Powering a Solenoid

Discussion in 'The Projects Forum' started by DrexelDooley, Nov 15, 2008.

  1. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Hello, I'm not confident in my ability to power circuitry and was wondering if what I am thinking makes sense. At this point I own no physical devices, but it is all just design and theory, and I'll need to buy the parts some time in December this year. (I do actually have a small micro controller board to control the whole thing right now though.)

    My problem consists of powering a tubular push type solenoid and a DC stepper motor. These are both on the small scale: I can't use more than 170mmx170mm area for my whole system. I only need about 10 lbs of force from the solenoid, and the stroke will likely be less than 10mm (I think but not positive): the thing will be used for kicking a golf ball. 50% duty cycle is what my mechanical engineer friend told me to we were planning on using for the solenoid, I really don't know why, but I'll take his word for it. I believe my battery will be about 10-12V DC. I've not worked with solenoids or motors before (other than basic motors used in an academic, follow the steps, environment.)

    I believe to get enough of a force out of a solenoid I will require more than 10-12V, so I've read about DC to DC boost converters which I think would be what I'd have to use. (Or charging capacitor banks to hundreds of voltages, but I don't know if I need such high voltages, nor how to do such a thing.) Once boosted, I suppose using a transistor (pmos I'd guess) would be used as a switch to source the voltage to the solenoid. From reading other posts, it seems I need to also use diodes in parallel to the solenoid to protect the circuitry. Also from reading other posts I saw advice to just use a solenoid driver chip, but I'm not sure if that is what I should use or not.

    So that's pretty much it. Any advice or help would be appreciated. In case you're wondering why I don't know any of this, I've been taught primarily in programming and logic design as a computer engineer, and now on this project I am having to also play the role as a electrical eng major with knowledge in power systems...

    Thanks in advance, and for at least reading.
    -Dooley
     
  2. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Dooley,

    This is not only an electrical problem, it is mechanical. You cannot start to create requirements for the solenoid or the electronics before you figure out the dynamics of your golf-ball kicking device. You would need to figure out what quantity of energy you need to hit the ball with. Then, you would be in the position to specify the kinetic energy required; hence the mass of the solenoid plunger and the velocity before contact.

    A solenoid driver is nothing more than a switch in form of a transistor (mosfet or BJT) and a freewheeling diode across the solenoid. The reason to increase the voltage is two fold: first, you would overcome the inductance faster, second, you can ultimately push more current through the winding. The average current must be less than the recommended current through the solenoid, or it will overheat. A dynamic model should be made first, then you will have the best possible idea.

    The only complex solenoid thing you can do is to have a feedback mechanism. There are ways to drive a characterized solenoid (force versus distance at constant I) open loop, and then there are ways to use a linear position sensor. I really don't know the application, so I am not sure what you will need.

    Steve
     
  3. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Thanks Steve,

    I think the dynamic model will be made soon enough. I think the device only needs to transfer like 1-3 J of energy in all (I don't need to hit the golf ball (46g) any more than 10m/s).

    But I was just curious if my description of the system makes sense, or is what the final system will likely look like in general. Such as ~10V battery, to DC-DC boost to X Volts, to mosfet or BJT transistor switch, to solenoid & freewheeling diode. By the way, is a freewheeling diode the same as a flywheel diode?
     
  4. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Dooley,

    You're welcome :)

    Yes, you may be giving 1-3 Joules, but how fast do you need to transfer the energy? For example, 3 Joules in 10mS is 300W of power.

    The reason I am inquiring is because the need for high voltage depends on the dynamics of your system. Like I said, you might not need to overcome the inductance faster or run the solenoid at very high current.

    Your system makes perfect sense. You will simply need a source, a switching regulator if you need, a MOSFET (much more efficient than BJT in this case).

    I recently designed a dc drive that uses a similar arrangement. Very simple PWM to a gate driver, with the diode across the motor.

    http://forum.allaboutcircuits.com/showthread.php?t=15711&page=4

    From wiki

    Steve
     
  5. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Just curious, if I were to need 300W of power, does DC-to-DC conversion allow for a scaling up of a ~10V battery up to the high voltage required to achieve this power (at least is it possible in for a small sized / light system.)
     
