# Alternate to Resistors in 12v LED arrays?

Discussion in 'General Electronics Chat' started by kcarring, Sep 6, 2011.

1. ### kcarring Thread Starter Member

Jan 22, 2011
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I have been reading through a good many great articles here, trying to sort out the current limiting situation with series LEDs, on 12v. My main objective is to have very powerful homemade LED boards running on solar charged batteries. There is no concern about them running "while charging takes place"... due to the fact that the solar system would be adding very little charge on a day where the building actually needed this light on (the building also has natural light). They could be shut off when a genset was used.

I've read several of Bill's great articles on using voltage regulation, and I have read a number of threads here.

MY main objective is to build arrays of 24 8mm LEDs, initially bright whites and later warm whites. The spec on the bright whites is 3.2v / 100mA / 110,000 MCD ea.

The solar system bank varies from 11.5v to 12.6v.

I was initially thinking I could use an LM350 to voltage regulate to 8v, and calculate resistors from there.

When I do the calculations, I am amazed at how much energy is wasted on heat! WOW. Then again... incandescant bulbs are LIKE HEATERS, so I guess it's not so bad afterall.

Is it possible to use tiny light bulbs in series instead of resistors, or does anyone know of any clever "use" for this energy?

If not, please take a stab at this theory:

Could I settle for arrays of 12 - voltage regulate to 7.2v, use 8.2 ohm 1/4 watt resistors instead, and

together, all resistors dissipate 492 mW
together, the diodes dissipate 3840 mW
total power dissipated by the array is 4332 mW
the array draws current of 600 mA from the source.

Is it possible to get that precise?

Does dropping the voltage that much lose a lot of energy within itself?

I kind of like this method too in that the lamps would work on two lithium cells as well. They are getting cheap now, you can get a 4000 mAh 3.6v battery shipping included for \$3.

Thanks

Apr 30, 2011
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Candela ratings are not very useful without considering the radiation angle. Convert to lumens to make comparisons to existing lighting.

3. ### Crispin Member

Jul 4, 2011
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Why not use the raw 12V and 4 LEDs in series. You'd be slightly under the 3.2V (3.0V each) though.
Or, use 3 in series with a 24ohm resistor. You'd have 8 strings of LEDs, 8 resistors.

4. ### iONic AAC Fanatic!

Nov 16, 2007
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Much more information is needed on your solar system. It seems odd or unconventional that your system only fluctuates between 11.5V and 12.6V throughout the entire day.
The power capability of the system would also be helpful as it would give us an idea on how to approach battery charging and voltage regulation.

You cannot use "tiny" light bulbs in series to limit the current as you need a precise resistance to regulate the current through the LED's. If you could find the right bulb with the right resistance you would probably use more power in the long run. Besides the light you get from this tiny light bulb will be insignificant and cost you way more than a resistor.

You won't be able to charge your batteries if you regulate to 7.2V. You would need to regulate to 8.2V to 8.4V to fully charge your batteries and regulate the current as well.
Your series resistance would need to be 15 - 20 ohms.

Did you crunch the numbers with the LED array current draw and the capacity of the batteries you mentioned. 4000mAH/600mA = 6.6hrs absolute max.

5. ### John P AAC Fanatic!

Oct 14, 2008
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I came in prepared to say you need some kind of regulator using an inductor, but KJ6EAD beat me to it. There are actually special LED control chips available for this job.

6. ### Crispin Member

Jul 4, 2011
88
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I think everyone is being polite and not shooting down my idea
Genuinely curious - is there anything wrong with the method? Accepted, there will be some fluctuation depending on battery voltage but given the details provided, I would not have thought it would be much.

- Curious

7. ### iONic AAC Fanatic!

Nov 16, 2007
1,420
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With no resistor? With those tight specs your LED's will be Dimming with the sun. So when a cloud rolls by the ambient light is reduce and so is the light from the LED's

Yes, but with a volt of swing (and we still don't know how tho OP's solar panel is managing that), what value of resistor should be used??

Not really shooting it down, well maybe the first of the two suggestions, just don't have enough data to make a good recommendation.

Apr 30, 2011
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There are lots of ways to limit the current to LEDs. A varying input voltage such as you have here pushes the logical choice towards active current regulation rather than a resistive approach. Operating from solar cells puts a premium on the most efficient, not the simplest or least expensive regulating method so that tends to rule out linear regulation schemes using transistors, op amps or adjustable regulators. With a large number of LEDs and a relatively low supply voltage, using a boost topology running 3 strings of 8 LEDs might be best.

