Hello there,

Recently i received a rather peculiar 120vac coil relay. It has what appear to be a 2mm red LED wired in series with a 50k resistor, and that placed across the coil of the relay.

Now if this was a DC relay and the polarity was marked, no problem. However, this is an AC coil relay 120vac, which has peaks of plus and minus 170v.

My problem is that i want to test the LED as to how it works with AC, but without blowing out the LED.
If i connect it to 120vac, the LED will get -170v across it when the peak goes negative, so it will blow out. That is apparently what happened with the first unit but i had no idea the design might be flawed (or not) so i just plugged it in and turned it on, and the LED does not light.

Now the second unit i dont want to blow out the LED, although it would be simple to replace. I would like to investigate this for the manufacturer of the relay so i can advise whether to fix it or not. If it really is flawed, it needs a design update. If it is not flawed then the first unit simply did not work and the second unit will work when plugged in, but i cant plug it in because if it is flawed it will blow and i wont have any other unit to test.

I guess the real question here is are there any LED's on the market that can handle plus and minus voltages like for AC, without adding a diode. IF there are, then this could be one of them.

So far the quick test i did was to connect 4vdc across the coil, first one polarity and then the reverse. It only lights (the 2nd unit that is) with one polarity and not the other. Unfortunately that's not definitive because if the LED is made for this, then it may not light in both directions, but it could be just a simple LED and of course then it would light in only one direction anyway.

So you see the sort of dilemma here, if the LED has built in protection then it will work, but if not it will blow out when connected to 120vac and it will be hard to explain to the manufacturer exactly what went wrong. Of course since i saw this one light, if it blows out then we know, but i'd like to keep this one intact in case i have to send it back to the manufacturer so they can see it work with most of the original parts and duplicate the fix/mod.

I thought this was an interesting problem too so that's also why i am posting, as well as to hear some ideas on a test procedure. Maybe look for reverse leakage, but something simpler would be nice and not sure if that will be definitive either.

Also note i dont want to bother the manufacturer for the part number of the LED although that may eventually have to happen. I'd like to solve this without having to do that.

Mucho thanks

 

Show us some clear pictures.

 

50kohm seems like plenty.....

 

you can put two led anti parallel

the currennt after breakdown will be limited by 50k but you often have surges and harmonics,
these can reach 1000v easily, and then you could see EMI events of some thousand volts, also damaging regular electronics, those with oldfashion 50hz transformer will be less affected but cheap switching supplies will blow out from these EMI events

Neon lamps are better suited they can take a lot overloading for a short time, can be used with resistor or also capacitor.

But when you do as instructed, the LED should not blow so easily, you can rectify first and put a small electrolytics + zener diode,
normally not much needed but it can deal with overvoltages

 

How do you know that it isn't the back emf from the relay that is killing the LED?

I just tested an LED with 33kΩ on 120VAC and it tested ok.

Try using a variac and gradually turn up and down the voltage.

 

How do you know that it isn't the back emf from the relay that is killing the LED?

I just tested an LED with 33kΩ on 120VAC and it tested ok.

Try using a variac and gradually turn up and down the voltage.

Hi,

Do you mean you connected an LED, a regular LED, in series with a 33k resistor and powered it with 120vac and it worked ok, as it lit up ok?
And you have no other components connected?

The LED on the first unit never lit, not even once, so i did not think it would be back emf. Back emf will light the LED briefly i believe, but i am not too concerned with that right now i am mainly concerned with the -170v that appears every half cycle.

 

Hi,

Do you mean you connected an LED, a regular LED, in series with a 33k resistor and powered it with 120vac and it worked ok, as it lit up ok?
And you have no other components connected?

Yes, that is what I did.

The LED on the first unit never lit, not even once, so i did not think it would be back emf. Back emf will light the LED briefly i believe, but i am not too concerned with that right now i am mainly concerned with the -170v that appears every half cycle.

I'm thinking, what happens when you remove AC power at the peak of the mains cycle, either at the positive peak or the negative peak? The inductive coil in the relay will kick back with a huge voltage spike.

 

Yes, that is what I did.

I'm thinking, what happens when you remove AC power at the peak of the mains cycle, either at the positive peak or the negative peak? The inductive coil in the relay will kick back with a huge voltage spike.

