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I am using a 12V relay to drive high voltage contactors. The contactor coil causes EMC when turned off. I am uncertain of which circuit I should use for reducing the EMC. Is there even any difference between option 1 and 2?

Secondly It was recommended that I should use a 24V zener diode. But I couldn't find any information on what current rating the zener diode and the normal diode should have. The operating current of the HV contactor coil is around 0.5A. Should the zener diode be able to handle the whole operating current? Is there a way to estimate it?

I had trouble finding a datasheet for the contactor it's name is AEV6505A from panasonic. So I based my info on a similar contactor I looked at the specs for the highest current contactor just to be safe.

EDIT! Would this zener diode work since it has a " Non Repetitive Surge Power Dissipation" of over 100W at 1ms. And the maximum wattage from the coil is only 0.5A*12V=6W?

EDIT 2! Thanks for all the answers!

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You really do need to find the right data sheet. \$\endgroup\$ – Andy aka Aug 12 '20 at 22:15
  • \$\begingroup\$ The contact current has nothing to do with the coil current. At only 12 volts normally a 1N4007 damper diode will work. It hard-clamps back-EMF. A parallel zener or MOV can clamp forward surges. Large contactor coils may fry the tiny 1N4148 diode. \$\endgroup\$ – user105652 Aug 12 '20 at 22:49
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    \$\begingroup\$ !. Pse advise ACTUAL COIL voltage or range. Is it 12V or "high voltage, other". 2. Both arrangements are equivalent. 3. CRUCIAL - do you mean EMC "when turned off" (ie when not on) , or EMC as they are turned off? The latter is more likely. 4. The diode and zener should be rated to carry the contactor current at the turnoff time. The zener should be able to dissipate energy stored in the inductor. 4. A diode alone will work - but take longer to release. The contactor holds on until the circulating current drops to below the holding current. Higher voltage dissipation path gives higher energy. \$\endgroup\$ – Russell McMahon Aug 12 '20 at 22:52
  • \$\begingroup\$ Okay, thanks! I ofcourse mean EMC as they are turned off. Using just a diode is not an option since I am afraid that can long term weld the contactors. I still can't seem to find a reliable way to decide the proper wattage of the zener diode long term. Is there a way to pick a current just to be on the safe side? How would one go to calculate the current that goes through the diodes? \$\endgroup\$ – Isak Söderlund Aug 13 '20 at 11:19
  • \$\begingroup\$ As asked above - Q1: what is the coil voltage - 12V as shown or ? || Q5: What is the contactor switching - voltage and current? || Q6: As SSG asked in his answer - what is the inductor energy storage? [Current and inductance allow this top be calculated. Datasheet should say]| .Using just a diode with polarity as shown will not weld the contacts. Why do you think it would. || Q7 I said the zener must dissipate (most of) the energy stored in the inductor. That answers your zener rating question. Why do you ignore this answer? If you do not know the energy then how can we know? \$\endgroup\$ – Russell McMahon Aug 13 '20 at 12:12
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High Voltage contactors need to switch as fast as mechanically possible to reducing the harmful effects of arc currents on the contact surfaces at extreme temperatures ~ 6000’K.

When a simple relay or inductor is clamped by a reverse diode, the same coil current travels thru the diode with the reverse voltage. When such clamping occurs the force on the coil persists and the decay time can be estimated by the L/R=T . Thus a low R conducting diode will force the relay to open slowly, exactly what you don’t want for a HV contactor.

Your figures A & B are identical because series elements are swappable.

However the addition of a Zener will dissipate more power in the diode, yet it will switch much faster. Because dI/dt=V/L the Zener voltage allows some drop in current due to the rise in voltage. Yet it is not very much voltage and power can be a problem for Zeners unless bulky.

Another solution is to allow much higher switching voltages by using an RC snubber to absorb the energy stored in the coil and increase the V=I*R transient across the switch within its allowable safe operating area (SOA). This gives the contactor or power relay the best possible solution for switching the contacts open and quenching the arc.

Then next possible solution to prevent damage is to use a large power snubber for flaming contactors to divert the arc energy to an external RC dummy load.

The specifics require you to define the actual VA being switched and inductive energy stored.

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    \$\begingroup\$ Nice answer. ................ \$\endgroup\$ – Russell McMahon Aug 13 '20 at 2:30
  • \$\begingroup\$ So how would one decide what wattage the diodes should handle. Lets assume that the operating current of the coil is 0.5 A then I would need diodes that can handle 12x0.5=6W of power. But the peak is for such a short time (single digit miliseconds) So it seems excessive to base it on that. \$\endgroup\$ – Isak Söderlund Aug 13 '20 at 12:14
  • \$\begingroup\$ 6W coil could easily be done by 1W Zener. If it was a 60V zener the surge of 30W @ 1ms might be 1 deg'C/W on a 5W Zener. So using duty cycle and pulse width, look up temp rise to choose if power rating is acceptable on thermal graph at end of datasheet. Higher the voltage , the less drag on coil , the longer the contacts live with arc quenching \$\endgroup\$ – Tony Stewart EE75 Aug 13 '20 at 12:34
  • \$\begingroup\$ If you model the coil L , I & R , you can simulate the relay speed. Here I have 3 different snubbers. The diode clamp testing at 50Hz stays stuck On, the Zener + diode is fast and RC snubber fastest. tinyurl.com/y5nmlqno I just used 50 Hz to measure relay latency turning off. \$\endgroup\$ – Tony Stewart EE75 Aug 13 '20 at 14:43

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