resistance values for LED's

Discussion in 'All Things Boats & Boating' started by Frosty, Apr 26, 2009.

  1. Frosty

    Frosty Previous Member

    Does anyone know the formula for working out the value for using a 3 V LED on a 12/14 volt system.

    I have found that it is a simple afair to change all my bridge guage lights to led's

    I can never use my last resistor because then I wont have a sample when buying new ones. Its time I knew how to do it.

    The last lot I bought was In Phuket with a girl behind the counter who just did it in her head. I have not since been able to find anyone male or female that can do it.
     
  2. masalai
    Joined: Oct 2007
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    Location: cruising, Australia

    masalai masalai

    I have seen some nice units - - - easier to buy them - - same size as the new-fangled downlights and bunnings have those assemblies to fix them through a panel - will be getting some soon so will let you know then.... also look at http://www.altronics.com.au/index.asp?area=item&id=A2499A
     
  3. Guest625101138

    Guest625101138 Previous Member

    LEDs turn on at about 2 to 3V depending on the colour and they have quite small internal resistance. The little panel mount ones will give reasonable light with 10mA.

    Making the assumptions that the internal resistance is zero and the voltage across the LED is 2V irrespective of current then on a 12V system the feed resistor will drop 10V.

    Resistor value is 10V/10mA = 1kohm.

    There is a table here that gives the characteristic voltage range:
    http://en.wikipedia.org/wiki/Led#Miniature_LEDs
    Most panel LEDs will give reasonable light on a 12V system with a 1kohm resistor.

    Rick W
     
  4. Frosty

    Frosty Previous Member

    Ok thanks for that Rik.

    Mas I am fitting 1 singular LED in place of the tiny capless bulbs in my instruments. Not to save power but to replace the bulb.

    Carefully soldering the resistor to the posative leg and then soldering them into the tiny bulb assembly plug that Teleflex sell as a complete assy that I cant get and why when it takes about 5 minutes to do this. The only prob is that the plug can go in 180 degrees so pos will be neg ---just take the plug out and replace 180- degrees.

    I have LED's all over the boat for generator fan to 220 V input warning, gas solenoid switch etc etc
    .
    Not interested in navigation lights what with gen and two engines I have power coming out of my ears.

    I love messing with LED's now they got white ---wow.


    Ive looked up the colour code which seems to be 151 at %5 if i am doing this right.

    That is brown-green-brown -gold??? how does that correspond to 1000ohms
     
  5. CDK
    Joined: Aug 2007
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    CDK retired engineer

    0 Black
    1 Brown
    2 Red
    3 Orange
    4 Yellow
    5 Green
    For the first two rings, use the numbers. The 3rd ring is the number of zeros to add. so brown-green-brown is 1-5-0 or 150 ohms.
    1K-ohm is brown-black-red (1-0-00), really quite simple.

    White and blue LED's are quite happy with 10 milli-amps, the green, yellow and red ones do better at 20 milli-amps. The proper resistor for these is 470 ohms (yellow-purple-brown).
     
  6. Frosty

    Frosty Previous Member

    So am I using the wrong value CDK. The ones im using appear to be 150HMS.

    You say I need 1000 ohms for a white LED?.
     
  7. pkoken
    Joined: Mar 2003
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    pkoken S/V Samadhi V

    Use a constant current driver! They are cheap and easy and your LEDs will last longer and be the same brightness across a range of voltages (typically 9-30v).

    Samadhi V is illuminated mostly with custom warm white LED fixtures I built.

    [​IMG]
     
  8. marshmat
    Joined: Apr 2005
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    Location: Ontario

    marshmat Senior Member

    Very nice, Phil. May I ask if you're willing to share anything more about the constant-current driver you're using? If I were doing something like that, I'd probably just put the LEDs on the output of a linear voltage regulator, something like an L7805 perhaps, with series resistors to control the current. I'm curious to see if you've worked out a better solution.
     
  9. pkoken
    Joined: Mar 2003
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    pkoken S/V Samadhi V

    They are buck drivers that I purchased over the internet, the LEDs are CREE. This was all more than 2 years ago, before we were cruising full time. I will try and find more detailed info, as I know it is around here... somewhere

    We haven't had any failures (yet). Each light cost me about $16 in parts, plus a stupid amount of labor soldering them together. Hence why I never considered selling them!

    There is a lot of satisfaction seeing a 1.2 amp halogen replaced by a 170 milliamp LED. We keep the boat lit up all the time and hardly notice the power use.
     
