Looking for a capacitor capable of driving a 12V / 7A valve for ~4 seconds?

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.

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

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.