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dc coil surge supression 11

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electricpete

Electrical
May 4, 2001
16,774
I was asked to provide a “quick” (i.e. not my real job…helping someone else out) general recommendation for dc coil suppression for relay coils powered from 125vdc. . There are about 30 different coils in the cabinet, all powered by 125VDC and all drawing about 0.25A or less. Various problems have been experienced that are believed attributable to voltage spikes from coil switching. Response time is not critical in this application… 1 sec delay would not hurt anything. It is important for relays to change state reliably and even more important not to short out the dc power supply if the surge suppression device fails.

I did a quick search and it has been discussed many times on eng-tips.
Also I found:

I’ve read through the above and formed my own conclusions, submitted for your comments.

I think the 2 most common discussed options are:
1 – flyback diode
2 – varistor.

The first link especially seems to push the option of varistors. They highlight a concern that flyback diode can make the coil be so sluggish that it might not even operate. Also apparently when it operates slowly, it’s output contacts can be degraded. I have to admit I have not heard much about these concerns before (other than time response).

I really don’t like varistors in this application. I think every time the coil switches open there can be fairly high current at high breakdown voltage drop and these things degrade. Sure there is a rating, but they use a little life every time they cycle. If they ever short circuit, life is not good.

So I like the flyback diode. Diodes can short, but then again I have an easy solution: put two diodes in series…. Makes me feel a lot better. Either diode can fail short and not a problem. Also I’m thinking I would put a resistance about equal to the coil resistance (R = 125V/0.25A = 500ohms) in series. That should tend to minimize concerns about effect of slow field collapse upon the relay discussed in the first link. I picked 1 times coil resistance since when the full coil current switches into flyback loop the voltage is limited to the original 25VDC voltage (if I had double the resistance I could have double the voltage).

In summary I am thinking about two reverse-biased diodes (during normal operation) and a resistor 1x coil resistance... all connected directly in parallel with the relay.

What do you think? Am I grossly overlooking anything?

By the way, any tips for diode selection?

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(2B)+(2B)' ?
 
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The other engineer seems to be interested in pursuing my idea but is stuck on the idea of using directional TVS unidirectional diodes (Littlefuse 5KP series). They are really being used as diode and not expected to break down. I’m not sure what it buys us, but he really wants to do it. Do you think that would that hurt anything that you can see?

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(2B)+(2B)' ?
 
If (?) I understand how the diodes work, they provide a path for the coils' 0.25A current to continue to flow (preventing high voltage from developing in the first place). As you noted, the release of the contacts will be delayed by the time it takes for the current to decay.

But I don't think that the diodes are 'tortured' in this sort of duty. All they see is the 0.25A current suddenly appearing and dying away. They should last 'forever'.

Stand by for others...
 
That TVS unidirectional diode is (obviously) only intended to conduct in reverse. They give power ratings related to a short pulse... nothing for continuous. Even though a 1-sec or so decay is not continuous, I'm thinking I personally would rather have a continuous rated diode (at least then I can look at the data sheet and verify it is more than 0.25A). What would your diode suggestion be?

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(2B)+(2B)' ?
 
I think you can forget about all expensive and exotic devices in this case. Your diode+resistor thinking is right on.

1N4007 plus resistor equal to coil resistance can never go wrong. Delay will be minimal and the diode's reverse voltage is more than adequate. Resistor wattage need not be high, but you may need to use resistors with same gague leads as the diode, for practical reasons.

Momentary power dissipation will be half the coil wattage for a very short time, around 30 W during the first milliseconds and then quickly dropping to zero. I would use a 2 W resistor and put resistor and diode in shrink tubing. Make sure the polarity can be seen on the package.

For high switching frequency, you may need to use a resistor with more watts.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Given that diodes are almost free, I'd splash out for 1A and maybe 400 PIV (the old 'double and round-up' rule). 1N4004 or better. Switching speed might be something to watch for. If the diode is slow, then 'bad things' might happen in that duration. The 1N4000-series are not known for being fast. Almost certainly a better choice out there.

