75 LEDs installed on a medallion

Discussion in 'The Projects Forum' started by flipper80, Feb 13, 2010.

  1. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    I am doing a school play with some kids. So I got this idea with large hip hop chains, lighting them up with leds as I got access to about 350 LEDs. They all are yellow, and the specs are 2V and 0.02mA.

    The idea came after I accidentally discovered how to light up leds, and managed to light up some leds in parallel here with a resistor between the V+ source and the first led in the array. Between the ground(- of battery) and the first led it's no resistor. This is based on a drawing I found on the net, and seem to correlate with info other places.

    But, here I find a lot of different and new info. As I understand it, parallels with one single resistor is a NO NO. Also I found a link here, led.linear1.org/led.wiz, that puts the resistors between the ground and the leds, not between V+ and the leds. Does it not matter where one put the resistor in relation to the LEDs and V+?

    So I am a bit confused now.. I have read several posts on this site, but still have a couple of questions.

    Specs on the Duracell homepage says the battery(mn1500) here will last for more than one hour at 1.08A. 75 LEDs multiplied with 0.02A is 1.5A. 2 AA batteries can then deliver enough current?

    Am I right, that IF I want to power up 75 LEDs in parallel with just one resistor, that resistor would have to lower the current fromn 5A to 1.5A?

    3V supply - 2 volt / 1.5 A = 0,67 Ohm. This means I would have to use a 1 Ohm resistor, or is it never a point using 1 Ohm resistors with LEDs due to the variation in usage of current and volts from LED to LED? What is then the minmum of resistance one should use with LEDs?

    I have tried to light up 18 LEDs in parallel with one 2,7 Ohm resistor(3v-2v/(0,02Ax18LEDs)=1/0,36=2,78 ohm). The resistor is placed next to the V+ source. The multimeter says the current is 90mA when measuring this circiuit near the ground. Near the V+, between the resistor and the + on the battery, it's about 115mA.

    From my understadning, the resistor is lowering the volt, from the ground to the resistor, so we get a higher current from the resistor to the +V? It still puzzles me as it looks like the resistor adds more current from nowhere between the resistor and V+. I understand it as current comes from the ground(minus pole of battery).

    Measuring current without LEDs, and only the resistor, display 470mA. If we subtract 470mA from the 100mA I got when measuring with LEDs, we got 370mA. This means the LEDs eat about 370 mA. This correlate with 20mA x 18 LEDs is 360mA. Or am I just doing cargo science now?

    I am puzzled by the leftover of 100 mA. Is this normal, or does it mean I should add more resistors? I wonder if it's possible to just add resistors until measurements looks good, instead of relying on math alone.

    The 2.7 ohm resistor gets hot after one minute or something, is this an indication something is wrong? The Voltage measurements also show 2,2V from ground to the resistor, that is more that the 2V each LED require. Is this normal?

    To lower the 100mA current, I put another 47 ohm resistor in serie with the 2.7 resistor. Current is lowered to around 20mA and volt to 1.8V between the resistor and ground. Between the V+ and ground, it's 3V. This somehow makes sense if the resistor lowers the volt, while the volt of the battery still is there. The lights are weaker, but still strong enough for use on a dark stage. I realize that this is the setup for one LED connected to a 3 V source(3V-2V/0,02A=50 Ohm), still I am surprised how well this works with 18 LEDs, still I want them brighter if it's possible in a safe way.

    I am trying to understand the concept that LEDs eat current. I have read "Electronics for dummies", but can't find an explaination there. So I hope those questions aren't too huge to answer here, and the leftover of 100mA puzzles me.

    I understand the reasoning for one resistor for each led, as forward voltage vary from led to led, but is it a way I can get away with less than one resistor to each led? Will it work for 10 minuttes without harming LEDs and batteries?

    A hip hop chain medallion can fit some 75 LEDs. Is it manageable to light them up with with 2 AA batteries? The area is a circle, measuring 30 centimeters in diameter, made up of hard paper, so LEDs will be easy to install through the paper, wiring it on the back, and adding a tree plate to cover the wiring and attach batteries to.

