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_ LED Lights: Criss - Cross light beam contamination/cancellation ?

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  • darryl
    replied
    In these particular led modules, the leds are not high-output types- more like an early version of the white led. Single white 3mm leds now are almost brighter than this entire array, and even they pale when compared to the newer versions such as made by Cree, etc. My opinion on this whole endeavor is that it will be a lot of work for an unsatisfactory machine light.

    Some months back I came across an led track light, MR-16, that works well. It runs from 12 volt, ac or dc, is internally regulated, has 3 emitters inside and a lens front which you can remove. This turns it into a flood vs a spot. Either way the light is a good color and it's bright. I mounted four of them on my mill, without the lenses, and I've been very happy with it. Without sockets to plug them into, I had to make the 12 volt connections directly to the pins, but that's a pretty minor detail. The light is bright white, no bluish tinge, and very smooth. By my guestimation, one of these will outshine five of these other led modules easily, and will do so using about 200 ma or so from the 12v supply. I placed four of these around my mill head. It's easily the best lighting arrangement I've made yet. I was well enough impressed that I bought another dozen. The lathe will get a 'light bar', both drill presses will get them as well, and the house is going to get a few in strategic areas for emergency lighting, or simply 'evening lighting'.

    They are normally a $20 thing, but were on sale at $5 a pop at Homely Despot.

    I have a 12v power supply rigged up in the shop, and my plan is to wire up all the machines with their own switch. There will be one light at the entrance to the shop, and an led indicator to show that the power pack is plugged in. It will be my choice to either unplug it, or leave it plugged in when I leave the shop, and the indicator at the doorway will be a reminder that it's still plugged in. the system will eventually include a 12v deep cycle battery with a two-way charger- ac and solar. Everything I need is here to put together a power-failure lighting system- it's just up to me to do it.

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  • mattthemuppet
    replied
    the point is though that these 3/5mm LED lights are simple, cheap and robust. They don't need any heatsinking as they don't produce any heat, they don't need any control electronics as the LEDs are carefully matched to the power source most people are happy to use and they'll probably last forever. My wife and I were given a couple of the really early LED torches (4x 5mm LEDs = a candle or two) for a wedding present. I thought we'd lost one but it turns out that we left it in the pocket of our tent when we packed it up. It was still on a year later, no worse for wear.

    Now when you compare that to a decent high power LED torch (I've built several) with it's thermal issues, complex drivers/ UIs and batteries that can explode if you fool around with them (or spot weld titanium if you short them, ahem) plus order of magnitude higher cost, you can see that there are reasons for those engineering choices. I'm not so much defending that style of torch as I personally can't stand them and I'm a tint snob, but for what they cost they do a pretty good job.

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  • vincemulhollon
    replied
    Originally posted by darryl View Post
    so the three AAAs in series becomes a voltage limited current source. Terrible way to drive the leds.
    The type of "engineer" who thought that cost savings move was a great idea, probably didn't use a milligram extra of heatsink. It has to work for five seconds in the store before purchase, and the customer will be ecstatic if it runs for a minute while flipping a circuit breaker back on, but try and run it for eight continuous hours and you may have a overheating problem. Just saying something electronically value engineered is probably thermally value engineered.

    Another funny way to look at it, is a really good flashlight costs far more than a screw in LED bulb at the store. So you're not going to be using a really good light source almost by definition if you're chopping up dollar store flashlights. And you could make a better heatsink, and a better active drive ckt, and better optics, but this is turning into that story about the silk purse outta a sows ear.

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  • mattthemuppet
    replied
    using flashlight heads sounds like a slightly complicated way of doing things, but if you have 8 of them and they should each be 3V (pretty much all 3 and 5mm LEDs are 3V), then wire 4 in series to get 12V and hook it up to one of your 12V wall warts. Then calculate the resistor you need to get ~20mA per LED or say 180-200mA overall (for the whole string) and you're done. If you want to keep things even simpler, wire the 2 strings in parallel and recalculate the resistor for ~400mA overall. At that current level per LED you're not going to get into any crazy problems. Wiring the lights into 12V strings will be much more efficient than wiring each light individually and burning off the voltage difference with a resistor.

    Personally, I think that you can get just as much and better quality light for similar money using high quality Cree or Nichia high power LEDs, optics and a dedicated LED driver, but that does require a bit more work (although no more money). I've made a ton of LED lights and they're really rather simple, plus you don't have the warm up times and cold weather issues that fluorescents are plagued with.

