High-powered LED bicycle light.

Discussion in 'The Projects Forum' started by Cowsandcowsandcows, Nov 27, 2011.

  1. Cowsandcowsandcows

    Thread Starter New Member

    Nov 27, 2011
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    First things first, I'm new to this forum, and relatively new to electronics, so I apologise if some of my questions are a bit basic.

    Now to the actual project. I'm trying to upgrade a large bike light from using a halogen bulb to using LEDs. It's running off a 4.8v NiMH battery pack with nothing in the circuit except the switch and the bulb. I've found these LEDs, but they apparently require a constant current supply. Maplin also sells constant current supplies, but the lowest voltage any of them operates at is 12v, which is too high for me. I've been reading up on LEDs on the internet, and from what I've read, if I were to use a variable voltage regulator to provide an LED with a voltage equal to its own voltage drop, it seems that it wouldn't need a series resistor to work without burning out. Firstly, am I correct in thinking that?

    Secondly, if I am correct, the Maplin website gives the "forward voltage" of the LEDs, is that the same as their voltage drop?

    Thirdly, do I really need a constant current supply if I can provide the LEDs with exactly the right voltage, or have I missed something?

    Thanks in advance
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    The idea is to not waste any battery energy as heat in a drop resistor and to make the battery last as long as possible. If your battery pack is 4.8V, you can wire two 2.4V (2.3V will do) LEDs in series with the battery and you will be ok.

    But a flashing circuit with a 50% duty cycle will make your light last twice as long and you will be more visible at night.
    In reality, you can buy such a bicycle light for under $5 so why bother to build one. I ride a bike a night and I use front and rear LED lights running off AA batteries.
     
  3. Cowsandcowsandcows

    Thread Starter New Member

    Nov 27, 2011
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    I'm building a light because I have to ride half a mile down an unlit road to get to and from work, and it's dark by the time I get out of work at this time of year. I need at least one constant light up front so I can see where I'm going, and the brighter the better. I already back up the halogen with an LED helmet light powered by two AAs, which is fine for making sure drivers see me, but doesn't really output enough light to see the road ahead.

    I might just go a couple of LEDs in series to be honest, from what I've seen, it seems to be possible to get a spectacular amount of light out of just one LED these days, two might well be enough. Is there some leniency in matching voltage drop to supply voltage then?

    Thanks
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    Go with two LEDs in series. No, you do not have to match the voltages. The LEDs will find their own operating voltages by themselves.
     
  5. Markd77

    Senior Member

    Sep 7, 2009
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    I'd recommend the ones premounted on the PCB. The PCB's are aluminium and are required to get a decent amount of heat transferred out from the LED.
    Rapid sells them a bit cheaper and they have free delivery until Christmas on orders over £10. They also have a range of optics.
    Best to wait until you know all the parts you might need before ordering.
    http://www.rapidonline.com/Electronic-Components/3W-High-luminosity-power-LED-stars-200094
    http://www.rapidonline.com/christmas/christmas-crackers
     
  6. bwack

    Active Member

    Nov 15, 2011
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    First. The biggest problem with halogen to led conversions is the heat transfer. Halogens lightsource get very hot and can take it, but the LED die on the other hand needs its heat drawn out of it. Therefor there may not be sufficient cooling in the housing that you have. Please take a pic and show us ?

    You won't get good control over the current when using a CV (constant voltage source). Look at the voltage-current curve of a LED. A small change in voltage will give you large change in current. Second, you are still burning off the same excess energi in the voltage regulator as if you where only using a resistor (but a resistor would be better because it keeps better control of the current).. I calculated roughly that you would need to burn off 0.7W (!) on the resistor, and you would get <70% efficiency (ignoring that the battery is 5.5V when fully charged. and then the efficiency and power loss is even worse).. You could classify these regulation methods as Linear regulators... If your battery was a lithium-ion (4.2V-3.0V) type, and you still wanted to use a linear regulator, you get a good efficiency if you use a Cree XPG-R5 regulated by the AMC7135. (better matching of Vf and battery voltage and gives good efficiency, low-dropout voltage (0.2V !) on the linear current regulator keeps itself in regulation as long as possible, a multiple of the chips will give you a multiple of 350mA current, and even 90% efficiency is possible if you match L-ion with Cree XM-L).

