LED lighting

Discussion in 'Tips, Tutorials & Tools' started by rmks2000, Nov 28, 2006.

  1. Zathros

    Zathros Guest

    Rather simplistic of me. You are 100% correct. I was thinking of led's in series running parallel to each other. 3 LED's conected in series, run parallel so that there are a total of nine (for example) three rows three a piece. At the Positive end have all LED's meet and then place a 16 OHM resistor (1 WATT) to a 12V source. n=Negative to ground. You'd pull .9 watts. To keep the voltage in check you could run a POT so the voltage could be controlled. The LED's would be in parallel to the pot but in series to each other. By lowering the voltage you could greatly extend the life of the LED's.
  2. SteveM

    SteveM Member

    No, you need a resistor for each of the 3 strings of series led's. Like this:

    Attached Files:

  3. Zathros

    Zathros Guest


    Those resistors won't be able to tell if there is one resistor per circuit of one connected to all three. You have a better chance of different actual resistor values with 3 resistors. We spent hours on parallel vs. series circuit electronics school. Here are a couple I pulled from the net. These would work fine. A voltage regulator circuit would give even more life., add a DC to DC converter for the best setup (longevity) and would handle small surges. Of course your free to do what you want.

    Attached Files:

  4. lizzienewell

    lizzienewell Member

    I have a lousy memory buffer and this passage exceeds my buffer in the first sentence. I'm been working alot with writing and so will try to rewrite it in a way that I can understand. I hope that this is helpful.

    Here is attempt one with a shift from passive to active sentence structure. It should get the ideas in a more workable order and reduce the number of words.

    The difference between supply voltage and the diode's forward voltage determines the resistor voltage, so resistor value determines current through the diode (V=IR). Check manufacture's recommendantions for current that should be flowing through the diode.

    Whoops still doesn't make sense to me. Here is attempt two.

    Check manufacture's recommendations for current that should be flowing through the diode. if you pump more current though the LED than recommended, the light will be brighter and the lifetime will be shorter. To determine resistor value, subtract the diode's foreward voltage from the supply voltage. Divide this number by "I."

    Err, I don't know what "I" means in the equation. I assume that it's a standard number of some sort. What is "I"?

  5. Bowdenja

    Bowdenja Active Member

    Dang........... I think philosophers have been pondering that question for centuries!:grin:

    Lizzies' right too much current going in, will eventually mean no light later............ usually when you really need it.:cry:

  6. lizzienewell

    lizzienewell Member

    Steve is the one who is right. I was translating what he wrote from mathmatician into I-gotta-degree-in-art.
  7. SteveM

    SteveM Member

    Yes, there is indeed a difference between 3 resistors and one. It has to do with the nature of diodes as non-linear devices.

    Lizzie, "I" is current, "V" is voltage, "R" is resistance. Ohm's Law is V=I*R so if you have a 3 volt supply and an LED with a 2 volt forward voltage and a 20mA operating current. You use Ohms law to calculate R in order to produce 1 volt at 20 mA, thus R=1/.020 = 50 ohms.
  8. Zathros

    Zathros Guest

    What ever floats your boat.

    There is no oscillation in this circuit, no frequency, no nuttin'. It is a simple circuit that can be over thought. I spend 8 years fixing TV's and the like. You can make circuits very complicated but simplest last the longest. If there were a frequency induced into this circuit maybe one could take your stand but to say it has to do with "the nature of diodes" is a non-statement. If you add a voltage regulator to the circuit I attached in my previous post you'd probably get 1000's of hours out of those LED's. We're talking about lighting paper models. This voltage regulator setup would be a cheap and easy way to go. (see attach.) This is a linear circuit. That is how that LED would act. The capacitance effect of a diode in an oscillating circuit would be a different story. (Yes, diodes can act as a "one way capacitor".) I can only speak from my experience and what I have seen in the 1000 or so TV's I have fixed and various other electronics. There are a lot of ways to do the same thing, so, "What ever floats your boat". I'm done.

    Attached Files:

  9. SteveM

    SteveM Member

    I am not referring to capacitance effects or oscillation, but simply the V-I curve of a diode is not linear. As I have said earlier, once the voltage across the diode achieves the forward voltage of the diode, it will conduct as much current as available even if it blows it self up. A resistor on the other hand has a linear relationship between the voltage across it and the current through it.

    Why is one resistor a bad idea? Take your circuit with only one resistor, tell me the current flowing through each of the 3 diode strings. Second, imagine one of those LED's blows out (opens), now calculate the current in the other 2 strings. Repeat for the circuit with 3 resistors. Now imagine you want some of those LEDs to be be blue and some red some small and some large. Unless the parallel strings have exactly the same forward voltage you will get some serious current imbalances resulting in shortened lifetimes.

