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series/parallel LED circuit voltage regulated.

huntxtrm

Feb 24, 2010
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Feb 24, 2010
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And also, 66 of them lm 317 is gonna be hard to house up in a console. I cant get away with less of them running more leds in parallel. that's why i was wanting to do a series parallel. But if it won't work, then it won't work, and I'll have to try something else. Input?
 

55pilot

Feb 23, 2010
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The problem with the two transistor current limiting circuit is that it is dependent on the current setting transistor's Vbe (Base to emitter voltage). Vbe not only changes from transistor to transistor, it changes a lot due to temperature. You can easily see a 1.5x increase in LED current over temperature. It may not be a big deal for some, but a deal killer for others.

The second issue is heat dissipation. If you are running 30mA and dropping 2V through the current limiting transistor, you are dropping 60mW, which is not a big deal. But if you are running 2 LEDs straight from "12V" (which can go as high as 14V continuously), you could be dropping as much as 7.5V or 220mW. A TO-220 transistor can handle that. A TO-92 will have a hard time and may fail. Using a TO-220 increases cost and physical size.

The problem with using small current limiting resistors is that you can see huge current differences. If you take 3 LEDs that nominally drop 3.4V, put them in series (10.2V), put a 27 ohm series resistor and drive it with 11V, your current will go up to over 50mA when the LEDs heat up and start dropping 3.2V.

The problem with using large current limiting resistors is that you drop a lot of voltage and power in the resistor, but you get good current control. In the above example, if you put only two LEDs and a 150 ohm resistor, your current will only go up from 28mA to 30.6mA. However you putting 0.14W in the resistor which will heat things up and now you need 50% more strings which represents more wiring and 50% more current out of your power supply.

What is the solution? There is no one solution that solves everything. You have to look at all these different issues and make compromises that fit your situation. By being consciously aware of the issues you are compromising on, you tend to arrive at a more optimal solution with fewer surprises.

---55p
 

huntxtrm

Feb 24, 2010
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Wow! way over my head. So your saying lm 317 won't handle it. 11 series circuts, 3 leds each?
 

Mitchekj

Jan 24, 2010
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I thought you were talking about 99 LEDs per side (33 parallel strings, 3 LEDs in series) for a total of 66 parallel strings, 3 LEDs in series each (198 total LEDs.) That's where the 66 LM317s came from... one on each string of 3 LEDs.

Pilot is talking about that all-discrete method I mentioned earlier, which would basically be two transistors and two resistors (instead of the LM317), and he brings up good points. Those points (and because of complexity, part count, etc.) are why I didn't draw it up for you and recommend you use it. :) The LM317 current regulating method sounded, to me, to be the most feasible method of doing this... but has its drawbacks, too (namely the voltage overhead issue.) The good side is that you'd have a constant current. Bad side is two components per LED string, and may have slight issues when the battery dips below 12V.

The adjust pins would be tied between the first LEDs' anodes, and the 62 ohm resistors... see the schematic I posted (dashboardLEDs.pdf)... that only shows the first 5 LED strings. Your application would have 66, not 5, and all white LEDs.
 

55pilot

Feb 23, 2010
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For your problem, I would recommend the KISS principle (Keep It Simple). Stick 2 LEDs in series with a 200 ohm 1/2W resistor (preferably 1W) and run from the battery voltage through a switch and a fuse.

When battery voltage is 12.5V (fully charged, engine off), and the LEDs are hot, you are looking at 31mA. At the other extreme, when the battery is down to 11V and the LEDs are relatively cool, your current will be down to 22mA.

You can just run a power wire all the way out. At regular intervals, strip the insulation and solder a resistor. Then shrink wrap over the exposed part of the resistor & wire and move on to the next one. You will end up with a long wire with resistors sticking out at regular intervals. Attach two LEDs in series to the resistors. Take a similar ground wire, strip and solder the other end of the second LED, shrink wrap, and repeat the process. You will end up with a string of 50 pairs of LEDs in parallel. You are looking at about 1.2A current in the leg (25mA typical * 50 pairs), so you can get away with a 24 gauge wire, though I would prefer 22 gauge. The work sequence I have described will work out only if you solder/shrink-wrap in the proper sequence. So think this through before starting.

