I have to assume that the concensus is that reverse biasing an LED will blow it?
Does any manufacturer "actually" say this?
I have to say that I haven't seen any LED datasheet that gives any specific information on this point!
However, is it not possible that at low enough reverse currents the LED acts like a low power Zener diode and "doesn't" blow but simply clamps it's voltage at the breakdown level?
With 240v AC and a 240k resistor, there would be a reverse current of approximately 1.4mA (240 x 1.4 peak voltage) and I can't really believe such a low level would damage the LED.
After all at reverse breakdown of about 5v for a red LED, dissipation is only about 7 - 8 mW.
However, if more current is set up to get higher brightness, the reverse current dissipation would increase dramatically. e.g. aiming for ~20mA would need about a 17k resistor but the reverse power would now be 100mW and I doubt any low power LED would survive that.
In addition, the dissipation of the current limit resistor would have to be uprated to at least 10W and get pretty hot.
I don't really see why there is such a negative slant to the replies on this thread as we are all content to use neons at line voltage with suitable in line resistors, so what's really the difference apart from the higher voltage drop across a neon???
Let's face it, if you're an idiot with mains voltages, you'll be just as dead either way!
Sandy
Does any manufacturer "actually" say this?
I have to say that I haven't seen any LED datasheet that gives any specific information on this point!
However, is it not possible that at low enough reverse currents the LED acts like a low power Zener diode and "doesn't" blow but simply clamps it's voltage at the breakdown level?
With 240v AC and a 240k resistor, there would be a reverse current of approximately 1.4mA (240 x 1.4 peak voltage) and I can't really believe such a low level would damage the LED.
After all at reverse breakdown of about 5v for a red LED, dissipation is only about 7 - 8 mW.
However, if more current is set up to get higher brightness, the reverse current dissipation would increase dramatically. e.g. aiming for ~20mA would need about a 17k resistor but the reverse power would now be 100mW and I doubt any low power LED would survive that.
In addition, the dissipation of the current limit resistor would have to be uprated to at least 10W and get pretty hot.
I don't really see why there is such a negative slant to the replies on this thread as we are all content to use neons at line voltage with suitable in line resistors, so what's really the difference apart from the higher voltage drop across a neon???
Let's face it, if you're an idiot with mains voltages, you'll be just as dead either way!
Sandy
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Same old song.
One that builds an appliance, whatever it may be, should know this!!!!
I have been lighting LEDs from mains through a capacitor, one resistor, and a zener diode (reversed, circuit from Elektor), for years without any problem.
So, why chase the LED???
All the best,
Max.
Same old song! How people is afraid around the world!
One that builds an appliance, whatever it may be, should know this!!!!
I have been lighting LEDs from mains through a capacitor, one resistor, and a zener diode (reversed, circuit from Elektor), for years without any problem.
So, why chase the LED???
All the best,
Max.
Now just try one series resistor, for the truth. Whatever way u use, is upto u.
A resistor parallel to the LED and series resistor is also a good design and has been used in one German product I have seen.
Believe me that resistor is 120 KOhms and is across 240volts AC.
Gajanan Phadte
A resistor parallel to the LED and series resistor is also a good design and has been used in one German product I have seen.
Believe me that resistor is 120 KOhms and is across 240volts AC.
Gajanan Phadte
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It works because you are obviously not exceeding a destructive reverse current .
As to your statement "we fit LED's in our 3 pin mains sockets", then if you fit them as described with no reverse protection then all I can say is that it's very bad practice, smacks of incompetent design, and leaves the indicator light open to failure at any time... and some person somewhere may be relying on that to warn of voltage present.
Many may work for years, how many more do not. The mains often has transients of 1kv and more on it and these could well contribute to a failure.
Sorry but you have left yourself "open" with this thread, particularly as you imply you manufacture the sockets with indicator.
It just read a little as though you might.
Nothing changes though... it's appalingly incompetent design in a commercial product that saves pennies and may cost lives.
What is the point of this thread? Something which is a bad idea can, under certain circumstances, seem to work OK. So what? Is this the freedom-fighters of poor engineering throwing off the yoke of good engineering, or just a minor rebel pointing out that some people can overstate their otherwise legitimate case?
Too much reverse current through an LED, like any diode, is not good and is likely to shorten its life so removing the main advantage of LEDs. Far too much reverse current will make it go bang. Someone making sockets with LEDs which are routinely reverse-biased had better make sure the LED manufacturer is happy with this, or the customers may find that the light fails earlier than an old-fashioned neon would. You can always make a lot of money from a cheaper product via poor engineering, but in the end the customer gets wise to this and prefers to pay more for quality.
Too much reverse current through an LED, like any diode, is not good and is likely to shorten its life so removing the main advantage of LEDs. Far too much reverse current will make it go bang. Someone making sockets with LEDs which are routinely reverse-biased had better make sure the LED manufacturer is happy with this, or the customers may find that the light fails earlier than an old-fashioned neon would. You can always make a lot of money from a cheaper product via poor engineering, but in the end the customer gets wise to this and prefers to pay more for quality.
It is a sort of a challenge. If u prefer, take it and find out. I have so many sockets in my apartment that r being used daily for 6 years and the LEDs in them r still glowing.
BEWARE:230 VOLTS AC CAN KILL U.
BEWARE:230 VOLTS AC CAN KILL U.
Oh, by the way, that yellow LED in the photo I attached still behaves like a normal LED.
Glows when forward biased and does not glow when reverse biased.
Glows when forward biased and does not glow when reverse biased.
There is no challenge...
240 volts AC equates to 680 volts peak to peak, or 340 volts peak. 120k limits current flow to 340/120000 which is just under 3ma. I'm sure most LED will survive that reverse current in practice.
