How to build a Power LED ring?

Thread Starter

q12x

Joined Sep 25, 2015
665
I decided today to write about this thing because it is over my head already.
My awesome and very trusty neon from my lamp, is burned.
(you can see the round neon on the table in the right corner)
(click on every image to enlarge)
IMG_20210723_201101.jpg
Now, I want to build a Power LED ring, to replace the round Neon.
My tests so far:
I lit up 1 power LED IMG_20210723_202852 Copy.jpg from my PSU and I got it to Full intensity (or very near Full intensity) for 8V@160mA
I then, very quickly build a capacitive dropper circuit and powered my LED to it. But the problem with this circuit, it does not let more than 10mA through the led. I've also noticed the low light intensity of the P LED (Power LED).
Sorry for my low resolution images but is quite understandable as it is. I have a very bad camera but is all that I have.
IMG_20210723_202307 Copy.jpg
The P LEDs, I cut them out from a standard Bulb LED light ALuminium disc.
I put the intact round AL disc in the picture to get the idea where I get them from.
The only problem I spent my time on, was the group of P LEDs in the right are Burned, but still working, and all the measurments I did to a point was on them. Then I realized they are just not good enough for the Illumination power that I wanted. So that was a very big downside in my experiment, very time consuming, a couple of days of testing and scratching my head, until I realized the real difference and also what I really want. Then, the group of the leds on the left (11 of them) are still taken from the burned AL discs, but they remained intact or very close to intact. They are VERY bright, especially compared with the burned ones. So that's what I want, I want very good illumination power !!!
Screenshot_7.jpg
Here is the data that I collected with my PSU for both categories: the more burned leds and the good leds:
IMG_20210723_204954.jpg IMG_20210723_205004.jpg IMG_20210723_205035.jpg
- After gathering all the information, I used my little knowledge that I have on calculating electronic components in circuit. I am not completely sure if those results I got there are correct or not. But it is the best I could do. Please take a look over those calculations and confirm to me, by saying "No, thats bad" (why and where if possible). But really, not those calculations are my issue here, but the circuit to build to properly power these P LEDs. So keep that in mind and dont get into the woods too far with the calculations. I only want a simple confirmation and that's it. If possible of course.
- Then, I thought about using Power Transistors in parallel to reach that 38.4W limit for only 1 P LED.
IMG_20210723_205049.jpg IMG_20210723_204904.jpg
But then I realized that the power transistors that I have, are limited on the Voltage side. This BD139 that I have is working up to 80V and not 240V that I need through the bridge rectifier. So the PT Power Transistors version failed very quickly after 2 days of thinking on the problem. It was a brilliant idea, but it failed. I also read that transistors in parallel like that need a less than 1ohm "balast resistors" on their Emitors to make a negative feedback and the transistor not to overheat and over conduct a surplus of current through it, and burn itself to death. At least I learned something interesting from all this experimentation of mine. Haha. I've also forget the filter capacitor after BR, now that I'm looking on it. Ive also thought on searching for real power transistors on ebay, like 300V@1A,50W, but I encounter another set of problems I will not mention at this point.
Possible circuits I am thinking to try in the future:
- I didn't try it yet, but I thought on using a SCR circuit:
This circuit I back engineered from a real SCR board from ebay some time ago. Not tested yet. !
IMG_20210723_204913.jpg
- Another very interesting idea is to take a normal NEW bulb led, cut its Aluminium disc and take the P LEDs from it, and re-arrange the P LEDs in my circle shape that I need. So instead of being a small circle, I will transform it into a large circle. Pretty much what I did so far with the burned P LEDs. But this time, using of course the same power supply that is coming with. I think this is a brilliant idea and I might make it. But I face a single and most important problem: How to stick those fragments of ALuminium with the P LEDs on them to the new ring shape? If I resolve this single issue, then is a GO. That's why I didn't start it yet.

- But now, I am so tired of trying, with so many failings on my head, I thought to expose my little madness here and hopefully, someone will give me an interesting Practical idea or solution to my problem.
Thank you very much for reading and hope to see some interesting answers from you.
 

LesJones

Joined Jan 8, 2017
3,497
Look up "angel eyes" on ebay. These are circular LED arrays used for car lighting and are available in different sizes. You may find one that fits your light fitting. These are powered from 12 volts DC so you would need a suitable wall wart to supply them.
Using a capacitive dropper will mean the LEDs will be at mains potential so you would need to make a cover so it was not possible to touch them. On the assumption that you mains is 240 volts 50 hz you would need a capacitor of about 2.2 uF to give a current of 160 mA. That would have to be increased depending on the number of LEDs in series. You would also need a low value series resistor to limit the surge current if you switched it on near the crest of the waveform.

