Power indicator LED in High Voltage Circuit

[Diver300]

I still puzzle how a single LED with a single current limiting resistor (100K, 0.25W) can work under high stress at mains voltage (240V 50Hz). On the CREE 5mm LED datasheet, the reversed breakdown voltage is only 5 volt, and during the non-conducting half cycle (no light producing), the LED receives reversed voltage as high as 339V (peak values of RMS 240V), without collapsing and without any diode protection.....only two components, one LED and one resistor. Fyi, I tested before a LED (only one time) in reversed biased at 15V from a desktop power supply unit and it was gone instantly!

https://www.electro-tech-online.com...-powered-from-230v.117495/page-5#post-1041139 will answer your question about the reverse voltage. In fact it is reverse current that the LED stands.

The reverse breakdown voltage of LEDs varies enormously. Some of the brighter ones are not rated for any reverse voltage. If you didn't have current limiting resistor, and breakdown was less than 15 V there will be a lot of current. Even with a current limiting resistor, the power may be a lot more than normal, and powerful LEDs will burn out very quickly if the max power is exceeded. There could well be other breakdown mechanisms in some LEDs.

 

[audioguru]

I saw a video a minute ago where a guy connected 244 9V batteries in series. They made a spectacular arc but made an LED dimmly blink for a moment. I betcha the LED became shorted.

 

[steven7890]

Thanks to Driver300 in sharing his knowledge of running an LED with only a resistor 100K across the mains voltage as an indicator. Yes, running LED with only a resistor under AC240V has shown working. My next question is how long it will last and what the failure mode is when it is down. Running an component beyong its maximum rating (one or more parameters) is not supported and warrant by its manufacturer. I have some neon bulb with resistor mains indicators around me running 7x24 for more than 6 years without visible degrading in their brightness and still going on. Will LED only with resistor indicators deliver the same level of performance? In total cost wise neon bulb appears to be 5% to 10% higher, but this fluctuates somehow dependent on the spot supply-demand at the moment of purchase

 

[Diver300]

In my experience, neon indicators can dim with time.

How long an LED will last when run beyond its rating is anybody's guess. I wouldn't do it in a commercial product.

 

[audioguru]

LEDs slowly get dimmer over some years. I have a clock radio that has been displaying the time continuously for about 34 years and its display is now so dim that I cannot see it in daylight.
NASA does not allow opto-isolators that use an LED in spacecraft because after some years the LED gets so dim the circuit fails to work.

I think name-brand (not cheap crap from ebay) 5mm LEDs are guaranteed for 50,000 hours (5.7 years) when their brightness drops to half at 20mA.

 

[steven7890]

Driver300: "How long an LED will last when run beyond its rating is anybody's guess. I wouldn't do it in a commercial product." --- agreed

audioguru: "NASA does not allow opto-isolators that use an LED in spacecraft......",
other than LED in opto-isolator, what are other light sources could be used in opto-isolator ?

 

[audioguru]

audioguru: "NASA does not allow opto-isolators that use an LED in spacecraft......",
other than LED in opto-isolator, what are other light sources could be used in opto-isolator ?

I guess better electrical insulation is used instead of opto-isolation.

 

[MrAl]

Hi,

I used to have a pdf of a study done on white LED's as to their half life. If the half life is specified at normal current (like 20ma) then the half life at half that current (like 10ma) is something like 5 or 10 times longer. So the usable life of the LED goes up very fast with a decrease in current. This is why it is good to use opto couplers at lower than rated current too as other types of LEDs are similar.

I also had a hard copy of a study done by HP years ago, back in the 80's, but not sure where it is now. It showed similar results and i think that was done with red LED's from that time period.

 

[ronsimpson]

Many white LEDs use phosphor to make white light. Most of the half life of a white LED comes for the phosphor not the LED.

 

[MrAl]

Hi,

It doesnt seem to matter much as both white and red show the same basic degradation. This is sort of obvious from the fact that optos age like this too, and follow the same basic pattern. The blue LEDs will age, green, yellow, etc., and the loss in the LED itself is not insignificant. The HP study showed regular LEDs loosing luminous output in almost the same way as the white ones do today. It is always related to the drive current in any type of LED.

