700mA LEDs in CCDriver

Discussion in 'The Projects Forum' started by rusk1y, Feb 17, 2013.

  1. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    Hi everyone,
    I built an aquarium canopy with LED lighting driven by 700mA Constant Current MeanWell LED Drivers. The LEDs i bought from ebay (i know they are cheap). They are rated at typical 3.8 Volts each bulb and mine only run about 3.6 on all of them. and at their rated 700mA.
    My MeanWell Driver gives out 700mA plus/minus 35mA(5%). I have two of them running all the same white bulbs. 10 in series on each. The drivers are connected to a mechanical timer and this is the second time the bulb burned out. They haven't been running for two months even (kinda sucks because everyone claims LED's can run for years). They are definitely not over heating because they are thermally compounded to three huge heat sinks (1.5" thick, 7" wide, 10" long) and also have two fans pushing air through the canopy with holes for air to escape on the opposite side. The wires i used to solder the LEDs are thick as well so i dont think there is any resistance causing heat in the setup.

    I'm pretty sure that it's just crappy LEDs (made in HK, boo!) but is it possible something else could be causing them to fail? any suggestions?

    Also, does anyone recommend any LEDs that are reputable and wont cost me an arm and a leg which fit my 700mA driver setup?

    The only spec sheet that came with the LEDs:
    [​IMG]

    The spec sheet of the CCDrivers (LPC-35-700):
    http://www.meanwell.com/search/LPC-35/default.htm
     
  2. #12

    Expert

    Nov 30, 2010
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    Everything you say seems valid, as in, "not stupid or even wrong".
    I'm going to read spec sheets for a while and append to this entry.
    Somebody else know where to get good quality 700 ma LEDs?

    Meanwell...have you measured the actual current? Did you know you don't have to run LEDs at their maximum current?
    (They last longer if you don't.)

    Here are some choices and they seem brighter than the ones you are used too.
    http://www.mouser.com/Optoelectroni...0wuo1Z1yzvdjcZ1z0wtjhZ1z0wumsZ1yzrybyZ1yzry3q
     
    Last edited: Feb 17, 2013
  3. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    Yes, i had a whole previous thread going on about having trouble with my measurements only to find out it was just a low battery in my DMultimeter. They dont go over 730mA. The Drivers i have are at 5% rated accuracy.

    I do know that now and i really wish i had thought about that before ordering my drivers. I will be educated on my next build i guess. But none the less, someone on this forum told me they ran their bulbs at the required 700mA on CC Drivers and they have been running smooth for a year+ now. Which is why i thought it unusual for mine to burn out in under a month (By the way, a second one went out this morning. I resoldered a bypass over the broken bulb hoping that was the end of that the first time, and here we are again :( LOL.

    These choices are pretty steep in price for me. Mouser is usually good with selection and pricing but per bulb is way too expensive it seems. Maybe i'm wrong too. idk.
     
  4. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    I emailed with the seller of the bulbs i currently have. I hope they send me some replacements for free. that would be awesome! We'll see if they send them out or not. Have my fingers crossed.
     
  5. #12

    Expert

    Nov 30, 2010
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    While you're figuring, figure 260 lumens per LED compared to 165 lumens for your first batch and adjust the price accordingly. You could get 10 cheap LEDs worth of lumens from 6.3 Cree brand LEDs.

    You could also put a 47 ohm, 1/2 watt resistor in parallel with each LED to run the cheap ones at 653 ma instead of 730 ma and get them to survive longer. The numbers are a little different with the Cree LEDs but the result will be very similar.
     
  6. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    So the seller replied. They are sending me out a few more free replacement bulbs. I might upgrade to CREE later if this reoccurs again too often or i run out of bulbs. :)
    Wouldn't adding a resistor to the DC side of a Constant Current LED Driver mess around with it's ability to provide the correct current? Wont it throw it off?
     
  7. #12

    Expert

    Nov 30, 2010
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    The constant current driver is going to provide constant current, period. The resistor simply routes some of the current around the LED.

    Example: Your LED has a forward voltage drop of 3.6 volts. It will change very little with a 10% change in current. A 47 ohm resistor in parallel with the led allows 3.6/47 amps (.077A) to flow without going through the LED. The remaining current of the 730 ma will go through the LED. 730ma-77ma = 653ma. The LED is now running at 93% of its rated maximum instead of 104% of its rated maximum current.
     
  8. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    I'm not sure how you got those calculations in the example. My light is using 10 LEDs wired in series not parallel. If the voltage output on the driver says it can handle from 9v-48v, how does the driver know how much voltage to supply? Is that just based on how many LEDs are connected in series?

    Anyways... lets say i have the 9v-48v source. I'm doing 10 LEDs wired series, What type of resistor would you suggest (ohms & wattage) and explain a little to me how you come up with the suggested resistor (show me the math so i understand it). I'll appreciate it thanks :) I'm mostly unsure of what number to plug into the equation first. How do i know how much voltage drop i need if my driver supplies that wide range of Volts?
     
