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I'm thinking of using a supercapacitor to power a device with low power consumption. I have one 1 F capacitor rated at 5.5 V and I plan to charge it using input voltage of 5 V.

What precautions should I take when charging it? I'm afraid that it will overload my power supply if I just connect it directly.

Any other tips would be appreciated too.

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  • \$\begingroup\$ When you say "power supply", I'm interpreting that as the dedicated power management you have on your board, not some bench PSU as some of the answers do. Which is it? \$\endgroup\$
    – Nick T
    Mar 12 '11 at 22:28
  • \$\begingroup\$ In my case, I'm referring to a bench power supply which has adjustable voltage, current limiting and should have short-circuit protection. Still, I'd like not to depend too much on the capabilities of the power supply. Instead, I'd like to limit intake current on the board side, so I could later switch to a simpler power supply. \$\endgroup\$
    – AndrejaKo
    Mar 13 '11 at 0:07
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I have had several students who have used super cap chargers. They provide a lot of nice features like variable current limit on the charging, the particular one that I linked to will also make it so you could use 2 caps with lower voltage on each one and it will make sure you don't over voltage one.

The only down side is the chips for these are VERY small and can be difficult to solder if you aren't very experienced.

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    \$\begingroup\$ A dedicated controller is going to offer a great mix of performance and features, likely at a price that's hard to match with discretes, and board area that's impossible. \$\endgroup\$
    – Nick T
    Mar 12 '11 at 18:18
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You are not only going to fry power supply, but also damage your supercap - it's internal connections cannot withstand high peak currents.

Charge it with some 0.1A current using current-limiting resistor in the simplest case.

Same applies for discharging - never short leads.

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  • \$\begingroup\$ Is there anything wrong with this answer? Except that 0.1 A would be specific to the type of capacitor used? \$\endgroup\$
    – AndrejaKo
    Mar 12 '11 at 21:03
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Buy a CC/CV power supply, I picked up a 3A one for £40. I've used them before and they won't damage themselves, even with a continuous short. You can set the CV control to about 5 volts and the CC control to the maximum charge current - 1 or 2 amps. Initially, current will stay high, but as the voltage rises the supply will drop the current automatically. Plus, the CV control ensures the cap won't ever be damaged by a high supply voltage.

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Step 1: Specify the constraints on the charging behavior that you want, based on the characteristics of the capacitor and/or your power supply. (e.g. current limit of not more than 0.2A or 0.5A or 0.1A or whatever)

Step 2: Build a circuit that will meet those characteristics. If the current is low, you can just do this with a resistor. It turns out if you do the math that charging a capacitor through a resistor from 0V to some fixed voltage, always wastes exactly 50% of the energy involved, so it's not hugely efficient, but it's not horribly inefficient either. If you need higher efficiency, you need a switching power supply of some kind. The ones appropriate for this application would produce nearly constant current into the capacitor with a voltage limit.

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If it were my power supply, I'd have one that can handle dead shorts with some sort current limiting, and not worry about it. This is inefficient, as power will be dissipated within the power supply.

If you're worried about inrush current harming the supply, put an inductor in series with the capacitor. Inductors resist changes in current, but eventually settle into acting like straight wire. You'll want something like shown here http://www.hammondmfg.com/153.htm with the laminated core and metal mounting bracket.

What size? To get a rough idea... The fundamental equation for inductors is dI/dt = V/L. rearrange this for dI = (V/L)dt and imagine capital Greek deltas instead of the infinitesimal lower case 'd's. The change in current is from zero to the initial inrush current. Let's put dt at 1 sec, since, although we're not trying to build a resonant circuit, the characteristic frequency for a circuit with 1F and 1H is 1 radians/sec. Suppose you get a nice hefty 1H coil. The initial voltage drop across the inductor is 5V. So we get 5 amps. In real life, the ESR of the capacitor and the internal resistance of the power supply, as well as its ability to deliver only so much current, means you probably wont' get that. To limit the current to say 1A means a 5H coil.

The least efficient way to limit current is with a resistor, or a current regulator such as an LM317. But this is likely much cheaper than a 5H coil, unless you already have a coil like that in a junk parts box (these were common in old 1960s/1970s TVs)

Supercapacitor charging circuits can get pretty involved for applications that harvest small amounts of energy, where not one microwatt should be wasted, where >90% efficiency in charging is vital. In some applications, the charging current might be only a small factor greater than leakage current. A good read with technical details: http://www.energyharvestingjournal.com/articles/using-a-supercapacitor-to-manage-your-power-00001921.asp

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    \$\begingroup\$ -1: Your discussion about inductance is correct from a theoretical standpoint, but useless from a practical standpoint, as any 1H inductors rated for 5A are huge custom magnetics. \$\endgroup\$
    – Jason S
    Mar 13 '11 at 4:31
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    \$\begingroup\$ Even from the theoretical stand point it means that when a circuit is pulling power, the inductor will want to provide constant current, which is not what the circuit will want. If the circuit is suddenly turned off, the inductor will want to keep delivering that current, which will cause the voltage to spike on the cap. \$\endgroup\$
    – Kellenjb
    Mar 13 '11 at 14:47

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