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  • Rosco-P
    replied
    Originally posted by iMisspell View Post
    The lighting in my basement sucks, ceiling is only 6'8" so the light will not spread and seeing at the mill is tough (right now im using a magnetic LED, want something better).
    Been wanting to get a "donut ring" of lights around the quill and seeing a post in the Tools thread got me motivated._
    What kind of lighting fixtures? Fluorescent tube? Bare bulb? If it's tubes, mounting the fixture up between the ceiling joists instead of parallel to them will improve the spread of the light. Also adding a diffuser will help as well as the diffuser now become a larger, softer source. Not a big fan of LED's, poor "quality" of emitted light, especially true in bottom of the barrel LEDs used in flashlights, still a heat source like a Tungsten bulb and 20 little emitters do not a coherent source of light make without reflector and lens. Having used (overly expensive) LED light panels in the film industry, I'll take a fluorescent, halogen or HMI light fixture any day.

    Leave a comment:


  • RichR
    replied
    Hi jhe.1973
    Am I right to think that for arriving at the value for the resistor, I can use an adjustable one to set the output voltage & then measure the value of the adjustable resistor to get what I need for a fixed one?
    You could, but you would be better off just calculating the value. The output voltage is based on the built in reference voltage
    (1.25V) and the ratio of two resistors. If you make R2 twice R1 you will get about 3.75V. Here is a link to an LM317 calculator:

    I recommend input and out capacitors to prevent oscillations. Don't make the values of R1 and R2 too large or the bias current
    from the reference pin will cause you problems.
    Last edited by RichR; 01-14-2014, 05:08 PM.

    Leave a comment:


  • darryl
    replied
    Ok, just for the heck of it I did a test. I cut open the body of the flashlight and wedged it open to get the module out. Hooked it to a power supply and ran the voltage up until the brightness stopped increasing linearly as the current rose. There's a sort of knee here, at about 350-375 ma. At 400ma, the voltage was 3.7, and at 300 ma it was 3.6 volts. From what I'm seeing, I wouldn't drive it above 300 ma, as the power dissipation rises with little increase in brightness.

    This is the same 9 led head as your picture shows, though the body on this one is a little different. It's likely the module is the same. It presses in from the front, but the ground connection is made by two wires that face forward on the module. This means that trying to press it out from the inside is likely to rip the wires off the pc board and wreck at least two of the leds.

    There's a definite beam to the light output. In my opinion it would be difficult to get an even illumination- you'll end up with six bright spots surrounded by shadow.

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  • darryl
    replied
    I was going to suggest using a variable resistor to get things just right, and then replace it with a fixed value- but these things aren't usually at hand. I was also going to suggest wiring the led 'heads' in series, but they have one lead connected to the case, which would mean isolating the mountings- then you have the possibility of a short at some time. But wiring three of them in series would make it so they require 9.6 volts to light to a good brilliance. That means your power supply would only be wasting 2.4 volts (if it maintained 12 volts under load). That would make the resistor required about 8.2 ohms, and it would only dissipate .7 watts. Wiring three heads in parallel would mean 8.8 volts lost, 900 ma flowing, and the resistor would be 10 ohms, but would need to be 10 watts. That's a fair amount of heat to get rid of, especially if it's all going to be compact.

    Just to take the example a little further- suppose you put four heads in series/parallel. That would mean 6.4 volts required, and 600 ma. From a 12v source, that would mean 5.6 volts lost, the resistor would be 9.3 ohms and would dissipate 3.4 watts or so- a 10 ohm, 5 watt resistor would be about right. The power supply would be happier delivering only .6 amps, and you would have less heat to worry about getting rid of.

    It's too bad the led circuit board in these lights can't be taken out easily so the ground connection could be isolated from the case. The best connection would be two strings of 3 heads in series, with the strings wired in parallel. That's a 600 ma draw and a 1 watt, 8.2 or 10 ohm resistor wired in series with each string, and a happy 12v power supply.

    Leave a comment:


  • jhe.1973
    replied
    Originally posted by ironmonger View Post

    Kind of looking like the halogen lamps would be simpler.

    paul
    Thanks Paul!

    You are right about halogen being simpler, but I like the cooler temperatures of the LEDs. Especially for a light that I will get up close & personal with when I am trying to see just what is going on at the cutting tip of a boring tool.

    I do a lot of fairly small stuff so I often need to get close.

    Leave a comment:


  • jhe.1973
    replied
    Originally posted by darryl View Post
    It really does depend on the power supply. Most wall warts are good for at least 300 ma, but that would only be enough current for one of those modules. Plus you wouldn't really know what the proper value for the resistor is until you put a load on the wart and then measure the voltage. This brings me back to the use of a regulator IC instead of resistors. You could use a 317 with a tab, and with a single 1/4 watt resistor you set the output voltage- and then drive up to 4 modules from it. The IC would mount to a heat sink, which could simply be the side of a case. If you want to drive up to 8 of them, then just use two regulators. You will be needing at least two amps from the power supply, so that narrows the field a lot if you're considering a wall wart.
    Thanks again darryl.

    Earlier this evening I was fiddling around in the shop & I found 2 - 12V warts that have a 1 amp output. I also looked up the 317 regulator & saw that my friend had me use a similar one for the built in adjustable power supply that regulates the transistor output of the EDM we built.