Les
 

Thread Starter

q12x

Joined Sep 25, 2015
665
That is a very interesting answer indeed. Thank you. I looked for "angel eyes" on ebay as you suggested mister @LesJones .
It is impossible to provide every detail of this project. I see that I failed to mention an important detail:
- the Outer diameter of the original Neon lamp = 112.4mm, Inner diameter =99.4mm and the thickness = 13mm.
So I searched for 112mm and the closest diameter I could find is 110mm. I suppose its ok. But the price... uhhhhh. Plus that these days I find out that from 01-07-2021 UE countries are charging VAT + local postage charges (2$) for every item that comes from non-UE countries, like china and it's ebay. I guess to stop people like me from importing the Famous virus from china. This news blow me away. All the small packages were Free until 01-07. Now we all must suffer.
These are all the diameters for "angel eyes" that I easily find: 60mm/70mm/80mm/90mm/100mm/110mm/120mm/130mm
I will think on your idea. It's a very good idea ! But a bit too expensive. Not much, but a bit.
Thank you for your answer.
 

ElectricSpidey

Joined Dec 2, 2017
1,806
Just for the record that is a fluorescent lamp, not neon, they can be replaced.

I don't have time to help you with the LED project, but I just wanted you to know about the circular lamp.
 

LesJones

Joined Jan 8, 2017
3,497
I have some similar lamps. The fluorescent tubes in them are about 118mm outside diameter. One point to look out for if you decide to order a spare tube is that there are two versions. One has 4 pins and one has 2 pins. (I think the 2 pin version must have the starter switch built in.) So make sure you get the correct version. Amazon have some angel eyes so I don't know if buying from them would avoid VAT charges. (I hope the UK does not introduce these charges on items from China.)

Les.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
Yes, you are right, mister @LesJones my fluorescent tube that is burned, has 4 connectors but 2 of them has a filter capacitor, so only 2 remains for connection with the wires.
(click on every image to enlarge)
IMG_20210724_121113.jpgIMG_20210724_121219.jpg
The wires has polarity colors on them (red for pos and black for neg) but I just measure them just now, and on DC I get 0V, every time I measure. But on AC I get a steady 50V. So it is a 50VAC supply inside. But I have no clue what current is providing. Isn't a way of checking an unknown current? Like in my case here?
-I did open it once and check for the room I have inside for my next board I will possibly make, and I know for sure it is an electronic board that is powering the lamp and not a coil transformer. But what kind of circuit is in there, I have no clue, im not that good. Haha.
-Your "Angel Eyes" are a very good idea and I totally like it, but it is an idea on the table. Like my many other ideas that stand on the table as well and wait their turn. So your idea will be put in practice after my other attempts. I want to build this thing as cheaply as I can, I know you understand me.
- I really like your idea with the "2.2 uF to give a current of 160 mA " but that must be a fix cap and not an electrolytic cap.
I have a bag of 100pcs of 100nf@600V specifically bought for this reason, to have spare capacitors for capacitive droppers I will build in the future. But at that time I didnt know how important the capacity is like you just point me at today. Damn... I am not sure I have such large capacitor value in my spares. I know for sure I have caps for very high voltages like 2kV - 6kV but they are low capacitive values.
I really-really wish I can build a power supply (very small, like the commercial ones) for these kind of P LEDs.
Another idea is to adapt the existent power supply that I get the 50V on its wires, to power these P LEDs... Hmmm? But I am not that good at this point, on adapting this kind of electronics. My brain stops after adding a BridgeRect and a filter capacitor after the50VAC wires. That's all I got. Haha.
Did you checked my calculations there? Are they any good? They are all bad? What can you say?
As always, I thank you from my hearth for all your help.
 

