# Question about LEDs Voltage and Current

Discussion in 'General Electronics Chat' started by flynnguy, Mar 13, 2009.

1. ### flynnguy Thread Starter New Member

Mar 13, 2009
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My goal is to light a string of LEDs efficiently. I talked with someone else who is doing something similar and came up with the attached image. I understand the number of LEDs I can light depends on their forward voltage but is this generally a good idea?

Also I realize that an excess of current can fry an LED, but what about excess voltage? IE, if I only hooked up 3 LEDs to line voltage (rectified and filtered) will this fry the LEDs? (Assuming I use the LM317 to limit current to the current specified for the LEDs) Or will the bleeder resistor and filter cap absorb this? Also should I be concerned about variations in line voltage?

Finally, say I wanted to add more LEDs, could I just use a step up transformer? Would this be a good idea or should I just start a second string?

I'd also apreciate any other tips that anyone might have as I'm mostly a novice when it comes to electronics. Thanks.

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3. ### SgtWookie Expert

Jul 17, 2007
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The circuit you posted will result in the immediate demise of the LM317 in a blinding flash, accompanied by a loud "BANG!"

You need to use a transformer to reduce the line voltage to a safe level, along with isolating your circuit from the mains. I suggest that you start out with a transformer that has a secondary rated for between 12v and 24v at 1A. You should use a fuse on the mains side connection rated for around 1/2 Ampere.

Then you use a full-wave rectifier bridge with relatively large capacitors to filter the output of the bridge.

Then comes the regulator. See National Semiconductor's LM117/LM317 datasheet for ideas:
http://www.national.com/ds/LM/LM117.pdf

Note that the LM117/LM317 has a maximum 40v differential between it's input and output terminals.

What are the specifications for your LEDs? What I'm asking for here is the typical Vf @ mA (forward voltage at a specific current). A quality supplier will give minimum Vf@mA, max Vf@mA, and typ Vf@mA among other specifications. Roughly 70% to 85% of the LEDs in a batch will fall close to the typical specification.

Last edited: Mar 13, 2009
4. ### beenthere Retired Moderator

Apr 20, 2004
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We are seeing a lot of these circuits where someone wants to run his circuit right off the line. This is inherently dangerous, and cannot be made safe to use under any circumstances. The use of a transformer is absolutely necessary to provide the user with safety.

No directly line-operated circuits are acceptable topics for discussion.

5. ### flynnguy Thread Starter New Member

Mar 13, 2009
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Ok, so what kind of transformer should I look for if I want to run say 40-50 LEDs figuring about 2-3V 20mA LEDs? Most of the ones I've seen only go up to about 12V. (At least of the wall wart variety) I definately want to do this the safe way which is why I'm posting here, thanks for the warning.

Apr 5, 2008
15,806
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Hello,

There are some led calcutators on the internet.
Here is an example.
http://ledcalc.com/

You can fill in :
the supply power,
the voltage of the led,
the current wanted and
the number of leds and
it will draw a little schematic.

Greetings,
bertus

7. ### flynnguy Thread Starter New Member

Mar 13, 2009
9
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Yes, thank you, I've seen this calculator and I know how to calculate for the resistor to use, the calculator page you sent me to only lets me put in a max number of Volts and max number of LEDs to 32, I was hoping to drive more than this, hence my question about the transformer.

8. ### SgtWookie Expert

Jul 17, 2007
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Well, the next question is; do you really want to build your own power supply, or just purchase one that's ready to go?

Marlin P. Jones & Associates carries a variety of supplies, wall-warts and plug transformers.
Here's a 24VAC @ 1.667A unregulated plug transformer for \$6.95+shipping:
http://www.mpja.com/prodinfo.asp?number=17561+PA
Here's a complete 24v @ 1.8A regulated switching supply for \$14.95:
http://www.mpja.com/prodinfo.asp?number=17426+PS
Here's a 15v@ 0.8A regulated switching supply for \$2.49: (you'll need to order a \$1.25 power cord for it, below)
http://www.mpja.com/prodinfo.asp?number=16790+PS

But to really nail down your power supply needs, it'll help a lot to get the specifications on your LEDs first. 2v to 3v @ 20mA is actually quite a variation. Are you planning on adding to the LEDs later, or is this going to be just a fixed amount for some kind of display?

