Can I use a Zener for 556 protection from relay coil back EMF?

Discussion in 'The Projects Forum' started by mrpict, Nov 19, 2012.

  1. mrpict

    Thread Starter New Member

    Nov 4, 2012
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    I have a beginner question please. I am trying to design a circuit to latch and unlatch a dual coil 4PDT relay with a SPDT toggle switch. The circuit will be battery operated so to save power I am using a latching relay.

    I intend using a 556 timer to provide two monostables to pulse the reset or set coils on the relay depending on the position of the toggle switch. I have got most of the way researching and building how I might possibly do this and have the timer circuit flashing LEDS with ~50mS pulses no problem.

    I am however unclear about how to protect the 556 from the relay coil back emf without degrading the switching speed of the relay. I have seen plenty of examples of using a single diode flyback in parallel with the coil but as I understand it this will delay collapse of the coil magnetic field and delay the speed and forcefulness of the armature operation. To keep switching speed snappy, would it be possible to use 36V zeners (D5 and D6) in the configuration shown? I am thinking that D1 and D2 should block 36V reversed (1N40004 good for 400V I think), and block any circuit developing through the 556. However the ground 0V rail is still connected to the -ve ends of the coils. Am I right in thinking that no voltage will develop here as far as the 556 is concerned (i.e. remains at 0v as the coil winds down). I will link to schematic of my proposed circuit here...

    http://pict.co.uk/556_RD_with_Relay.gif

    Apologies if this is a really dumb question, but if anyone can offer any learned clarification I would be most grateful.

    John
     
  2. ErnieM

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    Apr 24, 2011
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    Ya wanna know something funny? I've worked around and on several suppressor networks, either stand alone or in other circuits, all sold to several major relay manufacturers, designed by us or built to their prints, and the one common thing is every one uses a 1N4xxx diode and a 36 volt zener.

    Yeah, you not only reinvented the wheel, but you also made it the "standard" size.

    By all means use this scheme, you nailed it 100%.

    (all that assumes you have it connected correctly: I couldn't follow how the 555's were controlling the relay coils in your schematic.)
     
  3. mrpict

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    Nov 4, 2012
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    Thanks. My lucky day, though I wouldn't have known till you told me. Pretty sure the 556 wiring is OK. It seems happy enough with LEDs+resistors between the outputs on pins 5 and 9 and ground. Flash flash as I toggle toggle etc.

    Not sure I am completely understanding why I got it right though. Am I correct in thinking that as long as D1 or D2 are not reverse biased beyond their limit the 556 remains happy? i.e. I could use a 100v zener with a 1N4004 (400v peak inverse) but not with a 1N4001 (50v peak inverse). Also since D1 or D2 are not conducting in reverse there is no effect on the 0v ground rail as it is not forming a circuit that includes the coils (i.e. ground only connected to one end of the coils and no path to ground from the other end of the coil via the 556). Sorry for my struggling with this, but hoping someone can confirm my logic above is sound so I can learn and move on.
     
  4. THE_RB

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    The zener will only help the speed on relay release, not on engagement. Is that important? Also the speed increase won't be that much and is only really useful if your relay contacts are carrying large currents at the time of release.

    Also, zeners are notoriously unreliable, and you can probably get a very similar release speed just putting a resistor across the relay coil so that the back emf voltage is about 36v.
     
  5. mrpict

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    Nov 4, 2012
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    Speed of release maybe won't be important. I guess I am mainly interested in how to solve the "problem". The relay will be switching a buffered guitar signal at mA current levels, so maybe not important unless it mitigates any unclean bouncy disengagement. The resistor is an interesting idea. If I assume a 12v coil supply and 15 mA coil current (800R coil) then at 36v I would need resistance of 2400R. I presume this has to include the coil resistance so I figure 1600R in parallel with the coil would limit the maximum voltage to 36v. Is that how you would calculate it? Many thanks.
     
  6. THE_RB

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    You can't calculate it, you need a 'scope to see the back EMF voltage and adjust the resistor to suit.

    If the relay just switches audio then you should be fine with any normal diode (1N4148 or 1N4004 etc) across the coil, release speed does not matter and most commercial audo equipment that uses relays in the signal path just uses a standard diode on the relay coil. :)
     
  7. Ron H

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    Guys, you are missing something. The flyback voltage on the relay is negative (as you know). It will forward bias the series diode (D1 or D2, which have no purpose) and the collector-base junction of the pulldown NPN in the output stage of the 556. The zener will never conduct, because the voltage will never go low enough. I don't think it will damage the output stage, as the flyback current will never exceed the forward current of the relay.
    Zener flyback clamps only work when the driver goes open circuit. The 555/556 has a push-pull output driver.
    Regarding resistor flyback clamps, the same open-circuit driver principle applies. You CAN calculate the value. The initial current through the resistor will equal the relay forward current. The current will decay with a time constant of (Lcoil/(Rcoil+Rshunt)).
     
    Last edited: Nov 21, 2012
  8. ErnieM

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    Forget the scope you can calculate it (just as you have). The current thru the coil (basically an inductor) is continuous thru the turn off (meaning the value of current stays the same), so it just changes paths from the 555 to the resistor. Ohm's law applies, so if you want a 36V turn off voltage the resistor should be 36V/15mA=2400R, just as you calculated.

