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TBadertscher

Apr 14, 2015
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I am new to all of this and was wondering if there is a way to dump the energy from a fully charged capacitor to a battery. And what could be used to time this process automatically. Thank you in advance
 

Gryd3

Jun 25, 2014
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I am new to all of this and was wondering if there is a way to dump the energy from a fully charged capacitor to a battery. And what could be used to time this process automatically. Thank you in advance
You can't simply move energy from one place to another.
In order to get the energy from a capacitor to battery, you will need a charge controller. The charge controller must be able to accept a range of input voltage from the capacitor, and output a constant current to the battery.
As far as timing is concerned, it may be best to trigger this process when the capacitor reaches a certain voltage.

Regardless, what are you trying to do?
 

TBadertscher

Apr 14, 2015
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You can't simply move energy from one place to another.
In order to get the energy from a capacitor to battery, you will need a charge controller. The charge controller must be able to accept a range of input voltage from the capacitor, and output a constant current to the battery.
As far as timing is concerned, it may be best to trigger this process when the capacitor reaches a certain voltage.

Regardless, what are you trying to do?

Nothing in particular, just trying to understand on how all this works. Hypothetically, if I had a super capacitor of 3000 fareds and a max voltage of 2.7 volts and I supplied 2.7 volts to that capacitor, wouldn't that be 1/2(3000)(2.7^2) joules of energy? And then I was just wondering if you can transfer that energy in the fully charged capacitor to a battery? I could be missing major concepts here but just trying to learn.
 

Gryd3

Jun 25, 2014
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Nothing in particular, just trying to understand on how all this works. Hypothetically, if I had a super capacitor of 3000 fareds and a max voltage of 2.7 volts and I supplied 2.7 volts to that capacitor, wouldn't that be 1/2(3000)(2.7^2) joules of energy? And then I was just wondering if you can transfer that energy in the fully charged capacitor to a battery? I could be missing major concepts here but just trying to learn.
You *can*, but which battery?
A 1.5V battery will stop taking a charge from the capacitor when the voltage drops.
A 3V battery won't do much better.
So 'boost' type of switching converter would be used to output a higher voltage than what you put into it. This will allow more of the energy in the capacitor to be used.
This converter however, would lose energy... as no conversion can ever be done at 100%
The converter would also only be able to run down to a specific voltage.
Any more fine details beyond this I can't help with. Just know it can be done, with extra electronics.
 

TBadertscher

Apr 14, 2015
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You *can*, but which battery?
A 1.5V battery will stop taking a charge from the capacitor when the voltage drops.
A 3V battery won't do much better.
So 'boost' type of switching converter would be used to output a higher voltage than what you put into it. This will allow more of the energy in the capacitor to be used.
This converter however, would lose energy... as no conversion can ever be done at 100%
The converter would also only be able to run down to a specific voltage.
Any more fine details beyond this I can't help with. Just know it can be done, with extra electronics.
What about a very large battery like over 12 volts or something extremely high like 50 volts? does it change? Can you link some of these converters you are talking about. Are there some that are considered more efficient than others? Thanks again for your help
 

Gryd3

Jun 25, 2014
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What about a very large battery like over 12 volts or something extremely high like 50 volts? does it change? Can you link some of these converters you are talking about. Are there some that are considered more efficient than others? Thanks again for your help
https://www.google.ca/search?q=constant+current+boost
Same this with really high voltages.
These devices are meant to take a lower voltage and bump it up to a higher voltage... however, in doing so the current goes down.
So a 12V 1Ah battery will take considerably longer to charge than a 3V 3Ah battery with the same capacitor+boost setup.


And yes. Switch mode supplies and converters are much more efficient than linear converters. Off the top of my head, I think 80-90% is considered a higher end of what you normally get, but I could be wrong.
 

TBadertscher

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https://www.google.ca/search?q=constant+current+boost
Same this with really high voltages.
These devices are meant to take a lower voltage and bump it up to a higher voltage... however, in doing so the current goes down.
So a 12V 1Ah battery will take considerably longer to charge than a 3V 3Ah battery with the same capacitor+boost setup.


And yes. Switch mode supplies and converters are much more efficient than linear converters. Off the top of my head, I think 80-90% is considered a higher end of what you normally get, but I could be wrong.
  • So if I had a 60 volt battery. I couldn't just dump that energy straight to the battery multiple times until I achieve the Vmax of the battery. Is it necessary to use the boost type switching when you are continually adding that to the battery until its filled. And when you do transfer the energy, I presume the capacitor will be emptied unless of course the battery is full. Is any of this correct so you wouldn't lose current?
 

