Hvordan en "liten" LED lampe kan ødelegge en bryter (for tekn. interesserte)

[hkmod25]

Tok meg lov for å klippe/lime her en ganske god forklaring på et problem som en del av brukere har opplevd selv:

(Posted 17 January 2016 - 02:49 PM, http://forum.fibaro.com/index.php?/topic/20447-relay-switch-2x15kw-lights-not-turning-off/):

 

"For the technical minded: explanation of the "welding" and an attempt to explain why certain loads cause this, sometimes...

Mains voltage is AC. It has an amplitude that changes from 0 to 324 Volt in 5 milliseconds. It's a sine wave, so it does that again with a minus sign (- 324 Volt) after another 10 milliseconds. The cycle repeats 50 timers per second. When you switch on the light at 0, the capacitors in the LED charge with the rising voltage. That is "slow" compared with switching on at the peak. We say that ramping up has a low change of voltage per time, often represented by "dv/dt". A capacitor is essentially a dead short until it has charged. Switching on from 0 to 324 Volt in 0 seconds mathematically gives an infinute "dv/dt". So switching on the light at the peak gives a current that is only limited by the circuit resistance... That is: the copper wire in your house and to the distribution transformer, plus contact resistances everywhere, plus the resistance of the components in the lamp. If everything is very low, then you probably get around 1/2 Ohm so Ohm's law says the peak current is 324/0,5 = about 600 Ampere. It would really surprise me if you could get such a low resistance, in my house a dead short generates "only" 300 Ampere (I've measured the resistance. I have not shorted the supply...). A more realistic resistance would be "a few ohms" so let's take a nice figure of 3 Ohms. That is still 100 A. As a rule of thumb, relays get a difficult time when you get above 10 x rated current = 80 A (for a double relay). These numbers are all indicative. There is at least +/- 20 % of uncertainty in all of that. And measuring this current is difficult. It can't be done with a multimeter.

In a typical LED there are 2 capacitor types (so to spike), that contribute to the effect. The first capacitor, the main capacitor, holds the DC charge to make the supply work. The second capacitor is an EMI suppressor. Switching mode power supplies generate radio frequency noise. Anyone still listens to AM or shortwave radio? Then you'll know what I mean. To avoid turning the wires into antennas some circuitry is across the terminals of the lamp. Both types contribute to the inrush current effect. The amount of Watt and the design dictate the effect of the main capacitor. The number of lamps determines the current drawn by the EMI filter. To give an example: I have ONE type of LED, a 20 Watt floodlight, that would weld my relay contacts once every fourteen days. It doesn't do that because I have an NTC in series. The current drawn by the primary capacitor and the EMI filter is just to big. I don't have this situation, but I think the relays will have a difficult time if you have between 10 and 20 bulbs, even if the wattage is very low, because you get the sum of 10 - 20 EMI filters. BTW to get in the "safe zone" of the relay, I would limit the inrush current to the relay to 5 times its rated current (around 40 A for a double relay). So what you need is "a few ohms" in series with your load. I repeat: resistance IN SERIES, not in parallel! You want to INCREASE the total resistance, and lower the current.

This explains to me the randomness of the "welding". It depends on: How close you are switching on to the peak of the sine wave, the circuit resistance, the sum of the capacitive load(s). Not all LED lamps are equal, so you can't say "but it is only 20 Watt"!

Side note: if you have worked with dimmers or solid state relays, you may have noticed that they mention "zero-crossover switching" and "leading/trailing edge switching". This is related to the subject of unrush current. ZCS aims to switch on when the sine wave is near zero. With a capacitive load, this will be the most "gentle" way to do it. The FGD-212 auto calibration detects this and chooses the parameters to ramp up the voltage. But even this technology has its limits. I can't find the topic at the moment, but someone noted that with plus-minus 10 LEDs the FGD-212 would sometimes detect over-current at switch-on, and protect itself. These lamps were probably of a "high inrush current" type, but that is speculation! Because a double relay can't detect this situation, I bet this particular load is a fair candidate to make it stick... I'm not sure what to prefer... a stuck relay or a dimmer that doesn't turn on..."

[Karl Skapeland]

Hva med å skrive litt om problemet, istedetfor å tvinge oss til å lese tre svære bolker med tekst uten avsnitt?

[hkmod25]

Hva med å skrive litt om problemet, istedetfor å tvinge oss til å lese tre svære bolker med tekst uten avsnitt?

Bare se den første linje av teksten: "Explanation of the "welding" and an attempt to explain why certain loads cause this, sometimes."

Rele kontakter på en AV/PÅ bryter blir "sveiset" sammen...

[toreae]

Det er vel ikke noe nytt at en kapasitiv last kan ødelegge en kontakt ved tilkobling, like så mye som en induktiv last kan gjøre det ved frakobling. Men:

 

 A capacitor is essentially a dead short until it has charged. 

 

Rett og slett feil. Og dermed også det som kommer etter. En hver kondensator er født med en indremotstand. Denne indremotstanden vil være med på ¨å redusere strømmen ved tilkobling. Jeg tviler sterkt på at det er kondensatorer i en LED-pære som gir medfører 100 Amper ved tilkobling. 

 

Jeg vet ikke hvordan dagens LED-pærer er oppbygd. Men at ikke resistiv last kan gi problemer er ikke tvil om.