# Most Efficient way to drop voltage for LEDs

Discussion in 'The Projects Forum' started by jj_alukkas, Dec 8, 2011.

1. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
I need to light 3 x ½ W leds from a power supply which gives 28V. If I use the 3 leds in series, I have to use a 220Ω 2W dropping resistor in series on the line which uses more power than what the 3 LED's would use which is not worth using the LED's, so what would be the best possible and cheap way to light these led's without much loss?

2. ### bwack Active Member

Nov 15, 2011
109
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Hmm..Editted: my eyes are squinting today, you say three times 0.5W leds, i read 3.5W.
Anyway, the answer is to either use more of these leds, as many as it takes to get up to 28V total forward voltage, or use a buck regulator, which isn't at its best at such large difference but much better than using a resistor.

0.5W, lets say forward voltage is Vf = 3.1V .
Vf_total= 9.3V

28V / 3.1V = 9.0 , use 8 leds => Vf_total = 3.1*8 = 24.8V. Now you could use other less powerful leds if you wan't the same luminous output.

Last edited: Dec 8, 2011
3. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
No, you don't want them to add up to 28v exactly. You need to control the current, as the Vf of the LEDs will vary over temperature. The Vf might match at room temp, but as they heat up, the Vf decreases. This can lead to thermal runaway.

Joseph, you really need to give better specifications on your LEDs. Datasheet is most preferable. Otherwise, typical Vf @ If, and max If.
They ARE 500mW LEDs, correct?

4. ### bwack Active Member

Nov 15, 2011
109
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True, but I did back off one led from 9 to 8.

5. ### bwack Active Member

Nov 15, 2011
109
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Hi again I', pretty sure that thermal runaway is when you have some sort of thermal positive feedback, and typically for LEDs, diodes and BJTs you have that and Vf goes down, true, but the term thermal runaway I thought only was used for devices or strings connected in parallell, where one of them (if the thermal path is bad) will hog all the current (because of the lower Vf and hotter temp), and this will continue to do so because the difference in temp will increase more and more (runaway), but for a series of leds with a resistor I can't see how these will "runaway".

6. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
Yes, I think I must settle for ordinary hi intensity leds's instead of this, as 2 strings of 8 leds would be a lot better than wasting 2W for no purpose! Thank You

7. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
Sorry about that, I thought I could just give an outline of the situation. The led's on my tests ran fine at 3.1V @ 90mA, and I run them normally at these specs. Sorry, dont have a datasheet. And yes 500mW Leds

8. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
Even i do the same. I get the leds from the same batch, and always put atleast a small resistor even in case it doesnt need one and run them at small currents, like 14ma for normal hi intensity 25mA leds.

9. ### T.Jackson New Member

Nov 22, 2011
328
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All silicon technologies are NTC (Negative Temperature Coefficient) This means that as they begin to burn off electrons / produce heat, their internal dynamic resistance decreases. BJTs, MOSFETs, diodes, ICs, you name it. Thermal runaway is a condition when it is out the scope of being controlled, usually rendering the system inoperable.

10. ### T.Jackson New Member

Nov 22, 2011
328
14
A light bulb on the otherhand is PTC.

That's why if it is going to blow, then it usually does it when you first flick the switch.

11. ### bwack Active Member

Nov 15, 2011
109
10
Thanks. I can see now how thermal runaway may happen if the total Vf is too close to the input voltage, giving less or to little control over the current. Especially eight of these leds in series could give like 8x the change in Vf.

I (in a team) remember parallelling many mosfets in a design some years ago (the company wanted the Bill of Materials (BOM) as short as possible), and that they performed and balanced the heat very well (at 15A) because they tend to conduct less when they are hot (negative feedback), while bjt's conduct more when hot. Maybe i'm looking at this from the wrong angle... ? I can see how BJT's conduct more when Vf goes down, but in the mosfet, the n-channel created by the gate, will maybe conduct less, because heat makes it more likely for an electron to crash with the atoms (?)..

Last edited: Dec 8, 2011
12. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
Actually, MOSFETs have a PTC - one of the very few semiconductors that do. That's why it is so much easier to parallel them than bjt's/regular transistors.

13. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
You basically need a cheap buck switching regulator running as a current source. You could use something like an LM2575 set up for constant current. You could use national Semi's free software to design one. It selects all the parts for you and does the design. Since it runs CC mode, the Vf of the LEDs doesn't matter at all.
Here is a schematic of one:

• ###### LED.PDF
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Last edited: Dec 8, 2011
14. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
That was an interesting fact. I never noticed that, though I knew it would burn on usage.

15. ### jj_alukkas Thread Starter Well-Known Member

Jan 8, 2009
751
5
Thanks for this, it will certainly be useful for me in future projects, as in the current one, im looking to reduce the overall cost to make 8 units for my home use, so I chose to use smaller leds in more numbers to minimize the loss. But this one was certainly helpful! Thank You

16. ### sheldons Active Member

Oct 26, 2011
616
101
heres a way to drive leds(8 at a time,build as many units as your supply will drive(have built a few of these and no failures yet,been running for a few years now

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