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Relay Arc Supression Circuit

Discussion in 'Electronic Design' started by [email protected], Mar 14, 2006.

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  1. Guest

    I have the following circuit:

    | | |
    | | |
    330 Ohm Diode |
    | | |
    | | 5Vdc Relay Coil
    | | |
    LED 5.1V Zener |
    | | |
    | | |
    | | |
    To Micro----1kohm--+B C |Transistor
    | E |

    I'm concerned about reverse bias on the LED when the realy lets go. I
    measure about 7.5 Vdc across the LED when the relay lets go. I think
    the zener and the diode act as a good method to handle the coil
    magnetic field collapse and ensure fast armature motion but I do
    violoate the 5 Vdc LED reverse bias voltage. Any comments or
    suggestions for this circuit? I need to maintain the LEDs for
    visibility and troubleshooting. I'm hoping this thread discusses some
    general techniques for realy arc supression and addresses the LED in
    this specific example.

  2. Ian Stirling

    Ian Stirling Guest

    Why diode in series with the zener?

    Remove the diode and zener, place across the LED.

    Or actually, an ordinary diode across the LED.
    Turns off, and relay coil switches ends, and the 330 ohm resistor is
    used as a snubber to absorb the energy.
    Depends on the ratio between the 330 ohm and the relay coil impedence,
    and the Vce of the transistor.
  4. I would try this:
    | | |
    | | |
    330 Ohm Diode |
    | | |
    | +---------+ 5Vdc Relay Coil
    | | | |
    LED | 5.1V Zener |
    | | | |
    | | | |
    | | | |
    +-----+ +------------------------+
    To Micro----1kohm--+B C |Transistor
    | E |
  5. John_H

    John_H Guest

    Do you feel the Zener voltage is important? I only ever used simple diodes
    across the relay coil. The voltage is quite a bit smaller to collapse the
    magnetic field, but how long do you believe that collapse will take driven
    into 0.7V? If the zener is removed, the LED is protected. If you change to
    a 3.3V zener, your reverse voltage will hit about 5V - on the edge, but
    better. If you feel strongly that the zener is needed, I like Ian
    Stirling's idea of the doide across the LED which also gives you a snubber
  6. Guest

    The LED is supposed to be on when the transistor provides a ground
    showing that the relay is active. Wouldn't this connection reverse
    bias the zener (acutally it would foward bias it but zeners are
    normally used in reverse bias). So I don't think the LED would
    function properly.

  7. Guest

    If you have time visit:
    and go to page 3 "Other Arc Supression Methods."

    This is where I based the zener diode in series decision. Could you
    point me to a better reference? The article above will discuss the
    diode sereis resistor method you mention as well but concludes a zener
    is a better way to go for minimizing hold up time. I'm certain this
    type of circuit is done routinely so any other resources would be

    Thanks for your time!
  8. Guest

    Hmmm, I'm no expert at all. That said, enlighten me! I thought the
    rule of thumb was to target a zener voltage near the realy coil voltage
    (5Vdc). I think the gist is that this will effectively start the
    magnetic collapse sooner and provide a path for a 5.1 + 0.7 amount on
    energy absoprtion back to the rail. If you go to high, it never
    activates as the zener knee voltage is not met. If you go too low you
    don't get the benefit of the zener and it acts more like a solo diode.
    When both are in play, this aids in releasing the armature more
    quickly. The load on the contact side of the realy is another realy
    coil which drives an inductive laod.

    Thanks for your time!
  9. "For those applications that cannot tolerate lengthened hold-up time,
    a resistor may be placed in series with the diode. The resistor does,
    however, lessen the effectiveness of the diode and, usually, a
    compromise must be reached by trial and error."

    That is so sad.

    Best regards,
    Spehro Pefhany
  10. John_H

    John_H Guest

    Is holdup time an issue for you? We never had a need for 2 ms vs 9 ms.

    Would you mind providing a link to the relay you're using? (to help
    determine holdup time)

    Are you running a high voltage through the relay contacts such that contact
    arcing is your big concern?
  11. John_H

    John_H Guest

    By using a 3.3V Zener instead of 5.1V, you have a Zener + Diode reverse
    voltage maximum of about 5V instead of your measured 7.5V. The Zener plus
    diode at highest current gives you the max reverse voltage for your LED.

    If you want to use the Zener because you feel the reduced holdup is
    important, consider a higher voltage zener and either 1) a diode reversed
    across the LED or 2) a cap across the LED that just needs to keep the
    reverse voltage from getting to 5V during the coil discharge.
  12. John Fields

    John Fields Guest


    | |
    [300R] |
    | [COIL]
    [LED] |
    | |
    | |
    C |K
    µC>-------[1KR]----B [ZENER]
    E |
    | |

    Choose the Zener voltage to get the armature release speed you need
    commensurate with not zapping the transistor, and use the 1N4148 to
    keep from reverse biasing the LED.
  13. Guest

    You didn't show the cathodes of the diode or zener, so I made some

    I assumed that the cathode of the diode is at top, and cathode of zener
    is at bottom.

    So the LEd would be on when the transistor is on, through the forward
    biased zener.

    At turn off, the top diode is forward biased (while the inductance
    discharges through the two diodes), so the LED sees no more reverse
    voltage than than the top diode's forward drop.
  14. Guest

    Hmmmm, so the zener sinks the stored energy at relay releae to GND in
    this design. How would you recommend the sizing of the zener? I
    surmise it would be < Vce max of the transistor? Doesn't this present
    a similar problem as I have with the reverse bias on the LED? What I
    mean is, the zener will clamp to some V but suppose the stored energy
    is >> V then the transistor could be zapped anyway, right? And of
    course the diode doesn't relate to the stored energy in this design
    except for blocking any current during stored energy release.

