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diode for collapsing magnetic field

Discussion in 'General Electronics Discussion' started by Blank Stare, Oct 10, 2012.

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  1. Blank Stare

    Blank Stare

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    Oct 10, 2012
    OK, here's a very simplified drawing of what I believe is the sum of what we have discussed.

    Sure hope I got this right, because if not, I am definitely too dumb to play at this sport. :rolleyes:



    BTW - that 40 amp diode just bolts to ground, right?
     

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    Last edited: Oct 21, 2012
  2. CocaCola

    CocaCola

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    Apr 7, 2012
    I have not been following along fully, is the pump only supposed to run when both the remote switch and pressure switch are tripped? Because that is they way you have it drawn, neither switch will work on it's own, it's both or nothing... If it's supposed to work like that then you look good...

    But, if it's supposed to be an either or switch, put the pressure parallel to the remote switch, like in the attached...
     

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  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Blank Stare, and CocaCola, I agree with both of you.

    The polarity of the diode from hell across the pump should be pointing leftwards, as shown in CocaCola's version of your pic.

    For suppressing back EMF, the diode is connected so it doesn't conduct when the activating voltage is present. The back EMF has the opposite polarity, and makes the diode conduct in the direction of the arrow (for conventional current, that flows from positive to negative).

    This orientation is the same as for the diode (not the zener) across the relay coil. They appear to be pointing in opposite directions on your diagram, but that's because the positive and negative are on opposite sides comparing the relay coil with the pump.

    You can also test for safe power dissipation in each of the diodes. The 1N4001 will be alright but the big one and the zener diode dissipate significant power for a short time (< 0.1 seconds, probably) when the circuit de-energises. If it won't damage the pump, turn the circuit on and off say 50~100 times in quick succession and see whether those diodes are getting warm.
     
  4. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Make sure the 1N1190AR is solidly connected to ground as close to the motor's ground as possible. The mating surface must be flat and clean. No paint or grime. A layer of thermal compound would also be advisable prior to mounting it. Is the motor frame committed to ground?

    Chris
     
  5. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Kris, do you think this heat test is really necessary? The pump will never be operated like this during normal duty. ;)

    Chris
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    It's just a quick way to check that the diodes aren't being forced to dissipate too much energy while the magnetic field is collapsing. But it's not a very useful test - not for the zener, anyway. The instantaneous power dissipation limit could easily be exceeded and the zener could fail, without any heating being detected with that test. So perhaps there's no point.

    Steve, do you have an opinion here?
     
  7. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    On the other hand, if he cycled it that many times, in that short a period of time, it would be nearly proof positive that the problem was rectified. :D

    Chris
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I was thinking about this earlier. Clearly, if you have a normal diode and a zener back to back, the dissipation of the regular diode is going to be less than the Zener (same current, much lower voltage).

    I would be looking at the datasheet for the zener and seeing if it has peak current vs time graphs.

    Likewise I'd be checking the same for the diode if it's not rated for the full current.

    If either fail, they are likely to fail either short circuit or with obvious physical changes. The zener is most likely to fail first I suggest, leaving you with a simple reverse biased diode.

    If the Zener (or indeed the diode) can't handle the magnitude of the current immediately at switch-off, they may fail without getting noticeably warm.

    I would check the zener after "many cycles" and see if it is still acting as a zener.

    I also wonder if a tranzorb or similar device designed for transient high power pulses might not be a better idea. These act in a similar way to zener diodes, but are designed for the high peak power. In this case it's the power you want to handle, the voltage stability whilst doing so is not a real consideration.

    OK, Those were my idle thoughts that went through my mind yesterday or thereabouts whilst reading this thread. I'm not sure I've added much though.
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Agreed. Unfortunately, it's not going to be easy to work out how long the back EMF current pulse will be. I said less than 0.1 seconds but it's probably a lot less - less than 10 ms, maybe less than 5 ms. Mainly it depends on the inductance and mechanical characteristics of the big relay, and on the zener voltage. Of course you could measure it, but I don't think the OP has access to equipment to do that.

    I agree, I don't think power dissipation will be a problem for the diodes. The relay coil won't draw anything like 1A (at least I hope it won't!) and the meaty diode is rated for 40A (the load draws 30A). It's the zener I would worry about.