Last edited: Mar 13, 2009
9. ### flynnguy Thread Starter New Member

Mar 13, 2009
9
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It's for a grow light to start my seeds before I move them outside, and yes, I am growing legal plants like tomatoes, peppers, lettuce, spinach, etc...

According to what I've read, plants mostly need red and blue for photosynthesis so I was going to do about a 60%/40% mix of red and blue LEDs. I've talked to someone else doing something similar and while I did not see his circuit, I came up with what I originally posted though now that I read more in this thread, I'm sure he is probably using a transformer of some sort.

In looking at LEDs, I've noticed there was a wide range in the voltage requirements (seemingly ranging from 2-3V in what I've looked at) but I think I can limit myself to 20mA. Knowing what I know, I then figured I could just adjust the ammount of LEDs based on how many I could light. In other words, I thought the original circuit would limit the current to 20mA and then I would just put on the end however many LEDs I could light, figuring about 40 LEDs per string based on some rough calcs I did.

Ultimately I was hoping to get about 200-250 LEDs.

10. ### SgtWookie Expert

Jul 17, 2007
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OK, that makes more sense.

Red LEDs range from around 2.1v to 2.7v (the newer ones); blue range from around 3.2v to 4v.

You can mix them in a series string, just as long as you leave "headroom" for a current limiting resistor.

Let's say you went for that 15v@0.8A regulated switching supply for \$2.49.
If your current requirement for each series string of LEDs is 20mA, that means you could have 0.8A / 20mA = 800mA / 20mA = 40 strings of LEDs being powered by it.

Let's say you wanted to run the maximum number of LEDs that you could in each string; but you'd like to have an even number of red and blue LEDs.

I'll take a guess here that your red LEDs will have an average Vf of 2.4v @ 20mA, and your blue LEDs will have an average Vf of 3.6v @ 20mA, for a total of 6v.

15v / 6v = 2.5. You take the integer of the result; so 2 reds and 2 blues per string; they will drop a total of 12v across themselves leaving 3v to drop across a current limiting resistor. So, what value resistor will be needed?
R=E/I, or Resistance = Voltage / Current.
Since E = 3v and I = 20mA, R = 3 / 20mA = 3 / 0.02 = 150 Ohms.
We lucked out, 150 Ohms is a standard value. Were it not, we'd have to go to the next higher standard value, or use a couple of resistors in series or parallel to get the right value.
A table of standard resistor values is here:
http://www.logwell.com/tech/components/resistor_values.html

So, with that one supply, you could power 40 strings of 4 LEDs for 160 total with no problem.

You could actually power a few more if you didn't mind doing a little juggling with the LEDs from string to string.

From the above example, the blue LEDs were guesstimated to be 3.6v @ 20mA. You could power 4 of them in a single string, which would drop 14.4v (4 x 3.6=14.4), leaving 0.6v (15v-14.4v=0.6v) for the current limiting resistor to drop.
R=E/I, so R = 0.6/20mA=0.6/0.02 = 30 Ohms, another standard value.

I guesstimated the red LEDs to be 2.4v. 15v/2.4 = 6.25; the integer value is six.
15 - (6 x 2.4) = 14.4, again - so the current limiting resistor is again 30 Ohms.

So, for every three strings of 4 blue LEDs, you can use two strings of 6 red LEDs, and you'll wind up with 24 LEDs per that series of five strings; the total current draw for those 24 LEDs will be 100mA.
The supply will support up to eight of these series of five strings, for 192 LEDs total.

That in itself will represent quite a bit of soldering; 192 x 2 + 8 x 5 x 2 = 462 connections for the LEDs and resistors. It's a bit ambitous for a first project, but do-able. That's a lot of LEDs getting powered by only 12 Watts.

