Author Topic: I have an SMD LED which is 3mmx2.7mm which I cannot find, can you help? (Read 5442 times)

TT_Vert · « **on:** February 24, 2018, 05:54:45 am »

It it parts the item below. The forward voltage is 3V with a 20mA current draw and is powered by a 12V source. There are 48 of these on this board. I looked on digikey, mouser, etc. Seems these do not exist. Are these actually 3030s ? I measured about 10 times thinking I was going nuts.

Dave

Brumby · « **Reply #1 on:** February 24, 2018, 06:08:00 am »

Possibly 3528's

I found a datasheet where the white body is 3.2mm long x 2.8mm wide, but with the terminals included, it's 3.5mm long x 2.8mm wide.

If you only measured the white body to get your dimensions, then this puts you out by 0.2mm x 0.1mm.

Brumby · « **Reply #2 on:** February 24, 2018, 06:12:39 am »

Make that 2835's. They have the rectangular emitter area.

I've also found drawings with the external dimensions of two different data sheets, both for "2835 SMD LEDs":

One has: 2.74mm x 3.4mm
The other: 2.8mm x 3.5mm

TT_Vert · « **Reply #3 on:** February 24, 2018, 06:21:24 am »

I did measure at the body of the LED and with the terminals and It's definitely a max of 3.0mm . maybe 2.95mm w/o terminals, they don't protrude out much at all.

Dave

Brumby · « **Reply #4 on:** February 24, 2018, 06:26:54 am »

I threw 2830 into Google and came up with a couple of hits on Alibaba. One was a circular emitter and then there's this: https://hanhualed.en.alibaba.com/product/60267278940-801677488/shenzhen_led_diode_2830_snd_led_manufacturer_supplied_2835_smd_led_20_23lm.html

... but they seem to be 2835's

TT_Vert · « **Reply #5 on:** February 24, 2018, 06:29:31 am »

If they were 2835s they'd be 2.8x3.5mm at some point though right? What makes you think what i have may be the 2835s if they don't dimensionally match?

Thanks much,

Dave

Brumby · « **Reply #6 on:** February 24, 2018, 06:44:49 am »

Sorry if I wasn't clear - I meant the Alibaba ones that came up on a search for 2830 actually look like they are 2835's.

Not the ones you have.

Any idea on how thick your chips are?

Brumby · « **Reply #7 on:** February 24, 2018, 06:52:36 am »

With the dimensions you have given, the closest I have been able to get is the 3030. Found these on RS:
https://au.rs-online.com/web/c/displays-optoelectronics/leds-led-accessories/visible-leds/?searchTerm=3030%20smd%20led&h=s&sra=oss&redirect-relevancy-data=636F3D3126696E3D4931384E53656172636847656E65726963266C753D656E266D6D3D6D61746368616C6C7061727469616C26706D3D5E2E2A2426706F3D31313326736E3D592673743D43415443485F414C4C5F44454641554C542673633D592677633D4E4F4E45267573743D3330333020736D64206C6564267374613D3330333020736D64206C656426&applied-dimensions=4294439424,4294874375,4294873983

Probably time I step back and see if someone else can be more helpful.

TT_Vert · « **Reply #8 on:** February 24, 2018, 06:58:24 am »

Yeah I also found tons of 3030s. So is it possible these are just some one off or very hard to come by or obsolete chinese sourced LEDS? I did buy on Ebay from a US seller a few years back and looking even now the LED is unidentified in his ad.

https://www.ebay.com/itm/10X-White-48SMD-Panel-LED-Festoon-T10-BA9S-Adapter-Interior-Dome-Map-Light-Bulb/122603838385?hash=item1c8bc21bb1:g:2z8AAOSw4HlZbGCH:sc:USPSFirstClass!60107!US!-1

Dave

TT_Vert · « **Reply #9 on:** February 24, 2018, 06:59:07 am »

Quote from: Brumby on February 24, 2018, 06:44:49 am

Sorry if I wasn't clear - I meant the Alibaba ones that came up on a search for 2830 actually look like they are 2835's.

Not the ones you have.

Any idea on how thick your chips are?

Ahh understood. I'll check the thickness in the AM.

Dave

Brumby · « **Reply #10 on:** February 24, 2018, 11:02:30 am »

Quote from: TT_Vert on February 24, 2018, 06:58:24 am

So is it possible these are just some one off or very hard to come by or obsolete chinese sourced LEDS?

It could also be some LEDs that were manufacturer rejects because of their size (or other reasons as well) sold off at the Shenzhen markets.