    But why not use a switchmode regulator ? You say they need 12V. This is not true at all. What you need to find is a CC buck regulator that has a drop-out voltage lower than Vbat - Vf. The good thing with smitchmode regulators is not only the good efficiency, but the flexibility in input battery range..

    ..Ps. Vf = 2.4V ? Maybe if its a red LED!
    Ps.2 The X-ML's are actually not so much more expensive than the ones you linked to.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    You might look at the circuit used to power a string of LEDs using a solar-charged battery. I know it's more complicated than what you're looking for but might give you some ideas. And if you could scavenge the circuit from an old x-mas light set, you'd be good to go.

    That circuit, as I understand it, uses the IC to pulse the LED string directly from the 4.5 volt battery (4v nominal SLA) but varies the pulse width to limit the power so the LEDs aren't toasted. There's no wasteful resistor in series. You can connect as few as one LED, or up to 20 in parallel.
     
  8. Cowsandcowsandcows

    Thread Starter New Member

    Nov 27, 2011
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    @bwack, what I meant is that the only constant-current supplies Maplin seemed to have were 12v or above, I just managed to phrase it rather badly. What exactly is a switchmode regulator?

    I can post pics once I find my camera, but I'm not sure what I'm actually doing with the housing yet, I may end up not using bits of it.

    @Markd77, thanks for the link, I'm pretty sure I'm going with the PCB mounted ones, as I may well be rigging up some sort of heat sink to keep this thing cool.

    @Wayneh, I'm not sure I don't actually have an old set of LED christmas lights somewhere, that circuit might be an option if I can find them, especially if I do a rear light as well.

    Thanks for the help guys
     
  9. bwack

    Active Member

    Nov 15, 2011
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    Hi. I'll put up some links to some drivers for you :)
    I think wikipedia has a good description on what switched mode regulators/power supplies do: "... Like other types of power supplies, an SMPS (Switched-mode power supply) transfers power from a source like the electrical power grid to a load (e.g., a personal computer) while converting voltage and current characteristics". http://en.wikipedia.org/wiki/Switched-mode_power_supply .. For us this means transfering power from a battery to the led where the power in and out is the "same" (minus the loss), but the current and voltage configuration is different (remember the formula for power is P=voltage * current).
    Example: with one LED and a constant current regulator (switched mode).
    Lets say the leds forward voltage Vf=3.3V when driven at I_LED=1A .. The power dissipated by the led is then P_LED = 3.3V*1A = 3.3 W.
    Lets ignore the efficiency of the driver for a moment and say its ideal (100% efficient), the power drawn from the battery will then still be 3.3W, but since the voltage is 4.8V on the input, we put that together with the wattage into the formula for power: P=volt*current => 3.3W = 4.8V * current, and when you solve that eq. for input current you get 0.69A. You see, with the explaination from wikipedia ("current and voltage configuration is different") this makes sense, we have different voltage and current on both sides of the driver, but the power is still the same (ignoring the loss). Linear regulators on the other hand will simply burn off power to regulate the output, in some cases this is sufficient.

    links to some LED drivers for flashlights, but can well be used for cycling ofcourse:
    http://kaidomain.com/product/details.S002982
    or basically check this category: http://kaidomain.com/Category.224
    cutter-electronics has some very nice drivers by TaskLED for bicycle diy projects, but more expensive.
    http://cutter.com.au/products.php?cat=Taskled+Drivers
    Myself has used one from luxdrive called buck puck. http://cutter.com.au/products.php?cat=BuckPuck+-+Pins .. Only thing i can think of would be a problem for you is the 2V required drop out voltage which is too much for your setup.
     
