Circuit Challenge: Light Two LED's

[MrAl]

Hi,

We used to have a saying back in the 1980's:
"The hardest thing to do in electronics is to get a bulb to light up".

The reason for this is because it always seems so simple that we often overlook things that could go wrong.

This however is about lighting two LED's. Unfortunately, there are a limited number of parts available which i'll list here.
1. Bi color LED, one internal LED is blue 3.0v, one is green 2.5v, common cathode, so they are both in same package, and the package has three leads: one lead is CC, one is Blue anode, one is Green anode. (Also see side note far below).
2. Any number of 1N4007 diodes, but use the least number of diodes possible.
3. Any number of 82k, 1/2 watt resistors, but use the least number possible.

The application is as follows:
1. The LED will be used as an indicator for a power strip. The power strip has a single SPST switch on it that powers the six outlets when it is turned on, and does not power them when it is off. The power strip may be used with small wall warts or even a 500 watt appliance.
2. The 'blue' part of the LED is used to show that the power strip is plugged into the wall only.
3. The 'green' part of the LED is used to show that the switch is turned 'on' only.

So in a typical day of use it may goes as follows...
The strip is plugged in, the blue LED lights up because the switch is off, and nothing is plugged into the outlets yet. Later, something is plugged in, and then the switch is turned on, and then the green LED lights up.
Still later yet, the switch is turned off and the blue LED lights up again.

So the operation is fairly simple, but the part types are limited as above. The idea is to use as few parts of those above as possible.

There is a small side note about the bi color LED. If the blue LED is wired directly in parallel to the green LED (ie both anodes connected together) only the green LED will light up because it hogs all the current because the voltage is lower than the blue LED voltage, and is it low enough to really get all or most of the current.

This is actually something i may want to use myself too, not just an interesting problem.

 

[alec_t]

What is the mains supply voltage that feeds the power strip?

 

[MrAl]

Hi Alec,

Thanks for asking, i forgot to mention that. It is 120vac 60Hz. The 82k resistors are considered adequate for 1/2 wave operation of either LED. I would think that for 220vac mains the resistor values would increase, but i wont press this because i only need it for 120vac operation for the time being.

Also i forgot to mention that each LED must be protected against reverse voltage, even if there is a series diode, assuming that the diode reverse turn off characteristic is enough to blow the LED, and this usually requires a reverse diode across each internal LED (unless you can do it with one diode). All i really want to put forth though is that they have to be protected in some way against reverse polarity even if there is a series diode because a series diode can conduct for a very short time in the reverse direction. The usual method is a second reverse diode right across the LED, but there are two LEDs, but only one is on at a time too, so maybe one diode? Maybe not?

 

[Reloadron]

What is the mains supply voltage that feeds the power strip?

Since MrAl is in NJ, US I will venture a guess 120 VAC 60 Hz to the power strip. Might be fun to toss in other voltages like your Mains Alec. MrAl is likely busy this morning watching it snow and blizzard.

<EDIT> I see we are up and about and beat me with a post. </EDIT>

Ron

 

[MrAl]

Hi again,

Yes by about 2 minutes.

But let me explain why i would like to see a minimum parts design and also how yesterday i think i may have won the biggest and clumsiest dummy in the universe award, pending reply from all the other galaxies.

First, a min parts design because the space inside these power strips is limited so the less parts the better.

The biggest dummy award because, paying no attention to the known proverb from the 80's i mentioned, i quickly wired up a 'solution' to the problem, soldered everything, and found a big problem. Start counting. First, plugged it in, perfect, the blue led lit, then turned it on, perfect, the green led lit. Then, turned it off, plugged in a test 500 watt load, led goes out completely. Unplugged, still no led. Tested power, it's there, 500 watt load test power no problem. What happened was that in my haste there was a secondary current path through one of the diodes to the green LED when the load was plugged in, and that put a current much too high for the LED to handle so it blew out immediately, and took out the blue LED too. LED down count: 1.
Ok, so looked over the circuit, found the problem, drew new circuit. Got new LED, tested it, tested new diode set, all ok. Moved across the room to solder it all up again, then realized i left the meter on the other side of the room. Started across the room with LED in hand, something hanging on the wall caught the LED in my hand and flung it across the room. Where? Who knows. It's gone now and seems lost forever. LED down count: 2.
Ok, got still yet another LED which BTW was in the bottom box of the stack of boxes, so i had to move all the boxes first to get to that box. Test it, it's ok. Wired up new circuit. Plugged power strip in, blue lit up. Turned power on, green lit up and blue stayed lit (not perfect but what the heck). Plugged in load, load worked fine. Done? Nope. With load still plugged in, turned off switch. Green LED stayed lit up. Not exactly as bright as before, but still way too noticeable.
As you can see it was a pretty bad day yesterday

So as it stands now, when the strip is plugged in the blue lights up, and with the power off when the load is plugged in the green comes on fairly bright. When the switch is turned to 'on' the green gets a little brighter but not by too much. So it works in a way, but i got some unusual functionality i did not plan on.

