Parallel LED array.

stelleg151

Senior member
Sep 2, 2004
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This seemed a bit above the level of the usual questions in General Hardware, so I figure this would be a better place to ask.

I have some very basic knowledge about OHMs law, and DC circuitry in general, and am planning on building an LED array and want to clear one worry I have first.

The LEDs are rated to run at 20mA @ 3.3V(average). I will be hooking them up to a PSU's 3.3V line, with resistors inbetween. All 389 of them.

Now having 1 resistor for each LED seems to be recommended from Linear1's guide, but it seems to me that as long as the resistors can handle the power dissapation and have the right resistance then the LEDs can be run in parallel, say 21 per resistor. Because the difference in voltage will be reasonable small, the total power dissapation for a 1ohm resistor running 21 parallel 20mA LED's should stay below 1/4 (and probably 1/8) of a watt.

So basically it would both save me a lot of time and space in the project to just use one resistor per 21 LED's, and I am pretty sure it is possible, just thought I would run it by some of the Anandtech techies before I burn the house down by overloading a resistor.

To repeat myself/ sum up:
21 LEDs @ 20mA @ 3.3v.
3.3v DC line (more than enough Amps).
1ohm 1/4W resistor, regulating input current for all 21 LEDs, in parallel.

Thanks, and forgive my ignorance, Im actually a BA student.... bleh (in the process of switching to BSc though) ;)
 

Navid

Diamond Member
Jul 26, 2004
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I don't think that is such a good idea!

An LED is a diode (Light Emitting Diode). One of the characteristics of a diode is that after you overcome its forward knee bias voltage, its current changes very rapidly with voltage. In other words, if you change the voltage across a diode a little bit, its current can change a lot. That is the main reason you need a resistor in series with an LED. The series combination of the LED and the resistor will have an I/V characteristic that is less sensitive to voltage compared to just a diode. This will be more true the larger the resistor is.

Practically, the series resistor let's you control the amount of current that flows through the diode (LED).

But if you place diodes in parallel, you lose that control. You may think that half of the current will flow through each one of two diodes placed in parallel. That would be true only if the two diodes were perfectly identical. In practice no two components are ever identical due to imperfect manufacturing processes.

If you place diodes in parallel, you are running the risk of having one diode conduct most of the current (Thermal runaway). This can damage that diode (Overdrive).
In a less extreme scenario, you may end up with non-uniform lights coming from your LEDs.

Everything may work fine too. But, there is no guarantee. There will be a guarantee if you have an individual resistor with each diode since you can reasonably estimate and control the current of each diode.

This may be good to read.
This is the second page, which shows the point I was trying to make with a graph.
 

CrispyFried

Golden Member
May 3, 2005
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agreed, bad idea. what usually happens is one diode gets most of the current and burns out.. this allows more current through the rest of the leds and then another in turn will hog the current and burn out also. as more and more leds die the others get the current that the dead ones should have drawn, so even if the surviving ones do draw balanced current it winds up being way more than they should. repeat till they are all dead.

try it on a small scale and see.

try them in series. then you can run several with one resister. at 12v that allows 4 or 5 per resister.

some leds have built in resisters (pretty sure anyway), have you looked into them?
 

Mark R

Diamond Member
Oct 9, 1999
8,513
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You really, really want to use series strings of LEDs.

E.g. 3 or 4 in series, connected to the 12V line via 1 resistor for each string. This needs fewer parts, and is considerably more energy efficient - not to mention that the 12V line on a typical PSU is much stronger than the 3.3V

Unfortunately, you really do need 1 resistor per string - unless your LEDs are precisely matched for forward voltage and all mounted together so that they experience the same temperature.

Have you considered high-intensity LEDs? You could use medium-brightness LEDs like 'super-flux' LEDs (you'd probably need about 50 or 60 of these) or some high brightness LEDs like 'Luxeon III' - you'd need about 5 or 6 of these.
 

zachtos

Member
Sep 7, 2005
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check out my thread at reefcentral. I have created an array a few months ago that is used over a marine tank to grow coral. I'm guessing you are doing that too? anyways I have alot of explanation step by step and photos of what I did to construct it.
 