    The only reason people get away with circuits like this is that they use voltage sources that cannot supply enough current to blow out the LED's. But that is just shifting the problem into the supply, change the regulator to supply more current and you blow up the LED's. However, if the circuit is designed correctly you don't depend on the supply limiting the current, the circuit will operate safely regardless.

    And by the way, your most recent diagram is labelled wrong, the bottomost line is not -12VDC.

    And speaking of experience, I have an electrical engineering degree (Masters). I've built and designed circuits for 20 years. I have more than just a hobbyist's knowledge of how these devices work. Yes, there are many ways of accomplishing the same thing, there are also right ways and wrong ways, and just because it "works" doesn't mean it is not the wrong way.
  10. rmks2000

    rmks2000 Member

    I appreciate all the input this query generated. Now, I'm looking to upgrade my magnifier headpiece to replace the krypton bulbs with LEDs. The light assemblies are encased in plastic boxes riveted to each side of the headpiece and hold two AA batteries each. I've read an article on the net on how to glue an LED within an existing bulb's socket. This would allow me to retain the slide on/off switch. However, I'm not sure I have enough room in the case to add the requisite resistor. If that is the case, I could just use the LED without the bulb socket and add an exterior toggle switch. Can someone point me in the right direction as to what to look for in a low voltage non-illuminated switch?
  11. Zathros

    Zathros Guest

    That -12 v should be ground. Why not use a 10,000Volt source then you could power 1 billion LED's. If your using a nominal power supply that would not blow the circuit then you would have an efficient circuit. Why not use a 40 pound transformer to power your tiny paper model, heck, why not hook up a 40,000v volt generator. You made my point exactly without realizing it. Why don't you design a circuit for this chap using a nominal power supply that does the job and won't burn his house down. He wants to light up a paper model. By the way I have worked with many engineers as a machinists for 16 years and an electronic technician. I have learned it is what you produce not what you say. So let's see an efficient circuit. I have no doubt you can produce one. I believe he is looking for something battery powered. Remember KISS!!
  12. Amazyah

    Amazyah Senior Member


    Turning into a rough crowd in here.

  13. SteveM

    SteveM Member

    I already showed an efficient circuit (and I even showed it as battery powered). Just put one resistor in series with each of the series strings of LED's what is so bloody inefficient about that?

    And you didn't answer my questions about what happens if one of the LED's in your circuit opens up for some reason. The answer is the current in the other strings will increase by 50% very possibly blowing out an LED in each of the other strings. With my circuit, the other two strings are not affected at all and maintain the same current regardless of how many strings are in parallel.

    KISS indeed, keeping one resistor per LED or series string of LED's actually IS the simplest solution because it relieves you of having to know anything about the current supplying capacity of the voltge source. All you need to know is that it can provide more than you need. Without the resistors then you are dependant on knowing what the supply can deliver and that it will not overdrive your circuit. And that is important especially if you want to be able to substitute a DC adapter for a battery pack or D cells for AAA's.
  14. Zathros

    Zathros Guest

    Excellent, how about some values for the plebiscite. What I am saying is that with your wealth of knowledge, why not offer a complete circuit with values for a 3 string 6 string 9 string etc. set up. There are many people who don't know how to do these things. I am a technician, not an engineer. A chart showing the substitutions would be nice and helpful.. I could trudge thru this but if you placed a circuit at the beginning of this thread with some values this thread would have been a lot shorter. I belong to a few forums that show how to do DVD and Video copying/editing. When someone asks us something we answer it. If it is a question that has been answered a lot we refer them to the sticky on it. There is no question that your knowledge far exceeds mine. Share it. Post a circuit for people with a rudimentary knowledge and that will be helpful. It would be helpful if you included the variables the come with different color LED's. You have already shown you know how. Other people in this site seem very generous with their knowledge. It is better to give too much info than not enough. You might want to e-mail these guys and tell them their third example is the wrong way to do it.

  15. SteveM

    SteveM Member

    I am not trying to be stingey, I am not trying to keep the secret knowledge from you and all to myself. I have explained several times how to calculate the value of the resistor required. The reason I don't just give you a circuit with the values spelled out is because it depends on the LED you choose to use and the power supply you choose. That's why a gave the equation and tried several ways to explain how to use it.

    I feel that by giving you the equation and how to use it, rather that just a set of numbers, you will be empowered to use the parts you want rather than just the parts I tabulate.

    Want to use a 9v battery instead of 2AA's? knowing how to use the equation will let you choose the appropriate resistor.

    Why don't I just give you the proper setup for blue vs red led's? Because there are thousands of LED's available and they all have their own voltage and operating current. Tell me the LED you are using, its forward voltage, operating current and the power supply you want to use and either I can tell you what resistor to use or you could plug those numbers into the equation yourself. It is a very simple equation, not "rocket science".