 
Sorry, half the coil wattage is not 30 W, but 15 W. So 2 W resistor will be fine even with high cycling rate.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
With relay drop times between 100 and 300 milliseconds, even 1 ms in the diode doesn't matter. And the 1N400X are definitely much faster than that.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
My concern about the diode switching speed is that the coil current will try to continue even while the diode makes up its mind to forward conduct. During that time, the voltage may spike and possibly stress the diode. Peak voltage would be limited by the junction capacitance. And the diodes are rugged anyway. I doubt it's a real world problem.

But nothing directly to do with the "relay drop time".

 
No, it is not a problem. We use this type of protection in lots of installations in paper mills, steel works, power plants. Done it at least 40 years. Never had a problem.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Outstanding! That is fantastic and timely info.

This is not a high cycle rate application. I think we are going with 500ohm resistor fairly low wattage. And 1N4007 diode selection.

The other engineer pointed out that we don't really need those 2 series diodes to protect the power supply. If one diode should short circuit, the power supply is protected by the resistor which will limit currrent to 0.25A....

...but now the followup question: what happens when if the diode shorts and the resistor is now carrying 0.25A and protecting my power supply... and heating up since it is not sized for that 0.25A (we don't really want to use a big resistor rated for the full 0.25A / 30 watts since the physical weight will be supported by the electrical connections). Can we count on the resistor to fail gracefully and open circuit? Are there certain resistors that are expected to behave this way? We were thinking wirewound resistor might tend to open circuit easily.

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(2B)+(2B)' ?
 
You could use a small 7W 120V incandescent lamp for your resistor. Then it would light-up if the diode failed. A dim blink may be visible when the relay switches off. One drawback may be that lamp filaments are more fragile than some resistors.
 
I think "fusible resistors" are designed to open-circuit in a controlled fashion. For example this:

Maybe that would address the concern. Anyone know anything about these things?



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(2B)+(2B)' ?
 
It is not likely that a 1N4007 ever breaks down. And if it would, most resistors just open and do not catch fire. So it is only in the most critical applications that one needs to consider that failure mode.

Interesting to see that the fusible resistors withstand around 1 second at 15 W (the .25 W types) and the .5 W types six seconds before fusing. Lots of margin there.

It is not a bad idea to use the fusible resistors. Not that they are needed, but it doesn't hurt to use them.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
In some applications it can he better to have an 'obvious' failure that draws attention to itself, instead of a silent failure that quietly causes subtle problems.
 
We will go with the 0.5 watt resistor to for added comfort.

Good point about silent failures. If diode shorts and resistor then opens, I expect there would be no visual evidence. (We have seen visual evidence of diode short before, but only when accompanied by high current). The light-bulb has an advantage in that regard, but poses too much difficulty to install the way we were planning (supported by the leads from the device terminals). In the worst case, we end up missing some surge protection… the same as we are now… at least our modification has made the situation no worse. We will think about what strategy is viable for spot check of these circuits.


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(2B)+(2B)' ?
 
A TVS would be fine, but pricey.
The 1N4007/resistor is the better option.
Use a flame-proof resistor. Don't cover it up.
Makes a great visual indicator. :)
We had a whole cabinet full of these (>200 pcs.)
and they worked fine.
<als>
 
Re: Don't cover it up

There is transparent shrink tubing. Those little components need some physical protection in the sometimes harsh environment you get in a cabinet (pulling wires and removing/putting new ones).

But, as I said before, there is no real need to worry about the diode break-down. If you were using 1N4148 - yes. But never with 1N4007. Never.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Use a flame-proof resistor. Don't cover it up.
Makes a great visual indicator. :)
Can you explain what you mean by that 2nd sentence?
I haven't seen one of these but I was imagining it would melt open inside with no external discoloration. Have you ever seen one burnt open? (I might try that once we get them).

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(2B)+(2B)' ?
 
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