Current balancing LED strings

Discussion in 'The Projects Forum' started by bwack, Nov 28, 2011.

  1. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    Hello.
    Question in short if you don't want to read all of the below: I have a constant current source of 2.8A. I want to drive two LEDs in parallell. I'd need another regulator to control the two LEDs individually, but thats expensive. Could I some how regulate a current-mirror below the leds to balance the current ?? Matched pair transistors can be used for equalling the current trough strings of leds, but is it possible to balance the leds, say 40/60, 70/30 ?

    [​IMG]

    I'm working on (another) DIY bicycle light using high power leds. Since this is my second attempt I've noticed that when cycling in the dark forrest trails, I have a quad LED light with two wide optics and two spot optics. This is used to give some spill near the bike but also good throw in the center (spots). What I've notice is that if I decide to "dip" the beam down, the narrow light is too bright. The problem is that your nightvision is worse if the light in front of you makes your pupils close...

    So here is the dillemma. I could have used three spots and one wide optics, but then I saw an article on the runaway problem of parallel led-strings with constant current drivers. To prevent it they use matched pair sink transistors configured as current mirrors to ensure that the current in one of the led stirngs doesn't run away and blows the leds..

    Here is my question, can we control the balance of a current mirror?
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Yes, but that is not how you handle LEDs. You program the current you want, independently, so if one chain fails the other continues as it was.

    Even a well designed current mirror will provide this isolation. You don't want percentages, you want hard numbers on the currents that won't vary.

    LEDs, 555s, Flashers, and Light Chasers
     
  3. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    Just so i don't misunderstand: Do you say that I could divide the current 30/70 using a current mirror, but I can't make it adjustable ? Why is it not the way ? I'm not worried about one string failing..
    Edit: Thanks btw :)
     
  4. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    You want to implement a scheme that is fundamentally flawed it is your choice, but LED lighting doesn't work that way. You could, of course, program a µC to feed a current to control current.

    Here is the point you are missing. LEDs are not linear. If I feed half the current to a LED it will not be half as bright.

    LEDs also have an optimal current, usually the max current they will take. You can go over this number, they will still last a long time (unless it is out of the ballpark), but their life span will be shortened. Since it is measured in 10's to 100's of thousands of hours this may be no big deal.

    As for a simple current mirror, here you go. The temperature stability will be adversely affected a little with mismatches, it works best if all the number are the same.

    [​IMG]

    http://forum.allaboutcircuits.com/picture.php?albumid=152&pictureid=1670

    Use R2 and R3 to tweak the percentages. Your constant current source is basically a waste, since you needed a constant voltage source, the current mirror (or whatever) is already a constant current source.
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    To my mind, a "candelabra" circuit as proposed by the OP, in which one current driver powers a current splitter like this adds complexity for little advantage, unless the original source is something like a current-mode SMPS, which may not readily give a divided output. You might struggle to get the voltage drop of such a "balancer" down to a much lower value than would be lost simply by adding a low value resistor in series with each LED.

    If it is feasible, the use of a individual current drivers for each LED is more logical, as apart from dealing better with LED failure, it will keep the minimum voltage drop lower, improving low battery performance.
     
    Last edited: Nov 28, 2011
  6. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    No, I still wan't to use a constant current regulator, a buck with good efficiency and wide input voltage range, you won't get any of that with linear regulation .. and I know that intensity isn't linear to current, and a "linear" change in light that we observe is actually logarithmic, so if I'd like to step the light down i'd need to halve it or at least do some steps on a logarithmic scale.. (i did this on my pwm on the previous light to get 15 "equal"(as observed) steps of intensity) ok i see now that would be actually too much regulation and it is flawed.

    Herer is the schematic suggested for protecting strings of leds connected to a buck CC driver..

    http://www.ledsmagazine.com/features/6/2/2 Edit: see fig 3 ! :)

    I see you mention stability, but it is not necessary to get perfect 50/50 split of current, just to protect them. I could always go down on the total current fed to the circuit.

    Thanks Bill.. as you see I think you where right about the missing point afterall.. :) ..
     
  7. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    The amount of waste described in the link is vastly overstated.

    If I have 24VDC and a long chain of LEDs, I would use a resistor. If the number of LEDs in the chain were designed for the 24VDC, efficiency losses would be much less than 10%, which compared to a SMPS is equivalent, but one cheap part vs. a expensive circuit.

    There are lot of myths around LEDs, it is important to work the math.

    Lets go through some math. Assume the LEDs in question are white, which puts them around 3.6V drop.

    The example given in the link shows 5 LEDs in series. This means they will drop 18V. You need to regulate the current with the 6V overhead, 350ma. This works down to a 17.14Ω resistor at approx 2W.

    Of course, you add another LED it improves. R = 6.857Ω, resistor power is less than a watt. Overall efficiency is extremely high for something like this.

    The major problem I have from article is the design is not just bad, but major flawed. If you want to do what they show you need two independent SMPS supplies, and not try to skimp on the regulators. That design is exactly what I was talking about for cascade failure.
     
    Last edited: Nov 28, 2011
  8. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    @bill: I don't agree. (I do agree about doing the maths). If you see the link, the input voltage varies alot, and therefore in a linear regulator so does the efficiency alot. Sure if you use a 24VDC power supply, then you can match the leds to it. Second I'm going to use this on bicycle, max 1-2hours.. A L-ion battery varies from 4.2V full to under 3V low. I've done all the calculation, what I'm asking is if it is possible to control a make a current mirror adjustable, but I see that my idea would be more complex than useful.

