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Goosing a solenoid??

Discussion in 'Electronic Basics' started by Don A. Gilmore, May 5, 2005.

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  1. Hi guys:

    I have an application where I have 24 independently-operated push solenoids
    (very small tubulars) that are rated at 12 Vdc, continuous operation. To
    make them operate properly in my device, I need to deliver a 4x voltage
    pulse (48 volts) to the solenoid for a very short period (maybe 10 ms) then
    revert to 12 V to hold the solenoid extended while it is in use. This must
    occur every time it is fired. I will have both 12 V and 48 V sources
    available on the PCB and the circuit is controlled with CMOS.

    What is the simplest way to do this? Thanks for all replies.

    Don
    Kansas City
     
  2. Dave

    Dave Guest

    Hi Don,
    One solution is to charge a cap to 48V with a resistor that will drop
    36 volts when in series with the solenoid. Assuming the duty cycle will
    allow the cap to recharge.
    +48v
    |
    resistor
    |
    CAP-------solenoid
    | |
    | driver transistor
    | |
    | |
    common---------

    Dave
     
  3. Dave

    Dave Guest

    OOPS, my drawing with the keyboard leaves much to be desired..

    +48v
    |
    resistor
    |
    ----------solenoid
    | |
    CAP driver transistor
    | |
    | |
    common---------

    Dave
     
  4. Back in the day of wire matrix impact printers this sort of thing was
    very popular as the wire driver (called pick and hold). Sorry that I
    don't have time right now to do a more complete search, but here is an
    example of how it was done:
    http://www.allegromicro.com/datafile/archive/2962.pdf
     
  5. Don A. Gilmore said

    This is very similar to how automotive diesel injectors are driven.

    For optimal fuel control, one needs to open injectors very quickly
    with a fast current rise (often via a boost voltage), hold them
    open with a more convenient power source (vehicle battery voltage),
    and shut them down quickly via active low side clamping. They are
    typically current modulated at four current levels (two for
    opening, two more for holding). They are usually low impedance and
    will draw high currents if left at either voltage for very long.

    You speak of a boost voltage for a period of time and a hold
    voltage for the remainder. Is the impedance of these solenoids
    sufficient to limit the current to a managable level at 12V?

    We use two high side driven MOSFETS. One from the boost rail, the
    other from the battery rail. Each is modulated to affect the
    desired current levels.

    PFETS may the simplest?

    48V 12V
    ---------------------- -------
    | | | | | |
    | | | | | |
    | | R / | |
    --- Vz | --| NPN s s
    --- | |__| \ ___| PFET _| PFET
    | | | | / d d
    | | | --| PNP | |
    | | | \ | |
    |--------------------| V (Diode)
    | | | -
    | | | |
    R | -------------->>-
    | d L
    | _| O
    | s A
    | | D
    GND------------------------------->>--


    Maybe drive the 48V PFET with a one shot triggered off the rising
    edge of 12V PFET drive signal (not shown)

    There are many custom solutions for this topology. Here's one for a
    single voltage rail.

    http://www.onsemi.com/pub/Collateral/NCV7510.PDF

    Homer Simpson
    Spingfield
     
  6. Joerg

    Joerg Guest

    Hello John, Hello Don,
    A lot of these drivers will be discontinued because the impact printer
    appears to have followed the dinosaur. But you may be able to use a chip
    meant to drive a stepper motor. To crank maximum horsepower out of a
    stepper motor its coils must often also be peaked in a similar fashion.

    Regards, Joerg
     
  7. Homer.Simpson said

    Hopefully this will come out better. This is my first ASCII circuit.
    That's why it looks like crap. ;-)
     
  8. Mook Johnson

    Mook Johnson Guest

    A solenoid is actually a current driven device.

    you might try to make a closed current loop instead of a simple voltage on,
    off. While activiting L di/dt will cause high voltage (curent loop
    compliance) and will regulate the current to that required to maintain the
    device once set.
     