I think i have now found a solution but i'd like to see others ideas too if they care to share.
And it only took three LED's to find out
Also, i have not tested this solution yet either.

 

[MrAl]

Hello again,


Here's what i have so far. This is untested.

 

[alec_t]

You can save 2 components by eliminating the top 2 diodes.
If you want to have the blue LED go off when the green LED comes on, you could try this (which uses only 1 more component than your posted circuit) :-

 

[MrAl]

You can save 2 components by eliminating the top 2 diodes.
If you want to have the blue LED go off when the green LED comes on, you could try this (which uses only 1 more component than your posted circuit) :-
View attachment 90403

Hi there Alec,

Thanks for the suggestions.

But i think we may want to look at this a little more carefully. If i remove the top two diodes, then the anti parallel diode across the blue LED conducts when the the upper line line voltage goes negative, and then that causes the green LED to light up when the load is plugged in and the switch is off, but it's worse than that because then the only thing that limits the current through the green LED is the load resistance, which could easily be as low as 20 ohms. So we'd have a green 20ma LED in series with a diode and 20 ohm resistor being powered by 170v (at the peak). That's something like 8 amps through the LED.
This means at least one of the top two diodes has to stay. But unfortunately, they serve another purpose too. They eliminate half of the cycle of current from getting through the 82k resistor. That means the power in the 82k is half of what it would be with only one diode (and the switch on). So one upper diode will prevent over current to the green LED, and two keeps power dissipation in the 82k as low as possible.

Also, with the blue LED having a higher forward voltage than the green, when the switch is on the two upper diodes would conduct with about the same voltage drop if they were both conducting. That would mean if the diode voltage is 0.7 and the green LED is 2.5v and the blue is 3.0, then the total voltage for both LEDs and their diodes would be: green 2.5+0.7=3.2v, and blue 3.0+0.7=3.7, so again the total voltage difference is about 0.5v, which is enough to keep the green LED pretty much off. But the diode currents are not the same, so the green LED series diode voltage will be slightly less than the blue, so lets say it does down as low as 0.5v. Then the difference is 0.3v, but the current through the green LED is going to be very very low now, so i dont think it will light up very much (switch on of course). Do you agree with this?

I did not try this yet though, and i guess it would be interesting to simulate. I wanted to see what others ideas where first before i modify the power switch a third time (ha ha). Right now it is a little strange (a different circuit than shown) because the green light comes on when the load is plugged in, but at least i have a power light now when before i had none. So i can take my time about this now, and look over the possibilities. The next time i take it apart and rewire everything i want to get it right

 

[ccurtis]

I wonder if the antiparallel diode is necessary. Damage from reverse breakdown occurs from heat as I understand it, not from the avalanche itself. During breakdown, the heat dissipation in the LED is about 24 milliwatts thanks to the series resistor and that will only occur for about 200 ns, the recovery time of the series diode, for an average reverse heat dissipation of 125 microwatts over one cycle of 60 Hz (which is much less than the forward heat dissipation). I do understand that the beginning avalanche region is very localized (non-uniform) within the junction area, creating tiny "hot spots", but with so little heat, I wonder just how hot the spots get.

 

[MrAl]

I wonder if the antiparallel diode is necessary. Damage from reverse breakdown occurs from heat as I understand it, not from the avalanche itself. During breakdown, the heat dissipation in the LED is about 24 milliwatts thanks to the series resistor and that will only occur for about 200 ns, the recovery time of the series diode, for an average reverse heat dissipation of 125 microwatts over one cycle of 60 Hz (which is much less than the forward heat dissipation). I do understand that the beginning avalanche region is very localized (non-uniform) within the junction area, creating tiny "hot spots", but with so little heat, I wonder just how hot the spots get.

Hi,

Yes, very good point, and you may very well be right about that failure mode. But let me explain my position on this issue in more detail. My general approach to circuit design is if there is a question about a circuit that can not be answered then at the very least change the circuit so that the unanswerable question does not come up with the new circuit.