    So, this is coming back to me somewhat. He designed the electronics & I followed his schematics.

    Am I right to think that for arriving at the value for the resistor, I can use an adjustable one to set the output voltage & then measure the value of the adjustable resistor to get what I need for a fixed one?

    Leave a comment:


  • darryl
    replied
    It really does depend on the power supply. Most wall warts are good for at least 300 ma, but that would only be enough current for one of those modules. Plus you wouldn't really know what the proper value for the resistor is until you put a load on the wart and then measure the voltage. This brings me back to the use of a regulator IC instead of resistors. You could use a 317 with a tab, and with a single 1/4 watt resistor you set the output voltage- and then drive up to 4 modules from it. The IC would mount to a heat sink, which could simply be the side of a case. If you want to drive up to 8 of them, then just use two regulators. You will be needing at least two amps from the power supply, so that narrows the field a lot if you're considering a wall wart.

    Leave a comment:


  • ironmonger
    replied
    Originally posted by jhe.1973 View Post
    darryl: Thanks for the additional perspective.

    IIRC I gave my friend the 12 volt source idea 'cuz I already had one. Since then I have found plenty of lower voltage ones at thrift stores etc. and may use those instead. I even found a universal wart w/several voltages available just by selecting them w/a switch.

    Would I be safe to figure that doubling the size of the resistor to 94 - 100 ohms would drop the voltage enough to limit the current to what you suggest?

    I have a good understanding of electricity, but lack the solid state & digital application of it so thanks again for the help.

    Darryl was figuring about 30 mils per LED. That's 240 mils for the cluster of 8. That needs a series resistor of 50 ohms to set the current to 240 mils. That resistor at 12 volts needs to be at least 3 watts. If you increase the resistor to 100 ohms the current drops to 15 mil per lamp... if that works your home free. That also reduces the wattage rating to 1.44 calculated or a 2 watt resistor.

    If you find a 5 volt wall wart that can supply 500 mils or so, you would then need a 20 ohm resistor with a 1.2 watt rating or so.. 2 watt would be great. For each lamp of course.

    Kind of looking like the halogen lamps would be simpler.

    paul

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  • ironmonger
    replied
    Originally posted by darryl View Post
    Jim, I have the guts of one of those sitting on my desk. It looks to me like all the leds are wired in parallel, and I don't see a resistor. I think they are banking on the fact that a triple A will lose voltage right away when loaded that much, so the three AAAs in series becomes a voltage limited current source. Terrible way to drive the leds. If it's indeed drawing 600 ma, that's almost 70 ma per led- they will burn out and they do. I'd limit the current to something like 35 ma at the most, which means about 300 ma. You'll have a chance at them working for a decent length of time. A better way to drive them would be to supply 3.2 volts regulated to them. That would keep the current through any one of them within limits, even when they start to burn out. Otherwise as they burn out, the remaining ones will get higher and higher current and they will burn out in quick succession.
    Wow... good call Darryl. I never looked inside the one that I have. That's flakey.
    I went back to find the listing that had the loose lamp moduals.
    All the other ones that I have are Cree's or cree knock offs, and have dirvers built in.
    These are the single power level ones.

    see:

    although this is the same price:
    http://www.amazon.com/gp/product/B005E48K6I/ref=cm_cr_ryp_prd_ttl_sol_1

    paul

    Leave a comment:


  • jhe.1973
    replied
    Originally posted by darryl View Post
    Jim, I have the guts of one of those sitting on my desk. It looks to me like all the leds are wired in parallel, and I don't see a resistor. I think they are banking on the fact that a triple A will lose voltage right away when loaded that much, so the three AAAs in series becomes a voltage limited current source. Terrible way to drive the leds. If it's indeed drawing 600 ma, that's almost 70 ma per led- they will burn out and they do. I'd limit the current to something like 35 ma at the most, which means about 300 ma. You'll have a chance at them working for a decent length of time. A better way to drive them would be to supply 3.2 volts regulated to them. That would keep the current through any one of them within limits, even when they start to burn out. Otherwise as they burn out, the remaining ones will get higher and higher current and they will burn out in quick succession.
    darryl: Thanks for the additional perspective.

    IIRC I gave my friend the 12 volt source idea 'cuz I already had one. Since then I have found plenty of lower voltage ones at thrift stores etc. and may use those instead. I even found a universal wart w/several voltages available just by selecting them w/a switch.

    Would I be safe to figure that doubling the size of the resistor to 94 - 100 ohms would drop the voltage enough to limit the current to what you suggest?

    I have a good understanding of electricity, but lack the solid state & digital application of it so thanks again for the help.

    Leave a comment:


  • RandyZ
    replied
    Even if you did set up "standing waves", they would too short for you to see anyways.

    Leave a comment:

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