0
\$\begingroup\$

I'm switching on an LED strip power supply using an Arduino, but my relay has just started sticking. The relay switches the mains for a 5V 20A power supply. The relay board is this style.

The relay is apparently rated for 10A @ 250VAC, so I think should be fine for the continuous current, but I guess the inrush current when the power supply is first switched on is what's causing the relay contacts to fuse.

I have another 4 of these relay boards spare, so I'm wondering if I can either a) Somehow limit the inrush current to the power supply so that I can just use one of these boards, or b) Figure out the inrush current so I can use a more suitable relay.

If I go with b) - does anyone know of a relay that would be up to the task? Ideally 5v with the same package as the ones on those boards so I can do a straight swap, but I'm guessing the ones that can handle the higher inrush current are probably a bigger package, right?

Edit: Here's a quick schematic of the setup: Schematic

The reason I'm switching on the AC side instead of the 5V was really just for power saving, so that the power supply isn't on constantly, only when I want it on (when the Arduino signals D5 high).

The relay is sticking in the on position - the output stuck on closed and when the whole thing was unplugged (both USB and AC), I managed to give the relay a few taps and it eventually unstuck and opened up again.

\$\endgroup\$
  • 1
    \$\begingroup\$ Let's see the schematic. Is it sticking in what should be the energized state (possible mechanical damage) or in the deenergized state (possible coil/drive circuit damage)? \$\endgroup\$ – schadjo Apr 12 '18 at 19:41
  • 1
    \$\begingroup\$ Is there any reason for not switching this on the low-voltage side? The 5V power supply can also power the arduino if you do it that way. \$\endgroup\$ – Jonathan S. Apr 12 '18 at 20:42
  • \$\begingroup\$ I've included a schematic now. It's sticking in the energised state. The reason I'm switching the AC was just from an energy saving perspective, so that the power supply isn't always on. \$\endgroup\$ – Andrew Porritt Apr 12 '18 at 23:53
  • \$\begingroup\$ if the OP would switch low side the quiescent current of the supply always flows, to no good purpose. \$\endgroup\$ – dmb Jun 24 '18 at 19:04
1
\$\begingroup\$

I have faced this issue in past. An easy fix would be to move from relays to solid state switches. Use a standard triac with an optocoupler diac driver and you should be fine. I used BTB16 600BW and FOD420. If FOD420 is difficult to come by, MOC3021 can be used as well.

Since your output power requirements are less than 100 watts, the triac won't be supplying more than 1 A (in case of 110 V. For 220 VAC line, it will be even lower). I have tested the triac and it works comfortably without heat sink for currents up to 1 A.

You can follow this circuit:

triac

\$\endgroup\$
  • \$\begingroup\$ Thanks. I'm going to order a BTB16-600BW and MOC3021M off ebay and give this a try. For R14, what kind of wattage resistor would I need there? The current to the gate of the triac is quite low isn't it, so it shouldn't need to be a very big one, right? Also, looking at the datasheet for the FOD420 (page 6 - mouser.com/ds/2/149/FOD4216-195358.pdf) - it looks like they use a voltage divider instead of single series resistor. What's the difference between the two options? \$\endgroup\$ – Andrew Porritt Apr 13 '18 at 12:41
  • \$\begingroup\$ I have used 1206 package resistors as R14 and didn't see a failure till now (in over 1000 circuits working for almost 2 years). I think 1206 has a power rating of 0.25 watts. If you are going for through hole resistors, 0.25 watts will do fine. From the page you mentioned, 220 ohm is what I have used in place of 360 ohm resistor. Additional 330 ohm might be being used for preventing a trigger due to noise (perhaps). However I'd like someone more experienced to answer this part. That being said, my circuit works reliably. \$\endgroup\$ – Whiskeyjack Apr 13 '18 at 18:02
4
\$\begingroup\$

In general, relays stick because the contact open when the voltage is high and a spark is generated across the gap. This spark results in damages to the contact or welding. This can be dangerous in certain environments.

I don't know how often your circuit turns on the relay so check your specifications to make sure it can handle the switching. For AC applications we try to make the make/break point near the zero crossing.

Search for spark suppression on the internet you will find literature on how to design a simple RC circuit to reduce the spark. Care must be taken to select components which are rated for your application.

Update The value of the R and C across the open can be calculated based on a simple equation which has been derived through experience where C = (I square)/10 with I equal load current. The open voltage across the contact R = V/[10*I(1+(50/V))] . formula by CC Bates.

\$\endgroup\$
  • \$\begingroup\$ The relay doesn't switch often - it's going to be a case of turning the supply on and having it on for anywhere from several minutes up to a couple of hours, then off for a long time also (I.e. no rapid switching). Would something like this be appropriate? ebay.co.uk/itm/… And would it connect between the terminals of the relay (COM & NO)? I'm not sure how best to calculate the values that would be needed. \$\endgroup\$ – Andrew Porritt Apr 18 '18 at 15:59
  • \$\begingroup\$ +1 for identifying the voltage spike as the problem, rather than current. When you switch at the moment current flows, there is some stray inductance in the circuit, it is very small, but the sudden change of current means you can get considerable voltage transients. Putting a bit of capacitance in there counteracts the inductance, problem much reduced or solved. The very first google hit when you google as Y. Yee suggests gives you the circuit configuration and info. \$\endgroup\$ – dmb Jun 24 '18 at 19:19
0
\$\begingroup\$

There are relays with high inrush current capabilities upto 100A. For example check out the below relays which are 16A, 240AC rated (100A inrush) and comes in smaller size.

Pansonic https://www3.panasonic.biz/ac/ae/control/relay/power/dw/index.jsp

Omron http://omronfs.omron.com/en_US/ecb/products/pdf/en-g5rl.pdf

The other option would be switch to SSR, however you may need to understand their drawbacks like false trigger, heat etc.

\$\endgroup\$
0
\$\begingroup\$

You can reduce the current surge and prevent the contacts from welding by adding an "NTC inrush current limiter". This is probably being caused by a poorly designed LED power supply (normally there is one built-in).

Here is a training module on how to specify such a part (most likely they will specify a particular manufacturer, but you can search for others).

For it to be effective you have to allow enough "off" time for the NTC to cool in order for its resistance to go up so the next time the contacts close the current will be limited. They are designed to get quite hot at full load so the resistance drops to perhaps 1% of the cold resistance (and thus the losses are minimal when hot).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.