Best powersource, driver and wiring for leds- Mtbr.com
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  1. #1
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    Best powersource, driver and wiring for leds

    I am still thinking about the best way to power two leds in series with maximum efficiency. Well, actually, I am going to use 2*MC-E and 2*XP-E with two drivers. And I am still in between what to use. I can make a battery with 18650s. Let's take 2*18650 in series, it means 8.4V for fresh, 7.4V nominal and 5.5V minimum (2*2.75). On the other hand 2*XP-E in series must be around 6.6V. If I use a regular backpuck, I will not be able to drain the battery fully. If I use 3*18650, it means 12.6V for fresh, 11.1V nominal and 8.25V minimum. And efficiency will suck, as voltage drop is significant. What does that mean? Not to use Li-Ion 3.7V cells? And what to use then?

  2. #2
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    AMC7135 boards work fine with 2*18650 and 2 leds in series.

    But a step-down converter (and a battery pack with a higher voltage) would be the in the same efficiency range.

  3. #3
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    Quote Originally Posted by super-fast
    AMC7135 boards work fine with 2*18650 and 2 leds in series.

    But a step-down converter (and a battery pack with a higher voltage) would be the in the same efficiency range.
    I don't understand. What does it mean "work fine"? Sure, I can use many drivers and they will "work fine", but I want not fine, I want efficiently. I care about efficiency when I am going to power up ~26W. 80% is not enough.

  4. #4
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    Determine your circuit Vf.
    Determine your battery.
    Then someone could maybe point you in the direction of most efficient driver for that system.
    Efficiency is determined by those three variables, and a few more.
    Heat can also play havoc with efficiency. But we will just assume that you have the heat sinking under control.
    Sort two and someone can probably help with the third variable.

  5. #5
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    Quote Originally Posted by odtexas
    Determine your circuit Vf.
    Determine your battery.
    Then someone could maybe point you in the direction of most efficient driver for that system.
    Efficiency is determined by those three variables, and a few more.
    Heat can also play havoc with efficiency. But we will just assume that you have the heat sinking under control.
    Sort two and someone can probably help with the third variable.
    Heh. I thought I already did. Everyone here knows MC-E and XP-E, and everyone here knows 18650. I am just thinking about using something other than Li-Ion 18650, but not NiMh. There are not so many drivers to use, and mostly they have the same problem: voltage drop should be not much to archive efficiency > 90%.

  6. #6
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    Quote Originally Posted by Itess
    I am still thinking about the best way to power two leds in series with maximum efficiency. Well, actually, I am going to use 2*MC-E and 2*XP-E with two drivers. And I am still in between what to use. I can make a battery with 18650s. Let's take 2*18650 in series, it means 8.4V for fresh, 7.4V nominal and 5.5V minimum (2*2.75). On the other hand 2*XP-E in series must be around 6.6V. If I use a regular backpuck, I will not be able to drain the battery fully.
    OK with you on this bit so far....

    Quote Originally Posted by Itess
    If I use 3*18650, it means 12.6V for fresh, 11.1V nominal and 8.25V minimum. And efficiency will suck, as voltage drop is significant.
    Aha - now I'm lost...

    Generally buck regulators run at >90% efficiency as long at the battery voltage (supply) is more that ~2V above the Led (load) voltage. So no, your efficiency will not suck.

    If you were using a linear regulator or just a resistor, to drop the voltage - the efficiency would suck more as the voltage drop is dissipated as heat - but you would not be doing that.

    You can get buck regulators that will run with reasonable efficiency with 0.5V difference between battery and led, but that's beyond beer coaster level. You can also get buck/boost regulators that will allow you to increase the voltage between the supply and load (boost) and decrease the voltage between the supply and load (buck) but these are expensive and the boost is usually quite a bit less efficient (~50%)

    - Matt

  7. #7
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    Ttrying to figure out what question you are asking is the challenge here.
    US Navy research has reinvented cold fusion in a bottle it seems. That should give you the power without the voltage drop.
    Other than that I seem to be unable to help.
    Good luck.

  8. #8
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    Quote Originally Posted by odtexas
    Ttrying to figure out what question you are asking is the challenge here.
    US Navy research has reinvented cold fusion in a bottle it seems. That should give you the power without the voltage drop.
    Other than that I seem to be unable to help.
    Good luck.
    Looking at I see that I am wrong, I was under impression that I need to decrease input voltage a little to get more efficiency.

