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Snubber circuit for 30ma 240 to 9v transformer ?

MidAtlantian

May 12, 2015
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Hi,

While I can manage my way around digital circuits, I am pretty lost in the world of a.c... I am using a 9v. relay to turn on a 240vac - 9vac transformer from a digital circuit, and thought I should protect the relay's contacts against arcing, so I put a .033 µF 240v capacitor across the contacts. But then the transformer was partially powered while the contacts were open, and I measured 360 volts ! across the capacitor. So I know I've got a lot wrong.

Do I need to protect the contacts for the few milli-amp inductive load? If so, how?

Many thanks for any suggestions.

MidAtlantian
 

Harald Kapp

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Welcome to electronicspoint.

As long as the contacts of the relay are rated for the voltages and currents to be switched you need not worry about additional protection for the relay contacts.
A capacitor across the contacts is sometimes used for switching DC to reduce the risk of an electric arc when moving the contacts. But AC has regular zero crossings (periodic with mains frequency), so any developing arc is interrupted.

With an inductive load things are a bit different. Since the switching of the relay is typically not synchronized with mains, the inductor may develop a high flyback voltage when turned off due to the energy stored within the inductor which seeks a way to be released. In that case the high flyback voltage may produce an electric arc even across open contacts. In this case a protective element can be used. A capacitor is not a good choice, as you already noticed. A voltage dependent resistor (VDR) with a threshold above the mains voltage either across the contacts or across the inductor (the latter being the better solution in my view) will dissipate the excess energy from the inductor. When selecting the VDR, watch the ratings: some are rated fpr DC, others for AC. The rating must match your AC mains voltage (+ tolerances) to ensure the VDR is never active during normal operation.You should also use a fuse in series with the whole setup in case the VDR breaks down (on a second thought you should have a fuse anyway).

And don't forget to look at the VA rating, too, which is not just simply max. voltage*max. current!
 

MidAtlantian

May 12, 2015
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Halald, Thank you for the welcome and for a detailed, clear and usable response! You took what I vaguely knew and put it info a sensible structure. Of course, now I have a few more questions, but the end is in sight.

Can I assume that a VDR - which I'd never heard of - is the same as a varistor? I'd guess that I could use the smallest I can find; that even a few joules would be enough. Is that correct? And is 250v enough head-room, or is 275v better?

The only piece of the puzzle I am left with is how the voltage across the snubbing capacitor exceeded the peak voltage (about 310 ) on an rms of 220 ... by about 50 volts! Is that from the "kick" of the collapsing field at each of the a.c. cycles? All together it was certainly creating havoc on the digital side of the circuit.

There is one of your comments I do not understand: when you mention that the power (VA) is not just volts * amps - unless you are referring to peak voltages being higher than rms? But then I would haave thought that averages rather quickly.

And yes, I HAD forgotten about a fuse. Very good point! In fact, I'll need several, one for the longer running low power side of the circuit, and one for the few second bursts of high power.

Again, many thanks for your guidance.

( I hereby absolve you from any responsibility, even from any incorrect information you might give me, inadvertent or otherwise. You have Carte Blanche protection.)
 

Harald Kapp

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Can I assume that a VDR - which I'd never heard of - is the same as a varistor?
Yes, you can.

how the voltage across the snubbing capacitor exceeded the peak voltage (about 310 ) on an rms of 220 ... by about 50 volts
How do you measure the voltage across the capacitor? The peak voltage of an AC 220 V signal is 311 V (220 V * sqrt(2)). Depending on how you measure you may see this voltage or part of it and thus your measurement may indicate a voltage other than 220 V.

when you mention that the power (VA) is not just volts * amps
What I meant to express, sorry for not being that clear, is: A relay may have a max. voltage rating and a max. current rating. however, you may not be able to multiply these values to get the max. rating of the load to be switched. A rela's datasheet gives you I vs. V characteristics showing that often high voltages can be switched only at low currents and vice versa. And things are different for AC and DC due to the risk of arcing with DC.


( I hereby absolve you from any responsibility, even from any incorrect information you might give me, inadvertent or otherwise. You have Carte Blanche protection.)
Thanks a lot - take care!
 

MidAtlantian

May 12, 2015
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Hi Harald.

Thanks again for your responses. Yes I get your point about maximum current or voltge ratings and overall power ratings.

But I wasn't very clear about the "measured voltage" question. I measured 360, well over 50 volts above peak. How was I getting that?
 

Harald Kapp

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You may have see a resonance due to the LC circuit formed by capacitor and inductor.
 

MidAtlantian

May 12, 2015
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Thanks. Yes, RLC circuits: I've got only the vaguest idea of them: I can imagine a see-saw of energy being alternately stored in the capacitor and inductor, but how that actually plays out is not something I'd expect to learn without some serious study... and math.

In any case, I'm out of questions. Many thanks again for your help.

MidAtlantian
 
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