LED Array Help (Current regulation)

Discussion in 'The Projects Forum' started by 6FiRE9, Oct 9, 2009.

  1. 6FiRE9

    Thread Starter New Member

    Oct 5, 2009
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    So I'm new here and wish I would have found this site back when I was in school :) Anyways, I do have electronics training, but I've been out of school about 5-6 years and I'm very rusty.

    On to my project. Me and a friend, who also has basic electronics knowledge are trying to make a plant growing LED light. We have three sets of LED's that we plan on using and I was originally planning on using some LM317's with the voltage input coming from an old PC SMPS 12V rail.

    I have been working on the design, but keep hitting stumbling blocks.

    Say we had 5 of each LED to power.
    Blue Typ. Vf =3.4V Max current =300mA - 1W output
    UV Typ. Vf = 3.4V, Max current =300mA - 1W output
    Red Typ. Vf = 2.2V, Max current = 300mA - 1W output

    Now onto the "theory" that I think is in my head :)

    What if I set an LM317 to current regulating, and setup 3LED's in series under R1 with the ADJ reference coming from that string only.
    Then I take another parallel string with just two LED's in series and adjust the R2 on that line to compensate for the current needed. Would R2 not do it's job to control the current, as LED's are mainly concerned with the current??

    I have read the concerns on parallel circuits, but if R2 value is set properly and the current for both strings is even, wouldn't the regulator keep it that way?

    Love to hear your thoughts, and thanks ahead of time!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    When you use an LM317 as a current regulator, it drops a minimum of 3v across itself. It will also dissipate a good bit of power as heat. The more voltage it's dropping for a given current, the more power it will dissipate.
    Example: 3v @ 300mA = .9 Watts
    12v - (2x3.4v) = 5.2v; @ 300mA = 1.56 Watts
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    The LM317 is a voltage regulator when you connect R1 and R2 to the ADJ pin.
    But you don't want it as a voltage regulator, you want it as a current regulator where it uses only one resistor. The datasheet shows the resistor between the output and the ADJ pin and the regulated current output is the ADJ pin. For 300mA the resistor value should be (1.25V/300mA=) 4.17 ohms/1W.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    There are also other ways besides the LM317. 3V is a large chunk of voltage to loose powering up a chip.

    [​IMG]

    For 20ma this circuit will drop a minimum of 0.8V where it counts. The voltage goes up from there to match the current to 20ma to the LEDs. If the number of LEDs and transistors is substantial the dioes and resistor can be replaced with a LM317, which is a 1.25V regulator at the basic level.

    BTW, there are newer LEDs out there that use 350ma, or 700ma, or more. They are bright enough to replace house lamps, but are a bit more expensive. Costs are dropping fast though.
     
    Last edited: Oct 10, 2009
  5. hgmjr

    Moderator

    Jan 28, 2005
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    I like the efficiency of the approach of using a boost regulator such as the one shown in this link.

    hgmjr
     
  6. 6FiRE9

    Thread Starter New Member

    Oct 5, 2009
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    Thanks for the great info everyone. I do see the disadvantage of dropping 3V across the LM317 and the extra heat. That will also make 4-5 LM317 circuits to make this project come to life.

    One thing is that I would like to keep the circuit small, simple and inexpensive if possible. BillMarsden, I like the idea of constructing something out of transistors exc. but I'm worried about the power consumption and current.

    As I stated before these LED's are 300mA 1W and I need to power 15+ of them. The LED driver circuit looks like it's for 20mA LED's, and I'm not sure if it would work well in this application.

    Also the boost regulator requires a fair amount of additional circuitry, extra capacitors, diodes exc. Which I'm not completely against if it can run all of these LED's in one shot and not be too pricey.

    Any more suggestions and ideas would be welcomed.

    Cheers!
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    As with the 317, the transistors will disappate heat according to how much voltage they drop. Power transistors are cheap and the resistors are relatively low wattage since they drop very little voltage. You could use the approach in post #5 for efficiency, but I don't think it matters much.

    Wookie, you think the 317 by itself would still require the .1µF and 10µF filter caps on the output?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    No. When LM317's are used in current regulator mode (a single resistor from OUT to ADJ, current taken from ADJ) capacitors are not needed.
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    How about if they are used as regulators, but with out any resistors? As in a 1.25VDC regulator?
     
  10. 6FiRE9

    Thread Starter New Member

    Oct 5, 2009
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    But would that still regulate the current? Also would I have to worry about LED protection because of no current limiting resistor?
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    When they're used as voltage regulators, they require capacitors on the output; and if the input wire is more than a few inches long, one on the input as well.

    Remember, this goes along with the definitions of regulated current sources vs regulated voltage sources; current sources have very high impedance where voltage sources have very low impedance.
     
  12. 6FiRE9

    Thread Starter New Member

    Oct 5, 2009
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    One more question, lets say with the 12V input to the 317, this gives me only 9V out and I drop 1.25V across R1. That's 7.75V to the series of LED's, could I still properly drive 3x3.4V LED's since the current should still be regulated to 300mA, or will the current also drop because I don't have sufficient bias voltage?
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Ahh, where are you getting the 1.25v drop? Vref?
    An LM317 generally has a dropout voltage of around 1.7 from IN to OUT.
    Then there's the Vref from OUT to ADJ which is about 1.25v.
    That's how you get the 3v drop across the regulator when used as a current regulator. That is the minimum it will drop.

    So, with a 12v supply, you'll have 9v left. 9v/3.4v = 2.647 LEDs. Take the integer value, which is two 3.4v LEDs. If you tried to run three 3.4v 300mA LEDs in series on 9v, they would be very dim.
     
  14. 6FiRE9

    Thread Starter New Member

    Oct 5, 2009
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    My bad, I thought you guys meant that there was a 3V drop from in to out. Well that's good, but not ideal because now that means 8xLM317 circuits for 15 LED's. That's a bit excessive.

    Well I could use the negative voltage lines from the PSU for increased voltage. however I will need to check the amp ratings on those lines to make sure I still have 5A.

    Are there any chips that can do what the LM317 does with less voltage drop?

    Thanks again!
     
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