Minimum solenoid activation time?

Heey guys,

I have a problem.
I want to calculate the minimum time I have to activate a (latching)solenoid so that its final velocity is 0 m/s when it hits the inside.

Well I know the following things:
F= m*a
I know the force and the mass, so a can be calculated.
V = a*t.
the average speed after accelerating is V/2(assuming the acceleration is with a constant force/speed)

i know the strength of the spring(max. 0.8N)
with that I know the deceleration speed(f= m*a) again.

if also the deceleration is with a constant speed then I know the ratio between acceleration and deceleration.
the average speed of the acceleration is known, the deceleration is assumed linear and wished to end at 0 m/s and therefor has the same average speed as the acceleration period.

i know the distance of the plunger to pull inside.(9 mm)
S = 0.5at^2. where 0.5at can be taken as the speed.
s = v*t
Because we know the ratio between acceleration an deceleration we know how long the deceleration will take to hit 0 m/s again.
This would make s=v*t*(1+Ratio). 1 = acceleration time, ratio is de deceleration time.
s=v*t*(1+ratio) => S = 0.5at^2*(1+Ratio).

rearrange so that time becomes the subject: t=√(s/(a/2*(1+R))

This gives me te answer of 2.9ms while in reality I have to power the solenoid for atleast 11ms, I did not include the pull force when the plunger closes in to the solenoid. (this would make the power time even less).

neither did I include the inductance factor of the solenoid. And the friction(friction is almost none compared to the springs strength. pull force through the shaft is about <0.1N
I don't know if the plunger compresses air in the shaft and how much this force would be.

I want to know this so that I can calculate the minimum capacitance needed to power the solenoid.
for this I take the average voltage of the capacitor after 2-3T and take that as the power time. force is linear to the current through a coil and the current is linear to the voltage supplied so I can take the average of the capacitor.

But now the question:
What force do I forget that has such a huge effect on the power time.
with testing I tried using a power supply and I tried using 150uF capacitors because of their low resistance. best results with the 150uF capacitors. also tried capacitors parallel to the power supply but the power supply somehow reduced the performance except when I add a diode.

I hope someone can help me with this :)
If someone want my excel sheet they can get it. it also includes gravity when placed in vertical position.
nielskool
Asked Jul 28, 2014

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