    Last edited: Nov 28, 2011
  10. Markd77

    Senior Member

    Sep 7, 2009
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    A linear current regulator isn't too bad for this: a switching supply might be 80% efficient, but the linear one is about 70% most of the time (the battery voltage will be around 4.8V most of the time).
     
  11. bwack

    Active Member

    Nov 15, 2011
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    Yes not too bad, but lets look at the lumen (lm) pr watt for the LEDs mentioned where the wattage is the total used, and comparing using linear and buck regulators.

    Vbat=4.8V, linear regulator gives the following lm/W:

    linear regulation:
    - 3W led mentioned from maplin, @700mA: Vf=3.8V, 160lm, ( 60lm/W ) <-- this is only for the led itself.
    P = P_LED + P_driver = 0.7A*3.8V + (4.8V-3.8V)*0.7A = 0.7A*4.8V = 3.36W I see/remember now that linear regs have same current through as the led, so simple P = U_BAT*I_LED. And therefore 3.36W total use is also valid for the two leds below. Anyway, total lm/w is then 47.6lm/W

    XPG-R5 neutral white: @700mA: Vf=3.2V, 260 lm minimum, ( 116lm/W )
    lm/w = 77lm/W

    XM-L neutral white: @700mA: Vf=2.9V, 260lm minimum, ( 128lm / W, but
    more like 100lm/W at 1A ). Also 77lm/W.


    buck:
    please correct me if I'm wrong. Now lets say you find a >85% CC buck driver. the total power dissipation would be like this: P = P_LED/0.85 .

    3W led: P=3.13W => 51.11 lm/W .. abit better than the 47.6lm/W above, not much.. I feel ya.. but check this out:

    XPG-R5: P=2.64W => 98.5 lm/W .. this value of intensity jumps up quite abit.. the reason is the larger difference between battery and Vf.

    XM-L: P=2.38W => 109 lm/W .. yawsa!

    My vote for a Cree and a buck-reg is not random :) (260lm minimum versus 160lm typical(!)). So yeah, with the first led I wouldn't bother with something other than a linear regulator, but for the cree's the story is different. All data pulled from the datasheets at cree.
     
  12. Cowsandcowsandcows

    Thread Starter New Member

    Nov 27, 2011
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    @bwack

    Correct me if I'm wrong, but the basic idea of a switched-mode power supply seems to be that it works like a transformer, in that it changes the voltage and current, but the overall power remains the same. What it seems to be doing in this instance is stepping up the voltage to whatever level is necessary in order to produce a constant current, in this case 700mA, and still transfer the same amount of power.

    I assume the problem is that the LEDs can't withstand more than a certain amount of current due to the heating effect of that current, so the only way to get more power to flow through them is to increase the voltage.

    Have I understood the situation, or am I still missing something?
     
  13. bwack

    Active Member

    Nov 15, 2011
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    Yes.. Note that switched-mode psu's also comes as constant voltage sources too, and they come in two topologies/types: 1. step down (buck) regulator and step up (boost) regulators.
    If you go for switched-mode CC regulator you must select a "buck" type (battery voltage higher than LED Vf). If you have two leds, you could either parallell them and use a buck, or put the LEDs in series and use a boost driver. Buck is the most common used..