    I prefer to try to teach people how to do things themselves rather than just gve them the answer. I think it is better to teach the equation rather than the specific values. I believe in teaching yuo to fish rather than just giving you a fish. I think that ultimately, that is more generous.

    If the equation is confusing or I've explained it badly, I'm more than happy to try to clarify (as I just tried to do with Lizzy's question).
  16. Zathros

    Zathros Guest

    Please forgive me as I have worked with many engineers and I love squeezing them. They are like a big juicy orange's waiting to be juiced.:grin: You are absolutely correct and it's nice to know that you are a stickler for details. It may be useful to say when figuring out the total resistance of the circuit to use either Product over Sum or the Reciprocal method. If you don't know the Rt(ResistanceTotal),how could you figure out the It(CurrentTotal)? As far as explaining this stuff verbal representations of mathematical equations are difficult for many people. A "Sticky" is this forum on Electronic Circuit Designs for lighting Models would be interest and useful. I do not think I would be qualified to write such a thing, as a technician I only know how to put things together. Most people would not know how to figure out a circuit with just Ohm's Law unless there are examples. It is something that becomes obvious after you learn it. The things engineers take for granted are "rocket science" to the rest of us. I hope I have not offended you, as a fellow New Englander I assumed you are as thick skinned as the rest of us.
  17. lizzienewell

    lizzienewell Member

    Thanks. My highschool physics teacher taught me the formula years ago, but I forgot what the letters stand for.

    Here is what I understand of our discussion of two wiring layout options.

    1) Hook up each LED with its own resistor. Advantage: The LEDs will last longer, and the electronics can handle a bigger power supply. Disadvantage: requires more work in connecting each LED, and the stuff will take up more space in the model.

    2) Hook up several LED to one resistor. Advantage:The stuff will be more compact and require less effort to put together. Disadvantage: if one LED burns out they will probably all burn out. If the model small and has an onboard battery, it won't need to handle much power.

    I'm considering putting 3 LEDs on a model boat that is 5"x2" and has little space available for the electronics. The deep part of the hull is less than 1/4." I need to figure out if 3 resistors will fit. If I use one resistor, I might have room for a small battery onboard. The model would be disposable. The battery will go before the LEDs do.

    Not sure if I will try either. I'll have to go to the electronics supply store for the smallest LEDs available.

    I'm impressed with the wealth of knowledge and experience on this site. I'm also amused by putting in use for such a small(pun intended) issue. So how many cardmodelers does it take to change a lightbulb(err LED)? ;-)

  18. SteveM

    SteveM Member

    Okay, lets work an example: keep it simple, one resistor and 2 different LED's in series.

    Power supply will be a 9V battery.
    LED1 will have a forward voltage of V1
    LED2 will have a forward voltage of V2

    since they are in series, the current through them has to be the same so the LEDs have to be chosen so they both light at the same current (typically 10 to 20mA) but lets just call it "I".

    Now to find the resistance.

    The voltage across the resistor will be 9 - (V1+V2) and the current through it will also be "I".

    Ohm's law says V = I*R or R = V/I where V=9-(V1+V2)

    so R = (9-V1-V2)/I

    If you have other strings in parallel just repeat each string with no need to refer to the other strings.

    Now add up all the I's for each string and this is what your power supply will have to be rated for (or how long a battery will last)

    questions welcome. :)
  19. SteveM

    SteveM Member

    That was only for several parallel strings. If you connect say 3 LEDs in series with a single resistor, then if one burns out then it just shuts off and the other LEDs will be fine.

    Resistors can be pretty small too. here is a sample small LED. Forward voltage is 3.5V and maximum current of 30mA, so lets operate it at 20mA.
    Anyway if you want to connect 3 in series you will need a 12V battery. The resistor will be (12-10.5)/.020 = 150/2 = 75 ohms and the wattage is 75*.020 = .150 so an 1.8watt resistor is a little to small, you need to use a 1/4 Watt.

    Now try a parallel connection with 1 resistor per LED. We need more than 3.5 V so lets use 3 1.5V button cells for 4.5V. Voltage across each resistor is 1V and so R = 1V/.020A = 50ohms at .100W so a 1/8W resistor for each will work.

    This was just the first LED I pulled out of Google, I just selected it as an example. I hope this helps your design.
  20. Zathros

    Zathros Guest

    So a 9V batter powering a LED with a 2 volt voltage drop that needs .02ma would require a 350 ohm resistor. Since that would give max current a 390 ohm would give a little less current and longer LED life. COOOL!!! R=(Vs-Vl)/I, that was easy!! Thanks.

Share This Page