    My idea now, is to perhaps use 4 AA batteries, and put 4 LEDs in serie, each of them connected to 17 or 18 LEDs in parallel. Four 2v LEDs in series requires 4 LEDS x 2V, in total 6V, something that four 1,5V AA batteries in series can provide. Then I am planing to use as few as possible resistors, balancing between stability and resources(money and installing 350 resistors is..). I am also thinking about sorting out similar LEDs to be paralleled. I found a good way to measure LEDs in a breadboard circuit, described here: http://www.allaboutcircuits.com/vol_3/chpt_3/2.html, as measuring LEDs with my 10$ meter don't work very well(meter display a number for a short time, before going blank, and the number vary).

    Those LEDs won't be used for long on the stage, and will maybe be put into other more stable circuits for other purposes later, if they still work.

    Or is there other better solutions?

    The show only last for 5 minuttes or so, but it's imortant to not run risks of overheating or exploding anything here.

    Sorry for unclear english, as english not is my native language, but hope this still is understandable!
     
    Last edited: Feb 13, 2010
  2. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    You can buy LED strings at many hobby places complete with full instructions.

    [​IMG]

    This example is from Hobby King in Hong Kong. It does have reasonable shipping costs, if you are not located nearby.

    John
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    Putting LEDs in parallel is indeed a bad idea. If a few LEDs fail they can take the rest of them with them.

    Have you read this?

    LEDs, 555s, Flashers, and Light Chasers

    You will find we are quite willing to help out around here with any questions.

    How sure are you of the 2V drop across the LEDs? If so then they are of an older generation. Nothing wrong with them, but it affects the math.

    Generally you are better off with this configuration...

    [​IMG]

    There will be major adjustments in the resistor values and whatnot, but the base schematic give you the idea.

    Resistors are quite cheap, less than 2¢ each in quantity.

    Figure ½ of the LEDs will be used, so there will be 175 legs, if each leg is pulling 0.02A then you will need 175 X 0.02A, or 3.5amps. The AA batteries will not work.

    Like I said, we'll work with you. For now think car battery, even for 5 minutes. Is this possible?
     
    Last edited: Feb 13, 2010
  4. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    BTW, I was remiss in forgetting something. Welcome to AAC flipper80.

    Considering your one-time need, potential future needs, and need for simplicity, I believe an off-the-shelf solution is best. These LED strings, which are essentially bought by the yard, are tolerant to individual LED failures, are light weight, fool proof, and do not require car batteries to operate. Even small, electric models can carry enough for night flying. This video is a larger model (I didn't have one to share from my local club).

    The biggest problem our club has encountered is being arrested for creating public panic. It seems three members were flying advanced electric designs, like flying wings, and some local residents thought they were being invaded by UFO's. No one actually got arrested, but it did make the local newspaper.

    John
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Yep, welcome to AAC!

    Thinking about it. 75/2 is 38 legs, 38 X 0.02A is 0.76A. Better, but still a battery killer. Here is the odd part, if you double the voltage (12V) you can use half as many legs, which drops the current down to .38A. 24 volts is even better (which is 16 AAA batteries).

    Like I said, we can work with you to get you where you want.

    How hard is it to get parts where you're at?
     
  6. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    I agree on the power requirement, but not that it is a battery killer for a 5-minute show. 12V at 0.38A is just 4.56W for 5 minutes or 0.38 W-hours. A 1.5V AA NiMH is typically rated at 2000 mAH or more. 2X1.5V is 3.0 W-hours. No problem.

    John
     
  7. 2Tell

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    Nov 29, 2009
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  8. Audioguru

    New Member

    Dec 20, 2007
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    LEDs each have a different voltage. If they are connected in parallel then each must have the same voltage so flashlight manufacturers test and sort them so the ones in each flashlight are matched. They have thousands of LEDs to test and sort, you don't.

    You made many tests but you assumed your two AA battery cells always produced 3.0V. But they produce 3.0V only when brand new and only for a minute or two with a load. Their voltage drops as they run down as shown when you used your "3V" battery to light many parallel LEDs without a current-limiting resistor. Then you measured a fairly low current. So the battery voltage dropped to 2.0V but you didn't measure it.
     