  9. Bob Eldred

    Bob Eldred Guest

    The cleanest way to do this is with one or more microprocessors and FET
    drivers for the solenoids. The solenoids would be run on 48 volts. The
    microprocessor(s) would pulse the solenoids on for the required pull in time
    then go to pulse width modulation at one fourth duty cycle for the 12 Volt
    equivalent drive. There would be 24 input ports and 24 FET driver ports with
    the timing for the driver periods controlled by counter registers for each
    driver. A couple of PICs could easily do this. The PWM frequency could be
    20KHz. Futhermore, being coded, the pulse width, 48V time and other factors
    can be easily be adjusted to get the require performance, not easy with
    capacitors or other ideas.
    Bob
     
  10. Here are a couple more integrated drivers that almost fit your supplies:
    http://www.powerdesigners.com/InfoWeb/design_center/Appnotes_Archive/4032.pdf
    http://www.alldatasheet.com/datasheet-pdf/view/STMICROELECTRONICS/L6213.html
     
  11. I read in sci.electronics.design that Dave <>
    It's still mangled. Use Courier font and don't use tabs.

    It's also a circuit that slows down the operation of the relay. It's
    better to put the cap in parallel with the resistor. This simple trick
    works VERY WELL, and you don't need complex techniques. Use Courier
    font:

    +48 V----+----R----+---COIL------Switching device----- 0V
    | |
    | |
    `----C----'
    The resistor R should be three times the coil resistance, and RC should
    probably be around 50 ms for such small solenoids.

    This works because initially the cap ISN'T charged, so the 48 V gets
    directly to the coil. As the cap charges, the coil voltage drops to 12
    V. When the switch opens, the cap discharges through R.
     
  12. John Fields

    John Fields Guest

     
  13. Pat Ford

    Pat Ford Guest

    there are chips that do this already. look for peak and hold drivers for
    automotive fuel injectors.

    google can be your friend ( hint
    http://www.google.ca/search?hl=en&q=peak+and+hold+injector+drivers&meta=


    HTH
    Pat
     
  14. eromlignod

    eromlignod Guest

    that?


    I like it too. Thanks, John W. & Dave. I also like the idea of the
    solenoid driver chip, but it seems that they run three or four bucks
    apiece.

    The solenoids are 2 watts. So does this mean that I'll need at least a
    6-watt resistor (3 * 2W) for each solenoid circuit? No more than six
    of the solenoids ever fire simultaneously. So I guess the max. total
    heat loss would be

    (6 * 2W) + (6 * 6W) = 48 watts

    Am I all wet on this?

    Don
    Kansas City
     
  15. John Fields

    John Fields Guest

     
  16. eromlignod

    eromlignod Guest

     
  17. Pat Ford

    Pat Ford Guest

    NAtional semi samples them.
    Pat
     
  18. Rich Grise

    Rich Grise Guest

    I would recommend that when you _do_ get the solenoids, you actually
    measure their resistance. The "2 watt" spec sounds a little "loose"
    for me to want to base a series resistor on.

    Besides, I want to know. :)

    Good Luck!
    Rich
     
  19. Is is possible to connect it to 48V only, and switch from initial 100% duty
    cycle to something much lower duty cycle PWM, to drop to a current that is
    equal to 12V/Rsolenoid?
     
  20. I read in sci.electronics.design that eromlignod <>
    Not wet, but that would be true only if you really need to keep full
    rated voltage across them for more than a few seconds. Once the solenoid
    armature has come into position, you may well be able to hold it there
    with far less than 12 V. The spec may tell you. It depends on whether
    what the solenoid moves applies a restoring force. If there isn't much
    of a 'push-back' force, you might need only 3 or 4 V to hold position.
    That would cut the power down a lot.

    2 W and 12 V is 144/2 = 72 ohms. 3 V/72 ohms is 42 mA. 48 V times 42 mA
    is 2 W total for solenoid PLUS resistor.
     
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