The issue is really about failure *rate*. If we had 200 circuits connected 100 with a reverse diode and 100 without, how many of each would fail after 1 hour, 10 hours, 100 hours, 1000 hours, 10000 hours, 100000 hours ? I can not answer this question. Without that data, i can not make an informed decision to leave out the reverse diode. We would also have to think about the effects of an unfavorably timed line surge.

If this was just a hobby circuit wired up on a breadboard, i may take the chance. If the LED blows, just unplug it and plug in a new one. But this circuit is to be hard wired into a power strip, where there is little room to work in and installation takes about an hour or two. So replacing the LED and adding the two missing diodes could take up to two hours.

Your post does suggest that we should try it though. Perhaps wire up a single white LED with a diode and resistor and plug it in and leave it for days, if it lasts i guess. That would be interesting. The leakage current is very low too, so i dont think that would be a problem, except of course if there was a line surge. Line surges could be very high though, so it's difficult to predict what may happen. It could lead to what looks like random failures...ie no failure for years, then all of a sudden it blows out.
In our favor we have some LED capacitance which is probably a lot higher than the diodes capacitance. Unfortunately in order to swamp a 200v line surge and maintain a signal less than 5 volts, it would have to be 40 times higher than the diodes capacitance.

What else we dont know is the effect of a repeated high reverse voltage, even for a short time duration for each instance. What i dont know is if there is any accumulation effect over long time periods such as what we might see in standard relay contacts.

 

[alec_t]

Point taken re the top diodes.

 

[ccurtis]

The reliability and surges are reasonable concerns, Mr. Al. It would be interesting to see the result in practice, though, understandably, not so interesting that potentially a failed LED would have to be replaced.

 

[MrAl]

Hi again

Alec:
Actually that's how i blew out the first LED
The only difference is i had a diode also in series with the resistor, but that did not help because the current flow through the LED was not influenced much by the resistor current or lack of it.

ccurtis:
Yes, would be interesting to look at this and do some experiments.
One thing i do see now is that when the diode voltage reverses, it would probably be pretty low except if there was a surge at that point. That's because the time it reverses is near the zero crossing. With little capacitance in the circuit that means that the voltage should be low and thus the reverse current low. Maybe this is how some people get away with this idea. For me though, i feel more confident in the long term survival of the circuit with the reverse diode in there, and it's not that hard to add one more 5 cent diode to each LED.

 

[k7elp60]

MrAl,
I think I have a solution for you. I didn't figure the value of the resistor in series with the green LED as I am not sure of the current. After thinking about the circuit that is attached I think with a 5.1V zener diode in parallel with the 22uF capacitor would cover all possible failure problems.
Ned

 

[Tony Stewart]

Mr Al's solution works and is adequately protected.

ALL LEDs are rated for -5V and all will begin to fail with -15V, some will begin to fail with -10V.
However the LED current is ~ 1mA so it wont be very bright. Choose many 82K R's in parallel to increase brightness, each taking <0.2W full wave.

This is why Neon tubes are more popular than LEDs with 220K series R and 70V avalanche voltage for AC line detectors in extension cords.

 

[ccurtis]

The two-leaded, dual colored LEDs packaged and wired internally anti-parallel must necessarily experience brief periods of avalanche (if driven by, say, 10v or more through a series resistor) as the applied polarity is switched, given the capacitances and asymmetrical and finite switching times. I have wondered about the reliability of those if they are switched continuously to produce a mixed color.

 

[MrAl]

The two-leaded, dual colored LEDs packaged and wired internally anti-parallel must necessarily experience brief periods of avalanche (if driven by, say, 10v or more through a series resistor) as the applied polarity is switched, given the capacitances and asymmetrical and finite switching times. I have wondered about the reliability of those if they are switched continuously to produce a mixed color.

Hi,

That is another interesting question you brought up. This may be why some manufacturers are stating on the data sheet that the LED should not be pulsed.
It has been a common belief that pulsing is good because it allows a lower level of light in a very variable way. Simply change the 'on' period and the light can be dimmed to any level, rather than have to stick with set values of series resistance to change the brightness level which does not allow for continuously variable adjustment.
With the LED capacitance however the capacitance will have to be constantly charged (although not reversed in this case) and that means a higher peak current, even though it 's for a short time. But it will be a particular time period it takes to charge regardless of frequency, so increasing the frequency will increase the power dissipation just because of that capacitance and whatever series resistance internal to the diode.