    Edit: sorry I see they say a fixed 24VDC. You are right.
     
  9. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    Hi again Bill. I think I may have missed some of your post above, maybe it was edited in later or I failed to see it the first time (im sorry). I agree with you on this, the example they show is bad, why not use 6 leds instead and a resistor., but still even if the example is bad, thermal runaway is a problem with constant current sources. So one may ask if one really needs those for such large quantities of leds.. ? (and the answer is yes if you need to install leds in a plane for example where the batteryvoltage can vary alot)..

    Anyway I think I'll use two drivers instead, one for the near and one for the far. I could have a cheap linear driver, preferable amc7135 (or make one myself) for the near (wide) beam, and another (not thought of yet) driver for the throw... When thinking of it is nice to cut the high beem when I meet people in the woods to prevent blinding. Ok I drop this balancing idea. Thanks for the advice.
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Good luck!
     
  11. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    A trick I have used for matching is to order maybe 20 pieces on 2N2222 or 2N3904 devices in plastic To-92 from someplace (they are really cheap). Then I set up a little test circuit to measure the VBE at 1 mA and look for matched pairs. I usually find at least 5 pairs that match to within a few mV. Then when you build the circuit, Krazy glue the flat sides together on the pairs that need to track thermally.
     
  12. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Thinking like a designer: I would stack the two LEDs in series and redesign the top current source for 1.4A. Problem solved. I would use a PWM converter for allowing the current to be adjusted with high efficiency. You could even use a part like LM2576 set up as a current source.
     
  13. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    Thanks bountyhunter. The idea was to get some load regulation (in realtime by the user) between the two emitters using one and only one (relative expensive) constant current source. Think of it as the balance on a stereo amplifier, where the user can in realtime balance to his/hers needs.

    I did do some calculations and simulations on the idea of balancing some load between the two parallel strings, using an opamp to control a npn in series with the second led to regulate the current in that led, but _at_ the range I was looking for ( 25/75 load share), the losses at this range is so great that I could just aswell have dropped this idea and used a second inexpensive linear reg. instead when taking the cost and efficiency into consideration. For just 50/50 split its just more complex and adding losses. I'm not conserned about runaway either, and if we put them in series, problem solved as you say :)
    Edit: btw thanks for the LM2576 btw, never seen it before.
     
    Last edited: Nov 30, 2011
  14. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    Put a 25% tint on the glass lens where the LEDs are too bright in the near field.
     
  15. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Run them in series but put a small shunt (with pot adjust) around one LED to be able to dim it a touch. Use a very small fixed shunt on the other LED to make sure the user-adjustable LED is always a shade brighter at full current.
     
  16. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    If you mean be able to dim out either LED: yes, there's a way to do it. Use an unbalanced current sink (NPNs) at the bottom of each LED with emitter resistors ratioed to get the current you want. A dual gang pot attenutaor could have center position no attenuation (both LEDs full bright), and going left or right would adjust the emitter R value in each "reference" string which mirors over to the LED current to attenutate one or the other LED going left or right.. It's not that complicated.

    tell me what your power source is (voltage rails available) and it is an easy design. Also need to know the V drops across the LEDs.
     
    Last edited: Nov 30, 2011
  17. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    I find it difficult to see it without a diagram. This is what I tried to calc and simulate on.

    [​IMG]
    (current source should say {0.1A, 0.7A, 1,5A, 3A} sorry for the error)

    The current source can be set to different modes by a button. Lets just look at it as a current source for now, linear or not... Set it to 3A to drive two Cree XM-L's (rated for 3A max) to see the worst power dissipation.

    U1 keeps V- and V+ equal, and the voltage divider R4 and R5 makes a ratio between Vsense1 and 2 controlled by U1... If R4 is low then Vsense1 and 2 are equal. If the user adjust R4 to say 100k, then there is a 75/25 ratio I think... (I got about 1A in ID2 and 2A in ID1). The problem here is range. Sure we can balance to get equal currents, but to see a noticeable difference in light, I'll have to halve the current.. What makes this problematic is the rise and fall in the forward voltages Vf for the leds. What happens is that the currents ID1 increase and ID2 decrease, like we want.
    VD1+Vsense1 will rise, VD2+Vsense2 will fall but to compensate VCE,Q1 must rise, and I think you will have to dissipate 0.5 - 1W in Q1 just to do this, its not worth it.

    I'm planning to make a two XM-L light using two linear drivers and four parallell L-ion cells (protected 4P1S). Most of the runtime, the voltage will be between 3.7V and 3V, starting at 4.2V. Should give good average efficiency ?
    Edit: here is the datasheet for XM-L (neutral white) http://www.cree.com/products/xlamp_xml.asp if you want to have a look :) I appreciate all help from all of you.
     
    Last edited: Dec 1, 2011
  18. bwack

    Thread Starter Active Member

    Nov 15, 2011
    109
    10
    @bountyhunter Quick question, would the gang pot need to be really beefy to handle the dissipation ?
     
  19. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    The transistors certainly would be.

    Cost and complexity are reaching the point of a smaller buck puck for each LED.
     
  20. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    If the battery is two Li cells, the design is way too inefficient. Too much power diss in the transistors. I would use a cheap buck sw reg for each LED (set up as variable current source) on the top and have a dual pot controlling brightness.

    You could also set up some free running 555 timers with variable pulse width and use them to manually adjust the base drive to the transistors. That would reduce power dissipation and let you control brightness. Run them at 25 kHz so they don't whine. That would be a very cheap solution.
     
Loading...