Discussion in 'The Projects Forum' started by JoeJoe88, Dec 23, 2011.

1. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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sooo im hoping someone here will finally be able to end my suffering with this crap. Sadly i get the feeling my answer is super easy, and im just super stupid soo please pardon my ignorance if this is the case. Now to the issue, im trying to design an array consisting of 125 3 watt leds, ranging from 2.5 Volts up to 3.6 volts.The leds im using have a rated typical forward current of 700mA. Ive tried a few online array wizards, but id much rather have a understanding of how it actually came up with the blueprints, this way if i wanted to make more, or something failed, id be able to handle the issue better. So after all of that, I know ill have to group the leds in a series based on its current voltage rating, but i guess what i would like to know,

1.How do i go about calculating the specs of the driver i am to use (output voltage, output current)

2. If i am to use a driver do i need to use resistors or a regulated power supply?

3. If i wanted to connect 42 LEDS, all with a forward voltage of 3.6 and a forward current of 700mA, in a array, how much forward current is needed to power all the LEDs?

thank you in advance to anyone who takes the time help me with this, it will be greatly appreciated

Apr 5, 2008
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3. ### tracecom AAC Fanatic!

Apr 16, 2010
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The current required will simply be the number of LED's times the current required per LED, i.e., 125 times .7 A (which is 700 mA) equals 87.5 Amps, which is a very large current. And that's just for the LEDs; any current limiting resistors will add to that figure. The voltage is more complicated; it depends upon how you have the LEDs electrically arranged, which depends upon what voltage you plan to use for power. What do you have in mind as a power source?

ETA: As has been correctly pointed out in posts after this one, running large numbers of high current LED's in parallel is not good design. I did not mean to imply that it was. In addition, my statement above about the current limiting resistors adding to the required current was wrong. For the best answer, read SgtWookie's posts (numbers 7 and 8) in this thread.

Last edited: Dec 23, 2011
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4. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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thanks for the quick response, i think im starting to get it. i changed my location. i was unaware this was a international site, which is great.

5. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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thats the thing, i have no idea. lol
at first i thought about using a 24 volt led power supply, but it was only rated at like 2 amps. and as you made clear with ur helpful calculation, that that just wont cut it....im basing my design loosely off of a design i saw online..they claim to put 150 3watt LED (running at an assumed 700ma)
into one system that has one 3 prong plug that u can plug into one outlet...seems impossible. or am i mistaken?
i wanted to group the leds together with at least 15 to each group
im speaking strictly to the lay out of the actual diodes,not the wiring of the whole system
id like to string together the individual leds into a series of at least 12
so at this point i havent purchased any equipment, im still planning it all out
what kind of power supply would u suggest?
i thought about running mutiple power supplies to different outlets to avoid not having enough current since my breakers r only 20 amps each

6. ### tracecom AAC Fanatic!

Apr 16, 2010
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I may have posted before I fully understood the issue and what I said may be wrong. Let me work on some more calculations.

What I posted earlier is a worst case scenario.

I think the following would work. Suppose you decided you wanted to use a 24V supply (a couple of 12 batteries in series) for your 42 LED's that each have a Vf of 3.6V and a current requirement of 700mA. You could then put 6 LED's in series with a 3.6Ω 3 watt resistor, which would have a total power consumption of 16.8 watts at a 90% efficiency. Then, you could repeat that arrangement 7 times for a total of 42 LED's for a grand total of 117.6 watts. Because watts equals volts times amps, the total current required would be 117.6 / 24 = 4.9 A.

Last edited: Dec 23, 2011
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7. ### SgtWookie Expert

Jul 17, 2007
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Well, with 3W LEDs, you really should be using a switching current regulation scheme, as plain resistors or linear regulators would waste too much power.

When operating LEDs in series, they should all have the same current rating. If they have different current ratings, you need to use the lowest current rating. If the difference in current rating is significant, you would be wasting the light output potential of the higher current LEDs.

Here's where experimenting in a spreadsheet will help. You can create "what if" scenarios to help you determine what voltage would work best for your situation, and what resistors you might need.

I don't know how many you want to run in a series string. You could start out at an arbitrary voltage. 19V is common for regulated laptop supplies; and they can be cheap/handy to use for such things.

Take your poswer supply voltage, subtract 1v from it (if using a switching regulator, consult the documentation - but it's frequently 2v), then divide that remainder by the Vf of your LEDs; then round the result down (truncate fractional part) to an integer. That's how many you can operate in a series string.

