Minimum solenoid activation time?

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1

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=vt(1+Ratio). 1 = acceleration time, ratio is de deceleration time.

s=vt(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 15000uF capacitors because of their low resistance. best results with the 15000uF 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 :slight_smile:

If someone want my excel sheet they can get it. it also includes gravity when placed in vertical position.

2

I’m unsure what you mean by “so that its final velocity is 0 m/s when it hits the inside”. The inside of what ?

My usual idea is that the spring retains the plunger and when energized the magnetic force pushes the plunger out, fighting the spring. If no stop is hit, the plunger extends until the magnetic force = spring force, probably after some oscillations.

So the 1st of your problems seems to be that neither force, spring or magnetic, let alone the net force, is constant. A simple spring model is ;

F = k*x where k is the spring constant and x is the displacement. So in my solenoid the spring force increases w/distance and thus time.

The magnetic force is also not constant. It decreases w/displacement. You can use a freeware program, FEMM ver4.2, to calculate magnetic force vs displacement and/or current. Be aware it’s a static modeler, not one that computes the physics of moving objects. But you can easily iterate the static computations for various displacements.

Also you need to include the inductance as the magnetic force depends on current flow and the solenoid will be an RL circuit, w/it’s associated time constant. It gets worse as the inductance depends on the plunger position. FEMM can compute that part for you.

Lastly I would expect different times for the opening and closing.

3

need to calculate for a specific part number.

4

Mee_n_Mac:
I’m unsure what you mean by “so that its final velocity is 0 m/s when it hits the inside”. The inside of what ?

When the final velocity is 0 m/s at the exact moment it would hit the inside it means there is not too much energy put in the solenoid. it is to calculate the minimum capacitor needed. but for now the capacitor is not used yet with the calculations, for now I use a steady power supply of 10V.

//edit

the plunger is pulled in to the solenoid’s armature.

Mee_n_Mac:
My usual idea is that the spring retains the plunger and when energized the magnetic force pushes the plunger out, fighting the spring. If no stop is hit, the plunger extends until the magnetic force = spring force, probably after some oscillations.

To push the plunger out i would only have to power the solenoid for about 100uS.

I want to calculate the time i need to pull the plunger inside. the max force of the spring is 0.8N pressing the plunger outside.

Mee_n_Mac:
So the 1st of your problems seems to be that neither force, spring or magnetic, let alone the net force, is constant. A simple spring model is ;

F = k*x where k is the spring constant and x is the displacement. So in my solenoid the spring force increases w/distance and thus time.

With the calculations for now i used the worst case scenario for all.

max spring force(0.8N)

Mee_n_Mac:
The magnetic force is also not constant. It decreases w/displacement. You can use a freeware program, FEMM ver4.2, to calculate magnetic force vs displacement and/or current. Be aware it’s a static modeler, not one that computes the physics of moving objects. But you can easily iterate the static computations for various displacements.

also here i used the worst case scenario.

Minimum solenoid pull force 4.5N at 9mm stroke(10 V power supply)… increases if it gets inside the armature. (positive)

Mee_n_Mac:
Also you need to include the inductance as the magnetic force depends on current flow and the solenoid will be an RL circuit, w/it’s associated time constant. It gets worse as the inductance depends on the plunger position. FEMM can compute that part for you.

i didn't include this at first but later on i did. didn't really make a difference. because the force of the solenoid is linear with the current. and the current not drawn at the beginning will still happen at the end.

I used openmodelica to simulate the model with an inductor. but tahnks for the tip to use FEMM maybe it gives me new information.

Mee_n_Mac:
Lastly I would expect different times for the opening and closing.

True, the solenoid is a latching solenoid. inside there is a magnetic that has a 1.3N pull strength (when latched).

when the plunger exits the solenoid for more then 1mm is unlatches and the spring will be stronger then the magnet inside.

so powering the solenoid for more then 100uS is already enough to unlatch, also because the force is at its top when the plunger is inside.

Thanks for helping :slight_smile:

5

nielskool:

Mee_n_Mac:
Also you need to include the inductance as the magnetic force depends on current flow and the solenoid will be an RL circuit, w/it’s associated time constant. It gets worse as the inductance depends on the plunger position. FEMM can compute that part for you.

i didn’t include this at first but later on i did. didn’t really make a difference. because the force of the solenoid is linear with the current. and the current not drawn at the beginning will still happen at the end.

