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  • Electronic, electromagnetic help please...?

    I really should be able to do this but I seem to have an enduring brain fade right now..

    I am trying to use a double pole double throw relay to reverse the current through an inductive load, i.e. I want to use the relay as an 'H bridge'.

    OK, no problem connecting the relay contacts to the load and the supply but how do I protect the contacts from the inductive kick back? It sounds easy, just put a diode across the inductance, er no, the inductor has to handle a reversing polarity.

    Thanks for any assistance

    John

  • #2
    Use back to back zener diodes rated above the operating voltage by a few volts. It won't be as effective but it will greatly limit the back EMF.
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    • #3
      Originally posted by Evan
      Use back to back zener diodes rated above the operating voltage by a few volts.
      Thanks Evan, that should do it.

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      • #4
        John,

        Take a look at this link. Further doun the page is the layout for the diodes. I might add a small resistor in series with the diodes to limit the current when reversing the motor.

        linky
        Robin

        Happily working on my second million Gave up on the first

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        • #5
          You don't need a relay. All you need is a $5 double pole double throw toggle switch with a center off position.

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          • #6
            I suggest NOT using zeners.....

            Go with rdfeil's suggestion, which is 4 diodes, two at each end of the load.... One from ground to the load, one from load to plus. Same at other end. Poled with "bar end" to plus.

            That is totally standard for VFDs and all other sorts of devices that drive inductive loads.

            The problem with zeners is that they are power-limited.... you can cook them if you use the wrong type, and they require some "engineering design".

            The diodes are easy, effective, cheap, and all you need to do is get the diode current rating fairly close to the load current. ergular rectifier diodes are tough and take surges very well. Zeners.... not so much.
            1601

            Keep eye on ball.
            Hashim Khan

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            • #7
              Sorry, brain fade.
              Last edited by Evan; 10-18-2011, 02:22 AM.
              Free software for calculating bolt circles and similar: Click Here

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              • #8
                Originally posted by gary350
                You don't need a relay. All you need is a $5 double pole double throw toggle switch with a center off position.

                In this case I really need a relay, or a suitable H bridge IC which I do not have.

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                • #9
                  I did try a diode matrix today but after about 30 seconds one diode punched through then a couple of seconds after that one relay contact welded shut and the power supply blew its fuse.... unfortunately the diodes were salvage items and I have no idea what their specs might have been, they looked like rectifier not signal diodes.
                  Last edited by The Artful Bodger; 10-18-2011, 02:11 AM.

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                  • #10
                    You need diodes rated at least ten times the operating voltage and higher would be better. When the field collapses the voltage developed depends on the rate of collapse. With a mechanical switch it is a hard turnoff and the voltage developed will be very high. 1N4007 diodes have a 1000v rating and will handle several amps momentary current.
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                    • #11
                      An old trick used on stepper motors is to feed the winding through a resistor. The voltage is increased to give the proper winding current. The resistor slows the collapse rate and holds the voltage down. It is wasteful of power but it works.
                      Free software for calculating bolt circles and similar: Click Here

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                      • #12
                        OK, let's hear what your voltage and current are...... and what sort of load you have..... so far, no specific requirements have been mentioned.

                        It kinda sounds like the load may be more than the relays and/or diodes could handle, either current or voltage.... I suggest you give the particulars.

                        When choosing parts.......

                        Something to watch out for...... relays (and switches also) have AC AND DC maximum ratings..... sometimes only one is given, but the other "exists".

                        The DC rating is normally a LOT lower than the AC rating, either in current, voltage, or both.

                        To break a DC current , especially with an inductive load, takes a much tougher relay than an AC load does. The inductor does not change current quickly (it takes energy input or removal to do that) so when the relay or switch opens, the current continues, forming an arc at the load current.

                        if the contacts don't open far enough, the DC arc can be self-sustaining, very much like welding, and with a similar effect. Since it is DC, it never reverses direction, and so the driving voltage never goes to zero. Any time the voltage goes to zero, as with AC, it allows the arc to cool off and tends to stop it, again similar to welding.

                        A relay may, for instance, have a 30A rating on AC at 277V (like a P&B* "T-90" relay), but may be limited to only 28V at a similar current with DC. That's a pretty dramatic difference, but typical even of heavier-duty relays.**

                        I don't know if that even applies, since you didn't give the voltage, but it is worth mentioning.

                        BTW, with 4 diodes as described, the voltage rating does not have to be "that" high, since no matter what happens, voltage-wise, a diode is available to conduct to either ground or the power supply....That is what engineers call "clamping" the voltage. But for slower diodes, extra voltage rating is a safety margin. if your diodes are at a minimum of double the circuit voltage, you should be OK, higher generally does no harm.

                        * Potter and Brumfield is now part of "TE Connectivity", formerly TYCO

                        ** to get a higher DC "break" rating, you need "magnetic blowout" or similar special methods. I once got a T-90 type relay to break 30A at 75VDC with a magnet to "blow out" the arc, but the manufacturer wouldn't make it that way.
                        Last edited by J Tiers; 10-18-2011, 09:40 AM.
                        1601

                        Keep eye on ball.
                        Hashim Khan

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                        • #13
                          There is a component called a "transorb", a variety of zenier diode optimized for higher current switching.

                          http://www.statemaster.com/encyclopedia/Transorb

                          In my business, they are used on relays that carry a large current, like turbine engine starter relays.

                          prevents (like a diode) spiking the the rest of the harness when the start sequence ends, and the relay de energizes.

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                          • #14
                            One option, tough not frequently used, is to put a resistor in parallel with the load. It will absorb the spike, but the downside is that it consumes electricity all the time the load is energized.
                            Amount of experience is in direct proportion to the value of broken equipment.

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                            • #15
                              Originally posted by Thruthefence
                              There is a component called a "transorb", a variety of zenier diode optimized for higher current switching.

                              http://www.statemaster.com/encyclopedia/Transorb
                              The equivalent circuit of Transorbs are two Zeners wired in series back to back.

                              Chris

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