In this setup, you are wasting about 50% of the energy. If you were to use 3 LED strings and a control circuit for each string, you will bring the efficiency up to 66% and the power consumption down to 1.6A for the whole system versus 2.5A for this setup, but the whole thing will get much more expensive and complicated.

The reason I am suggesting 1W resistors is because if you have a problem and the non-heat-shrink side of the resistor gets shorted to the boat, you will see about 0.75W through the resistor. A 1W resistor will survive that. A 1/2W resistor will blow.

Good luck.

---55p

Edited to fix my stupid math mistake.
 
Last edited:

Mitchekj

Jan 24, 2010
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I'm tending to agree w/ Pilot here... simplicity may be the way to go. Though, I see the whole system (198 LEDs) only drawing about 3A... that's w/ 2 LEDs in series each leg, 99 parallel legs, at 30mA each.
 

55pilot

Feb 23, 2010
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I'm tending to agree w/ Pilot here... simplicity may be the way to go. Though, I see the whole system (198 LEDs) only drawing about 3A... that's w/ 2 LEDs in series each leg, 99 parallel legs, at 30mA each.
You are right. That is the last time I am posting something while on a 2 hour conference call. Next time I will stick to doodling!

---55p
 

abominableman

Mar 19, 2010
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Mar 19, 2010
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Hi everyone,

Check out the pdf attachment. Looks like a perfect, simple solution, but I do not know how to calculate the values for the resistors.

I'm also building an array of LEDs - 8 rows of 6 - powered by a 24V DC power supply.
I need 20mA through each leg
LED Forward voltage = 3.2V

Can anyone figure out the resistor values for me and for huntxtrm?

Thanks!!!
Dan

I found the diagram at the on Semiconductor website (can't post links yet) Here's what they say:
When the circuit operates properly and all the LEDs are
running, the three sense resistors have about 1.25 V across
them, which turns the transistor switches ‘on’. This connects
all three sense resistors back to the Vadj pin allowing the
proper current to go through each leg. If one string opens up,
the sense resistor for that leg won’t have any voltage across
it, turning ‘off’ the transistor and disconnecting its sense
resistor from the Vadj pin. Therefore, the other two LED
strings are unaffected by the fault. This same scheme can be
expanded to accommodate as many LED strings as needed.
 

Attachments

  • LM317_with_MPS2222.pdf
    53.7 KB · Views: 320

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
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Jan 21, 2010
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25,510
Rsense is determined by ohms law to drop 1.25V at the required LED current.

I would pick something like 4k7 for R4, 5, and 6 (etc). I doubt that it's particularly critical. It should be at least ten times Rsense, but that won't be an issue unless you want really small LED currents (fractions of a mA)

Note that for this circuit to work, all the strings must have the same number of LEDs in them.


edit: for further discussion of this solution, see here.
 
Last edited:

niftynev

Mar 20, 2010
22
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Mar 20, 2010
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12v led current limiting

You are correct in having just 3 leds in each string on 12v. At about 3.4v across each led, that leaves 12 - (3 x 3.4) = 1.8v spare. That's very little wastage of the available voltage. You'll save yourself a small fortune just by putting a 90 ohm resistor in series with each 3 led string, & not worrying about a 317 chip for every one! The 90 ohm resistor will happily drop that last 1.8 v for you, & it will be dissipating (wasting) just a tiny bit of power in the process.

r = v / i
1.8 / .02 = 90 ohms

I have to agree that you must keep all salt water away from the leds, otherwise they'll become leds without legs REAL quick. Probably a few kilos of neutral cure silastic will be required to seal the system.
 
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