 

[steven7890]

If the design of one LED + one resistor as mains indicators or pilot lights could not be accepted, due to the present of high reversed voltage (or current) exceeding its maximum rating stated on its manufacturer data sheet, then I can see using neon bulb with one resistor could be still the lowest cost and technically proven solution. LED remains a choice for low voltage, and neon bulb for higher voltage applications

 

[MrAl]

If the design of one LED + one resistor as mains indicators or pilot lights could not be accepted, due to the present of high reversed voltage (or current) exceeding its maximum rating stated on its manufacturer data sheet, then I can see using neon bulb with one resistor could be still the lowest cost and technically proven solution. LED remains a choice for low voltage, and neon bulb for higher voltage applications

Hi,

I have to disagree totally. That might have been the case 10 years ago, but i dont believe that is the case today because we have very high brightness LEDs that are cheap and long lasting. If you cant afford one diode and two resistors or two diodes and one resistor then might as well not build it

As far as "technically proven", i dont feel that is the right way to describe those pesky neon bulb indicators. They were fine in the days of yesteryear, but now we have better. It's almost like incandescent bulbs...yeah they worked, but they only emit 2 percent of the total light energy required to run them...might as well get rid of them now whenever possible.

 

[steven7890]

talking about power consumption of LED indicator (including resistor and other like diode) vs neon bulb (including only one resistor) powered at AC mains 240V 50Hz, sorry I dunt have good comparison infor to share here.

But I do have here one 5mm neon bulb running at 240V AC via a resistor 330K, 0.25W for days emitting orange light. I used Sanwa CD800a digital multimeter to check the voltage across the 330K resistor, and it was 163V AC, and the across neon bulb was 77 V AC. Therefore the AC current moving them was 163/330 = 0.5mA. I am not sure the power factor for neon bulb, but the higher consumption was 0.5mA x 240V = 0.12W. Of course using wattmeter to measure is the best.

Does anyone have single LED set up running at mains voltage and know its wattage ?

Also happen to find this video clip interesting to share:

 

[audioguru]

Many white LEDs use phosphor to make white light. Most of the half life of a white LED comes for the phosphor not the LED.

I disagree.
The "white" LED with a phosphor is actually a blue LED and the phosphor adds a yellowish color (you can see the yellow when the LED is not producing light) to produce white.
If the phosphor light fades then the color of the LED will shift to a very noticeable blue.

But I have never seen a white LED that has dimmed due to a long running time, only red ones.

 

[audioguru]

Does anyone have single LED set up running at mains voltage and know its wattage?

It is difficult to measure the power because a neon bulb are very different and and an LED are not linear.

A Neon bulb draws no current until the AC voltage rises to about 90V which is its "striking voltage" then draws current as its forward voltage drops to about 65V. When the AC voltage drops less than about 60V then the neon bulb turns off and draws no current until the next half-cycle voltage reaches about 90V again.

An LED has such a low voltage compared to the very high mains voltage that the LED conducts for almost the entire half-cycle that it has forward bias.

 

[audioguru]

The "commercial" about Neon vs LEDs shows that Neon is much better but did not show the short lifetime of Neon.
I would like to see a similar comparison commercial done by an LED manufacturer.

 

[MrAl]

Hello steven and audioguru,

It is difficult to measure
the power because a neon bulb are very different and and an LED are not
linear.

A Neon bulb draws no current until the AC voltage rises to about 90V which
is its "striking voltage" then draws current as its forward voltage drops
to about 65V. When the AC voltage drops less than about 60V then the neon
bulb turns off and draws no current until the next half-cycle voltage
reaches about 90V again.

An LED has such a low voltage compared to the very high mains voltage that
the LED conducts for almost the entire half-cycle that it has forward
bias.

That's an interesting observation. I was tempted to think of the neon as
linear myself.

With that information we can model the neon roughly as a device that is
open circuit until the line voltage reaches 90v, and then becomes a 65v DC
source which of course then draws current through the resistor. Knowing
the resistor value, we should then be able to calculate current and thus
the power lost in the resistor and in the neon bulb. It wont be perfect
but it should be good enough for a simple comparison. Ive done this below.

steven:
I have powered LED's at low current from the line. For a little more
brightness i have used 50k, and for more normal use 75k and 82k. 82k works
out very well because the resistor does not get warm even though the line
is 120vac and as audioguru said the LED is on almost the whole time. This
is when using a 1/2 watt resistor for physical durability and for it's
power rating.

