Custom LED light bulb and a sanity check

Discussion in 'The Projects Forum' started by rggreter, May 6, 2014.

  1. rggreter

    Thread Starter New Member

    May 6, 2014
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    I have a couple of rather unique applications that I am attempting to develop LED light bulbs for and I'm hoping that someone here might be able to help guide me in the right direction.

    With the phase out of incandescent light bulbs quite a few historic preservation groups are finding that the light bulbs used in their old equipment are simply no longer available. I've had some sample bulbs produced by a manufacturer in China and they failed miserably (sometimes explosively) when we tried them out. The power source for the application I am working on is a 1000W 34V DC steam powered dynamo with a 30 A output. The dyamo feeds into a breaker box which feeds several circuits. We run about 15 or so 15W 34V DC light bulbs (0.5 A each) on one of those circuits. I am looking to replace those bulbs with LED and this is where I've hit a road block. The Chinese manufacturer upgraded some of the components of the driver (40V for the IC, 50V for the capacitor) but as I said they simply didn't last. The dynamo speed (and hence voltage, etc) can swing pretty wildly and when we hit 42V the bulb failed pretty spectacularly.

    I have a number of components for the bulb and am going to try to figure this out on my own but I'm going in circles at this point. I've identified a driver (NSI50350ADT4G from ON Semiconductor http://www.onsemi.com/pub_link/Collateral/NSI50350AD-D.PDF) that I think will work just fine for this application and I've mocked up a bulb but the LEDs burned out as soon as I connected the power supply. I have a benchtop power supply (34-40V DC, 350W, 9.7A) that I use to test the mock-ups. The bulbs themselves are made of 8 separate boards with six 3528 SMD LEDs. I found a spec sheet that tells that the 3528 LEDs have a forward voltage of about 3.3V, forward current of 60mA, and dissipate 200mW. I'm not 100% sure but it seems the bulbs are wired as 8 parallel boards, with the six individual LEDs in series. I've attached a circuit diagram and I'm thinking I need to just find the right size resistor to make it work. If so, what size resistor do I need? I've done the math several times and I don't get the same answer twice. What am I doing wrong? What wattage would the resistor need to be?

    As a separate question closely related, I purchased a handful of 30W high power LEDs for a different style bulb running off the same power source. The bulb to be replaced is a 250W bulb and I will need 4 of the 30W LEDs to give the same amount of light. I can't even light two of these LEDs in series and before I try wiring them in parallel I need some advice on that as well. These LEDs operate on 32-34V DC and the limited info that I can find indicates a forward current of 1000mA and that the forward voltage is the same as the input voltage. I don't know if it's a bad translation or if the forward voltage is indeed the same as the input but that seems very suspicious to me. Does anyone have any thoughts on that? If wiring in parallel is the way to go, what size resistor would I need in that case.

    Appreciate the help!
     
  2. #12

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    That's a pretty smart chip you found. Can survive up to 55 volts :)
    (Can be designed to 50 volts max)

    That 32 to 34 volt LED MIGHT have its own current regulator. That would explain why Vf and Vmax are the same.

    I'm working on the math now.

    That dynamo is pretty spooky. Rated for 34 volts, but "when you hit 42 volts". I sure hope somebody that knows dynamos chimes in!

    There is some mention of, "adjustable" on the datasheet, but I am only seeing external pulse width modulation for that purpose. Did you see an adjustable chip in your searches?

    Many ways to skin this cat.
    I am posting the chip you found and a different one.
     
    Last edited: May 6, 2014
  3. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    Just for thinking:
    34 V supply - 4 V for reg. - 20 V for 6 LED's = 10 V for resistor @ .48 A = 21Ω or std. 20Ω @ 10 W. Reg only rated for.35 A, so split boards into 4 & 4 with 2 regs.adjusted for 240 mA ea. with 43Ω @ 5 W.ea.
    Or if possible, 2 ea. 330Ω Rs @ 1 W in parallel on each string, still 2 groups of 4 strings ea.
     
  4. #12

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    Would you be comfortable with changing the number of LEDs per series string?

    Forget comfortable. Can you do it? Will they fit the package? etc?

    What country are you in?
    If it's the U.S. I need a state.
     
    Last edited: May 6, 2014
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  5. #12

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    Here is one way to skin the cat:
     
    Last edited: May 7, 2014
  6. #12

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    Annnd...another way, but I don't like this scheme when there are many better ways to do it.
     
  7. rggreter

    Thread Starter New Member

    May 6, 2014
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    As they say, a picture is worth a thousand words. These are the bulb "innards" that I'm working with - as you can see I really can't modify the configuration of the LEDs. The other pic is of the 30W LEDs I bought on eBay. Hopefully you all recognize them and can offer some advice on those as well.

    If I understand Bernard's response correctly, I'll have a voltage drop of ~20V (3.3 Vf * 6 = 19.8V) across each of the eight boards. I'm not sure where the 4V for the regulator comes from (I didn't see it on the spec sheet) but I see how you arrive at the 10 volts yet to be dissipated. Since there are 8 parallel boards I assume the total current through the LEDs would be 60mA * 8 = 480mA, correct? However, since the regulator only delivers 350mA wouldn't it also be correct then to say that the resistor needed would be about 30Ω (10V/0.350mA = 28.6Ω)? Since LEDs are current driven devices, wouldn't it simply be a matter of the bulb being dimmer than what it would be if driven at 480mA?