  6. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    300W of power would be over a 30A draw on your battery, so it will depend on the battery. Is this an intermittent-duty contraption? If you need an hour of use, that's a 30A/h battery, which is about a 22lb battery (30-40 W/h / kg lead acid).

    If you need an intermittent 10mS pulse of 300W of power, then you do not need a 300W switching converter, unless it is a 100% duty cycle. So, if it were every second, then you'd need 1/100th of 300W, or a steady 3W converter. This is assuming 100% efficiency for ease of calculation. You would need to have the energy stored in a capacitor bank ( Joules stored = 1/2 * C * V^2 ). Also, ensuring the ESR of the capacitor is low-enough to supply the necessary current.

    Steve
     
    Last edited: Nov 18, 2008
  7. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Steve,
    I'm not sure what duty cycle is required, I've never worked with solenoids before. I am assuming that the project does not require 100% duty. I think I only have to kick every 3-4 secs or so, so I don't need the power for long (assuming a 10ms kick:) probably only need it long enough to charge it up, maybe even hold it for a fraction of a sec before kicking.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The solenoid's plunger weighs 4.3 oz, and the maximum weight of a golf ball is 1.62 oz. So, the plunger has around 2.69 times the mass of a golf ball.

    If you obtained a 4800 Ampere turn solenoid rated for 10.3v, the beginning of the plunger stroke would have about 25 lbs of force accelerating it. By the time the plunger had extended 1", it would still have around 4 lbs of force accelerating it. Let's just say it has an average force of around 10lbs for the first inch.
    Since f=ma, by the time the solenoid plunger has traveled 1 inch, it should be traveling at about 365 meters/second^2.

    If the plunger then struck a golf ball that had about 37% of the mass of the plunger, and the kinetic energy transfer was reasonably efficient, the ball should be travelling at least twice as fast as the plunger.

    Unless, of course, I've completely fouled up in my calculation somewhere. :confused:

    What I don't know is if the plunger has to be held in the starting position until the coil is saturated in order to realize the maximum thrust.

    You should be able to run the solenoid on a very low duty cycle, which would save power.

    I'd be inclined to see how well it worked with a SLA 12v battery from a UPS before considering adding a dc-dc boost converter to the mix; you'd be more efficient without it.
     
  9. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
    1
    Like I mentioned, the only good way to do this is by creating a dynamic model.

    I would think a model for a solenoid would not be simple. This is because inductance is a function of position, which means force is also a function of position. The mass of the plunger is idealized by newtons laws and put into the dynamic model.

    http://www.fibtex.lodz.pl/41_22_85.pdf

    May give you an idea of how to do it.

    Steve
     
  10. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Thanks Steve and Wookie, I appreciate the help. I think this will be put onto the back burner for now though as I have to write up a report and presentation for next Monday and Tuesday... Hopefully by then I will have more time on my hands to think about this and get a dynamic model.

    Thanks again,
    -Dooley
     
  11. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    Hello all (if anyone,) so term has started back up and I am still unsure about my solenoid/power problem, although I have made much progress in the micro controller / coding area.

    My teammates tell me that we need about 20J of energy, about 50W of power for the kicking of the golf ball. (I don't know their time assumption, but I am sure it is less than 1/2 a second for activating the solenoid to impact with ball.) They are unsure about the specifics of the solenoid and power supply to get. They tell me these values allow for plenty of extra power as well. This system will be small (smaller than shoe box,) so ideally if I can do without a bulky battery, or cut out DC-DC converter that would be nice. I believe they are going to try to get a solenoid * wiring that is rated at 120V (I think,) but as I said neither they, nor I, really know if what we are doing is the best idea as none of us have worked on solenoids or power systems before. I think we are looking at something like .75A for the current right now as well...

    Just curious if any of this sounds odd or contradictory before I end up going down a dead end path. Or perhaps any of your suggestions as to how you'd go about what I've described.
     
  12. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Fifty Watts at three quarter Amp means 67 Volts... Looks like you'll have to use a bulky battery or a DC-DC converter after all...
     
  13. DrexelDooley

    Thread Starter New Member

    Nov 15, 2008
    7
    0
    yeah, I had suspected as much...
     
Loading...