LesJones

Joined Jan 8, 2017
3,497
The capacitor across the two pins is not really a filter capacitor. It is part of the driver circuit that initially provides enough current through the heaters at the end of the tube to get it started. Once the gas in the tube is ionised the heaters remain hot due to the current flowing through the plasma. Mesuring the voltage from the driver circuit is not meaningful without a good lamp connected to it. Capacitors used in capacitive droppers should be Y rated for safety reasons. (They are designed so that if they fail they fail open circuit.) Your calculations do not seem to include the reactance of the capacitor. To do a better calculation of the required capacitor value we would need to know how many LEDs you will have in series. (So we know the approximate voltage across the string of LEDs)

Les.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
...To do a better calculation of the required capacitor value we would need to know how many LEDs you will have in series. (So we know the approximate voltage across the string of LEDs)
I have 11 Good leds and 28 Bad Leds (that are still working but considerable less Light power comparative with a good one). We will focus on the 10 Good LEDs. One will remain as reserves or for testing purposes.
One Good LED is giving me good illumination when powered with 8V @ 160mA. We will use 10 of them.
So, to continue your idea with "the approximate voltage across the string of LEDs", will be 80VDC @160mA. In total we have 12.8W.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
I didnt mention this test because it was a big fail for me and I got low morale from it.
I tried to power a Good led (8V@160mA) using this circuit.
This circuit works FINE with a normal 3mm,5mm or smd LED, so less than 10mA for a led consumer.
1627178499620.png
But these P LEDs are a big trouble for this type of circuit. All the resistors I put there, all got hot or burned.
I put in parallel as much as I could and still got them burned. If I remember right, it worked ok for under 10mA current per P Led. But the illumination was terrible. So, by trying to achieve it's optimal power, I burned/or damaged a couple of resistors.
So, on my PSU, I lower it to 8V*0.12A = 1W. I was happy with this current through it, because the illumination was acceptable. I wanted to test to understand what it takes. Apparently it takes every resistor that I have.
If R1 = a single 20k, if I remember right, R1 didn't burn. From my calculations, , 240V-8V=232/20000=0.01A=10mA; 8*0.01=0.08W=80mW. But the illumination was terrible.
If R1 = 10k/4 (it means four 10k in parallel)=2.5k=1W = R1 burned
If R1 = 12k/6 = 2k=1.5W = R1 burned
If R1 = 16k/8 = 2k=2W = R1 burned
After that, I concluded that my 250mW resistors may not be 250mW but something less, probably 200mW or even less. It's my only logical conclusion. So I got depress from such failings. Then I decided to write here on the forum.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
Good news, I managed to make my lamp work !!! Right now it is back in function.
But we will continue this conversation because I want to be able to construct by myself a simple power supply for a couple of P LEDs.
 

LesJones

Joined Jan 8, 2017
3,497
I don't know what "P LEDs" are. Can you post the data sheet for them ? 8 volts is an unusual voltage for an LED. White LEDs normally have a forward voltage of between 3.0 and 3.6 volts.
If you have 232 volts across a 2.5 K resistor that is 232/2500 = 0.0928 amps which is 0.0928 x 232 = 21.5 watts dissipated in the resistor
and 0.0928 x 8 = 0.74 watts in the LED. This is why dropping the voltage with a capacitor is much better as the current is 90 degrees out of phase with the voltage so no power is dissipated. If you chose a capacitor with a reactance of 2500 ohms at 50 hz the value of the capacitor would
be 1/(2 * π * f * X) (X is the reactance of the capacitor,) so C = 1/(2 * π *50 * 2500) = 1.273074474856779e-6 Farads = 1.27 uF

Les.
 
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Thread Starter

q12x

Joined Sep 25, 2015
665
I don't know what "P LEDs" are. Can you post the data sheet for them ?
"P LED" stands for Power LED. To make a clear distinction between them and the very low power usual 3mm,5mm and smd LEDs.
1627209779681.png
I truly don't know what to search for these types of P LEDs, to find their datasheets !!!
I'm !VERY! curious if you can find any datasheet.

8 volts is an unusual voltage for an LED. White LEDs normally have a forward voltage of between 3.0 and 3.6 volts.
Yes, you are referring to the very low power leds, 3mm,5mm and smd LEDs who have this 3.0 and 3.6 forward voltage.
8 volts is quite common voltage for a Power LED. And hold your stick, I just recently discover a NEW type of P LED that works for the incredible 32V @ 1.2mA (max). Truly fantastic. It is my first encounter with them. And I actually use both: these specific P LEDs and it's driver board, to construct my circular lamp.
I will probably make a new q12 artpage and show there how I build it, what I tested and so on. I hope I can manage to make it.
If you don't believe me, just tell me and I can make for you a movie with these super high voltages per LED.