    Thanks for your time.
  15. John Fields

    John Fields Guest

    To what article and to whom are you responding?

    Read this, From:

    "Summarize what you're following up.

    When you click "Reply" under "show options" to follow up an existing
    Google Groups includes the full article in quotes, with the cursor
    at the top
    of the article. Tempting though it is to just start typing your
    please STOP and do two things first. Look at the quoted text and
    remove parts
    that are irrelevant. Then, go to the BOTTOM of the article and start
    typing there.
    Doing this makes it much easier for your readers to get through your
    They'll have a reminder of the relevant text before your comment,
    but won't have to re-read the entire article. And if your reply
    appears on a site
    before the original article does, they'll get the gist of what
    you're talking about."
  16. Use what John Popelish told you. You obviously don't have a clue
    about diodes and zeners, so take the advice and stop questioning.
  17. Guest

    The 3.3 is my punt plan for the current artwork or to populate a 0 ohm
    resistor where the zener is.

    Here's what I'm not understanding...with the 5.1 zener and per my
    o'scope, it appears the worst case reverse bias on the LED is 7.5 as I
    aforementioned and I see about 3V across the series resistor, Rs. So
    this sums to ~10.5 Vdc. So does the 5V rail + 5.1 (zener) + 0.7
    (diode) create a voltage of about 10.8 reverse bias? This seems wrong
    to me as the 5V rail is fixed but empirically the 10.5 is close to the
    10.8. So, in my stupidity, I changed the value of Rs (reduced it as I
    could afford to have more current run through the LED in normal
    operation) but duh, the I-V charactereistics of the LED in reverse bias
    pretty much fix the voltage drop across Rs. So I think your statment
    about 5.1 + the diode drop at worst current must be what is yielding
    the ~10.5 Vdc I see across the LED and Rs. Is that correct? Will the
    diode/zener deviate that much from the nominal 5.1 + 0.7? Also, in
    your experience, if this relay swithces slowly, would a diode by itself
    be adeuqate? It only runs the coild of an offboard realy and is
    switching 24AC to this other realy. I know some of these are basic
    questions but I appreciate your input.

    Thanks again for your time.
  18. Guest

    RTFMed, thanks for that. I'd been using the reply at the bottom of the
    post and by default no text is included whatsoever. Sorry for the
    confustion I rarely use the web interface and usually use Knode to bad.
  19. John Fields

    John Fields Guest


    Go to:

    download the simulator, and run this:

    Version 4
    SHEET 1 880 680
    WIRE -384 176 -384 64
    WIRE -384 320 -384 176
    WIRE -384 416 -384 400
    WIRE -352 176 -384 176
    WIRE -208 176 -272 176
    WIRE -112 320 -112 288
    WIRE -112 416 -384 416
    WIRE -112 416 -112 400
    WIRE -48 288 -112 288
    WIRE 16 176 -144 176
    WIRE 80 288 32 288
    WIRE 144 176 80 176
    WIRE 144 176 144 144
    WIRE 144 192 144 176
    WIRE 144 240 144 192
    WIRE 144 416 -112 416
    WIRE 144 416 144 336
    WIRE 208 64 -384 64
    WIRE 208 144 144 144
    WIRE 272 192 144 192
    WIRE 272 272 272 192
    WIRE 272 416 144 416
    WIRE 272 416 272 336
    WIRE 272 448 272 416
    FLAG 272 448 0
    SYMBOL npn 80 240 R0
    SYMATTR InstName Q1
    SYMATTR Value 2N4401
    SYMBOL ind 192 48 R0
    SYMATTR InstName L1
    SYMATTR Value .1
    SYMATTR SpiceLine Rser=100
    SYMBOL zener 288 336 R180
    WINDOW 0 -42 32 Left 0
    WINDOW 3 -120 0 Left 0
    SYMATTR InstName D1
    SYMATTR Value BZX84C15L
    SYMBOL diode 16 192 R270
    WINDOW 0 32 32 VTop 0
    WINDOW 3 0 32 VBottom 0
    SYMATTR InstName D2
    SYMATTR Value 1N4148
    SYMBOL res -256 160 R90
    WINDOW 0 0 56 VBottom 0
    WINDOW 3 32 56 VTop 0
    SYMATTR InstName R1
    SYMATTR Value 300
    SYMBOL voltage -112 304 R0
    WINDOW 3 -152 153 Left 0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR Value PULSE(0 5 .1 1e-6 1e-6 .1 .2 3)
    SYMATTR InstName V1
    SYMBOL voltage -384 304 R0
    WINDOW 3 -6 144 Left 0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR Value 5
    SYMATTR InstName V3
    SYMBOL LED -208 192 R270
    WINDOW 0 72 32 VTop 0
    WINDOW 3 0 32 VBottom 0
    SYMATTR InstName D3
    SYMATTR Value QTLP690C
    SYMBOL res 48 272 R90
    WINDOW 0 0 56 VBottom 0
    WINDOW 3 32 56 VTop 0
    SYMATTR InstName R2
    SYMATTR Value 1000
    TEXT -418 506 Left 0 !.tran 1
  20. Guest

    RTFMed, thanks for that. I'd been using the reply at the bottom of the
    post and by default no text is included whatsoever. Sorry for the
    confustion I rarely use the web interface and usually use Knode to bad.
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