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Backvoltage from relays?

Discussion in 'Electronic Design' started by David Harper, Jul 10, 2003.

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  1. David Harper

    David Harper Guest

    In a standard relay, how much backvoltage can I expect to see across
    the coil terminals when the relay is turned off (and the magnetic
    field collapses)? For instance, say if the coil had a pickup voltage
    of 5V and a resistance of 200 Ohms, would there be any significant
    backvoltage induced?

    I've been told for solenoids, I should design for a potential
    backvoltage of three times the operating voltage. Since the
    resistance of the coil in the relay is so high, I was wondering if
    this rule of thumb would be applicable for relays also.

    Thanks in advance for any help!
    Dave
     
  2. CFoley1064

    CFoley1064 Guest

    In a standard relay, how much backvoltage can I expect to see across
    It might be better to figure some way to satisfy Mr. Lenz (such as putting an
    R-C snubber, zener, MOV or back-biased diode in parallel with the relay coil)
    than jacking up the V(ce) (sat). Relay and solenoid coil inductances and
    resistances vary significantly.
     
  3. CFoley1064

    CFoley1064 Guest

    V(ce)(max). Sorry.
     
  4. Ted Wilson

    Ted Wilson Guest

    V = L dI/dt, which means the faster you try to reduce the current, the more
    volts you get. The resistance of the coil doesn't have a lot to do with it,
    other than it sets the current you're switching.

    In practice, it's normal to put a diode across the coil to catch the
    voltage, (diode reverse biased by coil drive voltage), otherwise the voltage
    will rise to a level that enables the current to continue flowing and you'd
    be amazed how high that can be if you don't do something about it. (Years
    ago, I designed a pulse-induction metal detector where an air-cored coil of
    30 turns at 10" diameter was hit with pulses of about 0.5A and the inductive
    fly-back when the coil was turned off got up to around 800V!).

    Unless you are controling the rate of change of current in some way, (a bit
    pointless when a simple diode solves the problem so easily), you will end up
    with enough volts to break down the device driving the coil. Unless the
    driver has an adequate avalanche rating, that means it's going to 'pop'.

    Regards

    Ted
     
  5. There's not much point in calculating the exact value, as it will vary
    widely
    depending on uncontrollable factors, such as coil interlayer
    capacitance, coil Q, stray capacitance to ground, and more.

    Most engineers put a R-C, or diode, or varistor across the coil to damp
    the inductive kick. This is almost always 100% indicated, EXCEPT
    if you need the relay to release very quickly. The snubbers will extend the
    release time a bit.
     
  6. It depends on the relay you use, but in general, the case of the backvoltage
    being 3 times VCC is hardly ever true. Myself, I've seen relays inducing a
    voltage of more than 50 times VCC, so there is no general rule in this case
    and sometimes no transistors that can handle the voltage. Try using a diode
    antiparallel to the relay coil to get rid of the spikes the coil produces.
    If there is a need for the relay to switch off very fast and the current
    through the diode is high enough not to allow this, connect a zener diode
    in series with the first diode. This will raise the voltage in the zener
    circuit and result in slightly lower peak current. In this case however
    make sure that the other equipment can handle voltages more than VCC plus
    the zener spec voltage. If the relay is slow enough anyway and speed is not
    a concern, a capacitor can be connected to the coil, although this is
    usually not the preferred solution (beware of possible resonance etc).

    Dimitrij
     
  7. Ted Wilson

    Ted Wilson Guest

    It's not high current in the diode that causes slow switch off, it's due to
    the fact that the reverse voltage across the coil is limited to a forward
    diode drop, (approx 0.6V).

    Sinve V = L*dI/dt, dI/dt = V/L

    For a given value of L, the lower you make V, the lower will be the rate of
    change of current. That's the point of putting the zener in series with the
    diode since this increases V to Vf + Vz, and the rate of current decay is
    therefore increased accordingly.

    The peak current will be the current that was flowing in the coil at the
    moment you switch off, whatever configuration of diode or diode plus zener
    you use.

    Regards

    Ted Wilson
     
  8. David Harper

    David Harper Guest

    What about putting a cap in parallel with the relay's coil? Has that
    ever been done before? Once the 'input' current was shut off (and the
    cap charged), the cap's voltage would drop until the coil's dropoff
    threshold was reached. This alone would help decrease backvoltage, as
    the coil's field would already be weaker due to a slower voltage drop
    to the dropoff threshold... correct?

    The bigger payoff could be when the coil DID dropoff. The enduced
    backvoltage would be offset by the remaining charge on the cap... ?

    Just an idea, but would this work?

    Thanks in advance!
    Dave
     
  9. Without a snubbing or daming diode, the back voltage of an ordinary 5
    volt relay can easily exceed many kilovolts.

    It all depends on di/dt.

    Zero di/dt = infinite back voltage.

    --
    Many thanks,

    Don Lancaster
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    voice: (928)428-4073 email: fax 847-574-1462

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  10. One danger: An ordinary diode can significantly lengthen the relay
    dropout time.

    A zener or a MOV is a better choice for faster relay circuits.

    --
    Many thanks,

    Don Lancaster
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    voice: (928)428-4073 email: fax 847-574-1462

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  11. If it is underdamped, the relay could chatter badly.

    --
    Many thanks,

    Don Lancaster
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    voice: (928)428-4073 email: fax 847-574-1462

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
  12. CFoley1064

    CFoley1064 Guest

    What about putting a cap in parallel with the relay's coil? Has that
    Yes, but you put a resistor in series with the cap. The switching element
    ususlly doesn't like the feeling of a dead short at the moment of turn-on.

    Back in the days of relay logic, you started with the idea of how much current
    the switching relay contacts could handle, and then you chose fraction of that
    to select the resistor. You then just bumped up the value of the cap until the
    spark you saw at the switching relay just disappeared. Long ago and far
    away...
     
  13. Active8

    Active8 Guest

    i've run sims on electromagnetic clutches driven by MOSFET switches
    where it ran into the 4 kV range.

    i saw a circuit once where the diodes were connected across the switch
    transistor (which did the same thing) and in one case it looked like a
    Schottky. maybe it was a poor symbol for a Zener.

    what do you think? would a Schottky speed things up and release the
    relay faster or is it more likely they used a Zener?

    mike c
     
  14. Ted Wilson

    Ted Wilson Guest

    If the device was across the the coil driver, it would have been a zener,
    not a Schottky. Putting the zener there eliminates the need for a series
    diode required if you put the zener across the coil.

    Putting a Schottky across the driver would do nothing to supress inductive
    fly-back, except that it would then be a toss-up which broke down first -
    the Schottky or the driver.

    Regards

    Ted Wilson
     
  15. Active8

    Active8 Guest

    ah, sooo .......... makes sense now that think about it. i think i saw
    that "Zener across the driver set up" in a H-Bridge circuit.
     
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