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I am looking for a small form factor capacitor that is capable of 12V at 500 mA for approx 2 seconds. This is to open a small 12vdc air solenoid that is under about 100PSI of pressure for just 2 seconds. Then the capacitor could be recharged.

I have not been able to find a formula or calculator.

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    \$\begingroup\$ Capacitors are very bad at keeping a near-constant voltage at that drain unless they are very large. Figure out what the maximum and minimum voltages you can tolerate are. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 20 '15 at 21:01
  • \$\begingroup\$ 12V @ 0.5 Amps for 2 seconds? That's going to take a very large capacitor. \$\endgroup\$ – Dwayne Reid Feb 20 '15 at 21:30
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Do the math.

Energy needed by the solenoid: \$12 \text{V}\times 500\text{mA} \times 2\text{s} = 12\text{J}\$.

You can charge the cap to 12 V and a boost switcher will convert the cap's decreasing output voltage to 12 V. Let's say the boost switcher is 80% efficient and can operate down to 2 V. So the cap needs to provide \$\frac{12 \text{J}}{80\%} = 15 \text{J}\$.

After discharging the cap from 12V to 2V there will be a fraction of the energy left: \$\frac{(2 \text{V})^2}{(12 \text{V})^2} = 0.028\$

So the cap must hold a total energy of \$\frac{15 \text{J}}{1 - 0.028} = 15.5 \text{J}\$ at 12 V.

The minimal capacity needed is: \$2\times \frac{15.5 \text{J}}{(12 \text{V})^2} = 215 \text{mF}\$

So, you need at least a 220 mF 16 V cap. This assumes you have a boost converter that can produce 12 V at 500 mA from 2-12 V and is at least 80% efficient.

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    \$\begingroup\$ 6:1 is an awfully wide range over which to maintain 80% efficiency in a switching regulator. But fortunately, raising the minimum voltage to 6V (a 2:1 range, which is much more reasonable) only raises the capacitance required to 280 mF. \$\endgroup\$ – Dave Tweed Feb 20 '15 at 23:20
  • \$\begingroup\$ Thanks very much for all of the input. That was the formula that I was looking for. Fortunately, the air tank is tiny and during the 2 seconds which the solenoid valve is open the pressure will drop to ~0PSI so the power requirements to keep the solenoid open will decrease to approximately 2-3V toward the end. \$\endgroup\$ – kboyette7 Feb 21 '15 at 2:57
  • \$\begingroup\$ 220 mF = 220,000 uF. That's a BIG HONKING CAPACITOR, even by today's standards. \$\endgroup\$ – John R. Strohm Feb 21 '15 at 4:37
  • \$\begingroup\$ Yes, I've since discovered that size capacitor would be "BIG HONKING." :) My goal was to replace 2-9V Batteries with something smaller/lighter. Doesn't look like that is going to happen. I had assumed that since I only needed the power for 2 seconds that the resulting size would be smaller. That is what I get for assuming. Thanks again everyone for your help. (And yes, I'm aware that 2-9V batteries together are not 12v) the solenoid has a bit of tolerance. 18V for a few seconds will not damage the coils. In fact one 9v battery will work up to about 50 PSI. After 50PSI it draws too much. \$\endgroup\$ – kboyette7 Feb 21 '15 at 14:05
  • \$\begingroup\$ @Dave: I agree. I picked something to illustrate how the answer can be gotten just by doing the math from basic phyiscs. In reality, the boost converter would be more efficient in the beginning, then efficiency drop off as the voltage gets lower. Using 80% overall with your 12-6 V input range is quite realistic. \$\endgroup\$ – Olin Lathrop Feb 21 '15 at 14:20

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