Hi,

Ok thanks for clarifying the experiment. I should try that too i guess with a random LED.

I am not too interested in the inductive kick back right now though and here is why (although i wont forget that).
I already have a -170v reverse voltage to think about, and that is applied to the LED through a 50k resistor. THAT, in itself, i believe to be a bad thing without even going any higher like to say -1000v kick back.
So in short, if i solve the -170v reverse voltage "problem" then i solve the -1000v kick back too, most likely.
I also happen to know for sure that there is an inductive kick back although i made no attempt to measure it because of the above.
If i connect the coil to 4vdc backwards, when i remove the power source i see the LED light up for what must be 1/4 second or something.

Your experiment is interesting because that means the LED you are using can take -170v (reverse voltage) through a 50k resistor probably because of the low current, which must be around 120/50000=2.4 ma rms. If the reverse LED happens to 'zener' at 10v, that's only 12mw plus maybe another 5mw.

This operation surprises me a little because i've seen LED's blow out with as little as 9v reverse voltage, so maybe some LED's can take it and some not. I guess it partly depends on the reverse 'zener' voltage and maybe the capacitance in reverse. Maybe you could measure yours. I also hve to wonder how long it would work like this. There is usually a max reverse voltage spec too which is often 5v, and if we go above that then we go outside of the recommended operating conditions.

I hope i can try this too later or tomorrow some time.

There is also a secondary issue that arises here too though, and that is the power rating of the resistor.
When i do this i use two diodes: one in series, and one in parallel, so that the series diode limits the power dissipation in the resistor and the parallel protects the LED.
Without the series diode however and assuming 2v 'zenering' voltage and 2v forward voltage, we see almost the whole RMS voltage across the resistor which amounts to 278mw in the resistor, and this resistor happens to be a 250mw resistor. If we assume 10v reverse 'zener' voltage, then we have about 260mw in the resistor, which is still too high really. If we assume the whole 170v then i cant see the LED surviving.
Something to measure next i guess though.

 

Don't forget that you still have 50kΩ in series. So even if the diode breaks down, the breakdown current is limited to 3mA.
Is the reverse breakdown a temporary or permanent destructive breakdown?
Will the diode recover from a temporary 3mA reverse current?

 

It would be interesting to look at the voltage directly across the LED with an oscilloscope. (But using an isolating transformer for the test.)

Les.

 

Hello,

Usualy a led will break down at a reverse voltage of about 5 Volts.
Putting a diode anti-parallel to the led, will protect the led.

Bertus

 

It would be interesting to look at the voltage directly across the LED with an oscilloscope. (But using an isolating transformer for the test.)

Les.

Done that.
With green LED and 33kΩ series resistor, the peak reverse voltage is 100V.
So the diode looks like 50kΩ drawing 2mA.

LED still lights ok.

 

Hi MrChips,
I had always believed that even the tiniest current though the junction in the reverse direction would destroy the LED. I had expected that as soon as the voltage exceeded the reverse voltage rating it would behave like a zener diode (Around 5 volts or so.) or it would latch into a state with a very low voltage until the voltage was removed. (Or would be destroyed as soon as the voltage exceeded it's reverse voltage rating.

Les.

 

Don't forget that you still have 50kΩ in series. So even if the diode breaks down, the breakdown current is limited to 3mA.
Is the reverse breakdown a temporary or permanent destructive breakdown?
Will the diode recover from a temporary 3mA reverse current?

Hello again,

Yeah if the current is limited it leads us to think about the power in the LED then. There is a peak power point similar to max power tracking a solar array where when the voltage (across) is low the current is higher and when the voltage (across) is higher the current is lower, so there is a point at i think around 1/2 of the full voltage where the power is max.

Unfortunately, that is for a device with a perfect physical construction where the electrostatic field is the same across the entire cross section. If we have small areas that are different, we might see higher current flow in those small areas and thus the current is not distributed across the whole cross section and that leads to what some call "hot spots". It would be like having 100 resistors in parallel, but one resistor has a higher voltage across it so only that one resistor suffers, but after that blows out that leaves 99 and then another might blow next time.
The better analogy though i think is a comparison to a lightning storm. When a lightning bolt hits say a city block, it may take out one house, but the remaining houses are still ok. But then next time a lightning bolt hits, it takes out another house, etc. So the short term damage is small but not zero, and the long term damage is ultimate failure.
In the LED or other semiconductor, the lightning bolts would be analogous to tiny plasma streams that destroy minute parts of the semiconductor that may be hard to detect as being damaged. This might be detectable as electrical noise. Over time the damage is accumulative.

