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Whats a "snubber diode"?

Discussion in 'Electronic Basics' started by Steve Evans, Nov 27, 2004.

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  1. Steve Evans

    Steve Evans Guest

    Anyone know?

    tnx,

    steve
     
  2. I think this term is often used to describe the application where a
    diode allows a peak voltage (usually created by an inductor that has
    had its current source switched off) to be limited by conducting
    current into a capacitor that is precharged to a voltage just below
    the desired limit. Often that precharge is just what is left after a
    resistor has bled the effect of the previous pulse from the capacitor.
     
  3. Jamie

    Jamie Guest

    used to absorb the reverse (flyback) voltage release from a
    coil when energized source is removed quickly.
    the flyback voltage is in reverse of what went in and can get
    very high in level which will short out things.
    the trick is to place a diode across the coil connections.
    the Cathode is connected to the + side and Anode to the - side
    when energized, the diode does not conduct. when source is
    removed quickly, the release will generate high voltage in reverse
    polarity. at this point the diode will conduct and anything above the
    cut off voltage of the diode will get absorb in the diode and protect
    other voltage sensitive components.
     
  4. dB

    dB Guest


    The voltage isn't "absorbed" it is prevented from developing.

    (What do you mean when you use the word "flyback"? Do you know where
    the term came from?)


    I'll give you 3 points out of 10 for English, and 3 out of 10 for the
    clarity of your explanation.

    The same point is dealt with in this item from a forum:

    http://p199.ezboard.com/fbasicelectronicsbasics.showMessage?topicID=1705.topic
     
  5. Steve Evans

    Steve Evans Guest

    That makes sense. But the diverted engery has to be dissipated
    somewhere. If the diodes simply in antiparallel with the source, itll
    act as a short circuit on the back emf. Why doesn't that (the energy
    of that reverse pulse) destroy the diode? Someonne else said a cap and
    bleed resitor can be used to store and discharge the pulses
    harmlessly, but no one esle has verfified this. Can we have some
    clarification, please?
     
  6. If the coil were made of superconductor, this might well be the case,
    but in the more practical copper wound coils, the wire resistance
    consumes most of the energy.
    http://www.maxim-ic.com/appnotes.cfm/appnote_number/848
    http://focus.ti.com/lit/an/slup100/slup100.pdf
    http://powerelec.ece.utk.edu/pubs/pesc2002_sssi.pdf
     
  7. CBarn24050

    CBarn24050 Guest

    Subject: Re: Whats a "snubber diode"?
    There seems to be some confusion here, a flywheel diode is a diode connected
    across an inductor. It provides a path for the inductor current when the drive
    voltage is removed (switch off). If say there was 1 amp flowing and the drive
    was removed then the current (1 amp) would circulate through the coil and the
    diode, as there is resistance as well as the diodes forward voltage drop, the
    current will decay to zero.

    Sometimes diodes are used with snubber networks, typicaly accross transistors,
    igbt or thyristors. They are in series with the cap and allow the switch off
    pulse to pass into the cap,as normal, but prevent the enrgy flowing back into
    the transistor when it switches on again. A small resistor accross the cap
    dissipates the charge from the cap during the "off" time.
     
  8. john jardine

    john jardine Guest

    If you short circuit a charged up inductor then absolutely nothing happens!.
    On the other hand, a charged capacitor would destroy the shorting wire.
    ....
    Use a mechanical switch and switch off a relay coil without a diode across
    the coil. The relay discharges and drops out near instantly and you'll
    probably get a spark across the switch contacts as thousands of volts come
    back off the coil.
    Now put a ('flywheel')diode across the coil and again switch off the relay.
    The relay takes ages to discharge and drop out and there is no spark as only
    0.7V can come off the coil.
    Now put a 33V Zener diode (+ a reverse diode!) across the relay and switch
    it off. It drops out quite fast with a peak of 33V across the coil.
    Similar effect occurs if you put a diode in series with (say) a 100 ohm
    resistor. This time the discharging coil voltage can be quite high and
    depends on the inductance value.
    Put a capacitor across the relay coil and you've made a nice resonant
    circuit that takes many cycles to decay before the relay drops out. (it's
    the relay wire resistance that finally dissipates all the stored energy).
    Capacitor + resistor across the coil gives a very lossy 'damped' tuned
    circuit ' but a number of measurements and calcs then needed to allow a
    balancing act on the numbers.

