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I have 4 SSRs on my board with a steady state current rating of 2A and a peak current rating of 9A (Relays are CPC1976). This product is meant to be a commercial one and the relays will be powering bulbs (CFLs, Incandescent etc) upto 100W @ 240VAC

My main concern is short-circuits. A lot of CFLs that I've used in my house short circuit for a small time when they fuse. The short circuit is serious enough that the circuit breaker trips. This will obviously fry my relays.

I can easily include a fast-acting fuse but that will be have to be replaced everytime the lights fuse! I could consider a resettable fuse but most of them seem to be too slow to act to save a semiconductor device.

What are my options here to have a resettable solution yet save my relays from damage?

One idea I had was to do away with the relays and use opto-coupler + triacs instead. Triacs are cheap and I can easily find some that can handle large amounts of current (obviously a short circuit will be more than they can handle). These may survive such short circuits with the addition of a resettable fuse?

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  • \$\begingroup\$ Normal relays are pretty robust. \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2014 at 17:42
  • \$\begingroup\$ Do you mean electromechanical? \$\endgroup\$
    – Saad
    Commented Jul 18, 2014 at 17:43
  • \$\begingroup\$ Yes I do mean that \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2014 at 17:44
  • \$\begingroup\$ If there's no other option, I'm open to the idea. My main limitation in that case is I would rather avoid the noise and the limited turn-on cycles. Also, don't electro-mechanical relays also have a current rating? A short-circuit will obviously exceed. How is it that they will be more robust? \$\endgroup\$
    – Saad
    Commented Jul 18, 2014 at 17:47
  • \$\begingroup\$ Once a relay contact is closed the current can ramp up quite a way without necessarily destroying the contact and then the fuse/breaker blows saving the day for everything! \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2014 at 17:52

3 Answers 3

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You need to monitor the current using a low resistance in series with the load (bulb). Or possibly, the drop across the relay can be used to calculate the amount of current taken by the load.

Once the drop exceeds a threshold, the relay should be turned off. Make sure that the detection mechanism and relay is fast enough to allow for abrupt termination.

Otherwise, go for a fast blowing fuse in a holder; solid state stuff has low thermal mass. Replace the fuse while replacing the bulb.

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  • \$\begingroup\$ I actually like this approach but the issue is that the SSR will not switch off till the current waveform falls down to zero. The short time that it will take for the waveform will probably fry the SSR. \$\endgroup\$
    – Saad
    Commented Jul 19, 2014 at 13:14
  • \$\begingroup\$ Defends on how you detect the spike. You should remember that there will always be spikes even in the normal operation and I feel that SSR are designed and tested for that. Let's call your over-current situation as a spike. If detection system is a MCU that is looking at the spike (sampling a 50Hz or 60Hz waveform at a high rate by an MCU is not a big deal). If A.C is 50Hz, the maximum SSR "asked for turn-off" delay is time-period/4 assuming that at the peak of A.C waveform, the current was maximum (resistive load). Any SSR should be able to handle this stress. \$\endgroup\$
    – nvd
    Commented Jul 19, 2014 at 14:15
  • \$\begingroup\$ I feel the challenge is not to get false positives as MCU is doing fast sampling a take a short spike as a short. Thus, you need to low-pass filter (in analog or digital domain) the samples and take decision. Just make sure that the filter response is slow enough to allow a spike but capture a short circuit before SSR's back breaks. ;) You need a peak follower of the voltage across the sensing resistor. A peak follower consists of a diode followed by an RC network. It is a demodulator, the voltage waveform is amplitude modulated by the current. \$\endgroup\$
    – nvd
    Commented Jul 19, 2014 at 14:18
  • \$\begingroup\$ Sorry: "asked for turn-off" is "turn-off" above. \$\endgroup\$
    – nvd
    Commented Jul 19, 2014 at 14:22
  • \$\begingroup\$ could you please explain how the max. "turn-off" time is time-period/4? My concern is as follows: the mains outlet will be able to deliver a lot of current into the relay when there is a short circuit before the circuit breaker trips. I don't know much current will flow - but let's assume it's 100A, that 100A will be flowing through that "asked for turn-off" time period wouldn't it? \$\endgroup\$
    – Saad
    Commented Jul 19, 2014 at 14:28
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Normal relays are pretty good at surviving considerable current overloads providing the contact has already settled into its closed position. It's the closing and opening that is the killer for a lot of standard relays. I'm not saying all relays are the same so do read the data sheets and see what is said but, given you aren't using PWM via the SSR's normal relays should fit the bill.

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I did a DC current limit, current flows through a FET, and then a series resistor, which can turn on the ~0.6V Vbe of a bipolar transistor, which in turn switches off the FET. (there are some resistors involved too.) It's kinda a weird circuit, if you care about the off state. (since the bipolar needs to be on to shut the fet off.) If you can take the voltage drop, I think you could do something similar in AC... though again the off state might be turning on again off again each cycle. I'm not an AC/power guy.

This solution looks like it'd have to be another SSR in series that would turn off at over current. I'm thinking you might be able to turn off your existing SSR's with the same trick. Can you post a schematic of the circuit? (or maybe just the SSR piece of it.)

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