Since some data sheets reflect this i guess there is something to it. At 50 or 60Hz it may not be a problem though, especially when the drive wave is a sine rather than a true pulse. I cant say for sure of course because i have no data for this either.

I am happy that you came up with these ideas because they help us make a better circuit. Some external series resistance is a good idea to minimize these secondary effects.

 

[MrAl]

Mr Al's solution works and is adequately protected.

ALL LEDs are rated for -5V and all will begin to fail with -15V, some will begin to fail with -10V.
However the LED current is ~ 1mA so it wont be very bright. Choose many 82K R's in parallel to increase brightness, each taking <0.2W full wave.

This is why Neon tubes are more popular than LEDs with 220K series R and 70V avalanche voltage for AC line detectors in extension cords.

Hi Tony,

Well that used to be true, about the brightness, but today with the high brightness LEDs i think it is changing. 10 or 20 years ago i would not have considered doing this without using a series capacitor for increased drive for the LED(s). But today the brightness from a typical high brightness LED is quite impressive, especially when used as just an indicator which doesnt have to be as bright as a light source for illumination such as a flashlight. It is very surprising but it works really well. The green LED for example lights up so bright you could not ask for anything more or it would start to get annoying. The blue LED isnt quite as bright, but is clearly visible even with a 75 watt light shining on it from 5 feet away. In bright sunlight i dont know, i never tried that, but with the normal room lighting it is very very clear that the LED is on because it's just too bright to miss.

The type of LED i am using is actually a 5mm tri color LED, red, green, blue, but not using the red (bent lead over). The red is quite dim compared to the others, so i dont use that LED. That's my only complaint with these LEDs. Also, they have a diffused lens, something that works good for an indicator because you can see it from any angle (180 degrees solid angle).
I have used other LEDs too though, like a high brightness yellow 3mm, not diffused, but i see the diffused 5mm types are the best for indicators, as long as they are high brightness.

I also can use a 75k, 1/2 watt resistor, giving a little more current. Also, because of the upper two diodes, the resistor only conducts for 1/2 of the line cycle, so that means 1/2 the total power:
Full cycle power with 120vac line: 120^2/75000=0.2 watts
Half cycle power with 120vac line: 120^2/75000/2=0.1 watts
So you can see that i like to run the resistors at 1/5 of the power rating to keep them very very cool.

As a side note, the reason i am doing this with LEDs instead of neons is i am going for the reliability. For this application we might grow to rely on the lighting or not lighting of the indicator to tell us if the strip is turned on or not or plugged in or not. That means i am after extremely high reliability as well as the normal functionality.
I am also considering installing a back up system, where a small buzzer beeps once when the switch is turned on if there is power. This of course adds to the complexity but if it can be done cheaply i'll include that too.

 

[MrAl]

MrAl,
I think I have a solution for you. I didn't figure the value of the resistor in series with the green LED as I am not sure of the current. After thinking about the circuit that is attached I think with a 5.1V zener diode in parallel with the 22uF capacitor would cover all possible failure problems.
Ned

Hi Ned,

Thank you for the suggestion. What i am after here is simple construction and minimum parts count, and on top of that, high reliability. Thus i am going with parts that are known to have life spans on the order of 20 years or more. Resistors and diodes are high reliability devices when operated within limits, so it makes sense to stick with those components. Of course the circuit has to be done right too, and i have been playing around with this to get the right circuit
The zener is a pretty high reliability device too i think, so that may help somewhere too. I am still thinking about this circuit and not done yet really.

 

[k7elp60]

Hi Ned,

Thank you for the suggestion. What i am after here is simple construction and minimum parts count, and on top of that, high reliability. Thus i am going with parts that are known to have life spans on the order of 20 years or more. Resistors and diodes are high reliability devices when operated within limits, so it makes sense to stick with those components. Of course the circuit has to be done right too, and i have been playing around with this to get the right circuit
The zener is a pretty high reliability device too i think, so that may help somewhere too. I am still thinking about this circuit and not done yet really.

MrAl,
I think I finally understand what you are trying to do. I would eliminate the capacitor in my circuit and use a 3.3V zener diode. The reason I used a bridge rectifier is I think it removes the requirement for a diode to cancel the reverse polarity on the LED's. I have been building AC powered light strings for many years and I find that 1/2 wave or full wave rectifiers have treated me good. I did one last halloween that was powered by 120VAC and had 45 orange LED's in series with a full wave rectifier and a series resistor.
Ned