Kjelt · « **Reply #11 on:** February 24, 2018, 03:15:31 pm »

Quote from: Brumby on February 24, 2018, 11:02:30 am

It could also be some LEDs that were manufacturer rejects because of their size (or other reasons as well) sold off at the Shenzhen markets.

That would be very unlikely IMO, the plastick and metal base carriers are not made by the the led manufacturers themselves the same as that the ic housings are not made by semicon fabs. So on the intake these would have been rejected since in a 3030 manufacturing process it would cause a lot of problems.

My guess this is a niche product from one of the many chinese led manufacturers. Those small factories with their product range arre not so known in the rest of the world. I have some reels with leds where only chinese for me unreadable datasheet and catalogs exist. Some series are not even publicly for sale but are specially made on a mass contract for instance for some lighting manufacturer wanting to produce a series of a few million led lamps for the chinese or other market.

TT_Vert · « **Reply #12 on:** February 24, 2018, 06:16:49 pm »

Well that is good enough for me. I was trying to modify these lights I had to convert from 12 to 24V but it seems like more headache that it's worth. I could probably make the 3030s work but it's just not worth it.

Thanks for the insight guys, these little things really do confuse a newbie! I feel like I'm missing something and don't know what I don't know..
Dave

Kjelt · « **Reply #13 on:** February 24, 2018, 06:45:39 pm »

Quote from: TT_Vert on February 24, 2018, 06:16:49 pm

Well that is good enough for me. I was trying to modify these lights I had to convert from 12 to 24V but it seems like more headache that it's worth. I could probably make the 3030s work but it's just not worth it.

Thanks for the insight guys, these little things really do confuse a newbie! I feel like I'm missing something and don't know what I don't know..
Dave

What is the layout of the led string? If it is 12V they probably have only three leds or four in series and then many of these parallel. Are there resistors in the string? In theory you need to go from 4S12P to a 8S6P construction. Could be as easy as opening a pcb trace with a scalpel, cleaning the copper and soldering a few thin teflon wires.

Other perhaps better solution: place a high efficiency dc-dc converter in front.

TT_Vert · « **Reply #14 on:** February 24, 2018, 06:56:44 pm »

I will have to double check but it's a PCB w/ 6 rows of 8 LEDS. I am also trying to better understand how it is currently wired. There is one 8.25 ohm resistor on the positive side and one on the negative side. I was hoping to cut it in half and wire the two in parallel to make it 24V. I did this and got it to work on 24V doing this but it is getting very warm and I think drawing too much current. But i really don't know what i'm doing other than just experimenting

If you are interested in adding input I could get you the drawing I made of how I think it's wired. I could also tell you the current draw as it is in my config but every LED resistor parallel/series calculator I see does not show a recommended method as it is now even in its current configuration of two 8.25ohm resistors so I'm confused there as well.

Dave

drussell · « **Reply #15 on:** February 24, 2018, 09:20:35 pm »

Quote from: TT_Vert on February 24, 2018, 06:56:44 pm

I was hoping to cut it in half and wire the two in parallel to make it 24V.

That would make it 6V.

Why don't you just change the resistors or add some other kind of current regulator?

TT_Vert · « **Reply #16 on:** February 24, 2018, 09:33:10 pm »

I'm sorry I meant I wired the two in series to make it 24V.

It's lighting up a 3d printer and this PCB is on a custom bracket on the printhead (Extruder) and I don't have a bunch of space. I could add some type of current regulator but it'd have to be near the power supply. I did play w/ resistors but I'm still learning and not exactly sure what I'm doing but they were getting hot so I was clearly doing something wrong. I've been going through calculators and trying to understand what I"m doing and why it's having the effect it is but it's taking me some time. I am more of a hands on learner and really learn by doing and noting results to be honest. I did check the PCB again and this what I came up with (See attached pic). It's an 8x6 array of LEDS, the left 16 (left 2 rows) share the same ground and the right 16 share (right two rows) the same hot. THen the middle 16 are in series. Would this be a 3P? This is what I think it is but I'm not even remotely confident in that answer. Or perhaps S3P3 or 1S2P? The current draw @12Vdc is 130mA and forward voltage @12Vdc is 3.08V.

I'm also trying to find out how/why the 8.25ohm resistors work in this circuit, it just doesn't make sense to me.

Thanks again,

Dave

Kjelt · « **Reply #17 on:** February 25, 2018, 12:32:02 am »

It does not make much sense to me either.
Most white leds are slightly more than 3V, around 3,2-3.3V depending on the current.
If you have exactly 12VDC the string length will usually be 3 leds since 4 can be too much, certainly over time as the voltage slightly increases.
So that would make the PCB consist of 16 paralel strings of three leds in series.
Since you say there are only two resistors instead of 16 my guess is that each 8 paralel strings get one resistor.
So this would make it two times a 3S8P configuration. can you verify this?