All:
The line voltage is:
Vt(t)=170*sin(2*pi*f)

Solving this for Vt=90 we get:
T1=asin(9/17)/(2*pi*f)

Solving Vt(t) for Vt=65 looking for the solution between 90 and 180 degrees
we get:
T2=acos(13/34)/(2*pi*f)+1/(4*f)

Now that we have the turn on and turn off times per half cycle we can
calculate the power in the resistor and the power in the neon bulb.

The instantaneous current in the resistor is:
IR(t)=(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}

and so the instantaneous power in the resistor is:
PR(t)=IR(t)^2*R=(170*sin(2*f*pi*t)-65)^2/R, {T1<=t<=T2}

The average power then is:
Pavg=(1/T)*integral PR(t) dt [from t=T1 to t=T2], {T=1/(2*f)}

which for any frequency f comes out to:
Pavg=4249.0894/R

That is the power in the resistor alone.

Next, the voltage across the neon bulb is:
VN=65, {T1<=t<=T2}

and the current (from above) is:
IR(t)=(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}

so the instantaneous power in the neon bulb is:
PN(t)=IR(t)*VN=65*(170*sin(2*pi*f*t)-65)/R, {T1<=t<=T2}

Integrating from T1 to T2 and multiplying by 1/T where T=1/(f*2) we get:
Pavg=3286.9903/R

Adding the two we get the total power:
PT=7536.08/R

Calculating this for R=300k we get:
PT=25mw

and for R=100k we get:
PT=75mw

For comparison, my semi high brightness LEDs run at about 100mw but this can be reduced with ultra high brightness LEDs.


Without specifying the peak line voltage Vpk or the turn on voltage of the neon bulb VH or the turn off voltage of the neon bulb VL here is the rather long formula for the total power:

PT=
((4*VL^2-260*VL+2*Vpk^2)*acos(VL/Vpk)+(2*VL+260)*sqrt(Vpk^2-VL^2)-8*sqrt(Vpk-VL)*VL*
sqrt(VL+Vpk)+(2*pi-4*asin(VH/Vpk))*VL^2+(260*asin(VH/Vpk)-130*pi)*VL-8*sqrt(Vpk-VH)*sqrt(VH+Vpk)*VL-
2*Vpk^2*asin(VH/Vpk)+(2*VH+260)*sqrt(Vpk^2-VH^2)+pi*Vpk^2)/(4*pi*R)

The only variables are:
Vpk the peak line voltage (Vpk=170 for a 120vrms line),
VH the turn on voltage of the neon,
VL the turn off voltage of the neon,
R the series resistance (in Ohms).

The formula may be able to be simplified but that's it for now.

 

[steven7890]

MrAl,
Thanks for the precise calculation of power consumption for the neon bulb with resistor running at 120V AC. How about running at 240V mains? Can I simply multiply by 2 ?
Also I failed to catch how you get LED power consumption at 100mW

 

[MrAl]

Hello again steven,

Using the formula i gave in the previous post the total power with a 240vrms line would be:
PT_240=43616.72/R

The average current ratio changing from 120vrms to 240vrms is about 3.0707, so to get the same average current at 240 we'd need a resistor of about 3.1 times what we use at 120vrms. So lets make it an even 3.000 and go from there.

With R=900000 we get:
PT=48.5mw

and with R=300000 we get:
PT=145mw which may be too high.

You can check what resistors are normally used at 240vac and try them. I think the Ne2 bulbs take a 50k resistor at 120vac which would give 150mw, but i'd have to look this up. That would mean 290k at 240vac, but we should verify that the bulb can run ok with this value.

The "100mw" came from a quick estimate of the total LED plus resistor power when run from a 120vrms line.
The power is roughly:
PT=120^2/Rs

where Rs is the series resistor.
So for Rs=82k we get 176mw, and for Rs=100k we get 144mw.

 

[steven7890]

If I put 2 LED in reversed parallel and in series with a resistor 100K running at 240VAC, the total wattage will be 240^2/100K = 576mW. This looks high for me...need 1 watt rating resistor. Or use a higher resistor....
Thanks again to MrAl.....