calculating capacitor discharge

[d333av]

Hello everybody, im new here and could do with some help.
Ive got a problem where I think I can employ a large electrolytic capacitor to work the motor in a motorised valve. The 12 volt power supply from batteries and a solar panel come down a 2 core wire for some distance and empower a pump. This is quite a heavy duty affair but there has had to be a motorised valve fitted to prevent flow back when the pump is off. When the power to the pump is present I can easily tee off the power supply and connect it to the motorised valve to open the valve. But when the pump switches off there has be 12 volts present to shut the valve for 6 or 7 seconds. The current required to shut the valve at 12 volts is 60 milliamps. The valve automatically shuts the current off when fully shut and no power is then needed to keep it shut.
I got some 50 volt, 10 000 micro farad capacitors from an old stereo and built a circuit using diodes and a relay to switch the valve. The capacitors could only manage to shut the valve half way before they were drained.
I'm looking on the internet and i've seen some 12 volt, 1 farad capacitors for car audio systems. I fancy they might work but they're around £50 and i'm not sure they will work.
I've tried to get my head round the equations for calculating capacitors but i've been an auto electrician all my life and am too thick. I lose it in the differential bit.
Can anyone help advise me if they think this capacitor will work in theory? They're available in 1, 1.2, and 2 Farads. And the voltage in the system is 12v (disregarding higher voltage due to charging), at 60 milliamps for 6 to 7 seconds.
Thanks very much in advance for any help
dave

 

[JimB]

but there has had to be a motorised valve fitted to prevent flow back when the pump is off.

Does it have to be a motorised valve to prevent back-flow?
What about a simple all mechanical non-return valve?

JimB

 

[ericgibbs]

hi Dave,
A cheaper option would be a 12V 2AHr SLA battery and a simple transistor or FET delay circuit.

For those high value caps a 0.3Farad, a 12V starting voltage would decay to 10.5V in 7 seconds, into a 200R load resistance

A 1Farad down to 11.5V in 7secs.

Eric

 

[MikeMl]

CχΔV = Q = iχt

C = Capacitance in Farads
ΔV is the voltage difference from fully charged to ending voltage over the period t.
i is the discharge current in A.
t = discharge time in s.


C = iχt/Δv

Say you can tolerate 5V drop in 7s with a discharge current of 60mA.

C = 0.06χ7/5 = 0.084F = 84,000uF

 

[crutschow]

You wouldn't need a very large battery to provide the 60mA for 7 seconds. 10 AA or AAA NiMH batteries would easily provide that. You can trickle charge them to maintain their charge between uses.

 

[d333av]

HI again
some fantastic responses thank you all

I tried a simple solenoid but it didn't work because they become un reliable unless the supplied water pressure is at constant mains pressure (I don't know why). This system has variable and low pressure and it doesn't work.
Ive considered the battery circuit idea and im not convinced the pump is empowered long enough to reliably charge the battery.
Thanks to the two replies on capacitor info, I gather then that the worst possible voltage drop would be 5V
I think this would work and I think i'll give it a go for its simplicity
thanks again
dave

 

[dougy83]

CχΔV = Q = iχt

There shouldn't be any chi (χ) in the equation. I know you're using it as a symbol for multiplication, but it doesn't look like one.

Thanks to the two replies on capacitor info, I gather then that the worst possible voltage drop would be 5V

The voltage drop of 5V was just for an example of using the formula. You can plug in whatever the acceptable voltage drop of the capacitor for the solenoid to still be powered.

 

[d333av]

OK thanks ive saved the equations but whether i'll be able to get my head round them again when I next need them remains to be seen.
Well I bought a 1 farad from maplin today for £40 and it works brilliantly. It has its own LED voltage level indicator and if its to be believed then the voltage when the valve finished closing was exactly 11.5V. Well predicted Eric!
Thanks again
dave

 

[ericgibbs]

hi 333,
This image shows the formula used to determine the voltage at a given discharge time.

E