Brumby · « **Reply #18 on:** February 25, 2018, 01:00:18 am »

Following the PCB as best I can, I make it 2P-3S-8P ... like this:

(Sorry for the rough line work ... I'm not too crash hot drawing with a mouse.)

If so, then I think your connections are on the wrong side of the resistors ... but I can't see that clearly enough to be certain. The resistors will be for current limiting.

Brumby · « **Reply #19 on:** February 25, 2018, 01:14:05 am »

With so much parallel connection of the LEDs, the risk of uneven current distribution due to variances in each LED is significant. However, these variations tend to be much less when all the LEDs come from the same manufacturing run ... and that is how a lot of these cheap arrays can get away with it.

TT_Vert · « **Reply #20 on:** February 25, 2018, 01:20:01 am »

Yes Brumby, you are correct with your drawing. The connections are correct as i have another unmodified one which works fine. Care to explain how you get to the 2P3S8P?

Brumby · « **Reply #21 on:** February 25, 2018, 01:32:46 am »

Each row of 6 is as follows:
LED 1 and 2 are in parallel - 2P Call this group A
LED 3 and 4 are in parallel - 2P Call this group B
LED 5 and 6 are in parallel - 2P Call this group C

Groups A, B and C are in series - 3S .... so each row is 2P 3S

You then have 8 rows in parallel - so 2P 3S 8P

Brumby · « **Reply #22 on:** February 25, 2018, 01:36:23 am »

Do your power wires go directly to the LED rails down each side - or do they go through the resistors?

TT_Vert · « **Reply #23 on:** February 25, 2018, 01:38:42 am »

I will absorb this in a bit but answer your question the hot and ground go through resistor first

Brumby · « **Reply #24 on:** February 25, 2018, 02:10:11 am »

That's good. They should.

TT_Vert · « **Reply #25 on:** February 25, 2018, 06:33:25 am »

So after some trial and error I think I got to where I want to be. I noticed the original array drew about 140mA total. I played w/ resistors to that current draw w/ 24V input. I ended up at 22 $\$\Omega\$$ . If I understood more about calculating the values based on the series/parallel configuration I'd do the math. I'm still going to try but the way these are wired and the effect that has on resistance has me confused. Just as having one of the 3 parallel portions of the original wiring scheme w/o any current limiting resistor at all. It's not exactly pretty but it'll do the job. Now I just need to better understand why what I did is working.

Dave

TT_Vert · « **Reply #26 on:** February 25, 2018, 07:00:29 am »

I see what you saw before w/ that hot wire. I moved it over to the other side of the resistor so I could test the board. It was easier to connect my meter to a fixed wire rather than probe. Anyway I did get it working after some trial and error. I am trying to understand how these series/parallel configurations affect resistor calculations. I saw the original 12V array was drawing about 140mA. I ended up getting there at 24V w/ 22ohm resistor on each of the two arrays I have. I guess these two arrays are 2P3S4P if I'm understanding this correctly. Is there some easy way to calculate the resistance of these types of configurations? Do you calculate each group individually and then add them?

Thanks for all the help. I've included a pic before I cleaned it up some, it's not pretty but it'll do the job.

Brumby · « **Reply #27 on:** February 25, 2018, 07:30:09 am »

The topography you have worked out is exactly what I had sketched up for 24V.

The resistor in the middle was a neat move.

TT_Vert · « **Reply #28 on:** February 25, 2018, 07:37:58 am »

Thanks again, few questions to try to better educate myself.

Is there some easy way to calculate the resistance of these types of configurations? Do you calculate each group individually and then combine them?

Dave

Brumby · « **Reply #29 on:** February 25, 2018, 12:57:14 pm »

You need to work out the voltage across the LED matrix and the current they will draw - in total. You then subtract the LED matrix voltage from the supply voltage to give you the resistor voltage. You then take the total current and the resistor voltage and plug these into Ohm's Law to give you the resistance.

Let's say we have LEDs with an operating forward voltage drop of 3.5V and a current of 65mA

With the labels I used above:

* LED 1 has a voltage of 3.5V and a current of 65mA
* LED 1 and 2 in parallel (group A) have a voltage drop of 3.5V and a combined current of 65mA + 65mA = 130mA
* Groups A, B & C in series have a voltage drop of 3.5V + 3.5V + 3.5V = 10.5V and a combined current of 130mA. (The current going through group A is the same current that goes though group B and then group C)
* There are 8 sets of 10.5V, 130mA in parallel which gives 10.5V and 8 x 130mA = 1040mA = 1.04A

So, with a 12V supply, the resistor has to drop 12 - 10.5 = 1.5V with a current of 1.04A. In this example, the resistor would be 1.44 ohms.

