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The TRIAC actuator of a DC motor through a bridge rectifier.

I activate a DC motor from the mains through a TRIAC (BTA16) in series with a 10 Ohms resistor for the inrush current limitation, in series with a PTC thermal protection for the motor jamming protection and in series with a bridge rectifier needed to supply the dc motor with dc voltage, filtered by a 100 uF/400 V capacitor.

The TRIAC is triggered by a zero-crossing optotriac (MOC3043). The TRIAC is provided with a snubber of 100 Ohms and 100 nF. Hundreds of motors having this actuator have worked very well. It happened in some particular place that the motor once started didn't stop, but changing the place it keep working properly. It is the influence of the network of this particular place but I don't know which influence. I have added in series an inductance shifting the current with almost 30 degrees and the TRIAC worked very well.

Could some other inductive loads on this particular network have such a bad influence on the behaviour of the TRIAC? I mean a weak electrical network and some asynchronous motors around. Could you please give me any advice?

PS: I have attached a drawing but I was not able to post it. I hope that this description is good enough.

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  • \$\begingroup\$ Add a link of the drawing in your question and we'll post it for you. (Nice to see another fellow Romanian around here) \$\endgroup\$
    – m.Alin
    Commented Sep 27, 2012 at 19:57
  • \$\begingroup\$ The zero-crossing in a weak grid is not such a good idea, because it isn't always "strong enough" and the sudden current may cause "islanding", which is problematic for zero-crossing detection. The simplest of solutions could be an RC low-pass with a fairly small time-constant (depending on the actual effect and the desired cure) to smoothen out just a bit the waveform while not adding too much delay and fire the triac too late. Of course, there may be other problems, but this one seems like a cause for it. \$\endgroup\$
    – Vlad
    Commented Sep 27, 2012 at 20:23
  • \$\begingroup\$ I have uploaded the drawing at Google+. \$\endgroup\$
    – Mircea Sturzu
    Commented Sep 27, 2012 at 21:24

1 Answer 1

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Triacs aren't fit for switching DC. They basically consist of two transistors connected in such a way that they keep each other conducting once the triac is triggered. The only way to switch them off is to cut the current below the hold current level. In AC that's not a problem, since the current will become zero 100 times per second at 50 Hz, but in DC the current keeps flowing.

If it works most of the time it means that in these cases the current will indeed drop below the hold level.

Two ways to switch the triac off: interrupt the current with a normally-closed switch in series. Or have a switch parallel to the triac. Closing that switch will bypass the triac so that it doesn't get current anymore and will switch off. Open the switch again and the motor will switch off.

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  • \$\begingroup\$ [here is a link with the drawing: plus.google.com/#photos/107843529757978755623/albums] \$\endgroup\$
    – Mircea Sturzu
    Commented Sep 28, 2012 at 12:35
  • \$\begingroup\$ @Mircea - Link says "Mircea Sturzu doesn't have any albums", no picture to be seen. \$\endgroup\$
    – stevenvh
    Commented Sep 28, 2012 at 12:38
  • \$\begingroup\$ The electrical diagram is in the photos album, the only one. Thank you! \$\endgroup\$
    – Mircea Sturzu
    Commented Sep 28, 2012 at 12:41
  • \$\begingroup\$ @Mircea - Sorry, still nothing. "Mircea hasn't shared anything with you". I'm not really an active Google+ user, but I guess we'll have to share a circle. I have added you to my EE.SE circle. I guess you have to do something similar :-/, or make it public somehow. \$\endgroup\$
    – stevenvh
    Commented Sep 28, 2012 at 12:47
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    \$\begingroup\$ @MirceaSturzu: should someone find this topic in a few months or years, they won't be able to see your drawing. Either because of privacy settings or deletion. Please upload it on a public host such as imgur, for the sake of future generations :) \$\endgroup\$
    – Christoph B
    Commented Oct 28, 2012 at 15:57

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