9. ### Wendy Moderator

Mar 24, 2008
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The fact is, you must have an external current limit device. The absolute simplest way to do that is a resistor. Even a Constant Current SMPS involves using a sensing resistor in series with the LEDs, and while it tends to be much more efficient this is offset by the parts count. With a solar power supply this is probably the best way, given the electricity is in much shorter supply.

10. ### kcarring Thread Starter Member

Jan 22, 2011
38
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Thankyou for all the great replies!

I don't think I made myself clear --- I confused a couple of readers. It is a conventional solar system with lead acid storage... so... I mean that on a good day, the bank sits above 12.0v On the end of several "bad" sun days, the bank may drop as low as 11.5v - but that's about as low as I usually let it get - at that point - with no sun in the forecast - the genset may be fired up ... but when that happens, it's usualy not dark, and therefore the lamps would not need to be on... I could live with turning them off during genset charging. As for periods when solar charging is taking place, well... it's bright out, and therefore the indoor lamps are not needed. So when I say that the bank sits between 11.5 and 12.6 volts, that is to say the standing voltage of the bank, not including charging voltages. I confused matters by talking about lithium ion batteries - those are not the solar bank, those are just for portability.

I gave the number 3.2v.

That's a minimum. The "typical" rating for these LEDs is actually 3.6v, and when you run that through the calculator, things look a lot less wasteful.

Here's what that looks like (using 12v source, 100ma draw, 3.6v LEDs):

Solution 0: 3 x 8 array uses 24 LEDs exactly
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 12 ohms

The wizard says: In solution 0:
each 12 ohm resistor dissipates 120 mW
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 960 mW
together, the diodes dissipate 8640 mW
total power dissipated by the array is 9600 mW
the array draws current of 800 mA from the source.

With this calculation, less than a watt is shown wasted to heating resistors.

Initially, I had looked at this (using 12v source, 100ma draw, 3.2v LEDs):

Solution 0: 3 x 8 array uses 24 LEDs exactly
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 27 ohms

The wizard says:
each 27 ohm resistor dissipates 270 mW
the wizard thinks ½W resistors are needed for your application
together, all resistors dissipate 2160 mW
together, the diodes dissipate 7680 mW
total power dissipated by the array is 9840 mW
the array draws current of 800 mA from the source.

As shown here, over 2 watts in resistor "heat".

I suppose this is acceptable, and I am only left wondering:

What happens when the bank is at 12.6v?
What happens when the bank is at 11.5v?

Obviously the 11.5v isn't going to burn out the LEDs, and they will just be a bit dimmer...?

Would I be better off to design for 12.6v?

That would give me this (using 12.6v source, 100ma draw, 3.6v LEDs):

Solution 0: 3 x 8 array uses 24 LEDs exactly
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms
+----|>|----|>|----|>|---/\/\/----+ R = 33 ohms

The wizard says: In solution 0:
each 33 ohm resistor dissipates 330 mW
the wizard thinks 1W resistors are needed for your application
together, all resistors dissipate 2640 mW
together, the diodes dissipate 7680 mW
total power dissipated by the array is 10320 mW
the array draws current of 800 mA from the source.

And again, I'm back to over 2 watts of heat, or 20% wasted energy.

Maybe I am just way overanalyzing this? I guess the main thing is not burning out \$100 worth of LEDs prematurely, but I must admit, it'd also be nice to have the same light available under high and low potential of the bank. And if I seem overly concerned about a watt or two... well... solar energy kind of makes a person "at least attempt" to be power conscience, in every possible way, that's all.

I am reading Bills articles a second time through, they are very interesting and seem to elude to methods to have a safe working voltage for the lamps, and yet constant light, regardless of the standing voltage of the bank. What concerns me is the voltage drop of an LM317 or LM350... if the bank is at 11.5v and there is a 3v voltage drop... do I need to design the circuit for 8v, or so?

Thanks Everyone!
Sorry if I'm confusing!!!

Last edited: Sep 6, 2011

Apr 30, 2011
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12. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
A constant current source compensates for input voltage, a switching mode power supply even more so. It will feed the LEDs the same current no matter what the power supply voltage is.