If the LEDs ran at 3.2V and 20mA, the LED numbers would be: 9.6V @ 320mA. For a 12V supply, that works out as 12V - 9.6V = 2.4V @ 320mA which gives 7.5 ohms.

(I was drawing up a sketch to illustrate - but the program crashed when I saved it and corrupted the file.)

Brumby · « **Reply #30 on:** February 25, 2018, 01:01:02 pm »

The symmetry of this LED matrix makes calculation easy - but remember that direct paralleling of LEDs (as has been done here) can result in varying brightness across the LEDs due to unequal current sharing. This risk is much lower when they are all manufactured together.

Kjelt · « **Reply #31 on:** February 25, 2018, 01:06:19 pm »

Quote from: Brumby on February 25, 2018, 01:01:02 pm

direct paralleling of LEDs (as has been done here) can result in varying brightness across the LEDs due to unequal current sharing. This risk is much lower when they are all manufactured together.

I was worried for this a few years back and asked a specialist who insured me nothing could go wrong. The manufacturing process is so good at the moment that the differences in Vf for the individual leds are marginal but true what you wrote, use leds from the same bin.
What is important however is that you should actually use a current source for these kind of setups.

Brumby · « **Reply #32 on:** February 25, 2018, 01:29:33 pm »

Quote from: Kjelt on February 25, 2018, 01:06:19 pm

What is important however is that you should actually use a current source for these kind of setups.

Ideally, yes - but a resistor is cheap, easy and good enough for a lot of applications .... even if it is a bit power wasteful.

TT_Vert · « **Reply #33 on:** February 26, 2018, 03:33:56 am »

Thanks guys. I want to do this math but given i don't have a datasheet I'm trying to figure these specs out but I may be doing something wrong. I had read that you can measure your Vf by powering your LED at 12V w/ a 500 ohm resistor to get a 20mA current limit and measuring across the LED leads.. When I do this i get 2V. However when I put it in my diode tester on my meter It shows .875V. I then checked the diode testing specs of my meter and it states a forward DC current of ~1mA and reverse DC voltage of ~1.5V, Is it possible my meter is reading low because of the low 1mA current? Moving forward I decided to check the Vf of the SMD LED that I had been using for this project and I cannot even get a reading on my meter. Anyone care to chime in on what I'm doing wrong here, if anything?

Thanks much,

Dave

Brumby · « **Reply #34 on:** February 26, 2018, 05:28:45 am »

Quote from: TT_Vert on February 26, 2018, 03:33:56 am

Is it possible my meter is reading low because of the low 1mA current?

You are on the right track.

The actual voltage that drops across the LED when forward biased (V_f) changes according to the current flowing through it. This relationship also varies with temperature.

If you look up the datasheet for any given LED, it should show you a chart with current (I) vs forward voltage drop (V_f) ... and it will look something like this:

This image shows a comparison of two different LEDs, so you can see these graphs can vary ... and they do. (The three curves for each LED will be for 3 different temperatures.)

The important things to note are:

Low currents produce low V_f
Currents can skyrocket when you get past a certain point
Different LEDs have different curves
Temperature affects the curves

We use constant current sources or current limiting resistors to prevent excess current from flowing through LEDs ... because left alone, they will ~~try and~~ draw destructive currents if the supply voltage is high enough.

TT_Vert · « **Reply #35 on:** February 26, 2018, 05:36:35 am »

Yes I did notice that when playing around. I paid attention to current draws at a given voltage and actually wrote them down for my own personal knowledge. This particular datasheet shows a forward current of 2.8V even at 1mA. I don't know the specs of said LEDs I'm testing but I do know what my meter is saying which is confusing me.

Red LED i have. Meter shows .875V when on diode testing. Meter shows 2.0V when light is illuminated. Any idea the discrepancy? Could the curve be that steep below a certain current threshold (Say 1mV in this example?_
White SMD LED, I get nothing on meter w/a known good LED. Could this be because it's a white LED and a higher forward voltage? \\

I'm just trying to make heads or tails of what I'm seeing here so I can apply it moving foward.

Dave

Brumby · « **Reply #36 on:** February 26, 2018, 05:39:53 am »

Quote from: TT_Vert on February 26, 2018, 05:36:35 am

White SMD LED, I get nothing on meter w/a known good LED. Could this be because it's a white LED and a higher forward voltage?