Jeff7

Lifer
Jan 4, 2001
41,596
20
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I've got some pics (no writeup though, maybe in the coming months sometime...) here.
I wanted to make an LED growlight. But I eventually decided to go back to college, so money and time vanished from my inventory, so the project stopped.

I used Linear's LT1618 boost converter to make little power supplies for the LEDs, which may or may not have been necessary, but I did want a stable source of power. The little converters boost 2-18VDC to about 35.5VDC to power the LEDs. I'm not exactly sure what the max output of the LT1618 is, but it can power 5 1W LEDs, and at least 95 regular LEDs.
I have them in a few series arrays - about 19 LEDs per array, with the arrays in parallel to each other.
I did run the thing for a few weeks over some small plants. They didn't die, but they didn't grow much. It either was because of inadequate brightness, or maybe the lack of any IR LEDs. Once I'm out of college in a few years, well, LED tech will probably be a bit better, with lower prices. I'll have to ditch the loads of regular LEDs I've got now, and maybe go for some of the high-brightness things in red, with T 1-3/4 LEDs around the perimeter in blue, IR, and UV.
 

stelleg151

Senior member
Sep 2, 2004
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Thanks for the replies guys.

EDIT: Awesome stuff zachtos! Welcome to the forums. You have done more documented work with LED arrays than anywhere else I can find, so thanks a lot.
That is frustrating hearing of your LED failures. I saw a comparison somewhere that showed the output in light amount is actually not much reduced with a decrease in current, so I am considering doing a test run with a higher resistance resistor. I am also planning to buy from besthongkong, so I hope they arent just crap LED's, they are much cheaper than anywhere else I can find. It sounds like you are using the 5mm LEDs; I am going to try out the 10mm 12 degree 110000mcd LEDs. Youre project has convinced me to buy 50 or 100 first for testing, so while I am very sorry for your frustration, I thank you for being a pioneer. Anyone interested in LED arrays should see this guy's work: its awesome. I am also very impressed that you could solder 400 LED's in only 8 hours, that really gives me hope.

Again, thanks for your hard work and for documenting it so well.

That array calculator is why I thought I would ask before I started. The reason I thought parralel would be ok is because of this article : Text Basically it looks like he has them in series of two, but only one resistor for all 18.

I have looked at the higher intensity ones, although mine are fairly high intensity (135000 mcds max), and the ones like the Luxeons require heatsinks for each one, and I am looking at making a high density array, fairly evenly lit, for fairly cheap($.04 each). The reason I was hoping to opt for the one resistor per set was because I will be doing hundreds, but it sounds like I should just prepare for an even more absurd amound of soldering.

I looked into the voltage step up converters, and I figure I would need one for every row of my LED's so 34 of them in total, making it a bit too expensive of a solution, but thanks for the suggestion.

Thanks for the help guys, Im gonna go buy more solder.

Edit: Oh and one more question. Part of the reason I was looking at a computer powersupply as a powersupply(other than that I already have a couple) is that the reverse voltage for the LED's is rated at 5v @ 100micro amps. I figure the reverse current should be hooked up the -5v line on a PSU, correct me if I am confused here.

Thanks.
 

stelleg151

Senior member
Sep 2, 2004
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I have given it more thought, and I think this setup would work, and would be willing to spend the extra money on the DC power supply for significantly less work and a more efficient setup.

New Plan (edited)

The array theoretically should consume around 32 watts (confirmed only by linear1's calculator).

Most importantly: Any ideas on whether or not it will work?

A couple of worries about the 48v power supply. First, do I need to worry about fluctuations, or will the resistors be enough to stop any fluctuations? Should I put a capacitor to help regulate the voltage coming out of the PSU, or will Also, is 48v any more dangerous than 12v with the same current (should be just 20mA) running through? (not too worried about this, have been shocked by wall sockets a few times).
 

Navid

Diamond Member
Jul 26, 2004
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Originally posted by: stelleg151
I have given it more thought, and I think this setup would work, and would be willing to spend the extra money on the DC power supply for significantly less work and a more efficient setup.

New Plan

I thought you said the LED current was 20mA.

If that is correct and you have a 1100-Ohm resistor in series (from your diagram), you will end up with 1100 X 0.020 = 22V across each resistor. This leaves only 26V across the series combination of the LEDs.