    In my original configuration, I swired the regulator in series with the LED array and they promptly burned out. Would that be because without a resistor there simply was too much undissipated power or is there another flaw lurking in my circuit?

    As far at the high power LEDs go, if I follow the same logic I would have to dissipate nearly the entire 34V since the forward voltage is nearly the same as the input. Putting four 1000mA LED's in parallel would give a total current of 4A which would then require a 10Ω resistor (34V/4A = 8.5Ω). My question then is what size resistor would I need? Since they're wired in parallel would I use one 30W resistor? or would I need one resistor per LED?

    The dynamo's themselves are a particularly tempermental piece of hardware. They are in good shape for 73 year old units but if the governor gets a little sticky or if we get a little extra steam leaking through in the inlet valve they will rev up pretty quickly and start cranking out some extra volts.

    Thanks for your help everyone - I appreciate the advice and the warm welcome.
     
  8. #12

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    I stand by my calculations. I will list the reasons in a bit but right now I need to make it clear that LEDs don't play well together. They don't like sharing their current and the biggest current hog will burn itself out. After that, a domino effect. The cure is that every series string needs its own resistor and the lack of these sharing resistors is the evil lurking in the shadows.

    I used 3.5 volts for the regulator because that's where the regulation is at 90% on the graph, page 3 figure 2. I used 350 ma for the resistance only circuit because 1) that's where it was in the first place, 2) there is no law that says you have to run LEDs at the highest survivable current, and 3) you have a dynamo that is delivering higher voltage than the label says it should.

    I have to go right now. I'll be back in an hour or so.
     
  9. ronv

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    Nov 12, 2008
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    You kind of have a dilemma. Each string wants about 20 volts and you have 42 max so a lot of power is lost in the regulator. The one you found is nice but with al 8 boards in parallel at 350 ma (the current for the driver) the driver has to dissipate over 7 watts - it can't.:(
    The smaller one that #12 posted would work a ok, but needs one for each board. This may be the best deal. ;)
    The whole thing may be to hot. Just have to try it.
     
  10. #12

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    I can see from the photo in post #7 that you are going to have to take the whole thing apart to change anything. One resistor per board, one regulator per board, anything. This leaves me at a place called, "stuck".
     
  11. ronv

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    Your 30 watt leds are probably 10 3 watt leds in series. Vf of each about 3.2 to 3.4 volts at 1 amp. Finding a driver for this may be a trick. Most buck boost drivers stop at 36 volt input voltage. You can't just add a resistor because it needs to be 10 ohms when the dyno is 42 volts and almost nothing when it is at 34 volts.
     
  12. ronv

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    He could probably use a constant current driver for each lamp. Kind of expensive but it should work.
     
  13. #12

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    The 120 volt chip I posted would be right if using one per series string and have a lot more safety margin on over-voltage. Add a resistor to share the heat and it should work. Even better would be to design for the voltage available by using 8 or 9 LEDs per string.

    I feel that I made a mistake because I assumed that anybody that got circuit boards made had already gotten somebody that knows how, to design the circuit. Now I see that no accommodation was made for resistors or regulators to force current sharing and the LEDs were not installed in matched sets.
     
  14. Bernard

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    Looks like 8 boards with 6, 20 mA LEDs in a package. There should be a spec on the package.
    Are we sure that the 6 LED's on a board are really in series & not 2 parallel strings of 3 in series ?
    34 V - 7.5V [ ON seems to use 7.5V for specs ] = 26.5 V - 20 for LEDs =6.5 V / .35 =18.57 ohms or even 18 @ 5 W
    If module is really 12 V, then all bets off.
    This seems similar to the water turbine driven dynamo at Scotty's Castle in Death Valley.
     
    Last edited: May 7, 2014
  15. rggreter

    Thread Starter New Member

    May 6, 2014
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    I think Bernard is on to something here. I'll dissect one of these bulbs later tonight to confirm but it does make sense that they may indeed be 2 parallel sets of 3 in series. I do know for sure that the 8 boards are in parallel.

    I don't have any specs on the LEDs themselves since the Chinese manufacturer is not exactly forthcoming on information for their product. All I was able to determine is that they are 3528 SMD LEDs. I found a spec sheet online for a 3528 and that is where I got the 60mA from – do they come in other amperages?

    #12 – the manufacturer took an existing design of theirs and substituted higher voltage components on the driver. The LED array wasn’t designed for this particular use per se. I’ve attached a pic of one of the blown driver boards– the best that I can tell the original driver is similar to the circuit shown in the attached schematic. They supposedly had their engineer’s review the design and I really have no reason to doubt that they didn’t. The bulbs worked perfectly with my bench top power supply but when connected to such a crude power supply as the dynamo they didn’t stand a chance. So, now I am left with about a dozen finished bulbs and a box of parts to make more if I can figure this out. I am using the extra components from the run of samples that they produced for me to do this work and I’d rather not just throw them away. In terms of light output and size they are perfect for what we need.
     
  16. Bernard

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    WAY JUN Tech's 3528 SMD LED, PN: LED-1210WVC lists 50 mA & 100 mA @ 100μs pulse, 10% duty. Most of data taken @ 20 mA.
    Ni Tex shows 60 mA, with sep. pins for 3 cathodes, 1 for anode.
     
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