  9. #12

    Expert

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    The voltage is the 3.6 volts across each LED as reported by you in post #1, line 2.
    The fact that the LEDs are in series does not mean you are not allowed to put a resistor in parallel with each LED.
    The calculation came from Ohm's Law, V=IR
    The CURRENT supply does not decide a voltage. It decides a CURRENT. The LEDs decide how much voltage they use up to pass the current. That voltage is 3.6 volts each as reported by you in post #1, line 2.
    Post #5 says 47 ohms, 1/2 watt. That is the size of resistor you might connect in parallel with each LED.
    3.6 volts divided by 47 ohms, according to Ohm's Law, arrives at .0766 amps going through the resistor.

    Now, if you still believe you can not put a resistor on each LED, you can add the voltages together for 10 LEDs. 3.6 volts times 10 = 36 volts. 36 volts divided by 470 ohms equals .0766 amps. The first problem with this is that if you have an LED die and short it out to get the rest of them running, the current through the 470 ohm resistor will change. The second problem is that the wattage of the 470 ohm resistor will be 2.76 watts. You would have to buy a 470 ohm resistor of at least 5.5 watts instead of using (10) 8 cent resistors.
    The wattage is calculated with Watt's Law: P=IE
     
  10. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    so to understand how this is drawn out you are suggesting this,

    Res.--LED--Res.--LED--Res.--LED etc.?
     
  11. #12

    Expert

    Nov 30, 2010
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    I'm suggesting one resistor is to be put in parallel with each LED.
     
  12. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    Ohh... thanks for the drawing. i couldnt understand because i never seen that done before. im a noob sorry.
    So then wouldnt the voltage drop in ohms law equation be what your trying to get rid of and not all 3.6 volts?
     
    Last edited: Feb 19, 2013
  13. #12

    Expert

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    The parallel resistors do not get rid of any voltage. They use the existing voltage of each LED to divert some current around it.
     
  14. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    What about how you measured the wattage used. I am told I*I*R gives you the wattage resistor dissipates. When supplying current value why do you use .075A and not say .735-~.075=.66Amps? I know your answer is right because the result makes sense, just want to understand what current they equation needs. So its not the total current going through the resistor its the current the resistor will not let through that is applied to the equation?
     
  15. #12

    Expert

    Nov 30, 2010
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    Right.

    730 ma is coming out of the current driver. It will go through the circuit even if the current supplier has to apply 48 volts to shove 730 ma through the circuit. All of the 730 ma can go through the LEDs or it can find an alternate path. Adding a resistor provides an alternate path.

    Each LED has a breakover voltage of 3.6V. That 3.6 volts will be pretty stable at 90% or 110% of the intended current. 3.6V/47 ohms is 76.6 ma. P=IE so power in the resistor is .0766A times 3.6V and that is .2757 watts.

    It is a rule of design that you always make the resistor twice the size necessary to dissipate the heat. It makes them last a lot longer and avoids getting them so hot that the solder connections eventually crack. .2757 watts times 2 = .55 watts. Buy resistors rated at 1/2 watt or bigger. Either that or calculate a resistor that will stay under .250 watts if you want to adhere to the rule strictly.
    P=I*I*R
    P=E*E/R
    E*E/P =R
    3.6*3.6/.250 = 51.84 ohms.
    You could use a 51 ohm resistor or a 56 ohm resistor.
    A 51 ohm resistor will divert 3.6/51 amps (.0706A)
    A 56 ohm resistor will divert 3.6/56 amps (.0643A)

    The point of this exercise is to divert a little bit of the current so the LED will not be pushed past its maximum rating.

    Let's try 3.6V/30ma to find the resistor that will set the LED current at exactly the maximum...120 ohms.

    My opinion is that you should drive LEDs below their maximum. Thus I recommended 47 ohms. Any resistance value less than 120 ohms will keep the LEDs under maximum rated current. Half watt resistors cost 8 cents. You can provide for a much longer LED life for less than a dollar.
     
    Last edited: Feb 19, 2013
  16. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    This got a little confusing. How come a 120 ohm resistor only resists or diverts 30mA and a resistor with a smaller ohm rating at 47 diverts / resists more keeping it at 90% of 700mA?

    Also, is wattage calculated a different way when the resistor is in series with the LED?
     
  17. #12

    Expert

    Nov 30, 2010
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    A 47 ohm resistor does not resist current flow more than a 120 ohm resistor.
    It diverts more current flow that a 120 ohm resistor (in this example).

    Wattage is always calculated by using the voltage across something, the current through that something, and the resistance of that something. No other something is involved, only the something you are calculating the wattage for. If a resistor was in series with a bunch of somethings, the voltage across the resistor, the current through the resistor, and the resistance of the resistor would be used to calculate the heat wattage that is being inflicted on that resistor.
     
  18. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    So i just bought a set of 47 ohm 1/2 watt resistors. I am about to solder them in. How can i check for validity that the current going through the LED will be below 700mA with my multimeter (where would i connect the probes?), or will it not show it?

    I'm gonna get started, let me know meanwhile. :) Thanks!
     
  19. #12

    Expert

    Nov 30, 2010
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    To measure the current through an LED, you have to open one connection at the LED and put the amp meter in series with the LED.
     
  20. rusk1y

    Thread Starter Member

    Jan 9, 2013
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    I'll try that out tomorrow. Thank you for your help!
     
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