MIC1427 Driving a 9V solenoid @ 222.2mA

Discussion in 'The Projects Forum' started by MrJojo, Aug 26, 2013.

  1. MrJojo

    Thread Starter Member

    Jan 23, 2013
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    Hello all,

    I am making a simple two link robot (2 servos attached to each other by an arm) and I wanted to put in a braking system, so when the robot is shut down, the arms don't go limp and potentially damage itself or its surroundings. My plan is to have a solenoid's armature pushed into the spokes of the servos so the servo can't move when there is no voltage. When there is a voltage applied to the solenoid, it will contract and allow the servos to move.

    I currently have several MIC1427 chips laying around and I was thinking about using them to drive the solenoids. I was reading through, and I think the only thing I need to do is put a capacitor between Vs and GND (bottom of page 4). The solenoid takes 2W @ 9V (I=222.2mA, R=40.5Ω).

    I am not sure if the MIC1427 will do what I want. I know they are used to drive high capacitive loads. Therefore, will I have to make the solenoid look like a high capacitive load?

    Thanks,
    Matt

    EDIT:
    I found that the new IC for the MIC1427 is MIC4427.
     
    Last edited: Aug 26, 2013
  2. #12

    Expert

    Nov 30, 2010
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    That should work, and have quite a bit of safety margin. And, no, you don't have to make the load pretend it's capacitive. That's just the "worst case" scenario.

    However, you do need a protective diode across each solenoid to protect the driver chip from inductive reactions.
     
    Last edited: Aug 26, 2013
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  3. MrJojo

    Thread Starter Member

    Jan 23, 2013
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    #12,

    Thank you for your input. Attached is a quick diagram I drew with the diodes in the place where I think they should be. In real life, the push buttons would be something else and the batter is a 9V power supply. I just want to make sure I have the idea of the protective diode correctly.

    Thank you,
    Matt
     
  4. #12

    Expert

    Nov 30, 2010
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    No. The diodes go in parallel with the coil with their "plus" end pointed at the "plus" end of the battery.

    You see, when you shut off an inductor, it has to get rid of the energy in its magnetic field. That energy wants to keep flowing from the power supply, through the inductor, and into the switching chip. You put a diode from the switched end of the coil to the plus of the power supply and the diode redirects the energy back into the battery so it doesn't punch holes in the switching transistors.
     
  5. MaxHeadRoom

    Expert

    Jul 18, 2013
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    I think you missed, In his case the coil is sinking to -ve so it would be the reverse, from the switched end to common (-ve).
    Anodes to #1 of the coils K to 2.
    7 & 5 of the IC to terminal 2 of the coils.
    Max.
     
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  6. #12

    Expert

    Nov 30, 2010
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    Yep. You caught me not paying attention.

    Each diode still goes in parallel with each coil, but the plus end of the diode (with the stripe) is pointed at the driver chip.
     
  7. MrJojo

    Thread Starter Member

    Jan 23, 2013
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    I think of diodes as one way streets, where current can't go the wrong way. In my case, I thought the diode would have protected the MIC1427 chip from the current surge and directed it back to the battery. If thats the case, why wouldn't my original idea have worked?

    Max, will it look something like the attached document? Note, I changed some of the pin names.

    As for dealing with the inductor's power, is it getting dissipated in the diode-solenoid loop? It seems to me that when the power is being turned off to the solenoid, there is a current surge from the battery, and the diode there to direct it to power the solenoid again, but this time with less power...?

    Matt
     
  8. MaxHeadRoom

    Expert

    Jul 18, 2013
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    In your first dwg, the diodes would have been forward conduction and could operate the coils, but at turn off, the reverse voltage would be blocked by the diodes and have no where to go.
    Where do you get the power surge from the battery at switch off?
    With the correctly connected diodes across the coil, the BEMF spike is conducted through the diode, effectively shorting the coil IOW the current recirculates through the coil and diode only, this is why the down side of a reverse EMF diode is that in introduces a fraction of a delay in drop out until this BEMF is dissipated.
    In most cases this is not an issue.
    Max.
     
  9. MrJojo

    Thread Starter Member

    Jan 23, 2013
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    Max,

    I don't know much about inductors or solenoids, but I thought when the solenoid would "shut down", it would release the stored current in the inductor causing a voltage surge through the rest of the circuit. Kind of like in cartoons when someone hold a hose shut and there is the water build up and once the hose is released, the water shoots every (much like Tom and Jerry).

    Therefore, in my Tom and Jerry analogy, my original drawing would have kept the hose held shut, and once the circuit is powered on again, things would go bad... Where in the second drawing, this allows the voltage to dissipate thought the solenoid-diode loop, so there isn't the big rush of water when the circuit is turned on again.

    Dose this mean I'll have to look for a reverse EMF diode specifically? I did a quick search and found a wiki article on flyback diodes where it uses a 1N4007. I have a 1N4006, and that seems like it should be more than enough for what I need.

    Thanks for the help and advice,
    Matt
     
  10. MaxHeadRoom

    Expert

    Jul 18, 2013
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    But you don't want that surely?

    IN4002/7 is usually sufficient, if you do not suppress this BEMF somehow, it collapses back into the coil with the resultant spike that can be many 100's volts.
    You don't want it released back into your circuit at that voltage!.
    So the diode takes care of it.
    Max.
     
    Last edited: Aug 27, 2013
  11. #12

    Expert

    Nov 30, 2010
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    In your original drawing, the diodes in series with the inductors, the reactive pulse would just punch through the diodes. In theory, there is no limit to the voltage of an inductive reaction. energy = (1/2) capacitance x voltage squared. If there is pretty much no capacitance to shovel the energy into, the voltage approaches infinity. In real life, it can pass a thousand volts fairly easily. Basically, you can't fence in an inductive collapse. You have to give it a way to escape.

    Putting the diode in parallel with the inductor lets the energy circle through the inductor and diode until the resistance of the inductor wastes the energy as heat.
     
  12. MrJojo

    Thread Starter Member

    Jan 23, 2013
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    #12 and Max,

    Thank you for the help and better understanding of solenoids/inductors.

    Matt
     
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