If you use a switching current regulator for each string, you can theoretically use an unregulated power supply. However, regulated switching supplies are so cheap, compact and efficient, you might as well use one of those instead.

If you use resistors for current limiting, a regulated supply is highly recommended. Otherwise, you can wind up with fluctuations in light intensity.

You will need one current regulator or resistor per series string. If you try to operate LEDs in parallel with just one current regulator/resistor, you risk a sudden catastrophic failure of all strings.

That depends upon the voltage supply you're using.
Let's just say you're using a 24v supply.
For resistors: LED_Max = INT((24v-1v) / 3.6v)
For current regulators: LED_max = INT((24v-RegulatorDropoutV) / 3.6v)

Let's just go with resistors for now. (24v-1)/3.6 = 23/3.6= 6.388889 = 6 LEDs.
6*3.6 = 21.6v
24-21.6 = 2.4v left over for "headroom" - this voltage will need to be dropped by a resistor or active current limiter.

Let's just for the moment say we're going to be wasteful and use resistors.
Rlimit = 2.4v / 700mA = 3.428 Ohms. That is not a standard value of resistance. 3.3 Ohms IS, but it is too low; 2.4v/3.3 Ohms = 727.3mA, about 4% over. You could use that, but you would be over the maximum specification, and would shorten the life of the LEDs. A 3.6 Ohm resistor is a standard value. 2.4/3.6 = 667mA; 95.2% of the rated current, which is fine; it gives you an extra safety margin for resistor tolerance.

Then you need to calculate the power required for the resistor.
Power in Watts = Volts x Amperes; 2.4v x .667mA = 1.6 Watts.
For reliability's sake, we multiply that result by 1.6, winding up with 2.56 Watts. If you can't remember the 1.6 factor, then just double the actual power requirement.

2.56 Watts is not a standard power rating for resistors, but 3W is.
A quick search on Digikey for 3W 3.6 Ohms turns up this:
http://search.digikey.com/us/en/products/ERX-3SJ3R6/P3.6W-3BK-ND/36610

So, with 6 in a string, and you want 42 LEDs, that works out to be 7 strings even. How convenient; you're not always that lucky.

In this case, you would need a total of 700mA x 7 = 4.9 Amperes current.
For reliability, you would want a power supply that was rated at least 20% more than 4.9A; or 5.88 Amperes; 30%/6.37A would be better. Just round that up to 6A and 6.5A.

There is a supplier in my state that I like to use for supplies; they seem to be reasonable, and ship very quickly.
From their Powersupply1 page: http://www.mpja.com/PowerSupply1-Mfg/products/465/
this 150W 24V 6.5A supply would neatly fill the bill: http://www.mpja.com/24V-65A-150W-Power-Supply/productinfo/16032+PS/

Now, have a look at the efficiency; 24v total, of which 2.5 we're dropping across resistors in what is basically wasted power as heat. 7 * 1.68W = 11.76W, which really isn't all that bad considering the remainder of the array takes 105.35W, for a total of 117.11W. 105.35/117.11 = 89.96% efficient - it would be a lot worse at lower voltages.

You could also look at using different voltage supplies; that's when a spreadsheet really helps to explore different scenarios.

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8. ### SgtWookie Expert

Jul 17, 2007
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Gosh, I sure took a long time to write that - no one else had replied when I started.

That would be if you were operating them all in parallel on low voltage, which would obviously be a terriffic waste of power.

No, the resistors are in series with the LEDs; that would not increase the current requirement, only the power requirement (more voltage to drop across both the LED and resistor).

9. ### tracecom AAC Fanatic!

Apr 16, 2010
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Thanks SgtWookie. I was editing my last post and didn't read what you had said, but it looks like we came to the same conclusion. You just got there with a lot more certainty than I. It took a while for me to clear my head.

Dec 26, 2010
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If you are talking about an AC mains breaker of 20A rating on a 120V supply, any reasonably designed LED power supply should not get anywhere near the limit of your mains supply.

The mains supply is capable in principle of supplying 2.4 kilowatts, and you are talking about 375 watts worth of LEDs - even if you used a stupidly inefficient linear supply with generous resistor ballasting, it is hard to imagine this stretching to over a kilowatt. A good linear supply could need maybe 500W, and a switcher would be better.

Clearly you would need some kind of transformer (or isolating switch-mode converter), as otherwise it would be unsafe, and of a type not allowed for discussion on this forum. The transformation would allow the mains current to be less than the current flowing in the lower voltage supply fed to the LEDs.