Exactly right! This is one of the reasons why Dave has a 15V diode test range on the 121GW

Brumby · « **Reply #37 on:** February 26, 2018, 05:48:16 am »

Quote from: TT_Vert on February 26, 2018, 05:36:35 am

Red LED i have. Meter shows .875V when on diode testing. Meter shows 2.0V when light is illuminated. Any idea the discrepancy?

It comes down to the current used in each situation.

A meter will do a diode test at the lowest current possible. It is only trying to determine that the diode works as a diode (remember, LEDs are only one type of diode) - and it will give the voltage it sees when doing that test. When you actually drive an LED for it to light up, you are putting a significantly larger current through it.

Interestingly, LEDs can actually emit light at voltages far below those on the charts - but the level can be incredibly low. If I remember correctly, Dave did a video where he used an ultra sensitive device that could actually count individual photons - and I'm thinking he put an LED in the chamber and passed a spectacularly small current through it to get a really small - but measurable - amount of light out of it.

LEDs are not linear devices. Not by a long chalk.

TT_Vert · « **Reply #38 on:** February 26, 2018, 06:05:38 am »

so in essence the only accurate way to get Vf of an LED is in use if you don't have a datasheet? And then just guess on an If (Or start at 20mA). Or buy Dave's $250 meter which while I'd love it I just can't justify the cost at my current level.

Dave

drussell · « **Reply #39 on:** February 26, 2018, 06:41:48 am »

Quote from: TT_Vert on February 26, 2018, 06:05:38 am

so in essence the only accurate way to get Vf of an LED is in use if you don't have a datasheet? And then just guess on an If (Or start at 20mA).

You should always test and characterize your components...

Perhaps, you seem to forget that the LEDs are diodes? They behave as diodes. For best practice, you really need some kind of way of doing some sort of external current limiting rather than relying solely on the precise characteristics of each diode to set your operating parameters. Even a bit of resistance definitely helps on that...

Brumby · « **Reply #40 on:** February 26, 2018, 06:54:17 am »

Picking the right current is not always easy, but it's not all that difficult to get a V_f.

Let's say you have a 12V DC supply and you want to test some LEDs. Lets use a series resistor - but what value?

Pick a current. I'll go with your 20mA for this exercise.

Now, if you know the colour of your LED, you can look around and see what the typical V_f is for that colour and use that - but for a general figure to use in this exercise, let's use 2.5V.

Now, the resistor will have to drop 9.5V (12 - 2.5) when carrying 20mA

V = I x R
9.5 = 0.02 x R
R = 9.5/0.02
R = 475 ohms

So, grab a 470 ohm resistor and go for it.

Note: This will give you "pretty close" voltages, because the current will likely not be exactly the nominated value, but the numbers you get will be quite useful. You could also measure the voltage across the resistor and work out the exact current.

The other approach - and perhaps the best - is to get yourself a constant current circuit. With this, once you've set it for a particular current, your measurements will be more "correct".

Brumby · « **Reply #41 on:** February 26, 2018, 07:03:08 am »

Just looking back at that ... and I realised determining the maximum current is not easily done without some idea of its characteristics.

I had some 1W and 3w LEDs ... and working out the current for each was not staightforward. I worked backwards, slowly increasing the voltage, measuring it and the current to get the power they were rated for.

TT_Vert · « **Reply #42 on:** February 26, 2018, 03:41:06 pm »

but how do you really know what they are "rated for"? I do understand that basic math of calculating a resistor for an LED given supply and Vf but I don't know when you've got too much current. My power supply does show me current draw and voltage so i can watch these realtime but I don't know when I've gone too far aside from monitoring LED temp and brightness.

Dave

Brumby · « **Reply #43 on:** February 27, 2018, 12:05:26 am »

Good question.

Datasheets are the primary reference here, but if you don't have one - or at least some idea of the characteristics of the LED - then it becomes much harder.

I can't say I have the experience to guide you on this - perhaps someone better informed can offer something useful - but I would try doing the following:

1. I would leave brightness out of my initial investigation - unless you had a brightness figure from some source you could use.
2. I would plot my own V_f vs I curve and compare that with other LEDs
3. I would watch the temperature of the LED like a hawk. Letting the chip overheat and release the magic smoke wouldn't be helpful.

TT_Vert · « **Reply #44 on:** February 27, 2018, 05:43:57 am »

great thanks much

Dave


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Author Topic: I have an SMD LED which is 3mmx2.7mm which I cannot find, can you help? (Read 5442 times)

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