  10. BillyDoc
    Joined: May 2005
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    BillyDoc Senior Member

    Hi Frosty,

    Pkoken is absolutely right about using a constant current source or buck regulator (constant voltage source) for the LEDs. The buck regulators can be very simple to implement, just go to digikey.com and search for a "simple switcher" integrated circuit and download the data sheet for directions. You need to match your LED working voltage to the buck regulator output voltage. Then, you can light up quite a few LEDs from one regulator, just stay within the current capability of the regulator and pay attention to the heat dissipation of the regulator so you don't overheat it. Best to use a heat sink on it.

    On the other hand, if you want to use a resistor to drop the voltage here's how you go about figuring it out. You need to have some basic data for the LED to begin, specifically you need to know the "typical Forward Voltage Drop" and the "Typical current," both of which you can get from the data sheet for whatever LED you are using.

    Let me give you an example, assume a forward voltage drop of 1.2 volts, which is a common value for small LEDs, and a current of 0.020 amps (20 milliamps). Say you want to power these from a typical "12 volt" system, which may in fact get up to around 14 volts when you are charging.

    LEDs will work nicely over a pretty wide voltage range, so let's just assume 12 volts, exactly. When you have them on and are powering them with 14 volts you will shorten the life of the LED slightly, but probably won't live long enough to notice. So, you have 12 volts to work with, and the LED needs 1.2 volts. That means that your series resistor has to dissipate 12 minus 1.2 volts, or 10.8 volts. Now it is a basic rule of electricity that in a series circuit the current flowing through all the elements in series will be the same. We want 20 milliamps going through the LED, so 20 milliamps also has to go through the resistor in series. This is exactly the sort of problem that Ohm's law addresses, the relationship between current, usually symbolized with an "I," resistance, usually symbolized with an "R," and voltage, usually symbolized with an "E," for "Electromotive Force" . . . but I'm flexible, "V" will work too.

    Ohm's law stated is: "The current in a circuit is directly proportional to the voltage, and inversely proportional to the resistance." This is a very useful thing to know, by the way. You will be amazed how often you can haul it out and get an answer to something that seems impossible to determine with it! So, I'm going to belabor this just a bit.

    In mathematical symbols, Ohm's law is "E=IR." Now for the mathematically inclined that's all you have to know to apply it . . . but that certainly doesn't describe me! Instead I use a little mnemonic device I learned in navy service school (I was an "Interior Communications Electrician" on board the USS Hornet in my youth.) To picture the device draw a big "T" on a piece of paper. Put an "E" over the top of the "T," an "I" to the left of the "T," and a "R," to the right of the "T." Now say you know the volts (E) and the current (I) of some circuit and need to know the resistance (R). You look at your mnemonic and note that the "R" is on the bottom right, this is the unknown you want to find, so imagine an equals sign to the rest of the mnemonic and squint your eyes a bit and you see that the remaining "E" is left over the "I," so just like resolving any fraction you divide the "E" by the "I," which gives you "R" as the result. If you knew I and R, you would multiply them and get E. If you knew E and R you would divide E by R to get I. Handy little tool, isn't it!

    So, getting back to the example, we need to know what resistor to use with a LED that wants to operate on 1.2 volts and 20 milliamps (0.02 AMPS). We did our subtraction and know that we need to find a resistor (R) that will pass 20 milliamps at 10.8 volts . . . so, using ohms law we divide 10.8 by 0.02 (because it works in amps, not milliamps) and we get a value of 540 ohms for that resistor. 540 is not a particularily common resistor, but you don't need to be anal about this, just go to the next higher value (560 ohms) that is common. As an exercise you can now go back to ohms law and see how much current that would mean going through the LED (it will be slightly less than 20 milliamps)

    Sorry, we aren't done yet! Those details, you know.

    Now you need to know how BIG a resistor to use, in the sense of "how much heat can it dissipate." Again this is easy to calculate because we already have the two variables necessary to calculate "wattage." Watts are, by definition, amps times volts (we are talking about direct current here, alternating current is a little bit more complex, but not much) and we know both. That little 540 ohm resistor (I'm leaving the 560 ohm one for you) will need to dissipate 10.8 times 0.02 watts of heat, or 0.216 watts. Use a half watt resistor and you are more than covered.

    Now we get to look at something interesting. We calculated the wattage for our voltage dropping resistor and got 0.216, but what is it for the LED? Ahah! We know how to find that out! Just multiply the current times the voltage again, 0.02 times 1.2 this time! The result is: 0.024 watts.

    Wait a minute . . . we are using 0.024 watts of power to do something useful, make some light, and 0.216 watts for nothing at all???!! I don't like that. It offends my delicate sense of electrical husbandry. Now you see why using a resistor to drop the voltage isn't such a great idea, unless you have power to burn, literally.