    But that is a L/R time constant delay and will be slower then using a zener. A zener is the industry standard. It is used in military and aircraft applications where failure simple is not an option.

    But the zener scheme you initially chose is just fine.

    Keep the series diode (D3 or D5) with the resistor with the snubber for the resistor: it keeps your drive from trying to also power the resistor.

    D1 and D2 are going to keep the snubber flyback voltage from causing an over voltage on the 555 output stage, that's safe no matter what the output max voltage rating is, meaning I didn't look up the spec but would use those diodes that same way no matter what the spec sheet says.
     
  9. Ron H

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    Nope. See my previous post. The voltage isn't a problem, but the zeners and D1 and D2 do nothing.
     
  10. ErnieM

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    Yep. The voltage spikes to 36V on the 555 output. The Absolute Maximum Rating on Vcc is 18V. Nothing specific as to the output but I would keep it at or below Vcc.
     
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  11. Ron H

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    Do you mean -36V?
     
  12. Ron H

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    My previous comments were based on analysis, and LTspice simulation. However, both assumed that the equivalent circuit in the National LM555 datasheet is correct.
    I just now went out to my bench and ran hardware tests, driving a 10mA, 12V relay coil with both bipolar and CMOS 555s.
    There was no significant flyback voltage on either one.
     
    Last edited: Nov 22, 2012
  13. ErnieM

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    I just might. That schematic is not very clear to me, and since it's not my problem I am not redrawing it. If it does spike negative (and I thing you are correct sir) then it may do not very nice things to the 555 output even with that diode where it is.
     
  14. Ron H

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    See post #12.
     
  15. SgtWookie

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    John,
    Since you are using battery power, you really should use CMOS timers. In both bjt and CMOS timers there is a 3-resistor voltage divider; in the bjt version it's 15k; CMOS is 300k. You''ll go thru fewer batteries with CMOS. CMOS versions can sink up to 100mA and source up to 10mA. If your coils require more than about 2/3 that, you will need a driver circuit. You might consider using a logic level MOSFET, like an IRLD014 or IRLD024, which come in a handy 4-pin DIP package. Google their datasheets.
     
    Last edited: Nov 22, 2012
  16. mrpict

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    Nov 4, 2012
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    Leaving D1 and D2 in was initially an oversight on my part. I have another application that requires pulses on the same output regardless of toggle switch position. In that configuration the outputs from each timer are joined together. The diodes (D1/D2) were there to stop the pulse from one timer shorting to ground via the other output which would be low.

    This confuses me. I was under the impression that D1/D2 would be reverse biased by the back EMF when the coil is turned off. With the zener scenario If D1 or D2 can't conduct, how does the 556 output see the 36V? What am I missing? Leakage current through the diode?

    Regards,
    John
     
  17. mrpict

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    Nov 4, 2012
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    Great point. I actually came to this realisation yesterday... I was measuring realtime current consumption using CMOS TLC556 versions (~1mA) in the prototype and reading the datasheet for the LM556 to check output source capabilities. I'll need more than the 10mA available for the relay so I tried adding a 2N7000 mosfet to the output and that seemed to work but the switching is a tad slower (source to ground, timer out to gate, drain to LED and other end of LED to +ve). I am thinking there may be some capacitance in the D1 to gate connection delaying it turning off. I'll try tying gate and source together with a resistor tomorrow. Regards, John
     
  18. THE_RB

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    I didn't look at the circuit. ;) And of course assumed the relay was being driven in a typical fashion by a single switch (transistor or FET). My apologies.

    I'll argue this one having actually done that tuning operation using a 'scope. The relay coil is not a typical inductor, it is very lossy on a back EMF spike at high speeds as its armature is solid iron.

    Uisng the logical formula when the relay coil is deenergised with no snubber (ie infinite ohms) it would generate infinite volts, but due to the very poor behaviour of the relay coil as a high speed inductor a 12v relay coil makes only about 150v or so. A well chosen resistor only needs to be about 5% to 10% of the relay DC current to snub that down to a usable value say <40v. This was common practive on relay banks when i worked in industry in the early 80's on equipment that was older than that. In those days using zeners on relay coils was considered unreliable and preference was to use good voltage rated transistors to drive the coil, and R or RC snubbers on the coil.
     
  19. Ron H

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    When the 556 output goes low, the relay coil voltage instantly drops to around -0.7V (due to D1/D2). The coil current cannot change instantly, so it continues to flow through the diode D1 or D2. The part that's hard to understand here is that, even though the 555 pulldown transistor will normally sink current, in this case it sources current from ground until the coil current decays. The presence of D1/D2 does nothing, because it is always forward biased, or at zero current.
    See the attached simulation. I verified this in hardware. I got similar results with a CMOS 555 in the hardware test.
    As you can see, a snubber will do nothing, because the 555 does the snubbing. The down side is, I don't see any way to speed up the relay turn-off, short of adding a transistor. The 2N7000 should allow you to do that.
     
  20. Ron H

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    I won't argue with your experience. I'll just tell you mine (from 10 minutes ago).
    I added a 2N7000 between the 555 and the relay in my test circuit. The relay is a small unit with a 12V, 10mA coil. I snubbed it with 1kΩ, and the drain voltage peaked at about 21V, very close to Vcc+(Icoil*Rsnub).
     
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