You say you have 28 LEDs in each row (your diagram). So, there will be 26V/28=0.93V across each LED. Is that the voltage your LED operates at?
 

Navid

Diamond Member
Jul 26, 2004
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The current of the negative terminal of the supply is going to be exactly equal to the current of the positive terminal. Except its direction. The current leaves the positive terminal of the source and enters the negative terminal of the source.

This is assuming you are using a supply with one positive terminal and one negative terminal. If this is like a computer power supply, which has one ground but multiple positive terminals (+12V, +5V, ....), the current of the negative terminal will be equal to the total current of all the positive terminals.

If this is a power supply that you have bought, the negative terminal (ground) can definitely handle the current that is within the limit of the + terminals.
 

Navid

Diamond Member
Jul 26, 2004
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Originally posted by: stelleg151

Also, is 48v any more dangerous than 12v with the same current (should be just 20mA) running through? (not too worried about this, have been shocked by wall sockets a few times).

No, the current will be equal to the current of one LED multiplied by the number of rows.

So, if you have 34 rows, and if you want each LED to have 20mA, the total current of the power supply will be 34 X 20mA =680mA.
 

stelleg151

Senior member
Sep 2, 2004
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Originally posted by: Navid
Originally posted by: stelleg151
I have given it more thought, and I think this setup would work, and would be willing to spend the extra money on the DC power supply for significantly less work and a more efficient setup.

New Plan

I thought you said the LED current was 20mA.

If that is correct and you have a 1100-Ohm resistor in series (from your diagram), you will end up with 1100 X 0.020 = 22V across each resistor. This leaves only 26V across the series combination of the LEDs.

You say you have 28 LEDs in each row (your diagram). So, there will be 26V/28=0.93V across each LED. Is that the voltage your LED operates at?

Sorry, my information there wasnt very clear. It is supposed to say one 100ohm resistor per row. Thus giving 100x.02 = 2v. Leaving 46v accross the LEDs. I also am an idiot and forgot to write that I'm putting 14LED's per row, not 28. Thus, we have 46v/14 = 3.29v, which is exactly right.

I am considering moving up to maybe 150ohm resistors to make sure that as the LEDs heat up they dont burn themselves out by letting too much current through(not exactly sure how this works, but from what others have put it seems to happen). I have also seen comparisons putting slightly undercurrented LED's together and they still were nearly at original brightness. However, I will decide on the resistor value for certain once I have done more testing.

Edit: New Plan (fixed)

Edit2: thanks for your help Navid, Im still a bit early in the learning/planning phases.
 

Mark R

Diamond Member
Oct 9, 1999
8,513
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That latest plan looks excellent.

Just bear in mind that sometimes the forward voltage of LEDs varies a bit between (and even within) batches.

It's fine designing your system for 100 Ohm resistors, but ideally you should measure the current through each string and adjust the resistor as necessary so each string gets the same current, and so is the same brightness. This is more important the lower the voltage you are leaving across the resistor.

The other catch is that if your power supply is unregulated then using a low value resistor can mean that very minor power fluctuations can turn into big brightness changes on the LEDs. If you can't have any flicker then either use a regulated PSU, or chop an LED from each string and use a correspondingly larger resistor.

The other thing to remember is that cheap LEDs aren't designed for use in hot environments (this is especially the case with white LEDs). The centre of your LED array will get hot unless you ensure plenty of space for air to circulate. (That's why I like luxeons - get a flat aluminium plate (to act as a heatsink) and glue them on with arctic epoxy - job done).
 

Navid

Diamond Member
Jul 26, 2004
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Originally posted by: stelleg151
I am considering moving up to maybe 150ohm resistors to make sure that as the LEDs heat up they dont burn themselves out by letting too much current through(not exactly sure how this works, but from what others have put it seems to happen).

Thermal runaway (constant voltage across the semiconductor):

You directly place a DC voltage across a semiconductor (like a diode) to bias it with no considerable resistance is series.
The diode turns on.
It warms up.
A semiconductor conducts more as it warms up (a difference between a semiconductor and a conductor). Or, you can say the effective resistance of a semiconductor drops as it warms up.
Now that the diode conducts more current, it warms up even more.
Now that it has warmed up more, it conducts even more.
.
.
.
.