  9. #9
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    Quote Originally Posted by Itess
    I don't understand. What does it mean "work fine"? Sure, I can use many drivers and they will "work fine", but I want not fine, I want efficiently. I care about efficiency when I am going to power up ~26W. 80% is not enough.
    You don't seem to really understand the numbers about batteries and leds. For example when the voltage of the li-ion cells have reached 3V there isn't any energy left in them.

    A buck converter is more efficient when the difference between input and output voltage is smaller, but there needs to be enough difference between input and output otherwise the output won't be completely regulated.

    A AMC7135 doesn't have a typical efficiency, it is a LDO regulator. The input voltage should stay just just above the led voltage for proper operation. A good li-ion pack with a led with a low forward voltage (like a MCE) works perfect with the AMC7135 chips. The efficiency is around 90%. A buck driver wouldn't be more efficient.

  10. #10
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    Sorry - I think I am starting to get the idea

    Sorry to Itess - I think that I'm now understanding the question.... to paraphrase(I think);

    - the efficiency of a buck is a function of V(in) - V(out) [the bigger the difference the lower the efficiency]
    - the efficiency of a buck is a function of V(out) [the higher V(out) the higher the efficiency]
    - the cost/weight of a battery is a function of voltage [higher volts mean more cells and packaging costs weight]
    - the cost/weight of a battery is a function of capacity
    - the capacity of a battery is a function of current draw
    - the light output, drive current and voltage is a function of emitter configuration [series or parallel]
    - the light output is a function of the emitter device selection and how hard you drive it
    - the weight of heatsinking is a function of how hard you drive your emitter

    ....what is the best optimisation of these functions to run 2* XP-E and 2* MC-E

    If you were trying for the most efficient buck for the two MC-Es then I think you would be far better off running the 8 devices in the two MC-E all in series for a combined Vf of 8*3V3 (@500ma) = 26.4V allowing 2V for the buck this would be a minimum V(in) =28.4V
    using Li-Ion at a EOL of 3V this is 10 cells

    But I am guessing that the buck will be very efficient ~95% at battery EOL
    maybe over 90% at battery SOL even though the difference between Vin and Vout at SOL is 15V, the higher V(out) should dominate.

    But,But - 10* 1000mAhr cells will weigh more than 5* 2000mAhr cells due to additional packaging so maybe it is better to go for lower efficiency and lower weight,

    But,But,But - you will porobably get more of the 1000mAhr capacity out of a cell at a draw of 500ma than you will out of a 2000mAhr cell at a draw of 1A. Also the I^R losses will be much lower at 500mA.

    ....it all looks like a slightly complex optimisation problem, we know all the variables - I'm sure someone could model this???

    - Matt

  11. #11
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    Quote Originally Posted by mattD110td5x
    ....what is the best optimisation of these functions to run 2* XP-E and 2* MC-E
    Thanks! Yes, that was the question.

  12. #12
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    @Matt, choosing a battery and a driver for a bikelight isn't rocket science, so don't try to use rocket science to solve the problem. A difference of a few percent in efficiency can't be noticed, the only way to find out is measure everything and calculate the efficiency.
    Using the right numbers makes a calculation a lot more usefull. Nobody is going to build a li-ion battery pack for a big bikelight with smaller cells then 18650s, because that would be really stupid. 18650 li-ion cells are the most high tech li-ion cells available and they are also the cheapest.

    @Itess, do you want to have multiple levels? Difference between the XPE and MC-E in drive currents? There are so many options that would suit you when you would be happy with just a single light level. But with more levels it's becoming a bit more interesting to find the best solution.

  13. #13
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    Quote Originally Posted by super-fast
    @Itess, do you want to have multiple levels? Difference between the XPE and MC-E in drive currents? There are so many options that would suit you when you would be happy with just a single light level. But with more levels it's becoming a bit more interesting to find the best solution.
    I am thinking on using 2 drivers: one for MC-Es and one for XP-Es. For XP-Es I think one level is enough, but for MC-Es it would be ridiculous to drive them on maximum all the time. And wiring the MC-Es is still the question of 2 parallel leds in series or 4 series of two parallel dies. So, I have four variables:
    • wiring of MC-Es
    • battery configuration to use, 3*18650 or 4*18650
    • driver to use (I am between bFlex and hipFlex)
    • efficiency of the second driver, not to generate much heat, because it will use the same powersource as the first driver

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