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Protecting DC Output SSRs against inductive kickback

Discussion in 'General Electronics Discussion' started by istvanb, Jan 25, 2013.

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  1. istvanb

    istvanb

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    Dec 10, 2012
    Hi,

    I understand why do we need to protect an SSR against inductive kickback and I also understand how the flyback diode works (http://en.wikipedia.org/wiki/Flyback_diode). I have attached a picture about the common configuration.

    Now imagine the diode connected parallel to the SSR and NOT the relay coil. I was told its a good protection as well. I am somewhat concerned since I have never seen such a circuit, but at the same time I can not prove if I am right or wrong. (see picture2). i dont think this circuit makes any sense...

    Please help me out with an explanation!

    thanks a lot!
     

    Attached Files:

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    Last edited: Jan 25, 2013
  2. GreenGiant

    GreenGiant

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    Feb 9, 2012
    You would actually want to place the diode like you have it in the second picture.

    Make sure that the diode is rated for the same or higher reverse voltage/current than the relay you have
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The second method requires that the voltage source is capable of absorbing the spike. It may not be able to do that if it can't sink current (imagine it has a diode in series with it)
     
  4. istvanb

    istvanb

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    Dec 10, 2012
    Well, I dont think so. Or else manufacturers have to change their documentation since I have seen the first method in dozens of them and have never seen the second.
     
  5. istvanb

    istvanb

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    Dec 10, 2012
    So lets assume the power supply can not sink the current. What would happen to the SSR/the diode in the second configuration?
     
  6. BobK

    BobK

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    Jan 5, 2010
    The current from the indcutor has to find a path. This may include blowing out semiconductors.

    Bob
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    To expand on Bob's comment, *something* will conduct in a manner that it is not supposed to.

    You may get an arc somewhere, a diode may conduct in the reverse direction, or something else in the circuit will conduct to prevent the voltage reaching a (theoretically) infinite value. One of these things (or a variant of them) is always going to happen. If the device that breaks over is not designed to do so then you are likely to cause some amount of cumulative damage until it eventually fails.

    If there is a capacitor in the circuit, this impulse will be limited in amplitude, and may not cause damage if the rise in voltage can be tolerated. Circuits also have inductances, and they may work for or against you. It gets very complex to define the behavior.

    But it does depend on your power source and the magnitude of the energy contained in the pulse compared with the capacitance.

    Because there is always some capacitance (even in wiring) the theoretical infinite voltage is always limited to a lower value.

    If your source of power can sink current, then the voltage rise is going to be limited by the transient response of the power source and its impedance. Any inductance or capacitance will also need to be factored in.
     
  8. Starbuckin

    Starbuckin

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    Jan 22, 2013
    I agree with istvanb...

    The diode is conducting in REVERSE, becoming a near short circuit for the back emf/inductive kickback of the relay coil NOT the power supply. Even though the diode is energized it is ONLY the relay that "sees" the short (or low impedance load, I should say)... This is because current can flow in only ONE direction through the diode and it is the OPPOSITE of the normal current path of the power supply.

    It does not matter which circuit you wire up (picture one or picture 2) as far as the power supply goes, it will not conduct extra current...

    The power supply sees practically an OPEN circuit whether the diode is on or off does not matter. The diode is conducting ONLY reverse current produced by the inverse voltage created by the collapsing magnetic field of the coil the instant it is switched off...

    I am an Associate Electronics Engineer and that is my take on it.

    I have NEVER seen a reverse biased diode in the second picture configuration, only reverse biased DIRECTLY across the relay coil itself... That is the ONLY logical way to absorb the inductive kick-back...
     
  9. Starbuckin

    Starbuckin

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    Jan 22, 2013
    O.K. I have reviewed the 2nd picture circuit in more detail...

    I agree with you Steve, that the output circuit in the power supply or (voltage source) WILL indeed have to sink the current produced by the back emf and and diode turning on because the current MUST flow through the power supply because it is in the circuit. Current must flow through everything in the circuit for any current to flow at all...

    However, since the current will be in reverse direction of the voltage source, the power supply output circuit is simply acting as a resistor, the power supply itself produces no current it is only absorbing it.

    Most power supplies I have seen would not be able to do this. You would basically be subjecting the power supply to a huge voltage that is opposite of the supply itself, DIRECTLY across the output leads. Not to mention that the circuit itself is very low impedance (High Voltage D.C. Pulse with a forward biased diode and a low D.C. resistance coil wired strait to the output of the power supply). This would not be a good thing for the power supply, in my opinion...

    I stand by my original statement that putting the diode directly across the relay coil is the ONLY logical way to absorb the back emf.

    Most systems have tons of other circuits hooked to the output leads of the power supply that is powering a relay circuit similar to the one above. ALL of the other circuits would also be subjected to the huge D.C. pulse of the back emf and would also have to either sink the current or deal with the voltage, if circuit #2 were used.

    Why would anyone want to use #2 when a simple reverse biased diode across the coil that is chosen right would easily take care of the back emf?

    Also #1 limits the effects of the inductive kick-back to ONLY the coil and the diode. As long as the diode conducts quick enough and can handle the current, the rest of the circuit never knows the back emf is there and the problem is contained right there at the relay coil...
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    The effect of #1 is to allow current to flow through the coil longer, and to make the switch-off slower.

    #2 may be a misguided attempt to allow switching to happen faster.

    A better way to do this is to place a zener in series with the diode (as in #1). The losses through this will be much larger and the energy in the circuit will be dissipated faster (naturally the zener will be doing most of the dissipation!)

    The zener voltage should be chosen so the peak voltage is still safe for the switching element.

    Since we know that the maximum current through the diode will be the operating current of the coil, we can also choose a series resistor to drop a "safe" voltage. This will be slightly slower that using a zener, but resistors capable of higher power instantaneous loads may be easier/cheaper to find.
     
  11. istvanb

    istvanb

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    Dec 10, 2012
    Thanks for all of you for your ideas.
     
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