Back in the 80's, i also learned that when the voltage goes too high there is a small amount of damage that accumulates over time, and that could lead to early failure. There was no good explanation given back then though like the plasma explanation.

Now you are reporting 100v reverse voltage across the LED itself. IF that is the peak, then with a peak line of 170v that means 70v across the 50k resistor and that's a current of 1.4ma peak, which is around 1ma rms.
100v is interesting because i didnt think it could get that high. The small dimensions of an LED die would make me think that there should be lots of arcing inside the LED just because of the sheer distance inside the package. I have t wonder if we are seeing some sort of anomaly. I say this because normally we would think that the LED would start to conduct in reverse, and that would keep the voltage lower as long as there was a significant series resistor. So i have to think that there could be significant differences in LED makes. I also have to wonder how long the LED would last under that kind of excitation. I would think it would be related to how extreme the "hot spots" are during conduction and this would be related to the homogeneousness of the die. So in short, we'd have to see how long the LED would last like this.

It's also kind of funny, the solution that would always work with no doubt is to use a two lead bi color LED, although the resistor would then have to be made a 1/2 watt unit instead of a 1/4 watt unit.

What else is interesting is that the LED on the first relay is blown out. How did it get like that

 

Hello,

Usualy a led will break down at a reverse voltage of about 5 Volts.
Putting a diode anti-parallel to the led, will protect the led.

Bertus

Hi,

Yes that's what i always did too, so i was puzzled when i saw this. In fact, when i first saw the very first relay i just assumed that there was a diode hidden in the package that could not be seen too easily so i didnt pay too much attention to it. Then i saw that the LED was blown out and so i started to look closer.

I also use a series diode too because that keeps the average current lower so the resistor can have a lower power rating like 1/4 watt instead of 1/2 watt.(120vac systems).

 

Hi MrChips,
I had always believed that even the tiniest current though the junction in the reverse direction would destroy the LED. I had expected that as soon as the voltage exceeded the reverse voltage rating it would behave like a zener diode (Around 5 volts or so.) or it would latch into a state with a very low voltage until the voltage was removed. (Or would be destroyed as soon as the voltage exceeded it's reverse voltage rating.

Les.

Hi,

I think the damage is minimal with one occurrence, but it is also accumulative so over time it fails early. I suppose that's the best case though, as the worst case is total failure right away. That's probably why we always protect against this kind of thing.

 

I have left the LED and resistor circuit connected to 120VAC mains for as long as you wish.
(Hope I am not violating forum rules.)

 

Just to make sure that I don't run afoul with forum rules, I have a RED LED, 33kΩ series resistor and isolated 150VDC power supply connected.

No breakdown so far. No appreciable current measured.

Switched back to GREEN LED.
0.4mA measured.

Will leave it on indefinitely.

 

Just to make sure that I don't run afoul with forum rules, I have a RED LED, 33kΩ series resistor and isolated 150VDC power supply connected.

No breakdown so far. No appreciable current measured.

Switched back to GREEN LED.
0.4mA measured.

Will leave it on indefinitely.

Hi,

Oh yes good idea, 150vdc isolated is safe and does the test well i think.

The red LED must be dropping the entire 150vdc then, that's amazing.
The green LED must be dropping about 136.8v then, also amazing.

Do you have any idea what the age of these two LED's could be? At least from the day of purchase.
I am wondering now if modern LEDs are made better, but i have also read now that it may take some time for it to actually fail and that would be a time that is much shorter than the typical life at such low forward current.

Oh wait, there is no forward current so that might change the results. One of the problems with LEDs is decreased light output. We might have to use isolated AC, and connect a control sample (same value and exact part number resistor, same exact part number LED) up to the same with a reverse diode. Unfortunately this could take a while, maybe even a week for example or more.

This got more interesting with your tests so far Mr Chips. I think most of us thought it would fail much sooner.

Isolated supplies are ok with the terms of service right?

 

My guess are the RED and GREEN LEDs tested are at least 10 years old.

BLUE LED dies after about 80VDC.

I have newer stock that was purchased two years ago.