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Is it possible for a Capacitor to supply a servo motor with enough power for it to close the servo when power is disconnected? I've got an application where I need to close a servo valve that takes 7 Amps (12 Volts) to run and it takes 4 seconds to close the valve. Voltage can vary between 13 and 7 Volts.

Any idea what Capacitor size I need to look at. I tried a 40V 5100uF capacitor that was sold as a battery replacement for a motorcycle but it discharges before the valve even moves.

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  • \$\begingroup\$ 7A ⋅ 4s / 13V = 2.15F \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 29 '16 at 18:35
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    \$\begingroup\$ Much higher capacitance needed since you can only discharge from 13 --> 7 V so... 7 A * 4 s / 6 V = 4.6F \$\endgroup\$ – Jack Creasey Dec 29 '16 at 18:41
  • \$\begingroup\$ How do you know it was fully charged and what voltage did you charge it to? \$\endgroup\$ – vicatcu Dec 29 '16 at 18:46
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    \$\begingroup\$ Possible? Yes. Practical? Eh... not so much. You'd probably be better off adding a small battery to provide shutdown functionality on power loss. \$\endgroup\$ – CHendrix Dec 29 '16 at 19:41
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    \$\begingroup\$ Thanks! I think I'll look into a small battery. That seems like a more viable solution in this issue \$\endgroup\$ – Dean Walker Dec 29 '16 at 20:07
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Q = C * V Charge stored in Coulombs is Q Rated capacitance in Farads is C Voltage across the plates in Volts is V

Divide both sides by a period of time and interpret V as the change in Voltage across the plates over that period of time and... Q / t = I = C * V / t Current is I in Amps if the is time in seconds.

More precisely, and since things vary continuously in the real world, I = C * dV/dt

The time derivative of Voltage is dV/dt.

Rearrange these with the usual algebra to answer your question.

C = I * t / V = 7A * 4s / (13V - 7V)... as a coarse estimate.

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If you could work with a higher voltage, it might be feasible. You might be able to find a switching power supply that would convert rectified 120 VAC to 9 VDC. That would make your capacitor voltage about 150 volts. If the motor would open the valve in 4 seconds at 9 volts and 7 amps, that would be 9 X 7 X 4 = 252 joules of energy. Say 300 joules to account for losses. Energy stored in the capacitor is 1/2 X C X V squared. So C = (2 X 300) / (150^2 - 11^2) figuring the power supply quits when the input voltage drops to 11 volts. That would mean that you would need about 30,000 microfarads.

If you could do the same thing from 40 volts to 9 volts, you would need about 80 of the 5100 uF capacitors.

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    \$\begingroup\$ For the OP (or others reading this) inrush current limiting would be required if you put 30,000uF on rectified mains. \$\endgroup\$ – mkeith Dec 29 '16 at 22:49

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