Case in point, a 3.6V LED at 20ma (this is just an example) uses .072W. If a SMPS regulator is 80% efficient (a reasonable number) then you will use 0.09W. If the source power supply is 12VDC then the current from the power supply will be 7.5ma. An added bonus is there will be much less heat generated than using a resistor (a direct consequence of efficiency).

Last edited: Sep 8, 2011
13. ### kcarring Thread Starter Member

Jan 22, 2011
38
0
Thanks Bill, I think you are leading me where I want to go. Of course there is a lot of ways to do anything, but I think this method could be particularly educational and interesting.

I found this document:

http://cache.national.com/ds/LM/LM1577.pdf

I will read up on this, and look into it more. Thanks again.

14. ### colinb Active Member

Jun 15, 2011
351
35
The answer is no to using light bulbs as if they were resistors. This is because the tungsten filament has a positive temperature coefficient of resistance and a much higher resistance when it is fully on and hot than when it is cold. This means it would not adhere to Ohm's law as you would require for your LED current limiting calculations.

As other posters have mentioned, a switching regulator is ideal. A constant-current regulator for the series LED string is the optimal type, since LED forward voltage is then inconsequential. However, a constant-voltage switching regulator would work too; you would just set its output to the sum of the LED's Vf and then you don't need any series resistor to reach appropriate string voltage.

Apr 30, 2011
1,425
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The 1577 is an obsolete product. National has an online calculator application for designing LED drivers. Using a voltage regulator, especially without any limiting resistor, is not ideal either since the dynamic resistance of the LEDs exhibits a negative temperature coefficient effect. As the LEDs heat up, the current through them increases which raises the temperature, etc.

Last edited: Sep 8, 2011
16. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
LEDs do not typically run away thermally, they exhibit some of the effects you describe, but not to the degree you describe. Do the math, I have. The more voltage dropped across the limiting resistor the more likely the effect will be negligible to non-existent.

I suspect people who try to limit current with voltage are the ones who have run into this. I've stated this many times, you do not try to run an LED without a limiting resistor or equivalent. It is a beginners mistake.

With a constant current source it doesn't exist at all.

17. ### iONic AAC Fanatic!

Nov 16, 2007
1,420
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As Bill stated, trying to regulate current through an LED by the adjustment of the voltage is not the way to go as the heating of the LED will cause an increase in current in the LED's (An unwanted side-effect that can lead to premature LED death over the long haul).

18. ### kcarring Thread Starter Member

Jan 22, 2011
38
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Once again, thanks for the input and clarification. Getting back to Bill's blog article, referring to the current source diagram...

In the case of the LEDs I have (8mm Bright White Strawhats 3.6v typ. / 100ma) , and, given the battery bank drops to possibly (extreme case) 11.0v, AND given the 3v potential drop of the LM317T, should I then design the series LEDs config. for 8v? To clarify my ignorance here... if I have (3) LEDs in series, then 3.6v x 3 = 10.8v. On a bad day, if the battery bank was as low as 11v, and I lose 3v across the LM317T... then I'd have 8 to work with, no? 2 LEDs in series would require 7.2v. Then in the simple case of using one LM317T for just (2) LEDs in series, I'd design for 200mA current?

I suppose if my thinking is close... then I'd extend the design to accomodate the current potential of the LM317T, what is it, about 1a? So I might have 4 parallel legs of (2) series LEDs, and design for 800mA?

I'm only trying to understand how this works... I'm sorry if I appear to lack intelligence here!

19. ### Wendy Moderator

Mar 24, 2008
20,766
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This would work, but it would be hot! Really hot, as in boiling water. It is an analog regulator. I would use it to test concepts out, but in the long term I would continue to look for a SMPS design.

I often recommend transistors for these kind of jobs where efficiency doesn't matter. There are few resistor designs that a designed for heatsinking, but they tend to be expensive. There are many more transistor designs meant for heat sinking, they are controllable with external devices, and tend to be much, much cheaper. They also tend to run very hot, but can be designed for better heat dissipation. For 1A to 4A designs that is the way I'd go. Again, this would be for trial setups, if you decide this is the permanent way to go the designs are quite robust.

We have had several threads where people are trying to design an economical SMPS current regulator. The commercial models tend to run \$10-20. You can probably build one for \$5-\$10 or so. Parts count is the major bugaboo IMO. One of the other guys, Tom66, has a particularly long thread where this is discussed.

As you have noticed, there are several LED driver ICs out there. IMO this is a very good option.