This regenerative process keeps going on until the diode breaks down.


How to avoid thermal runaway (make the semiconductor voltage an opposite function of its current):

You place a resistor in series with the semiconductor (diode).
The diode turns on.
It warms up.
As a result, the effective resistance of the diode drops.
As a result, a larger portion of the constant supply voltage drops across the resistor and a smaller portion of the supply voltage drops across the diode.
The lower voltage across the diode decreases its current.
The decrease in current reduces the temperature of the diode.

This is a stabilizing process as opposed to the one described in the previous paragraph. So, the series resistor creates a negative feedback that avoids thermal runaway.
 

stelleg151

Senior member
Sep 2, 2004
822
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Originally posted by: Mark R
That latest plan looks excellent.

Just bear in mind that sometimes the forward voltage of LEDs varies a bit between (and even within) batches.

It's fine designing your system for 100 Ohm resistors, but ideally you should measure the current through each string and adjust the resistor as necessary so each string gets the same current, and so is the same brightness. This is more important the lower the voltage you are leaving across the resistor.

The other catch is that if your power supply is unregulated then using a low value resistor can mean that very minor power fluctuations can turn into big brightness changes on the LEDs. If you can't have any flicker then either use a regulated PSU, or chop an LED from each string and use a correspondingly larger resistor.

The other thing to remember is that cheap LEDs aren't designed for use in hot environments (this is especially the case with white LEDs). The centre of your LED array will get hot unless you ensure plenty of space for air to circulate. (That's why I like luxeons - get a flat aluminium plate (to act as a heatsink) and glue them on with arctic epoxy - job done).

Good points. I definately plan on checking the currents often, and will buy a small batch to test for a few weeks before I go big. I thought about the regulated power supply issue, and since I will probably be going for the cheapest 48w power supply I can find, so its possible that one I get wont be regulated. I have been looking for 48v regulators, but have only found ones that run around 12v. Any ideas anyone?

As for cooling, I admit that LEDs with heatsinks are very cool(yay for puns). However, I think for this high density high volume project its a bit out of my budget. I dont think it will even matter, though, as I already have 2 120mmx38mm fans ready to suck ducted air straight accross the LEDs, which I think will keep them cool enough, especially considering that I will probably be undercurrenting them to help prevent failure.

Quick and Dirty Ducting Diagram
 

stelleg151

Senior member
Sep 2, 2004
822
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Originally posted by: Navid
Originally posted by: stelleg151
I am considering moving up to maybe 150ohm resistors to make sure that as the LEDs heat up they dont burn themselves out by letting too much current through(not exactly sure how this works, but from what others have put it seems to happen).

Thermal runaway (constant voltage across the semiconductor):

You directly place a DC voltage across a semiconductor (like a diode) to bias it with no considerable resistance is series.
The diode turns on.
It warms up.
A semiconductor conducts more as it warms up (a difference between a semiconductor and a conductor). Or, you can say the effective resistance of a semiconductor drops as it warms up.
Now that the diode conducts more current, it warms up even more.
Now that it has warmed up more, it conducts even more.
.
.
.
.

This regenerative process keeps going on until the diode breaks down.


How to avoid thermal runaway (make the semiconductor voltage an opposite function of its current):

You place a resistor in series with the semiconductor (diode).
The diode turns on.
It warms up.
As a result, the effective resistance of the diode drops.
As a result, a larger portion of the constant supply voltage drops across the resistor and a smaller portion of the supply voltage drops across the diode.
The lower voltage across the diode decreases its current.
The decrease in current reduces the temperature of the diode.

This is a stabilizing process as opposed to the one described in the previous paragraph. So, the series resistor creates a negative feedback that avoids thermal runaway.

Awesome, thanks a lot man, I had a hunch it had to do with reduced resistance with increased heat, but nice to have a real definition and name.


Edit: Another Diagram of LED Layout, almost perfect 1:1 pixel:mm^2 ratio, but stupid me forgot how paint draws a circle, so its a 9:11 pixel:mm^2 ratio.:disgust:

9:11 Pixel Ratio LED Array Diagram

Edit 2: [warning for circuitry ignorance] : It doesnt matter which side of the LEDs the resistor is on does it?