Gryd3

Jun 25, 2014
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  • So if I had a 60 volt battery. I couldn't just dump that energy straight to the battery multiple times until I achieve the Vmax of the battery. Is it necessary to use the boost type switching when you are continually adding that to the battery until its filled. And when you do transfer the energy, I presume the capacitor will be emptied unless of course the battery is full. Is any of this correct so you wouldn't lose current?
Well. You would still need a boost converter and here's why:
The voltage of the battery and capacitor are both 0.
The voltage of the capacitor goes up to 30V.
The capacitor and battery are connected together and a few amps begin to flow into the battery gradually raising the voltage.
The capacitor at this point is currently loosing voltage, for sake of argument, the battery and capacitor are now both at 15V.
The capacitor is no longer charging the battery, because the voltage in the capacitor is no longer higher... (like water flowing down-hill)
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 21V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 25V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 27V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 29V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 30V.
You disconnect the capacitor and repeat the process. Although this time the battery stays at 30V.

So, if you repeat the process multiple times, you 'may' eventually get the battery to be charged to what the capacitor is, but never more.
Using a boost converter will raise the voltage higher, which will ensure that the current flows into the battery.
Additionally, when the voltage is very different like at the beginning, you risk damage to the battery, as the current flow can be quite high.
 

TBadertscher

Apr 14, 2015
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Well. You would still need a boost converter and here's why:
The voltage of the battery and capacitor are both 0.
The voltage of the capacitor goes up to 30V.
The capacitor and battery are connected together and a few amps begin to flow into the battery gradually raising the voltage.
The capacitor at this point is currently loosing voltage, for sake of argument, the battery and capacitor are now both at 15V.
The capacitor is no longer charging the battery, because the voltage in the capacitor is no longer higher... (like water flowing down-hill)
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 21V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 25V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 27V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 29V.
You disconnect the capacitor and repeat the process. Although this time the battery goes up to 30V.
You disconnect the capacitor and repeat the process. Although this time the battery stays at 30V.

So, if you repeat the process multiple times, you 'may' eventually get the battery to be charged to what the capacitor is, but never more.
Using a boost converter will raise the voltage higher, which will ensure that the current flows into the battery.
Additionally, when the voltage is very different like at the beginning, you risk damage to the battery, as the current flow can be quite high.


Okay, that makes sense now. Thank you for the explanation. Now can you use one boost converter for multiple capacitors or is there a way to connect multiple capacitors together(like batteries) and then use a boost converter?
 

Gryd3

Jun 25, 2014
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Almost all components can be connected in series or parallel.
There are some exceptions of course ;)
When you connect Capacitors in parallel (ie. all of ther + is connected together, and all of the - is connected together, then you end up with a capacitor of a higher capacity)
When you connect Capacitors in series (ie. you string them together like XMas lights. The + of 1 goes to the - of 2, the + of 2 goes to the - of 3... and so on. Then you get a capacitor of a higher voltage)

It can be tricky though... and I'm sorry, but I won't be covering many more details in series and parallel as It's a large topic.
 

TBadertscher

Apr 14, 2015
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Almost all components can be connected in series or parallel.
There are some exceptions of course ;)
When you connect Capacitors in parallel (ie. all of ther + is connected together, and all of the - is connected together, then you end up with a capacitor of a higher capacity)
When you connect Capacitors in series (ie. you string them together like XMas lights. The + of 1 goes to the - of 2, the + of 2 goes to the - of 3... and so on. Then you get a capacitor of a higher voltage)

It can be tricky though... and I'm sorry, but I won't be covering many more details in series and parallel as It's a large topic.
I am familiar with that from high school, it's just every example was with batteries so I wasn't sure. Thank you for your help, If I have any more questions I will report back to you. Haha
 

TBadertscher

Apr 14, 2015
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I am back with another question, after the energy "transfered" from the capacitor to the battery using a boost converter, i presume the capacitor is immediately emptied and does not have to discharge?
 

Gryd3

Jun 25, 2014
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I am back with another question, after the energy "transfered" from the capacitor to the battery using a boost converter, i presume the capacitor is immediately emptied and does not have to discharge?
The capacitor will discharge down to the lowest voltage that the boost converter can operate on.
Anything left over will self-discharge if given enough time.
 

poor mystic

Apr 8, 2011
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Even with that arrangement, the capacitor is prevented from complete discharge. The output 1N004 will prevent discharge below about 0.6V no matter how long you wait. So there's always a charge left on the capacitor.

(PS
that's because the diode won't conduct below 0.6V.)
 

Colin Mitchell

Aug 31, 2014
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"Even with that arrangement, the capacitor is prevented from complete discharge."
Who said anything about fully discharging the capacitor ??????
 

Gryd3

Jun 25, 2014
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"Even with that arrangement, the capacitor is prevented from complete discharge."
Who said anything about fully discharging the capacitor ??????
Chill out. You didn't actually say ANYTHING about the circuit or explain to the op how it works.
Mystic was merely pointing out a characteristic of the device's operation.

Additionally, it does not actually charge only from the capacitor, it simply uses the capacitor to help charge from from another supply that isn't quite high enough. It's merely using the capacitor to bump up the other charging voltage.
 

TBadertscher

Apr 14, 2015
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The capacitor will discharge down to the lowest voltage that the boost converter can operate on.
Anything left over will self-discharge if given enough time.
What happens to the current in this case? Does a fully charge capacitor even have enough current for when charging the battery despite using a boost converter?
 
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