Or, did you really want to go for more LEDs?

11. ### flynnguy Thread Starter New Member

Mar 13, 2009
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Thanks, lots of good information there. This won't be my first project, I've done some microcontroller work before and soldered many things but never really designed anything. (Other than the microcontroller stuff but that was more programming than Electrical stuff)

I suppose that's a good place to start, what about the efficiency of these power supplies, it was my understanding that wall warts can be inefficient and for something that I was planning to leave plugged in all day and running for several hours 24/7, I'd like an efficient power supply.

I suppose if I want more strings, I just have to find a power supply with a higher current rating? Also in hooking this up, do I need any sort of regulator, filter caps or whatever? As I said, I've done some microcontroller work and I typically use a wall wart and then a 7805 circuit to get a nice 5V but again, most of these projects aren't running for long periods of time and I guess that's what concerns me most. Thanks for all your help!

12. ### SgtWookie Expert

Jul 17, 2007
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According to the information on MPJA's website for the 15v 0.8A supply, it's a switching power supply. Those are generally between 80% and 94% efficient. It's not likely that you'll be able to improve on that very much.
Yes.
You could use a filter cap across the output if you wish, but you probably won't need it. That supply already has a regulator built in.
Efficiency should always be a concern.

A wall wart with a 7805 isn't nearly as efficient as a switching power supply; much of the power is dissipated in the regulator itself.

For example, if you're powering a 7805 from a 10v wall wart, and you have a 50 Ohm load through which 100mA of current is flowing, that's 1/2 Watt.
At the same time, the regulator is dissipating 10v - 5v = 5v x 100mA = 1/2 Watt. This isn't counting the nominal 5mA current flowing from the GND pin, either; that's another 25mW's worth of inefficiency; so over 50% of the power is wasted in the regulator alone.

Note that the Vf @ current values I used were realistic, yet hypothetical. You will have to re-calculate as I did in order to get the proper resistor values.

13. ### flynnguy Thread Starter New Member

Mar 13, 2009
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Thanks again for your help, I think I understand all your calculations and realize I'll have to recalculate once I figure out what LEDs I'm using. (which is turning out to be harder than I originally thought to find LEDs in the wavelengths I need)

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15. ### flynnguy Thread Starter New Member

Mar 13, 2009
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Ok, I'm a little further than I was before as I found some LEDs and now have real specs!!! I just have a few questions to help me determine which power supply to use to light how many LEDs...

I'm thinking of using a 55% Red, 35% blue, 10% white mix. I'm going to use 2 wavelengths for each of the red and blue, 660nm, 645nm for the red and they are both 1.8V@20mA so that's easy. Then for the blue, I'm using 430nm and 470nm at 4.9V@20mA and 3.4V@20mA. Then for the white, I have 3.3V@20mA.

Ok, I get the above part and I'll start here as it seems like as good a place as any. So we get 40 strings.

Ok, I think I understand this but I'm having trouble adapting it to my circumstances. As far as I understand it I figured:
55% * Red V + 35% * Blue V + 10% * White V
.55*1.8 + .35*8.3V + .1*3.3 = 4.225V
Then 15/4.225 = 3.55 or 3 LEDs per string which doesn't exactly work out right.

Of course I realize that the two different voltages for the blue are probably messing up my calculations quite a bit and I don't have enough to do one of each on a string and still keep my ratios right. If I follow your logic in the rest of your post, I should be able to do the following:

15 strings of red, 4 strings of white, 12 strings of 430nM and 9 strings of 470nm which keeps it at 40 strings and 58% Red, 7% white and 35% blue which is about what I originally was shooting for.