    I assume the problem is that the LEDs can't withstand more than a certain amount of current due to the heating effect of that current, so the only way to get more power to flow through them is to increase the voltage.
    For my bikelight I used:
    - four cree's XPG-R5's in series (total Vf=13.2V)
    - a 5 cell L-ion battery pack (18.5V (actually 21V full charge and 15V cut-off (empty / a pcb inside the batterypack cuts it to protect it over discharge (dangerous))).
    - a CC buck driver (buck puck luxdrive) at 1A (about 90% efficient at that level).
    Initially I had planned to use a 4cell L-ion, but then I wouldn't be able to fully use the whole battery capacity because my driver required 2V drop-out voltage ( drop_out = Vin - Vload = Vbat_empty - Vleds = 4*3V - 13.2V = a negative value, but should be 2V). I could have done it all the same, but then I would waste/never-be-able-to-use maybe 20% of the battery capacity, and also it was nice with the 5th sell for added runtime (and it fitted so snugly in the my enclosure too). :)

    Just to give you some idea of what I went through when designing my light. Today I probably would use a cheap linear regulator with a good match between leds and battery. :)

    EDIT:
    Yes, ultimately it will break if the temp inside the led (the chips Tj junction temp) exceeds 150'C. If you heatsink is too small, then that is the limiting factor for max current, but if you use the largest heatsink in the world (and maybe also a peltier ?? hehe) then it is the max current allowed set by the manufacturer that is limitting (because there is a or several thermal resistances between the led's chip and the heatsink that will rise the Tj temp), and also remember that the led is a semiconductor, so a too high voltage will break it before it heats up too..
     
    Last edited: Nov 29, 2011
  14. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    These LEDs are the basis of many blinding flashlights.

    I recommend them. What you will want to drive them is a buck/boost constant current supply, so when your battery is low, it still keeps the constant current through the LED, with NO RESISTORS.

    If you try to use a resistor to limit the LED, you'll be wasting a lot of battery power as heat in the resistor, because the LED is drawing over an amp of power. (compared to 20mA of your head mounted LED light).

    If you have generators on both wheel hubs, you should be able to keep the battery somewhat charged if you have 2 of these lights on. To run two and keep the battery full, you'll need to pedal an additional 1/10th of a horsepower. That's roughly 15-20% more pedal power to keep these lighting up the road in front of you, and they will Certainly do that. You'll have roughly the output of a 60W incandescent bulb (without focus) shining your way. Add a lens and you will be able to see a few hundred feet.

    --ETA: If you were generating the power for the halogen lamp, then you will have a lighter load for more (1.5-2x) brightness if you buy the star mounted LEDs with a lens.
     
    Last edited: Nov 30, 2011
  15. bwack

    Active Member

    Nov 15, 2011
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    Hi thatoneguy, lets put this into an example using two of these in parallell (one would be too dim, atleast for me), and looking at expected runtime for an ok buck driver (85% efficient), and linear driver or resistor ( ~70%). Lets say that his battery is 4.8V 2400mAh (if they where fresh). Thats 11.5Whr (watt hours). Using the LED you recommend at If=0.8A (the max recommended) parallelled to 1.6A.
    - buck: Ptot = 3.4V*1.6A/0.85 = 6.4W. runtime= 11.5Whr/6.4W= 1.78hr
    - linear reg. or resistor = Ptot = 4.8V*1.6A =7.7W. runtime=1.5hr (1h 30min). Not too shabby?

    (The battery capacity is probably not realistic because it is not a fresh battery).
     
    Last edited: Nov 30, 2011
  16. iONic

    AAC Fanatic!

    Nov 16, 2007
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    Just thought I's toss a like to a very professional LED bike Light that utilizes the ATTiny13 Programmable IC And a triple Luxeon Rebel LED.

    Bike Light
     
  17. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    The difference between lathe/mills proejcts, and dremel and hacksaw is shocking. That light looks like something that would be sold for $125, +charger and spare battery!
     
  18. wayneh

    Expert

    Sep 9, 2010
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    I'd love to have one of those.
     
  19. Cowsandcowsandcows

    Thread Starter New Member

    Nov 27, 2011
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    Right, looks like my design has changed. I'm now looking for one of these buck regulators, and some of those Cree LEDs, does anyone know of a supplier for either of them in the UK?

    Thanks a lot, guys!
     
  20. bwack

    Active Member

    Nov 15, 2011
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