Back EMF-diode direction?

Discussion in 'General Electronics Chat' started by somlioy, Jun 17, 2011.

  1. somlioy

    Thread Starter Member

    Jul 28, 2010
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    So, I've been doing a bit research about stepper-drivers. I've seen some circuits have the back-EMF diode between the coil and V+ (cathode towards V+), some have it between the coil and ground (cathode towards coil) and some have both.

    What's the most "correct" way to do it and why?

    Also, would it be smart to use a pull-up or pull-down resistor on the base of my transistor (TIP122) ?
     
  2. #12

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    Nov 30, 2010
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    Which TIP122?
     
  3. t_n_k

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    The diode must effectively be in parallel with the coil. The diode cathode goes that end of the coil which is the more positive when energized.
     
  4. SgtWookie

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    Jul 17, 2007
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    As far as the reverse-EMF diodes, you can place them either way; from the coil to ground or coil to +V. The reverse-EMF will likely be stopped more quickly if you go from the coil to ground, as the power supply is then what is completing the circuit. If you go from the coil to +V, then when the polarity across the coil reverses when the circuit is opened by the transistor turning off, it's basically just the Vf of the diode and parasitic resistance of the coil that causes the current to decay.

    As far as a pull-down resistor - it couldn't hurt. Use a value of 3 to 10 times what your base current limiting resistor is.
     
  5. t_n_k

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    I've never been convinced that a diode from say the switch end of the coil to ground actually does anything.

    My reasoning is ....

    When the switch element (say transistor) is turned off the coil back emf will try to maintain the current in the coil. The back emf is of opposite polarity to the energizing voltage, so the diode must have the anode at the switch end of the coil (say the collector) for the current to "commutate" to the diode and thereby continue flowing in the coil.

    It might be worth drawing some pictures to make sure we are talking about the same circuit topology.
     
  6. t_n_k

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    This is what I mean ...
     
  7. Ron H

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    Considering an inductor to +V, the other end to a switch, and the other end of the switch to ground:
    The only time a diode across the switch will conduct is when the voltage across the switch tries to swing negative. This only happens when the inductor resonates with capacitance on the switch node, causing the voltage to ring. By this time, the switch may already be damaged due to the high voltage swing that precedes the undershoot. The diode does nothing to protect the switch from overvoltage.
     
  8. Ron H

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    It looks like your transistor is not staying in saturation. I think you need more base drive (or more coil resistance) to illustrate what would happen with a real relay driver.
     
  9. t_n_k

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    Thanks Ron - you're right. I've redone the simulation to better (?) illustrate the result. I also put a different transistor & diode type with a higher reverse breakdown.

    It's the "nasty" behavior of the second case that is of interest to me.
     
  10. The Electrician

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    Oct 9, 2007
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    It doesn't do anything unless the diode experiences reverse voltage breakdown.

    What should be used in that position is a zener diode to clamp the voltage across the switch to a safe value.

    In your simulation you show a 15 volt supply voltage. A 30 volt zener across the switch would limit the voltage across the switch to 30 volts and would limit the voltage across the coil to 15 volts. The same result could be achieved by connecting a 15 volt zener across the coil (in series with a non-zener diode to prevent the zener from conducting in the forward direction). In both cases, the voltage across the switch would be clamped to 30 volts.

    However, connecting the zener across the switch results in more dissipation in the zener than connecting it across the coil. An interesting article describing this effect was published in one of the magazines many years ago.
     
  11. t_n_k

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    Thanks Electrician - do you maybe have a link to the article?
     
  12. The Electrician

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    I scanned it from a library copy of EEE Magazine, Jan 1968

    Post an email address, or PM me one, and I'll send it to you.

    It's an interesting phenomenon that is obvious once you think of it, but not often mentioned.

    If you put a zener across the inductor and turn off the switch, the zener must dissipate the energy in the coil, and in a time determined by the volt-seconds the zener allows, and the discharge current flows only in the inductor.

    But, if the zener is across the switch (a 15 volt zener in this case), the same voltage appears across the inductor during its discharge, but the current also flows in the power supply, and energy is drawn from the power supply as well as the inductor.

    If the zener (or other clamp device) is placed across the switch, and as the clamp voltage is reduced until it's just a little more than the supply voltage, the inductor takes longer to discharge and the clamp has to dissipate more energy.

    Whereas, if the clamp is directly across the coil, the energy it has to dissipate is just the energy stored in the inductor no matter what the clamp voltage is. Of course, if the clamp voltage is low, the inductor takes longer to discharge. If the coil is a relay coil, a low clamp voltage will make the relay slower to open.
     
    Last edited: Jun 18, 2011
    t_n_k likes this.
  13. ErnieM

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    The Electrician is spot on in his analysis.

    The company I work for sells a line of "suppressors," basically the diodes that go into relays for several major relay manufactures. I also designed a time delay controller for use inside a relay "inspired" by one manufactures earlier design (he used a 4000 gate and I used a PIC).

    What many relay manufacturers use inside the relay is a diode and a zener. Some will just use a diode but the combo is perhaps a better choice as it turns off faster.

    It looks like this:

    [​IMG]

    We package up a 1000 volt rectifier and a 36V zener.
     
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