Let's try the maths. Personally I probably might use something like a set of series strings of LEDs arranged for operation at 24V. You might get five or six LEDs per string depending on their voltage and whether they used resistive ballast or an active current source.

Each string would take 700mA, so you might end up with 125/5 = 25 strings using a total of 25*0.7A = 17.5A at 24V, =420W. A modest power supply efficiency of 70% would require 420W/0.7 = 600W at the input, so from a 120V supply you would draw 600W/120V = 5A.

This is for a very basic linear supply using resistors. A good modern switched-mode system would do much better.

Edit: As usual, the younger and less decrepit have done a more comprehensive job. Never mind!

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11. ### tracecom AAC Fanatic!

Apr 16, 2010
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Well, I beat the Sgt to the first post, but unfortunately, what I posted was only half thought through, and while I was trying to get my act together, he came to the rescue (as usual.)

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Dec 26, 2010
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I just hope that the point has now been taken that no reasonable design using N LEDs each requiring 0.7A would require N*0.7A at the AC mains input.

You would never want to run such a thing as a parallel network straight off the mains. I shudder to imagine the scale of the dropping resistors required to do that. Perhaps about 10kW worth of tungsten lamps might fit the bill, but then the LED lamps would be rather redundant!

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13. ### tracecom AAC Fanatic!

Apr 16, 2010
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I have edited my first post in this thread (number 3) to add corrections. I agree that powering LEDs directly from the AC mains is a terrible idea, and I don't think I suggested that. I certainly didn't intend to.

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Dec 26, 2010
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I didn't suppose you meant that, but it seems possible that the OP was thinking along those lines.

It also may be that he is not familiar with the principles of transformers (or SMPSs) changing current and voltage relationships, and of how further voltage versus current trade-offs could be made by rearranging elements in different series and parallel groupings.

Perhaps this is not the case: if so I may be being unfair, but in my opinion anyone who can't get these ideas clearly in mind is not ready to plan a job on this level.

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15. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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i appreciate all of you for taking the time to comment on this thread and help me find a solution...i find it great that something like this exist for people to come together and share knowledge for free. thanks to all of you
im now starting to finally find the answers i seek...
and dont worry
i have no intention on plugging the LEDs into the ac
thanks for looking out tho

16. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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do u care to elaborate? these resistor methods seem functional, but i would like to go for maximum efficiency...
is there a way to wire 126 3 watt LEDs, with different foward voltage ratings, from the same power supply/driver?

17. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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i am not familiar with the knowledge needed to complete this project. i was hoping someone on here would be able to help or least point me in the right direction. i would like to learn about the "principles of transformers (or SMPSs) changing current and voltage relationships,......and voltage versus current trade-offs" so i can have the ability to design an array for specific needs. please help

18. ### Wendy Moderator

Mar 24, 2008
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We regularly run across this with grow lamps, I have quite a few schematics drawn for the occasion.

This article deals with much lower power LEDs, but goes over the basics.

LEDs, 555s, Flashers, and Light Chasers

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19. ### JoeJoe88 Thread Starter New Member

Dec 23, 2011
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ive looked over your article, but im unsure how to translate that into my situation with the higher current needs. is there a link for the schematics that you speak of?

20. ### SgtWookie Expert

Jul 17, 2007
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JoeJoe,
Sorry we seem to be a bit slow in responding. The board is not interactive, and lots of people are wanting help. We try to help most everyone that comes in, but things usually don't get resolved in one single day.

I went through a rather lengthy monologue on how to calculate for # of LEDs in series, and resistors for the "left over" voltage to limit the current.

If the LEDs are of the same current rating, you can actually more or less mix and match them in a given string; it frequently works out that a higher voltage (like 24v, 28v, 32v, etc) works out to be more efficient, as you can make the forward voltages all total up to be nearly the same.

Now, you said that you have have 3 Watt LEDs, and apparently they are all rated for 700mA.
You have said that you have 125 LEDs total.
You have said that the voltage ranges from 2.5v to 3.6v. It's this latter specification that is the item that needs to be worked out.

Let's separate them by color or part number or whatever distinguishes them from one another - and a count for each category. Then give the forward voltage (Vf) range for each type/color/etc., and what the typical Vf is. That way, we can start getting a better idea for what might work best for you.

One thing I forgot to mention in my previous post, is that the power supply that I provided a link to is adjustable; about ±10% of the rated supply voltage; this is indicated in the datasheet: