LED Headlights

Discussion in 'The Projects Forum' started by Coleman, Nov 28, 2008.

  1. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    I'm building a couple LED headlights for my eBike. The bike runs off a 24v LiFePo battery, so I figured, why use a separate battery pack for lights? A 500w motor is hardly going to notice the extra drain from a couple 5w leds.

    The problem I've run into is finding an appropriate voltage regulator. I originally bought one that was supposed to accept input voltage of 14 - 30v DC and output 12v DC. Turns out they changed the specs on the thing and didn't update the website till later; now it only accepts 14-16v input. So I returned it, and haven't been able to find a regulator to do the job.

    System Specs:
    Battery: 24v 20ah - actual voltage 24.5 - 27.5v DC (Cyclone eBike kit)
    Lights: two 5W 12v MR16 LED (p/n EO-EL16-51W3-1 from theledlight.com)

    I'm looking for a regulator that will output 12v DC @ 1a (or two regulators @ .5a). Or if that's just impossible, I could get a 24v regulator and run the two lights in series. That wouldn't allow them to be switched individually, but I suppose I could live with that.
     
  2. SgtWookie

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    Jul 17, 2007
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    Can't find your supplied part number on that site.

    However, since they're 5W at 12v, that's 416mA current (I=P/E).

    You could use a couple of LM317T regulators wired as constant current sources.
    R1 = Vref/Iout; Vref is nominally 1.25 but may be anywhere from 1.2 to 1.3; 10mA >= Iout >= 1.5A
    A 3 Ohm 10W resistor between the OUT and ADJ terminals should get you pretty close to 416mA. Connect the IN terminal to +24v, take your current from the ADJ terminal. Connect the other side of your LED to ground.

    You'll need a good-sized heat sink on each LM317T. Not an efficient solution, but cheap and easy to build.

    A much better solution would be a DC-DC supply, or switching "buck" converter. Much more complicated though.
     
    Last edited: Nov 28, 2008
  3. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    Thanks, that's helpful info.

    You can find the lights I'm using here. It's the 2nd and 3rd from the bottom of the page (10deg and 38deg lens being the only difference between the two - that should give more light in more places when I ride).

    Where would you recommend getting the parts you mentioned? I'm not quite sure why you would set them up as "constant current" rather than "constant voltage" - it's my understanding that LEDs require exact voltage inputs, but the current is self-regulated according to the demands of the LED. I'm always interested in knowing more about electronics, so if that's wrong please explain.

    Also, why would the DC-DC supply be better, and in what sense would it be complicated? Can you show me an example?

    Thanks again.
     
  4. SgtWookie

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    Jul 17, 2007
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    Digikey is a good place for small orders. They'll ship small orders via 1st class mail, which saves you money.

    LED's are current controlled devices, not voltage controlled devices. From looking at the datasheet, they recommend 9v to 13v (rms) as an input voltage; with a 5w rating.
    So, if you put about 416mA through it, you should measure 12v across it.

    LEDs have a pretty consistent voltage drop across them for a given current. However, it takes a very closely regulated voltage to get a consistent current flow, and thus consistent brightness intensity.

    Those LEDs must have a current limiting resistor already in them to allow the variation in supply voltage that they do.

    The circuit I proposed requires (for each LED) one LM317T, one 3 Ohm 10W resistor, and a heat sink. About 1/2 the power in the circuit will be wasted as heat in the regulator and resistor.

    A synchronous buck regulator would be very efficient (up to around 93%) but will require several ICs, two MOSFETs, an inductor, and perhaps a dozen resistors & capacitors - per LED. The trick here is that the inductor's current is kept nearly constant by alternately switching the supply current on via the upper MOSFET to increase current, and then shut off the current supply and use the lower MOSFET as a "flywheel diode" to keep the current flowing through the load via the inductor while it's magnetic field collapses.
    A portion of an example circuit is contained in Microchip's datasheet for the MCP14628 - 2A Synchronous Buck Power MOSFET Driver (attached).

    It's an intermediate level type thing. If you're a beginner to electronics, stick with the LM317T solution for now.
     
  5. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    Thanks again for the help. Your right, I don't feel ready to take on a project as complex as building my own, but what do you think about using this ready-made Buck-Puck or Power Puck. These specs seem promising - input up to 32vdc, various constant-current output options. And efficiency around 85%. If you think it would work, how would you recommend wiring it?
     
  6. SgtWookie

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    You could use a single 350mA unit per headlight, but your LED won't be as bright as it could be.

    You need about 420mA current for that 5W output. Running it with 350mA would be about 83% brightness.

    As far as wiring, both datasheets contain wiring information. Yours would be more simple, because you would only have one LED per module.

    Don't try to run both in parallel using a 700mA module; if one of your LEDs burned out, the other would burn out VERY quickly due to 700mA being rammed through it.
     
  7. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    That all makes sense. The curious thing is why their current doesn't better match the requirements of the light. On the description page, it says "All 350mA models have been designed to power one or two 1 Watt Luxeons" and "All 700mA models will handle 1 5-Watt Luxeon" So either:
    1) They're wrong, or
    2) A 5W Luxeon is somehow different from a 5W Edison LED, or
    3) There is no difference, and my 5W Edison LED will run just fine on 700ma, the same as a 5W Luxeon

    Maybe you can shed some "light" on it... I'll also contact the seller and see if they have experience with it.
     
  8. SgtWookie

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    Your Edison light must have a current limiting resistor built in. Otherwise, it would not operate across such a wide voltage range (9v-13v, max 15v).

    The power consumed in the internal current limiting resistor is being wasted as heat.

    5W Luxeon LEDs do not have internal current limiting resistors. If you attempt to operate them without a constant current source, or a current limited source, you will burn them up very quickly.
     
  9. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    I'll try to find out from the seller if that's true. If it is, doesn't that mean the 700ma puck would work fine?
     
  10. Wendy

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    Just curious Wook, can't you use a simple PWM circuit with a 50% duty cycle to drop the 24V to 12?
     
  11. SgtWookie

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    Not without using an inductor as a current limiter, and feedback to ensure the current doesn't exceed maximum.

    With LEDs, as you increase the current over the maximum continuous current, the allowable duty cycle rapidly decreases.

    It takes a relatively small change in voltage for a very large increase in current. As the LED heats up, it's Vf decreases, causing a further increase in current.

    One way to think of our OP's LED is an 8.2v Zener diode in series with an 11.4 Ohm resistor. If subjected to full battery voltage, current flow would be about 1.5A, and power dissipation in the resistor would be > 25 Watts, where it's designed for about 2 Watts. Power dissipation in the LED would be over 12 Watts, where normally it would be around 3.5 Watts.
     
  12. Wendy

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    I was thinking strictly as a voltage regulator, without the regulation. More of a divider. Get it down to 12VDC without all the waste heat.
     
  13. Coleman

    Thread Starter New Member

    Nov 28, 2008
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    Well if that's all that's needed then I could just run the two lights in series directly from the battery. Wait, is that really all there is to it?
     
  14. Wendy

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    If these are LEDs there will need to be resistors, unless their built in.
     
  15. SgtWookie

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    Nope, not without a limiter; and nowadays the most efficient way to do that is to use an inductor.

    Try it yourself. :) Use a regular LED with a current limiting resistor calculated for 6v, and then drive it with a 50% duty cycle 555 timer running at 1kHz-40kHz. See how long it lasts. Normal lifespan should be around 100,000 hours. I'd be suprised if an LED would take that kind of abuse for more than a few hours.
     
  16. Wendy

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    A simple capacitor would average most of it, I would think. Minor ripple isn't a problem, and inductors aren't absolutely required for switchers. I'm thinking KISS, that and alternate ways to do this. If you have a hard voltage (such as the 24VDC), PWM converstion ought to be pretty stable. I'll be running some experiments in the not too far future on this, it is a thought I've had for a while.

    Do these LEDs require an external resistor?
     
    Last edited: Nov 29, 2008
  17. SgtWookie

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    Sure, you could run two of them in series - as long as your total voltage on the electrical system never exceeds 26v. 15 volts is specified as the absolute maximum, but you don't want to go there; it will stress your LEDs and shorten their lives drastically.

    But, if you run two in series, you will have "all or nothing". You won't be able to turn just one on. If one should fail, you'll be in darkness.

    Carry spare bulbs for emergencies.
     
  18. SgtWookie

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    Nope, sorry. A capacitor "looks" like a dead short when it's discharged. It would rapidly charge as high as it could, then taper off - sort of like a sponge sucking up water.

    An inductor is the inverse function of a capacitor. Current flow begins slowly and builds gradually as the magnetic field builds. The current flow doesn't "like" to be cut off; it's like slamming a truck into a concrete wall. Reverse EMF (voltage) across a coil can get extremely high very quickly as the magnetic field collapses. But an inductor can be used to keep a relatively constant current flowing through a load.
     
  19. Wendy

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    I'll have to agree with Wookie on this one, a switching regulator is the way to go. We were just discussing details on how to do it. That and it lets you run the LEDs in parallel, redundancy is your friend on equipment that your life may depend on.

    Switching regulators are extremely efficient, so it would be great for your battery life too, ½ A at 12V is ¼A at 24.
     
  20. SgtWookie

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    You'd have to use switched capacitors, and they're only used for very low current levels (perhaps up to 50mA); and normally used to boost voltage, not reduce it.

    That is not indicated in the datasheet.

    They do recommend their own driver circuits for it - but their driver circuits are only for 120vac and 220vac.

    All they give is a recommended voltage range (9-13), max voltage (15), and wattage (5w). I had to deduce the rest.
     
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