  9. Wendy

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    Mar 24, 2008
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    I just did, it is seriously flawed. You can not use 1Ω resistors to limit current, the program is suffering from a serious lack of common sense. It is a pretty basic bug.

    Don't believe me? Try these parameters...

    Battery 6V
    Vf 2V
    Current 20ma
    Number of LEDs in your array 75

    Following AG's advice, try dropping the battery voltage to 5.8V, you'll get more realistic results.

    Ever been around kids? That battery will be toast long before the show. :D
     
  10. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    Thanks for reply!


    The 2 volt drop across the LEDs is something I am not sure I did mention? Sorry for beeing unclear, and I see my question is a bit confusing.. The current dropped 380mA across the LEDs, by measuring the difference in current by including the LEDs versus going straight to the resistor, bypassing the LEDs. It looked like the LEDs did eat some 380mA of current, but not sure if the current/electrons have disappeared into light and heat, or if it's related to volt.


    The voltage did not vary the same way, but will have to test this again once I have purchased a new fuse to the meter. Blew it a few minutes ago. No surprise really, as I was too curious to see the numbers of measurements from pole to pole on the battery. A whopping 5.5A and something new learned...


    What I am trying to understand is the Ohms law and how voltage and currents drop across the circuit, so guess I will have to some math and more measuring to hopefull detect a pattern here.


    The link you provided, was one of the first articles I found on this site, and very helpfull!


    After checking your suggestion, I am thinking about a 75/3 solution, 25 legs with 3 LEDs in series, as more LEDs in series are favorable and more stable? Also this requires fewer resistors, 25 instead of 38.


    3 LEDs, 2V each, requires 6V in total(1.5V X 4 AA alkaline batteries)? Each leg would consume 6V and all the legs 0.02A X 25 = 0.5 A. Resistors for each leg would be 6V – 6V / 0,5 A is ZERO? Sorry if the question is silly. Is the calculating here suggesting that the power supply is too low in voltage? Another AA battery makes the supply 7.5V. 7.5-6V/0.5A=3 Ohm. This suggest a 3.3 Ohm resistor at each leg with 7.5V supply?


    A 75/2 solution requires 4V at each leg. This is within the range of what three AA batteries can provide? One alkaline AA battery can do 1.08A in one hour according to the Duracell website? 4.5V-4V/0.76A=0,66 Ohms. 1 Ohm resistors at each leg, or can I throw in 10 Ohm resistors to make the cicuit a bit safer?


    The problem with car batteries is that the players on the stage is wearing the chains around their neck. As said, just 5 minuttes, and only requirment is that the batteries don't explode.


    I found a dealer nearby that seems to got cheap resistors if I buy 100 of the same kind, they also seems to have common parts in electronics.
     
  11. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    The voltage varied measuring between the ground and V+ and between the ground and the resistor. This correlate with what I have read, so doubt it's the battery that got weaker meanwhile. Sorry for unclear explaination. I did use a current-limiting resistor all the time. Between the ground and the V+, it's allways 3V. Between the resistors, the voltage rise as I get nearer to the V+.
     
  12. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    Hello, and thanks for reply!


    I did not knew about those strings, thanks for the tip. They can be handy if everything else fails here.


    I don't plan to get in the local newspaper, but it's inspiring to learn what one can do with LEDs :) You got a point with off the shelf solution, but I have some time left over here, and want to try to see how far I can get on my own, and perhaps learn something new.
     
  13. Audioguru

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    Dec 20, 2007
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    You don't understand that the voltage from a battery drops as it is used.
    The AA Duracell alkaline cell starts at 1.5V. If its current is 0.5A then it is less than 1.5V after less than 1 minute. The curves in the datasheet shows its voltage dropping.

    You talked about using three 2V LEDs, a 3 ohm current-limiting resistor and a 7.5V (5 AA cells) battery.
    When the five AA cells battery is new and is 7.5V then the current is 0.5A. The Duracell datasheet shows the battery lasts about 1 hour with the LEDs dimming the entire time. At 1 hour the LEDs don't just suddenly shut off, they are so dim that you can barely see them. At half an hour they might also be too dim to be useful.

