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I am trying to figure out a system to charge a series capacitor bank without too much current going through the capacitors. I had the idea of putting a battery in parallel with the series capacitor back to divide the current, but since the current through the capacitors is \$i=C\dfrac{dV}{dt}\$, and the battery would have a close to constant voltage making \$\dfrac{dV}{dt}=0\$. Would this effectively divide the current or would it all pass though the capacitors because of the constant voltage of the battery?

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The rest of this is just for application specific reason and not as urgent as the main question above. Thanks for looking =)

The reason I am trying to do this is to cut the cost of buying more super capacitors when they are $50 each because another method would be to add more capacitors in series to raise the voltage level and reduce the current. Are there any cheaper alternatives?

I have 1 kW of power being converted down to whatever voltage level I need. The capacitors I am looking at are the Maxwell K series. They have a voltage rating of 2.7V each, with C=1500F and a current rating of 97A. The cost of 4 is the upper limit of my budget.

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  • \$\begingroup\$ Have you considered what happens when you try to drain them at a current much higher than the battery can provide? \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 18 '14 at 4:57
  • \$\begingroup\$ I think your best option is to use a current source such as what is used to power LEDs. These things use a DC-DC converter so the efficiency is high enough, and you can precisley control the maximum current that your battery is sourcing. \$\endgroup\$ – Vladimir Cravero Jun 18 '14 at 7:09
  • \$\begingroup\$ There are a number of different approaches. Do a search for "capacitor balancing" or "super capacitor balancing" and you'll find plenty of discussions of different solutions. \$\endgroup\$ – JimmyB Jun 18 '14 at 8:34
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    \$\begingroup\$ There are too many unknowns. WHY limit current? What is capacitor bank used for (in general terms). A battery across the capacitors will discharge into them. etc. A fuller description of what you are trying to do and not how you are trying to do it will help get you a good answer and more rapidly. \$\endgroup\$ – Russell McMahon Jun 18 '14 at 10:58
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An efficient way to charge high capacitance capacitors with a small current is to use an inductor. i.e., you see the technique all the time in voltage converters. i.e., charge an inductor field by placing a voltage across it, thus building up the current in the inductor. Then, place the inductor in series with the capacitor that you want to charge. I.e., high speed switching. The inductance in the inductor will limit the current, while the energy stored in the inductor field will cause the voltage of the inductor to change to meet the voltage in the capacitor that you want to charge. i.e., at the point where you stop charging the inductor, and place it in series with the capacitor, the inductor becomes a current source. You probably remember hearing in your first AC electronics course that capacitors resist a change in voltage, and inductors resist a change in current. These characteristics make them a perfect combination for solving your problem...

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Here is a Linear Technology site where you can make use of an IC selection tool.

http://www.linear.com/products/led_driver_ics

This tool will help you select a current regulator that may meet your needs of constant and selectable charging current. The LT3795 is an example. I believe it can be programmed to provide either constant current or constant voltage, so the current set feature you need is available.

There are many devices like this that are designed to provide a constant current to strings of LEDs, with high efficiency, and many suppliers that offer them. Many manufacturers offer small, pre assembled, demonstration boards that you can purchase to prove your concept. Also, if you only need a few, just buy a few demonstration boards, and you have already built sub-circuits… just think of the demonstration board as a hybrid module.

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