy.
Not taking into account switching energy or ohmic losses etc etc etc this is approximately 300% energy gain.
Surely I am missing somethig here.
What is it I am missing.
Thanks fo the help
Hi,
If you connect 3 capacitors of 3.333F each in series, you get an equivalent of 1.111F, ok? (NOT 3.333F what you considered).
That is, you get back your input energy you invested in charging up the 3 parallel 3.333F capacitors. So no free lunch here, unfortunately.
rgds, Gyula
gyulasun said:
Quote
...If you connect 3 capacitors of 3.333F each in series, you get an equivalent of 1.111F, ok? (NOT 3.333F what you considered).
www.allaboutcircuits.com/vol_1/chpt_13/4.html
The 'Web page above has the concept down right. Capacitors hold a charge, not resist electron flow like a coil or resistor.
Suppose you had several capacitors with an equal number of relays sitting around having two sets of contacts, one normally open and one normally closed.
Use one set to charge the capacitors in parallel from a power supply. All the capacitors will have the same voltage.
Wire the other set to discharge in series when the first set is opposirely opened/closed. All voltages in the capacitor string will add with their combined charge(s), right?
I have in storage in another part of my home town a U.S. patent that describes the same thing using electrical components only. That's why I kept a copy. Elegant and clever. I can't get to it without hours of unstacking and searching, though. Unfortunately.
--Lee
the_big_m_in_ok
Hi Lee,
Do you probably mean this patent, US 7085123 ? ( see a free pdf copy here: http://www.pat2pdf.org/ )
This patent's circuit seems at least to double the working time of a battery, this sounds like a COP of two... However the power needed for operating the switches is not included in the test? (though it could be made very small)
rgds, Gyula
Quote from: the_big_m_in_ok on May 26, 2009, 03:43:00 PM
gyulasun said:
www.allaboutcircuits.com/vol_1/chpt_13/4.html
The 'Web page above has the concept down right. Capacitors hold a charge, not resist electron flow like a coil or resistor.
Suppose you had several capacitors with an equal number of relays sitting around having two sets of contacts, one normally open and one normally closed.
Use one set to charge the capacitors in parallel from a power supply. All the capacitors will have the same voltage.
Wire the other set to discharge in series when the first set is opposirely opened/closed. All voltages in the capacitor string will add with their combined charge(s), right?
I have in storage in another part of my home town a U.S. patent that describes the same thing using electrical components only. That's why I kept a copy. Elegant and clever. I can't get to it without hours of unstacking and searching, though. Unfortunately.
--Lee
the_big_m_in_ok
gyulasun said:
Quote
Do you probably mean this patent, US 7085123 ? ( see a free pdf copy here: http://www.pat2pdf.org/ )
Hey Gyula,
>>No, I was thinking of an earlier patent, but I did look up that one on
www.google/advanced_patent_search .
I think you're correct in the operation. More power in the long run. But special coils like tesla flat circular/bifilar coils might improve efficiency more.
Quote
This patent's circuit seems at least to double the working time of a battery, this sounds like a COP of two... However the power needed for operating the switches is not included in the test? (though it could be made very small)
>>I didn't notice the switches at first. ... Now that I've looked, the resistance should be about the only loss in the system. I only looked at the first drawing; as straightforward as it seems to me, testing for such a small reduction might be difficult?
Would I be correct?
--Lee
Quote from: the_big_m_in_ok on May 27, 2009, 03:30:05 PM
...
>>I didn't notice the switches at first. ... Now that I've looked, the resistance should be about the only loss in the system. I only looked at the first drawing; as straightforward as it seems to me, testing for such a small reduction might be difficult?
Would I be correct?
--Lee
Hi Lee,
Yes the losses would be the contact resistances of the switches and the equivalent series resistances of the supercapacitor and that of the battery.
The super capacitor should be a bipolar type to accept the reverse voltage too.
Sorry I cannot get what you mean by: "testing for such a small reduction might be difficult" ? what is the small reduction?
rgds, Gyula
gyulasun said:
Quote
The super capacitor should be a bipolar type to accept the reverse voltage too.
>>Hey Gyula,
Right. Attaching the terminals to the wrong polarity on a large capacitor might be a disaster! Tesla thought the capacitor was the same as "electrical dynamite!"
Quote
"Sorry I cannot get what you mean by: "testing for such a small reduction might be difficult" ? what is the small reduction?"
>>I meant in contact resistance. It, and the wire/cable resistance, should be as low as practical. It's usually too small to be worth measuring in most systems.
--Lee