If you have 232 volts across a 2.5 K resistor that is 232/2500 = 0.0928 amps which is 0.0928 x 232 = 21.5 watts dissipated in the resistor
I am very glad you posted your way of doing the math, because it is confirming that I am not wrong. I did indeed come to similar results but I thought I miss something and didnt trust quite well my results. But seeing you with very close results than mine, that is ohoa, very good for me to see. Again, im not used to do this type of calculations every single day, but I will start from now on a bit more serious. Thanks to you !
My calculation was:240/0.120 = 2000=2k; 240-8=232*0.12=27.8W heat dissipation over R1. I thought, wow, that is way too much, i must had miscalculating something. And I thought the experimental way is the proper way, so this is why I burned so many resistors. Well, I didnt let them actually burn, but I saw clear smoke coming from them, those in parallel I mention earlier. The smoke of knowledge !
So my calculation was correct. Imagine that. Hmmm. The question is... what can I put instead of a FIST of resistors in parallel? And that is for 1 single P LED. I imagine, the power will dissipate on the other P LEDs if Iput more than 1, so i think the resistor value will start to decrease, but here is my imagination at work and not my incredible math skill.

I need another help from you! And is a very important one for me. This is more a mechanical/chemical question. My solution of catching all the P LEDs in my lamp, is very practical and it works very good. BUT, it is extremely FRAGILE. My entire "doing" is very fragile. I need you to give me an idea of a viscous flowing/pouring substance, I don't care what it is, but it has to have 3 very important properties: 1-Heat resistant, 2-solid as a rock, 3-light. The temperature should be up to 300 Celsius (550 F). The light part is debatable, it can be heavy, but not very heavy. Ideally light or very light. The most important part is the temperature. If you tested stuff like this, or you happen to know, please give me a good product, substance name (hopefully very cheap).
Also, do you happen to know what is the black isolator material they pour over an IC or transistor? It's very hard and very heat resistant case I mean. Ebonite is the material used perhaps?
Thank you very much!
 
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LesJones

Joined Jan 8, 2017
3,497
LEDs are not suitable for use at 300 C. I have just had a look at the data sheet for 5050 type SMD LEDs and the absolute maximum temperature rating is 95 C. (I would not run any LEDs at more than 50C.) Many commercial LED arrays are made on a type of PCB that has an aluminium substrate. (Not glass fibre or SRBP.) This is to provide heatsinking for the LEDs
This is how I mounted 12 1 watt LEDs for a ring light for macro photography. The LEDs are mounted to 3mm thick aluminium with heat sink compound to reduce the thermal resistance.
IMG_0363b.JPG
You need to provide the datasheet on the LEDs THAT YOU ARE USING.

Les.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
OHO! I LIKE THAT ! Haha. Very nice ! Now we're talking !
I know you want the datasheet, I want it too, but understand that I took the leds from an already built LED Bulb and reshaped its leds position. I have to make for you a video because clearly you dont believe a word Im telling you here.
This is the type of LED bulb I used, the most common and commercial one possible: 1627216177447.png
Inside, on its ALuminium heatsink had 8 LEDs. I dessasamble it and reshape it. But it is still the same circuit and components from this specific LED bulb. Which is impossible to get any datasheet on its leds. I got the datasheet on the IC is using and its driver circuit... but absolutly nothing about its leds.
I will make you a movie and hopefully I can demonstrate to you that crazy high voltage they are working at.
 

Thread Starter

q12x

Joined Sep 25, 2015
665
This is the light Bulb I re-shaped to fit my larger circle.
082G IC_20210724_1.jpg082G IC_20210724_2.jpg082G IC_20210724_3.jpg
Those yellow dots are the Power LEDs I am talking about.
If I test 1 of them I get it is working with 32V@1.2mA. They are 8 in total, it is how it was designed.
In the end I used the exact circuit you see here intact, to power the exact number (8) LEDs on my final circle. I will post it soon.

I searched for this IC and I found a very similar datasheet, but not the exact one for it. Check my circuit and check the circuit in the datasheet, you will see they are the same, even the datasheet IC name is not exact with what I have here on this board.
https://pdf1.alldatasheet.com/datasheet-pdf/view/1178098/FS/FC2082G.html
this one:1627218660379.png
Unfortunately they dont provide a real example with values for the compoents, but the circuit itself is exact as mine.
 
