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I have a circuit that uses a TRIAC that controls when 120VAC passes to the output. Never really had any problems until recently when I noticed that I had continuity through the TRIAC without any power going to the gate. I have 100 other boards that don't have this problem. So my main question is, is it normal for a TRIAC to fail closed and if so, how does this happen? Manufacturing error perhaps? My knowledge on TRIACs is pretty limited. The only reason I am so concerned is because I use this TRIAC to control power to a heating element. Without any control on the gate the temperature just skyrockets.

Oh, the TRIAC I'm using is: MAC4DLM.

Schematic: https://ibb.co/f4QPAk enter image description here Thank you!

Derek

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  • \$\begingroup\$ What does the datasheet say? Unless it gives you a failure mode guarantee, anything can happen \$\endgroup\$ – PlasmaHH May 4 '17 at 13:36
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    \$\begingroup\$ It's more likely to fail-conducting for any semiconductor than fail-open. That is because overheating makes the device conduct better, and that leads to more overheating, and ultimately to a short-circuit between the bond electrode and the substrate electrode. That's the moment when current limiting through the load kicks in, and prevents melting of the bond wire. \$\endgroup\$ – Janka May 4 '17 at 13:36
  • \$\begingroup\$ You assume it's the TRIAC but you show no proof. There could be another components on the board causing the problem, you show no schematic so it is anyone's guess. \$\endgroup\$ – Bimpelrekkie May 4 '17 at 13:36
  • \$\begingroup\$ How a device fails depends on the circumstances. If the device fails open then that usually interrupts the current and no more damage is done assuming no more current flows. If it fails short, too much current can flow which can cause additional damage to for example the bondwires. It could then fail open. Lots of ifs and buts, no way to be sure what is more likely to happen. \$\endgroup\$ – Bimpelrekkie May 4 '17 at 13:39
  • \$\begingroup\$ Oh, you better add a thermofuse to your device to avoid that thermal runaway. Any kitchen appliance with a heater has it. \$\endgroup\$ – Janka May 4 '17 at 13:39
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Fail-closed is the usual failure mode for power semiconductors, triacs, diodes etc. An overcurrent event will tend to overheat the semiconductor, which melts everything into a highly doped and therefore conductive mess. An overvoltage event will tend to punch through insulation layers, with much the same result.

This is why you always use a fuse in the input to a bridge rectifier supplied instrument. If one diode fails, it's likely to go short, which will then take out its opposite number, also short, shorting the input.

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  • \$\begingroup\$ Fuse probably would not help in this case though, since presumably the heaters are on for a while. Fuse would never notice then being on forever. \$\endgroup\$ – Trevor_G May 4 '17 at 13:57
  • \$\begingroup\$ Right, this is all good information. Now I'm just scratching my head thinking about how to protect from the heaters staying on indefinitely. I didn't know that TRIACs had to ability to fail closed so I never thought that I would see this type of scenario. \$\endgroup\$ – Derek Lyons May 4 '17 at 14:03
  • \$\begingroup\$ Use a thermofuse. They are cheap, safe and simple to apply. \$\endgroup\$ – Janka May 4 '17 at 14:39
  • \$\begingroup\$ resistors fail open. not sure if that's a counter-point or a hint... \$\endgroup\$ – dandavis May 4 '17 at 14:52
  • \$\begingroup\$ @dandavis resistors are not semiconductors, neither are rice puddings, and rice puddings don't fail closed. Your point is ... ? \$\endgroup\$ – Neil_UK May 4 '17 at 15:29
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If the condition of a failure of one component has a follow on undesirable result, then you have to prevent that undesired result. In this case, you are controlling a heater, so the undesired result is excessive temperature. The best way to prevent that from occurring is by using a thermal cutoff. This should always be installed in heating appliances anyway. I'm speaking with 35 years of experience in product safety. The fuse is not the solution. The thermal protector can be a one-shot, self-resetting, or trip and hold type, the selection would be based on whether or not the product is attended during use or whether it would continue to operate in the fault condition without any user intervention. Other sensors should be provided to alert someone of the device's inability to maintain temperature control.

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  • \$\begingroup\$ +1. Common industrial control is to incorporate a "loop-break" alarm. If the output goes to 100% or 0% and the feedback doesn't start to trend appropriately then a secondary protection device switches off. This could be a relay feeding all the OPs heaters. The relay only switches in power-up and fault conditions so it doesn't suffer contact wear as it would if it was switching the heaters for temperature regulation. \$\endgroup\$ – Transistor Dec 12 '17 at 14:33
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That really depends on the rest of the circuit and the supply.

Initially power conductors usually fail shorted as others have mentioned. What happens after that will be one of three things...

  1. The output will be driven continually with whatever internal or external consequence that might be.

  2. Some other part will be over driven and will fail. This could make it worse or better...

  3. The original part will literally burn or melt open.

When designing control circuits that drive loads that can cause unfortunate consequences if let in the wrong state, it is critical to ensure that any design is "SINGLE-FAULT-SAFE". Totally fault safe is not possible, but your system should at LEAST be able to stop driving for any SINGLE fault.

Whether that be by complex hardware/software, series gates, fuses etc. depends on the design and the consequences of failure.

ADDITION

In this particular case you could add a 2nd Triac in series with the current one, perhaps on the low side. That will save you if one fails, but you really need to detect that one or the other has failed and stop driving or it will only delay the problem till the second one fails. Feedback is your friend.

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  • \$\begingroup\$ adding a second semiconductor which can have the same failure mode seems like a bad idea. I'd recommend a fuse over a second thyristor. \$\endgroup\$ – akohlsmith May 4 '17 at 19:49
  • \$\begingroup\$ @akohlsmith fuse will not work... the heater can be on for a while under normal operations anyway \$\endgroup\$ – Trevor_G May 4 '17 at 20:00
  • \$\begingroup\$ How would a fuse not work? Can you elaborate? \$\endgroup\$ – akohlsmith May 5 '17 at 22:09
  • \$\begingroup\$ @akohlsmith, if the mode of the circuit is to turn on the heaters for a longer duration, the fuse needs to be large enough to handle on all the time. Sine the failure is "on all the time" the fuse won't know the difference. \$\endgroup\$ – Trevor_G May 6 '17 at 12:53
  • \$\begingroup\$ Ahh; you want a maximum-on time. A fuse can still work here, if you size the fuse so that its trip time at heater load is longer than the longest on-time you will accept. This gets tricky if the fuse environment has varying temperature though. \$\endgroup\$ – akohlsmith May 13 '17 at 15:43

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