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Help required with placement of diodes on electromechanical vane display

Discussion in 'General Electronics Discussion' started by mightyconfused, Apr 9, 2013.

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  1. mightyconfused

    mightyconfused

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    0
    Apr 9, 2013
    Hi Guys,

    Relatively new electronics newbie here - so thanks for your help :)

    Recently I purchased several electromechanical seven segment vane displays for a project.
    As you can see in the photo, each segment flips out or back by a magnetic field generated through the coil.

    Each segment has only 1 coil. By changing the direction of the current impulse through the coil, you can change if the segment flips out or back.

    [​IMG]

    Each coil has a resistance of 50 Ohm and requires ~ 24 V impulse to drive.

    I am planning to use SN754410 chips to drive the impulse (with the logic coming from an Arduino + shift register). (SN754410 data sheet: http://docs-asia.electrocomponents.com/webdocs/0b87/0900766b80b878b5.pdf).

    After doing some reading I realise that when the current across the coil is turned off, there can be quite a high reverse 'breakdown' current that I need to protect my chips against.

    I am quite confused about the placement of diodes between the SN754410 and the coils to protect agains this (??)kickback current. I initially thought the diodes would be in series with the coil, but some sites suggest they should be in parallel. But if they are in parallel with the coil - how does that work when I reverse the polarity.

    Can someone please give me a guiding hand. This is well over my simple head.

    Thanks
     
    Last edited by a moderator: Apr 9, 2013
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,838
    Jan 21, 2010
    You can't put the diodes across the coil if you're using (say) an H bridge to reverse the voltage across the coil.

    In this case (an H Bridge) you can place a diode in parallel with the switching transistors or you can use a snubber

    There's an example of what I'm talking about here.
     
  3. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Unless I am misunderstanding something, the diodes are there in the chip.

    Bob
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,497
    2,838
    Jan 21, 2010
    Correct, you are not missing anything.

    mightyconfused, you don't need any more diodes :)
     
  5. mightyconfused

    mightyconfused

    6
    0
    Apr 9, 2013
    Thanks guys. I was under the impression that the internal diodes were not 'kickback' diodes. Given that it is 24V driving the coil, I thought that the kickback voltage might have been higher than the 36V the SN754410 is rated for.

    I'll go with you guys. It all does my head in. It's more confusing than learning a new language!!
     
  6. mightyconfused

    mightyconfused

    6
    0
    Apr 9, 2013
    Thanks for your advice so far guys. But I've become mighty confused again and would appreciate your help.

    I've gone ahead and created a simple circuit to a single coil/segment on one of the displays. After learning Fritzing, I've created a little diagram here to show you my circuit.

    H-Bridge_bb.jpg

    And my basic Arduino code to flip out/flip in the single segment:

    Code:
    int segmentOUT = 4;
    int segmentIN = 5;
    
    // the setup routine runs once when you press reset:
    void setup() {                
      // initialize the digital pin as an output.
      pinMode(segmentOUT, OUTPUT);     
      pinMode(segmentIN, OUTPUT);  
    }
    
    // the loop routine runs over and over again forever:
    void loop() {
      
      //Sent a 50ms pulse to put segment to OUT position
      digitalWrite(segmentOUT, HIGH);
      delay(200);
      digitalWrite(segmentOUT, LOW);
      //Delay 2s
      delay(2000);
    
      //Sent a 200ms pulse to put segment to IN position
      digitalWrite(segmentIN, HIGH);
      delay(200);
      digitalWrite(segmentIN, LOW);
      //Delay 2s
      delay(2000);
    }
    
    However - it's not working as expected!!

    The external power supply is a 24V/1A power pack. There is limited data on the actual display devices (bought of eBay second hand from Latvia), but I believe they require 20-30V for a (minimum) 50ms impulse to fire. If I directly connect the 24V + and - directly to a coils terminal, I can flip out the segment (and with reversing the polarity, flip in the segment). So that aspect of things appears ok.

    If I replace the coil in the circuit above with my multimeter, I can see the voltage alternate between +24V and -24V with a 2 second delay as expected by the program.

    However, if connected directly to the coil, there is no flipping of segments.

    Can anybody suggest where I have gone wrong/what I am missing?? The only strange thing I can note is that if I turn on the external power supply while the Arduino is already up, the onboard LED initially fades for a second as though there has been a power drain (it then stays on normally).
     
  7. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Deleted, I was wrong here.

    Bob
     
    Last edited: Apr 10, 2013
  8. mightyconfused

    mightyconfused

    6
    0
    Apr 9, 2013
    Addit: I should also point out that I missed putting a wire between the power supply + rail and pin 8 of the chip in the circuit. It is there in real life
     
  9. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
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    May 8, 2012
    The Vcc and display lines in your image make no sense. Are they really wired like that?

    Chris
     
  10. mightyconfused

    mightyconfused

    6
    0
    Apr 9, 2013
    Hi CDrive,

    Can you explain what you mean. I haven't heard the term display line.

    My understanding in terms of VCC to the chip is that:

    - Pin 1 requires 5V (high) to enable the bridge for motor 1 (or coil, in my case) - taking 5V off the arduino
    - Pin 16 requires 5V for IC Logic Power - taking 5V off the arduino
    - Pin 8 is power to the coil (as noted above, I forgot to include this in my diagram - but it runs from the 24V power supply + rail)

    As I said, very much a newbie, so please correct anything you can see wrong.
     
  11. mightyconfused

    mightyconfused

    6
    0
    Apr 9, 2013
    With some help, I managed to figure out the problem was because I didn't have the grounds connected between the Arduino and 24V power supply. Once I connected the two ground rails, the circuit started working.

    I've updated my diagram to show (and also fixed up my drawing errors previously - like forgetting the lead to pin8).

    H-Bridge_bb_working.jpg

    Thanks all for your help. If you can see any further errors in my design, please let me know.
     
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