    Good advice Steve. If it's not possible to confirm that the zener is operating within its data sheet specifications for power vs. time, because the back EMF pulse time is not known, then I would cycle the thing as many times as I could be bothered with, and check that the zener is still zenering. For example, connect a neon bulb with a 10k series resistor, straight across the zener; it should blink each time the relay is de-energised.

    That's true. The PDF application note that I linked to only suggested a zener diode, but a tranzorb-type surge suppressor might be better. Perhaps tranzorbs didn't exist when the app note was written. There's no date marked on the app note. Grrr.

    Perhaps 100V zener voltage is unnecessarily high. The app note suggests a 24V zener. A higher zener voltage means higher power dissipation, but allows the magnetic field to collapse more quickly, so the energy should be roughly constant regardless of the voltage; I suggested a high voltage because that will give the quickest release of the relay.

    It's good to hear that your general ideas agree with mine. I always value your opinion.
     
  10. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    This earlier quote relates the maximum power theorem and to my statement below.

    If we take a relay coil example of 12V/100Ω = 1.44W then the back EMF, regardless of the voltage magnitude, can't output power greater than this. To do so would be the holy grail of over unity.

    Chris
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I was thinking the current through the diode(s) was higher for some reason. For a relay it shouldn't be a problem.

    One day I might try a few of these methods and measure the result. It does seem interesting
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, maybe. WHat matters is not power in Vs power out, but energy in vs energy out.

    the current can never exceed 120mA, but the power could be several watts (for a 100V zener, you could have 120mA @ 100V -- briefly. That's a peak power of 1.44W, for a 24V zener it would be 0.34W

    What remains constant is the energy in Joules that is stored in the inductor that must be dissipated. The higher the peak power, the faster the current through the inductor falls.
     
  13. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Steve, I must confess that my initial reaction was to reiterate my over unity position. I'm glad I read this a couple of times though, which put the brakes on that. It appears that I'm mixing apples and oranges. Yes, you're dead balls on the mark,...Tech term! :D
    Joules are what we're dealing with here.

    Thanks for correcting me. ;)
    Chris
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes there's been a bit of confusion on the thread because there are two inductive loads. There's a heavy pump, rated at 30A at 12VDC, to which we are suggesting adding the 40A stud mount diode suggested by CDRIVE I think, connected in reverse, to suppress any back EMF that the pump might generate.

    That pump is switched by the contacts of a big relay, with a coil rated at 12V. We don't know the coil current or resistance, or its inductance. OP, do you have any information on the relay? Would be interesting to know.

    I've suggested considering a PDF (linked in an earlier post) that says that relays open less quickly and cleanly if the back EMF is suppressed by just a diode, since there is relatively little energy dissipated so the magnetic field in the relay takes longer to collapse. So that's the reason for adding a zener in series with the back-EMF diode across the relay COIL.

    Yes. I think for quickest relay contact opening, you want something that can absorb a lot of energy over a short time period. D'oh, you might say. But I think it's the ability of a suppressor to consume a lot of power for a short time that's needed. I would be interested to hear your results.
     
    Last edited: Oct 22, 2012
  15. screwball

    screwball

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    Jan 9, 2012
    Thanks, thats the one ;)
     
  16. The Electrician

    The Electrician

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    Jul 6, 2012
    Blank Stare, you should read these:

    http://relays.te.com/appnotes/app_pdfs/13c3203.pdf

    http://relays.te.com/appnotes/app_pdfs/diagnostics.pdf

    http://relays.te.com/appnotes/app_pdfs/13c3311.pdf

    http://relays.te.com/appnotes/app_pdfs/13c3236.pdf

    The reason I asked if you still have any of the relays that failed open, is because I find it difficult to see how a relay could fail open as the result of contact erosion due to turning off an inductive load. One of those app notes mentions that the turn on inrush current for a motor may be 600% of the running current. In the case of your motor with a 20A run current, the starting surge might be 120 amps. I would expect this to cause the relay contacts to weld together, which would lead to a failed short condition, not failed open.

    I wonder if a 120 amp startup current could melt the wire connecting to the armature contact. To get an open failure otherwise would seem to require contact erosion to proceed to a point where so much material had been lost from the contacts that even when the armature is seated, there would be no contact.

