# How to charge a supercapacitor FAST?

Discussion in 'The Projects Forum' started by lxlxlx, Mar 23, 2011.

1. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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Hi all,

I'm working on a design project that needs me to charge a supercapacitor as fast as possible. The source is shown in the attachment. It's a LM7805 regulator (powered from a 9v battery) in series with a 120 ohm resistor. The supercapacitor is a 5.4V, 0.5F. I need to charge it to around 3.5V as fast as possible. Any method would work as long as all the energy comes from the given power source. Any advice?

Thanks!

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
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Does it matter what happens after the cap gets to 3.5V?
Does the 120Ω resistor have to stay? Can you bypass it with a lower resistance during charging?

3. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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It doesn't matter. The cap will be used later as a power source in a different circuit.

The source (including the 120ohm resistor) cannot be changed, only the two terminals on the right side are accessible.

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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Yeah, but does it matter if the cap continues to charge after it reaches 3.5V?

5. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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That would be fine.

No, only the two terminals in the circuit are accessible.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
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Is this homework?

7. ### Papabravo Expert

Feb 24, 2006
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The answer is more current. The voltage is given by the time integral of the current divided by the capacitance. For a given amount of current it will take a fixed amount of time to get to 3.5V. To cut the time you need to increase the current. More is better, so why do you have the 120 ohm resistor in the path? It is only preventing you from accomplishing your objective.

8. ### SgtWookie Expert

Jul 17, 2007
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I'd say the 120 Ohm resistor is in there to limit max regulator output current to 40mA.

It would take around 2 hours to charge a 0.5F cap to ~3.5v from a 5v source via that resistor. But, if one charged a 1F cap (or a couple of 0.5F caps in parallel) from that circuit until they reached nearly 5v (which would take a very long time) then they could be used to charge another 0.5F cap via a low or variable resistance, like a light bulb filament.

If 1F's worth of capacitance had 5v on it, and you added another 0.5F cap in parallel that was fully discharged, you'd wind up with 5v*2/3 = 3.3333...V on all three caps. That's pretty close to 3.5v.

[eta]
Whoops, I goofed - not 7,220 seconds (2 hours) but 72.2 seconds.

Last edited: Mar 24, 2011
9. ### Ron H AAC Fanatic!

Apr 14, 2005
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The actual charge time will be about 72 seconds. I can't think of any way to decrease that time, unless, as Wookie described, you keep a 1F cap that was charged from the previous cycle of power ON, and connect that in parallel with the 0.5F cap when you want to use it.

Last edited: Mar 24, 2011
10. ### Papabravo Expert

Feb 24, 2006
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Well yes that is true, but the original question was how to charge the super cap as fast as possible. The answer is still more current and you can get that from a 9V battery with the regulator configured as a current source and a smaller resistor. This may lead to other problems but that is the answer to the original question.

Apr 14, 2005
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12. ### Papabravo Expert

Feb 24, 2006
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Well post #5 is by lxlxlx so now I'm really confused.

13. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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the 120 ohm resistor is fixed. it can't be removed. is there any way to boost the current, but lower the voltage? no external sources are allowed.

14. ### beenthere Retired Moderator

Apr 20, 2004
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Ohm's law has it that I = E/R. The current is determined by the voltage divided by the resistance. If the resistor is fixed at 120 ohms, the lowering the voltage can only give less current.

15. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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That's not what I meant. I want to connect some active components to the source circuit that produces a lower voltage but higher current, while maintaining the same power.

16. ### SgtWookie Expert

Jul 17, 2007
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Adding caps to total ~1F on the output and let them charge ~7 minutes before trying to charge your 0.5F cap using a low-value resistor is the only way I can think of to speed the charging. Any circuit you try to add to it will just dissipate power as heat.

17. ### Ron H AAC Fanatic!

Apr 14, 2005
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If you read ALL of post #5, it has my questions and his answers. That's why I referenced post #5 instead of post #4.

18. ### Papabravo Expert

Feb 24, 2006
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I guess I'm just a dumb graduate hillbilly cuz ah dun cee wha yo anser has to wi' mine

19. ### lxlxlx Thread Starter New Member

Mar 23, 2011
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That won't work. As other member said more current will result in faster charging. I wonder if there's anything to boost the current that doesn't use any external source?

20. ### SgtWookie Expert

Jul 17, 2007
22,183
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Well, I missed something pretty obvious.

When T=0, the cap is completely discharged and current begins flowing through the 120 Ohm resistor, 5v/120 Ohms= 41.666...mA. That's the highest current flow that you will obtain.

As the voltage on the capacitor increases, the current through the resistor decreases.
If V(C1) = 1, then current through R1 is (5v-1v)/120 = 4v/120 = 33.333...mA.
By the time V(C1) = 3.5v, R1 current is (5v-3.5)/120 = 1.5/120 = 12.5mA.

So, what you need is some kind of boost circuit that keeps the voltage across R1 as high as possible for maximum current flow.