    The answer is the buck regulator. These are nifty little critters that can change that 0.216 watt loss into a roughly 0.011 watt loss (95% efficiency). Check out the "simple switcher" data sheets! A constant current source has some serious advantages over a resistor (the current stays constant as the voltage varies), but is also resistive and does not help with the power loss issue.

    BillyDoc
     
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  11. pkoken
    Joined: Mar 2003
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    pkoken S/V Samadhi V

    Wow, not often I am right :) Gotta tell my wife about it!

    Using resistors isn't too bad unless you start to talk about equalizing the batteries. We have AGM onboard Samadhi so that isn't a real issue. Our Buck regulators are attached to the brass lamp housings with a dollop of polyurethane- That's all the heatsink they need, the lights don't even get warm.

    It's really nice to have a constant quality of light regardless of your battery voltage! Batteries low? - Same light! Genset running and voltage high while we charge? - Same light! You come to realize how much the color temperature of halogen changes across a couple of volts once you throw a few LEDs alongside.

    We have also used CCFL (light too cool) and CFL (OK, but I prefer LED in the same warm white 3000k color temperature)

    BTW, BillyDoc sounds like one of my electronic instructors from USN Tech School @ Pensacola ;) That was a few years back though...
     
  12. BillyDoc
    Joined: May 2005
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    BillyDoc Senior Member

    That's a good guess Phil, I was an instructor at the Service School in San Diego for a while . . . back in 1961. But I did work on the navy base here in Pensacola for several years, at the Naval Aerospace Medical Research Laboratory as a research scientist.

    I've always loved to "inconvenience" electrons, ever since I "rescued" an ancient battery operated radio from my grandmother's chicken coop and powered it up. It had a plug with six or eight connections on it, so I just arbitrarily mated all the black ones together then all the white ones, put on a typical AC plug and punched it into the wall socket.

    It was spectacular! Sparks everywhere, smoke, flames . . .and then the main fuse went. I was about nine years old at the time and was thoroughly hooked! I've been playing with the stuff ever since. My grandmother got gray hair for some reason.

    BillyDoc
     
  13. pkoken
    Joined: Mar 2003
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    pkoken S/V Samadhi V

    I was in tech school all the way back in 1989... well it seems like a long time ago to me!

    These days I get to be my own captain - at least as long as the Admiral lets me keep my job :)

    Anyway, back to the original post - If you are replacing instrument bulbs, then I HIGHLY recommend taking a look at the replacements available from http://www.superbrightleds.com/1157.htm as they will drop right in with no soldering or resistor matching.

    [​IMG] [​IMG]
     
  14. Fanie
    Joined: Oct 2007
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    Location: Colonial "Sick Africa"

    Fanie Fanie

    Making a constant current LED is a wee-wee.

    The attached diagram has everything you need to build one. All the components are off the shelf el cheapo's. The circuit is also simple enough so you don't need a PCB or veroboard, you can just solder the components together.

    The sypply can be up to about 40V. The 317 allows only 1V2 between pins ADJ and Out, so when the 1V2 is reached, conductivity is being limited and the In / Out voltage drop increase to keep the output current constant.

    The current limit is set with the resistor. Using Ohm's law you can calculate the current required.

    Note that the higher the input voltage and the higher the current the hotter the 317 will run. At 12V input and 30mA the TO92 package should be ok.

    You can add more LED's in series, BUT the total voltage drop per LED, plus the 1.2V of the 317, plus the 0.8V of the 4007 must not exceed the supply voltage or the current will drop.

    ie If you have 3 LED's in series = 9V (if the drop is 3V/LED) + 1V2 + 0.8V = 11V. All this will work well but if for instance the 12V battery goes flat to below 11V the LED intensity is going to fall off fast. At 10V the LED would make almost zero light, while only 2 LED's in series would still run happily down to around 8V and a single LED down to around 5V.

    Don't put 220V on it, if the smoke comes out it is stuffed, getting it fixed by putting the smoke back is very very difficult. Trust me.

    The 2nd pic is a TO220 LM317, up to 1A5 but up to 500mA without heatsink.
    The TO92 package if you look at the flat face legs down the pins are from right to left In, Out, ADJ. The TO92 is a smaller half-round device.
     

    Attached Files:


  15. marshmat
    Joined: Apr 2005
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    marshmat Senior Member

    Aha! A linear voltage regulator, just as I would have expected (although wired in regulated-current configuration).... and here I thought everyone was coming up with some really fancy custom stuff ;)

    Would anyone be able to explain what, if anything, would be the problem with using a constant-voltage regulator (such as an L78xx) to fix the voltage across the LED at, say, 3 V regardless of whether the supply voltage is 3 or 30 V? These things only waste a couple of milliamps and tend to be very reliable. I'm wondering if anyone who's tried them for high-output LEDs has found any problems.
     
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