    Examination of the internals of a failed open relay would allow you to determine this.

    I suppose another possibility is that the coil had become open. Do you know or remember whether you could hear the noise of the armature moving when you applied 12 volts to the coil of one of the failed open relays?
     
  17. The Electrician

    The Electrician

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    You don't need a 40 amp diode for your purpose. The 20 amp motor current will only flow in the suppression diode for a few milliseconds at turnoff. Most rectifier diodes can withstand a half-cycle (that's half a cycle of the power line frequency, 8.3 milliseconds) many times their steady state rating. For example, a 1N5400 diode can withstand a 200 amp surge for 8.3 milliseconds; and according to figure 3 of the datasheet, it can withstand a 20 amp surge for about 1/2 second (30 cycles).

    http://www.fairchildsemi.com/ds/1N/1N5408.pdf

    But I think the best device for transient suppression in your situation is a transorb:

    http://www.vishay.com/docs/88301/15ke.pdf

    I'd use about a 15 volt bi-directional part. If you can't find a bidirectional part, just use two of the regular parts connected back to back, or use a single 15 volt transorb with a 1N5400 diode back to back.

    Transorbs are better than an ordinary zener for this purpose. Transorbs are actually zeners, but they are designed to withstand large surges.
     
  18. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That sounds good. You don't think there will be much electrical energy generated at turn-off by the pump as it runs down? I guess the run-down would be pretty short and not able to generate much power?

    Are you suggesting this across the pump?

    How can you estimate the amount of energy that the pump will generate at turn-off due to its inductive behaviour and its run-down? For surge durations short enough that the burst of heat can be dissipated, the tranzorb's limits would be in terms of peak instantaneous power, and amount of energy, but the only dissipation rating stated seems to be 1500W peak dissipation using a standard 10/1000 us surge waveform.

    Thanks for your input on this thread :)
     
    Last edited: Oct 22, 2012
  19. duke37

    duke37

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    Jan 9, 2011
    There seems to be siome confusion between back EMF and inductive current.

    The motor will have some inductance and when switched off, the current will continue, decreasing, for a time.

    The 12V motor will produce perhaps 11V back EMF when running, the difference driving the current through the windings. Whem switched off, the motor will slow and the back EMF will decrease. This should not cause any problems.

    Any diode across the motor will only need to pass the inductive current for a very short time as has been said, it will have nothing to do with the back EMF.
     
  20. The Electrician

    The Electrician

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    Immediately after the relay opens the motor armature will still be rotating. The motor will act as a generator (assuming it's a permanent magnet motor), generating a voltage just about equal to the applied voltage. The polarity of this voltage will match the applied 12 volts--in other words, it will be positive at the positive terminal of the motor. This voltage therefore will produce essentially no voltage across the relay contacts, and it will therefore cause no arcing at the relay contacts and doesn't need to be suppressed.

    On the other hand, the armature does possess some inductance. When the relay contacts open, that inductance will produce a negative voltage spike at the positive terminal of the motor. Its duration will depend on the distributed capacitance of the armature winding and other parasitics.

    So, when the relay contacts open there will be a very short, relatively high voltage, negative going spike of voltage at the positive terminal of the motor and this spike will cause arcing at the relay contacts if it isn't suppressed. The relatively low voltage due to the spinning of the armature can be ignored.

    To estimate the energy of the voltage spike we would need to know the apparent inductance of the armature; it varies somewhat as the armature turns, The "run-down" energy doesn't need to be suppressed as I explained above.

    The OP's motor is fairly large--20 amps at 12 volts is 240 watts which suggests to me a motor of about 1/4 horsepower. I don't have a DC motor that large, but I do have a small motor and I measured its armature inductance as about 1 mH. If the OP's motor had a 1 mH armature inductance, the energy stored in the inductance would be 1/2 * L * I^2 = 1/2 * .001*400 = .2 joules of energy at a rate less than 1500 watts, well within the capability of a transorb (assuming the time constant of the inductance isn't too long; a measurement really should be made, but I would guess it would be ok). Even with an inductance of several millihenries a single transorb of the size in the datasheet I linked could handle the spike. Larger transorbs are available, or several lower voltage units of the ones I referenced could be connected in series for more energy handling capability.
     
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