I used openmodelica to simulate the model with an inductor. but tahnks for the tip to use FEMM maybe it gives me new information.

What's the coil resistance and inductance ? I find it hard to believe their time constant isn't at least a couple of msec.
6

1.94 Ohm + 200mOhm for the driver.

8mH when plunger is out of the armature.

pulling the plunger inside takes 11mS with a steady 10V power supply

plunger weights 14.2g

time constant would be something near 4mS.

although the time constant is higher than the time to unlatch there will still be a current through the coil… but less, but enough to unlatch apparently.

for the latching it has a greater impact then i expected… if i calculate 3mS and the time constant of the inductor is already 4ms… hmm then i will also have to calculate the average current through the inductor which i did with the capacitor in a later stage… hmmm i am going to do that now :slight_smile:

//edit

3ms on the 4ms time constant would give a 30% average current through put of the max. so this comes very close to my difference factor between reality and theory :D… i will continue with this and come back with my new calculation results :slight_smile: thanks… dammnn feel bit stupid for assuming the inductor wasn’t a big deal:P

//edit 2

oke i included the inductor current through put and get 7mS now instead of the 3ms without.

still 3-4 mS off… hmmm(30+%)

very little friction or force deviations have a great influence now. with 0.4N extra friction the result is what i expect.

but i don’t think it is 0.4N friction.

it can be a bit from misreading the graph of the solenoid and maybe some deviation in the solenoid. but i just want an asnwer a bit closer to what i expect. But i will try to find that tomorrow.

7

Have you looked at the rise time of your 10v supply on an o-scope ? How are you determining/setting time t = 0 ?

8

The power supply i had was actually very slow… so i used over sized capacitors to simulate an ideal power supply( high current, low resistance) 25000uF. parallel I’ve put the power supply with a diode. without the diode the power supply drained the current somehow…

the solenoids were controlled by a motor driver.

The on time was experimentally set to 10.5mS which was just enough to latch the solenoid.

9

And you’ve verified that the driver output goes to 10v PDF and doesn’t droop or ring during the 10 msec ?

Is there a link to the solenoid ?

10

jup i verified that with the oscilloscope.

used solenoid: http://www.ledex.com/ltr2/access.php?fi … B14M-L.pdf

//edit

i probably failed with the oscilloscope… because i notice that the driver limits the current to 3.3Ampss…

3.3 amps is reached after 2Tau. the remaining time has 66% of the current so that is another loss of 34% during 1.4mS a bit less because the inductor is still “charging”

still 2 mS off

11

nielskool:
still 2 mS off

Well .... is that inductance measured or by spec ? I'd expect at least a 20% tolerance vs spec. Also what is the stroke you're measuring ... is that the aforementioned 9 mm ?

Also flat faced cs conical ?

My Excel sim is showing (I think) different results from yours.

12

nielskool:
jup i verified that with the oscilloscope.

used solenoid: http://www.ledex.com/ltr2/access.php?fi … B14M-L.pdf

//edit

I probably failed with the oscilloscope… because i notice that the driver limits the current to 3.3Ampss…

3.3 amps is reached after 2Tau. the remaining time has 66% of the current so that is another loss of 34% during 1.4mS a bit less because the inductor is still “charging”

still 2 mS off

:angelic-innocent: The inductance was calculated… using the coil turns and the by hand measured radius and length of the coil. with a core permiability of 1(air) :angelic-innocent:… yea will probably be in there :stuck_out_tongue:

Solenoid used is conical.

Remeasuring the stroke gave me 11.2mm. should be 9mm, probably because i have an experimental setup right now without the final weight attached.

What are the results of your excel sim?

//edit

i just noticed something wrong in my first post. i didn’t use 150uF capacitors but 15000uF capacitors.

to fully pull the plunger inside a minimum of 4700uF was needed.

13

oke i verified my calculations today. i think that my formula is right now. my calculations result is a bit higher then the reality but this is alright. i think that is because the increasing force while closing in.(which i didn’t take into consideration with the calculation).