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Minimum solenoid activation time?

  1. Jul 28, 2014 #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=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.
     
  2. jcsd
  3. Jul 28, 2014 #2

    Baluncore

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    A silent solenoid?
    I very much doubt you will be able to control the final arrival velocity. As the armature begins to move, the magnetic circuit changes it's characteristics. This change will be magnified as it approaches the shortest path length at something like 1 / gap2 , which hits infinity at zero.

    I see no way you can overcome that effect electrically through the main coil because it will have a high inductance. Using a separate coil, a bit like a Thomson repulsion coil to stop the armature is a possibility, but the activation would need to know exactly when to discharge the brake capacitors. Thomson coils are used to blow out the arcs that form in electrical contactors.
    See: Electrodynamic suspension at http://en.wikipedia.org/wiki/Thomson_jumping_ring
    Google 'thomson jumping ring'

    A mechanically absorbent stop, such as trapped air in a leaky rubber cup might be made to work. That will counter the armature with a pressure proportional to 1 / gap. The elastic material will need to handle the high temperature of the compressed air.
     
  4. Jul 29, 2014 #3
    It is not about the velocity being zero at impact but knowing the minimum capacitance of the capacitor that is used to drive the solenoid.
    i know the force increases, but I take the worst case scenario because i can't integrate the characteristic in my formula. With the worst case i am sure it reaches the inside(at least that was what i thought)
    It is about getting the plunger inside with minimal power consumption.

    Thanks anyway :)
     
    Last edited: Jul 29, 2014
  5. Jul 29, 2014 #4

    Baluncore

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    The hardest thing is getting the armature to start to move. Once it starts, you won't be able to stop it.

    Use a DC supply to gradually increase the solenoid current until the armature starts to move.
    That will tell you the minimum current needed, if given plenty of time.
    If you can set a higher current it will happen sooner.
    Set the current to twice the minimum, then make the solenoid connection. It will be much faster.

    The time to close the solenoid will be a function of the rise time of the current, i, and that will be a function of the solenoid inductance, L, and the capacitor initial voltage, V. Where V = L * di / dt
    The capacitor, C, must have sufficient charge, Q, to build up sufficient current. C = Q / V

    You know the minimum current needed to close the solenoid armature given sufficient time.
    The energy stored in the solenoid inductance will then be; E = ½ * L * I2.
    That energy will all have come from the capacitor where E = ½ * C * V2.

    When you equate those two equations you will get a minimum C and V2 pair needed to initiate movement of the armature in a solenoid having an inductance of L when open.
    That analysis gives the minimum C and V2 as it ignores losses due to coil resistance.
     
  6. Jul 29, 2014 #5
    hmm let me take a look at that.. and try to add that to my current calculations. I'll be back :)

    //edit
    The energy placed in in the inductor before it starts moving the armature is about 0.017J compared to 0.15 i know i need in the capacitor.

    So that is about 12%.
    although when the inductor is fully saturated then the energy stored in there is about 0.05J increasing even more when the plunger gets inside the solenoid.

    But now the question.... this energy is not lost right? its in the solenoid and can still be used when it flows out of the inductor after the capacitor is emptied, right?

    //extra note
    even when a steady power supply is used the calculations are still off.
     
    Last edited: Jul 29, 2014
  7. Jul 29, 2014 #6

    Baluncore

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    A flyback diode on the inductor could recover some of the energy back into the capacitor.

    To select the right capacitor you will need to specify:
    1. Resistance of the solenoid coil?
    2. Minimum coil current needed to close the solenoid?
    3. Inductance of your solenoid coil when open?
    4. Voltage rating of the solenoid coil?

    Without those fundamental parameters you will be unlikely to get further help.
     
  8. Jul 29, 2014 #7
    1. 1.94 Ω + 200mΩ for the driver
    2. 1.45 A at 9mm stroke(horizontal position)
    3. 8 mH
    4. 100% duty: 3V, 50% 4V, 25% 6V, 10% 10V.

    extra:
    plunger weight: 14.2g
    spring strength:(max. 0.8N)

    I would like to charge the capacitor to about 8V or 10V.

    The force of the solenoid with 10V at 9mm stroke is 4.5N

    //edit

    so i found out that the inductor did have a greater impact than estimated. (when solenoid is 11mS powered)
    the current limitation by the inductor is about 35% average. This included in the calculations made that the result became 8mS instead of 3, so i am jus 2-3 mS off now. i will try to find the remaining difference tomorrow. by adding 0.4N friction to the 0.2 i already have the result becomes right... but i dont think it is friction now. maybe air compression or misreading the graph of the datasheet.
     
    Last edited: Jul 29, 2014
  9. Jul 29, 2014 #8

    Baluncore

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    Critical parameters;
    R = 2.15 ohm
    I = 1.45 amp
    L = 8. mH
    V = 10. volt
    Minimum inductive energy needed;
    E = ½ * L * I2 = 8.41 mJ
    Equals capacitive energy;
    E = ½ * C * V2
    So minimum capacitance is;
    C = 2 * E / V2 = 168.2 uF
    The LC Resonant frequency;
    F = 1 / ( 2 * Pi * √( L * C ) )
    F = 137.2 Hz
    The period of the LC resonance;
    T = 2 * Pi * √( L * C )
    T = 7.29 ms
    Quarter cycle current risetime;
    Tr = T / 4
    Tr = 1.822 ms
    Average resistive power loss;
    W = I2 * R / 2
    W = 2.26 watt = 2.22 J/s
    Average resistive energy loss;
    Er = Tr * W
    Er = 4.12 mJ
    Total energy required;
    Et = E + Er = 8.41 + 4.2
    Et = 12.6 mJ
    So more capacitance is needed;
    C = 2 * E / V2 = 2 * 12.6 / 100 mF
    C = 250.6 uF
    Tr will be greater, … so use next standard value;
    C = 330. uF

    Will a 2 ms current pulse of 1.45 A be sufficient to pull in the solenoid ?
     
  10. Jul 30, 2014 #9
    i think that you forget the inefficiency of the solenoid it self. the energy put in to the solenoid is not all energy that is used to pull in the plunger, that's why i use the force from the datasheet.

    a pulse of 2mS 1.5Amp did not work to pull the plunger inside.

    I have now included the inductors current throughput and it seems that this almost solved my problem.
    the inductance is calculated by hand and can be ±20% inaccurate. so i guess it is solved now.
    thanks for the help :)

    i just noticed something wrong in my first post. i didn;'t use 150uF capacitors buto 15000uF capacitors.
    to fully pull the plunger inside a minimum of 4700uF was needed.

    //edit
    my working result:
     

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    Last edited: Jul 30, 2014
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