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LesJones

Joined Jan 8, 2017
3,497
IF the data sheet is the correct one for the IC on your board then from the datasheet and the value of R3 and R4 the LED current will be 15.3 mA. Using all the existing components is the best solution as we know the circuit was deigned to drive the LEDs correctly. You main problem now is providing heat sinking for the LEDs. I don't know where you can obtain aluminium substrate printed circuit board to make the circular PCB to fit your lamp.

Les.
 

timm27

Joined Dec 11, 2020
7
"P LED" stands for Power LED. To make a clear distinction between them and the very low power usual 3mm,5mm and smd LEDs.
View attachment 244368
I truly don't know what to search for these types of P LEDs, to find their datasheets !!!
I'm !VERY! curious if you can find any datasheet.


Yes, you are referring to the very low power leds, 3mm,5mm and smd LEDs who have this 3.0 and 3.6 forward voltage.
8 volts is quite common voltage for a Power LED. And hold your stick, I just recently discover a NEW type of P LED that works for the incredible 32V @ 1.2mA (max). Truly fantastic. It is my first encounter with them. And I actually use both: these specific P LEDs and it's driver board, to construct my circular lamp.
I will probably make a new q12 artpage and show there how I build it, what I tested and so on. I hope I can manage to make it.
If you don't believe me, just tell me and I can make for you a movie with these super high voltages per LED.


I am very glad you posted your way of doing the math, because it is confirming that I am not wrong. I did indeed come to similar results but I thought I miss something and didnt trust quite well my results. But seeing you with very close results than mine, that is ohoa, very good for me to see. Again, im not used to do this type of calculations every single day, but I will start from now on a bit more serious. Thanks to you !
My calculation was:240/0.120 = 2000=2k; 240-8=232*0.12=27.8W heat dissipation over R1. I thought, wow, that is way too much, i must had miscalculating something. And I thought the experimental way is the proper way, so this is why I burned so many resistors. Well, I didnt let them actually burn, but I saw clear smoke coming from them, those in parallel I mention earlier. The smoke of knowledge !
So my calculation was correct. Imagine that. Hmmm. The question is... what can I put instead of a FIST of resistors in parallel? And that is for 1 single P LED. I imagine, the power will dissipate on the other P LEDs if Iput more than 1, so i think the resistor value will start to decrease, but here is my imagination at work and not my incredible math skill.

I need another help from you! And is a very important one for me. This is more a mechanical/chemical question. My solution of catching all the P LEDs in my lamp, is very practical and it works very good. BUT, it is extremely FRAGILE. My entire "doing" is very fragile. I need you to give me an idea of a viscous flowing/pouring substance, I don't care what it is, but it has to have 3 very important properties: 1-Heat resistant, 2-solid as a rock, 3-light. The temperature should be up to 300 Celsius (550 F). The light part is debatable, it can be heavy, but not very heavy. Ideally light or very light. The most important part is the temperature. If you tested stuff like this, or you happen to know, please give me a good product, substance name (hopefully very cheap).
Also, do you happen to know what is the black isolator material they pour over an IC or transistor? It's very hard and very heat resistant case I mean. Ebonite is the material used perhaps?
Thank you very much!
An 8 volt LED may be 2 or three LED chips in series perhaps with a current limiter resistor. A 32 volt LED is almost certainly a COB (chip-on-board) device - which is constructed from a chain of individual LEDs integrated on to a substrate perhaps with a driver circuit. 'LED filaments' are constructed in a similar way, but typically the driver is built into the lamp base. These devices are typically designed and built for a specific manufacturer and typically don't have commercially available data-sheets. But there are an enormous variety of devices available quite cheaply - look on eBay, searching with 'COB LED'. Typically sellers give enough information to be useful, but if you want a drive circuit you will have to find a suitable part to buy or design for yourself. I would be very wary of designing a transformerless design to run off 230 - or even 110 volt mains. Commercial designs are built in huge volumes and the designs are typically will tested, and the effect on safety for any single component failure will have been checked.
 

LesJones

Joined Jan 8, 2017
3,497
IF THE DATASHEET SUPPLIED IN POST#15 BY THE TS is the correct one for the IC on his original board then the LED current works out to 15.3 mA so the total power to the 8 LEDs would be 15.3mA x 8 volts x 8 = 979 mW. This seems to be too low for the lamp that he describes. (I would expect 5 to 10 watts.) The current he quotes in post #1 of 160 mA makes more sense. It would mean the lamp from which they were removed from was just over 10 watts. I think he has the wrong datasheet for the IC on the board.

Les.
 
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