    Idea is that with an inductor the stored energy can be removed (discharged)
    by allowing it to develop a voltage across some kind of load hence lose
    it's stored energy as heat.
    Bigger the discharge load resistance, then bigger the voltage, then bigger
    the power loss, then quicker the discharge. Put a short circuit across the
    inductor and a big current would try to flow but that same current will also
    be charging up the inductor so nothing actually happens. Put an open circuit
    across the inductor and the voltage screams upwards until something gives.
    Tiny current flowing but high voltage and power dissipation hence fast
    discharge.

    The only inductor formula worth noting are ...

    Stored Watt seconds(Joules)=1/2 x Inductance Value x [current through it
    ^2].

    Amps per second though inductor = V across inductor / Inductor value.

    [snip 8" of tedious relay coil example calcs]

    These simple calcs are useful for test purposes. Generally it's easier and
    much more accurate, just to use Spice.

    regards
    john
     
  9. Steve Evans

    Steve Evans Guest

    On Sun, 28 Nov 2004 21:52:57 -0000, "john jardine"

    [snip]

    Tnx, john (and others). Thats a pretty comprehensive answer, i guess.
    I'll ponder on it for a while. I must say your first bold
    pronouncement about coils discharging harmlessly and caps desroying
    had me confused, but your explanation of these pheonomena is of
    considerable help!
     
  10. Somebody check me on this.

    AIUI a cap stores voltage and an inductor stores current, so the result is
    that when you discharge a cap you can't see a voltage higher than what you
    had put into it, and when you discharge an inductor you can't see a current
    higher than what it carried before discharging.

    If the cap sees zero resistance then it produces a large current. If the
    inductor sees infinite resistance then it produces a large voltage. Both of
    these conditions must be avoided.
     
  11. Steve Evans

    Steve Evans Guest

    IIrc,. they both store *charge*.
     
  12. Actually the inductor stores a magnetic field, which is converted back to a
    current. People say inductors store current because there is a direct
    relationship, but there is no direct relationship to charge. Charge is a
    function of the number of electrons (6 * 10^18 electrons per coloumb I
    think), but that parameter is not characteristic to the stored energy in
    either device, and measurement of columbs only takes you farther away from
    the parameters that matter.

    One big and one small cap/inductor in parallel/series both have the same
    voltage/current but not the same charge within them.
     
  13. Active8

    Active8 Guest

    Nope. Faraday's Law states that a time varying flux (which is caused
    by the "field") induces an EMF.

    d(Phi)
    EMF = - -------
    dt

    That flux can't drop to zero instantaneously, but you can get a
    pretty respectable voltage, anyway.
    it is in a cap because it takes work to get that charge in there.
    Yeah, like C and V.
     
  14. Mike, can you reconcile that please with the numerous references that say
    it's a current and not a voltage that is proportional to the rate of change
    in a magnetic field? I don't see how it can be both.
     
  15. Active8

    Active8 Guest

    Show me a law of physics that states such a proportion
    mathematically. How are you going to get a current in a transformer
    secondary if it's open? Same thing with a generator or motor - the
    EMF or back EMF is what we talk about.
     
  16. Obvious. Should have seen that. My physics is very rusty.

    But I asked (3 of my messages back) if I was correct in stating that the
    snubber can't see a current higher than what the coil carried before it was
    turned off. What about that?
     
  17. Active8

    Active8 Guest

    Ok. That would have been in reply to someone else, but lemme see...
    no, I don't see how you could, but I might be brain farting here.
     
  18. More accurately, capacitors store energy proportional to voltage
    squared, and inductors store energy proportional to current squared.
    For fixed inductances and capacitances, I think these are good
    generalizations.

    If the magnetic path can change (think relay armatures moving,
    solenoids with moving plungers, etc.) or if the current changes from
    all the turns to some of the turns (or charges through one winding and
    discharges through another winding) the current can change.

    Likewise, if capacitor plate spacing can change, voltage can change.

    There are somewhat unusual conditions but not at all unheard of.
     
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