And some won't over many years.
Now put a transient of 3 or 4 kv on the line for a micro second or two. That may cause a spike of 30 ma or more in the LED as a reverse bias current. Will it survive, or has the life been shortened. Have you tested it at that ?
240 volts AC equates to 680 volts peak to peak, or 340 volts peak. 120k limits current flow to 340/120000 which is just under 3ma. I'm sure most LED will survive that reverse current in practice.
And some won't over many years.
Now put a transient of 3 or 4 kv on the line for a micro second or two. That may cause a spike of 30 ma or more in the LED as a reverse bias current. Will it survive, or has the life been shortened. Have you tested it at that ?
I'm no EE, or even an Electrician, but I thought electricity at the wall is only 1/2 cycle. So it's more like pulsed DC than actual AC(?) The power coming into my home is 240v AC. This goes to my breaker box where the two voltage carrying wires each feed one raill that then connects to a breaker and then the ground. Is this not correct?
What kills the LED is heat, not reverse current. If the current is limited by the resistor - under reverse bias conditions - so that (reverse current x reverse breakdown voltage) does not exceed the device dissipation, it will not be harmed.
Since reverse breakdown voltage is larger than the forward conduction voltage, the allowable current under reverse breakdown is proportionally smaller (for the same power dissipation) which may lead people to believe any reverse current is harmful.
I do agree that on mains applications there is a lot of unpredictability about what peak voltages may occur, so much more 'headroom' should be allowed for.
Since reverse breakdown voltage is larger than the forward conduction voltage, the allowable current under reverse breakdown is proportionally smaller (for the same power dissipation) which may lead people to believe any reverse current is harmful.
I do agree that on mains applications there is a lot of unpredictability about what peak voltages may occur, so much more 'headroom' should be allowed for.
The mains is sinusoidal, and usually to quite a high level of purity.
If your interested in how it's generated and distributed then this is good. Read the bit about single phase loads which applies to domestic supplies.
Three-phase electric power - Wikipedia, the free encyclopedia
As far as I was aware, ALL mains voltage in the States is 120v 60Hz A.C.
However on checking here;
Mains power around the world - Wikipedia, the free encyclopedia
it would appear that under SPECIFIC conditions 240v A.C. is available.
I've got to say that that is quite scary.
Gos help you if you manage to connect a lower voltage device to the higher voltage rail
Sandy
Hello Sandy,
Yes, 240 is available, but not at every outlet, and where it is available, the receptacle is different, so you can't get 240 going into a 120 plug 😉
Best, Barry
Yes, 240 is available, but not at every outlet, and where it is available, the receptacle is different, so you can't get 240 going into a 120 plug 😉
Best, Barry
Assuming that you're on 240v mains, the current with 120k in series is only 2mA, so maybe the LED is happy to handle this in reverse conduction? Remember though that you are dissapating nearly 0.5W in the resistor, so a quater-watt part won't be up to the job.
It would be interesting to see the waveform across the LED in reverse bias to see just how high it is going. Could be the LED is acting like a zener in reverse bias conditions and so limiting the voltage across it.
The current you are using is pretty small so that might be why the LED isnt blowing.
The current you are using is pretty small so that might be why the LED isnt blowing.
Out of curiosity, I looked into the issue.
A book I have a plot showing the leakage current starts around 5V for a red GaP LED. The current increases exponential to the voltage and the increase is slower than true Zener diodes. The voltage varies a lot within a lot by samples so picking an LED to substitute a Zener isn't a good idea.
The destruction of LED comes from the over current not from the over voltage, as others stated already, and ~2 mA seems fine for your LEDs.
What happens is that the brightness of LED degrade down faster, compared to normal usage. There are many papers on this issue but recent example is:
Defect generation in InGaN/GaN light-emitting diodes
under forward and reverse electrical stresses
X.A. Cao *, P.M. Sandvik, S.F. LeBoeuf, S.D. Arthur
Microelectronics Reliability 43 (2003) 1987–1991
which demonstrates slow degradation of the LED but mainly in the low current region. The authors suggest that the reverse voltage causes some physical change which leads to final defect eventually. There were almost no noticeable difference in the current-brightness plot at higher current region with reverse bias stress.
Dear Gajanan, do you have a reference to the description of the "theory" you disproved? I couldn't find a reference saying that a red ~ yellow LED would fail immediately if a reverse current of 2 mA was applied.
Please let me know for mistakes I made here. All the criticisms are welcome!
Best regards,
Satoru
A book I have a plot showing the leakage current starts around 5V for a red GaP LED. The current increases exponential to the voltage and the increase is slower than true Zener diodes. The voltage varies a lot within a lot by samples so picking an LED to substitute a Zener isn't a good idea.
The destruction of LED comes from the over current not from the over voltage, as others stated already, and ~2 mA seems fine for your LEDs.
What happens is that the brightness of LED degrade down faster, compared to normal usage. There are many papers on this issue but recent example is:
Defect generation in InGaN/GaN light-emitting diodes
under forward and reverse electrical stresses
X.A. Cao *, P.M. Sandvik, S.F. LeBoeuf, S.D. Arthur
Microelectronics Reliability 43 (2003) 1987–1991
which demonstrates slow degradation of the LED but mainly in the low current region. The authors suggest that the reverse voltage causes some physical change which leads to final defect eventually. There were almost no noticeable difference in the current-brightness plot at higher current region with reverse bias stress.
Dear Gajanan, do you have a reference to the description of the "theory" you disproved? I couldn't find a reference saying that a red ~ yellow LED would fail immediately if a reverse current of 2 mA was applied.
Please let me know for mistakes I made here. All the criticisms are welcome!
Best regards,
Satoru
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