    If you use a battery with a higher voltage then you can limit the current to the LEDs with a current regulator circuit. Then the LEDs will not dim, they will be bright for an entire hour.
     
  14. Wendy

    Moderator

    Mar 24, 2008
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    OK, 100 part of the same kind will get you there. But first we need to know what the Vf (dropping voltage) the LEDs use is.

    Are you commited to the 6V power supply (be it 4 AA batteries or 4 D cells)? If so we can go with that.

    Assuming it is really 2V dropping voltage (you need to verify this!) then 2 LEDs in series will drop 4V together. This leaves 2V (6V - 4V) for the resistor. Ohm's Law is:

    E=IR, or Voltage = Current X Resistance

    So, 2V ÷ 0.02A = 100Ω.

    The voltage you measure across the 100Ω resistance is the real current being fed into the LEDs on that chain. So if you measure 1.9V, it means the LEDs are getting 19ma.

    Need a schematic?

    38 chains at 20 ma work out to 0.76A.
     
  15. Audioguru

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    Dec 20, 2007
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    You forgot to say that an alkaline battery begins to die the moment it is first used and keeps dying until its voltage is so low that it doesn't power your project properly anymore.
     
  16. Wendy

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    He also said he want 5 minutes use, which can be done. Personally I'd want more, but to each his own.
     
  17. Audioguru

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    But during the 5 minutes the LEDs are very bright at the beginning then they dim, and dim more, then dim more and maybe then are not bright enough.
     
  18. Wendy

    Moderator

    Mar 24, 2008
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    Your point being? :D I still recommend a car battery, or a plug in, or D cells, but it is the OPs call. Those are going to be hot batteries.
     
  19. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    Thanks for decent replies. A good night sleep also help clear up what's going on here.

    Your way of calculate resistors indeed makes sense, and undertanding this Ohm law better now.

    I suspect the specs for LEDs here, at 20mA is a maximum value, and that those LEDs are effective at 10 to 15mA, too. 18 LEDs in parallel sharing only 20mA(3V supply and a single 50 Ohm resistor) are ok lit, more than dim. Tried one LED, 3V and 100 Ohm, giving about 10mA, and the difference isn't that huge from 20mA. I am getting light spots in my eyes after staring into it(6 degrees, clear water glass LEDs).

    So a setup with 76/2, 10mA in each leg, draining 0,38A could work too, or even a 75/3, draining 0,25A? It would be safer, too, help avoiding the popcorn effect you mention in your article?

    D type batteries is fine if better. As long the players can move freely, it does not matter what kind of source they use. But this leaves out car battery and wall plug. But what about a 9V E block alkaline? The size of 2 AA batteries, but 9 VOLTS! Can that work here?

    Is it a limit how thin the wire should be here? I checked limits of AWG wires, and looks like at least 24 is required, with limits of 0,577A for power transmissions. I got a 32 wire here, that is pretty easy to wrap around the leds and quickly solder, and worked with 18 LEDs, but looks like a bad choice according to the charts. Will a 32 wire still do ok for 10 minuttes?

    Yes, would like to see a schematic of 72/2 if it don't take much timeto make one.

    I am not sure how to measure the forward voltage of the LEDs, but after some reading, my guess is that one have to measure voltage before and after the LED. Before it's 0,04, and after, it's 2,07. At the V+, the battery, it's 2.99V. The circuit got a 52,6 Ohm resistor. 2,07-0,04 is 2,03 Vf?

    I have uploaded a drawing of the circuit and where voltage was measured, in case this don't make sense. If it's a better or another proper way to measure Vf I would be happy to know. The multimeter don't work very well for this.

    What is the minimum of Ohms in resistors for LEDs for good protection, if one can ask such a question?
     
  20. flipper80

    Thread Starter New Member

    Feb 13, 2010
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    Ok, now I got what you are talking about, and will have this in mind.
     
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