Now for the resistor values for each string:
Red: 1.8V*8 per string = 14.4V
15V-14.4V = .6V
so R = .6/.02=30 ohms
White: 3.3V*4=13.2
15V-13.2V = 1.8V
R = 1.8/.02 = 90 ohms (3 x 30 ohm since I'm already using a 30 above)
430nm: 4.9V*3 = 14.7V
15V-14.7 = 0.3
R = .3/.02 = 15 ohms
470nm: 3.4V*4 = 13.6V
15V-13.6V = 1.4
R = 1.4/.02 = 70 ohms (and since I can't seem to find a 70ohm, use a 75 ohm)

So the end result would be 15 strings of red, each with a 30ohm resistor, 4 strings of white, each with 3, 30ohm resistors, 12 strings of 430nm blue, each with a 15 ohm resistor and 9 strings of 470 blue, each with a 75ohm resistor.

So am I on the right track?

16. ### SgtWookie Expert

Jul 17, 2007
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55% red, 660nm, 645nm at 1.8V@20mA
35% blue, 430nm at 4.9V@20mA, 470nm at 3.4V@20mA.
10% white, 3.3V@20mA.

Here you go:

11 strings consisting of: (2)430nmBLU, (1)470nmBLU, 91 Ohm resistor (88% efficient)
11 strings consisting of: (1)430nmBLU, (2)470nmBLU, (1)RED (alternate 660/645nm), 75 Ohm resistor (90% efficient)
7 strings consisting of: (3)WHT, (1)660nmRED, (1)645nmRED, 75 Ohm resistor (90% efficient)
11 strings consisting of: (4)660nmRED, (4)645nmRED, 30 Ohm resistor (96% efficient)

430nm: 33 (blue, 16.5%)
470nm: 33 (blue, 16.5%)
645nm: 56 (red, 28.2%)
660nm: 57 (red, 28.3%)
White: 21 (10.5%)
Total: 200 LEDs (100%)
91 Ohm resistors: 11
75 Ohm resistors: 18
30 Ohm resistors: 11

It's not exactly your ratios, but it's adjusted for efficiency. Those blue LEDs (particularly the 430nm) were efficiency killers; that's why 11 of the blue strings had a single red LED stuck in there.

You don't want to get closer than about 0.5v to the power supply's voltage, even if it is regulated. Variations in the regulation could cause significant overcurrent, which will kill LEDs quickly.

Last edited: Mar 18, 2009
17. ### flynnguy Thread Starter New Member

Mar 13, 2009
9
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Just so I know, the main problem with my original calculations was the 430nnM coming out to 14.7V?

I know what you mean about the 430nM blue's, they are also the most expensive of the bunch. I'm actually considering removing them or even just reducing them. But thanks, you've given me a lot of good information to play with.
-Chris

18. ### SgtWookie Expert

Jul 17, 2007
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Yes, you really don't want to run them that "close" to the power rail. Even a small supply voltage fluctuation would cause a large increase in current.

I whipped up a spreadsheet in Excel to come up with the LED combinations that I recommended. It's not pretty nor suitable for general use; just a tool to simplify the calculations. If the LED specs you've given me are good, and the tolerances of the LEDs are reasonably good, it should be OK.

Well, I'm no plant pathologist. Go with the frequencies that have been reported to work the best.

19. ### flynnguy Thread Starter New Member

Mar 13, 2009
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It's still a lot of theory at this point and aside from the inefficiency, they are also a lot more expensive than the other blue. (~\$55 vs ~\$19 for 36)

Suppose I decide to do less strings, can I just eliminate them and keep the same resistor/number values? If I understand this correctly, the extra current just won't get used because the LEDs aren't pulling it, right?

20. ### SgtWookie Expert

Jul 17, 2007
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Cost is almost always an issue.

You should plan on buying around 10% more LEDs than are actually required. This will save you time and money when some of them fail or are damaged during assembly; you won't have to place a 2nd order and pay more shipping/handling charges. Besides, I can virtually guarantee you that some of the LEDs will have a Vf that is significantly higher or lower than the typical.

If you are going to eliminate entire strings, that's fine - the current limiting resistors in the other strings will assure proper current flow through the individual strings.

If you are going to change the mix of LEDs in individual strings, the current limiting resistors will have to be recalculated.

For example, if you decided to eliminate the blue 430nm LEDs from your array, the first two strings would have to be re-examined and recalculated.