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LED wiring

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Posted by Anonymous on Friday, June 25, 2004 2:14 PM
HMMMMMMMMMMMMMM.....................

Lets talk....

I get you them for 50 cents a piece...

or

you could just go here:

http://www.besthongkong.com/

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Posted by Anonymous on Friday, June 25, 2004 12:16 PM
I would like to know where you found white leds for 24 cents. I have several projects waiting for the price of white LEDs to come down.
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Posted by Anonymous on Friday, June 25, 2004 12:14 PM
Angelo, I would like to know where to buy white LEDs for 24 cents apiece. I have lots of applications, but not at $1 each!
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Posted by Anonymous on Thursday, June 24, 2004 6:52 PM
Lionelsoni
I used the configuration that you explained to me.

wire (a)---ledpair----ledpair---ledpair---resistor----wire(b)

The resistors are:
5% tolerance 330 ohm 1/2 watt

The leds are:
pure white 11000 mcd (the higher the number the brighter)
foward voltage 3.5

transformer running at full power at 10 volts.

I soldered the connections before i hooked to electric.

I shut the lights off and the six led lit up the whole room. The room is 10x 20.
Really bright.

Thanks again for all the info Lionelsoni!!
Angelo
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Posted by lionelsoni on Thursday, June 24, 2004 9:10 AM
Great! Let us know all the details--voltage, resistance, etc.

Bob Nelson

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Posted by Anonymous on Wednesday, June 23, 2004 4:21 PM
SUCCESS!!!!

I got it, it works and with just the right ammount of light. I got the picture but couldn't figure out how to get it posted. I am sure you know what it would look like already. I will try to figure out how to post the finished product, when it is done.

Thanks Lionelsoni.


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Posted by lionelsoni on Friday, June 18, 2004 10:27 AM
I'll bet they are not actually "white" LEDs. The color of an LED is produced in the semiconductor die buried in the plastic. The plastic itself is often colored, but just so that you can tell what the color is even when it is off. So many colored LEDs are made with clear plastic. White LEDs are fairly new and less likely to be found in old computers (although I realize that "old" for a computer is not as old as "old" for other stuff). The fact that they have 3 leads probably means that they have two colors, probably red and green, and are actually two LEDs in one package. They might be useful for a train signal, if you can figure out which lead is which. The most common arrangement is for the two LEDs to share a common cathode lead. You can get yellow from such a dual LED by turning on both red and green at the same time, but you have to adjust the relative brightness to get a good yellow.

Bob Nelson

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Posted by Anonymous on Friday, June 18, 2004 8:57 AM
Thanks lionelsoni for your response. Now another question...I came across some white LED's in some old computers my office had laying around. These have 3 wires coming out of the LED, how would I go about using this in my engine for lighting that only has 2 wires?

Thanks,
MarkB
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Posted by Anonymous on Thursday, June 17, 2004 9:53 AM
Thanks Lionelsoni,

I haven't gotten the leds or the resistors yet, but hopefully they'll be there when I get home from work.
I got it for sure this time .

Angelo
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Posted by lionelsoni on Wednesday, June 16, 2004 12:05 PM
I wi***his forum editor were wysiwyg (what you see is what you get) because it seems to have garbled your second picture. However, the first one looks right to me.

I'm going to give the circuit drawing a try:

...............|----|<|----|......|----|<|----|......|----|<|----|
(A wire)----|...........|------|............|------|...........|----Resistor----(B wire)
...............|----|>|----|......|----|>|----|......|----|>|----|

Try to ignore the periods (.....). They're there just to try to hold the space open.

Bob Nelson

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Posted by Anonymous on Wednesday, June 16, 2004 10:01 AM
Hey lionelsoni,

I just had a thought, when you said that I can wire three pairs of LEDs does that mean that they go in a chain or seperate?

(A wire)-----led pair ------led pair ------led pair------resistor------(B wire)
or

___Led pair
(A wire)-----Led pair ------Resistor----- (b wire)
------Led pair


Thank you
Angelo
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Posted by lionelsoni on Wednesday, June 16, 2004 9:45 AM
The short lead, or the one nearest the flat on the base of the LED, is the cathode. The end of the diode with the stripe around it is the cathode. The arrowhead in the diode symbol on paper points to the cathode. The other terminal in each instance is the anode.

You can wire all three components in series, with both the LED and the diode pointing the same way. Or you can wire the diode back-to-back across the LED, cathodes to anodes, and then put the resistor in series with that pair. In neither case does it matter which end of the resulting circuit connects to which of the wires feeding the lamp.

For headlights, I like to use the series circuit, with a capacitor (about 100 microfarads) across the LED, to make it turn on and off less abruptly, like an incandescent headlight. The negative terminal of the capacitor goes to the cathode, the positive to the anode.

Bob Nelson

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Posted by Anonymous on Wednesday, June 16, 2004 5:58 AM
If I wanted to use an LED inside an engine for lighting instead of the regular light, how do I figure out which wire is the ground and which one is power? Also, do I only need a resistor and a diode along with the LED for the circuit? I have the "Easy Electronics Projects for Toy Trains" book currently checked out from the library and from what they show that is all you need for an LED circuit. Is this correct?
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Posted by Anonymous on Friday, June 11, 2004 3:09 PM
I paid thirteen of that for shipping and insurance. I got them for about ten cents each. Sounds like I'll doing some trimming!! Thanks for the idea. I got some more in blue to put in between the station supports. I am just waiting on them now. This led thing may go a lot futher than chandelires.
Thanks for the info Lionelsoni,

Angelo
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Posted by ChiefEagles on Friday, June 11, 2004 1:56 PM
Yes Angelo., toy train passenger cars. Pull down the voltage so your engines are straining due to the wattage and the voltage drop and not the load. Disconnected all my passenger car lights and the F3's fly around the track.

 God bless TCA 05-58541   Benefactor Member of the NRA,  Member of the American Legion,   Retired Boss Hog of Roseyville Laugh,   KC&D QualifiedCowboy       

              

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Posted by lionelsoni on Friday, June 11, 2004 1:21 PM
Angelo, that sounds like a pretty good price.

It might be useful to know that you can do a bit of cutting on LEDs without hurting them, if you don't cut into the electrical insides. In particular, you can cut off the flange around the base if it is in the way, either mechanically or cosmetically. You can also cut off the rounded tip and poli***he flat end with toothpaste, to make something that looks more like a headlight or signal light. Also, you might want to try roughing up the surface of your chandelier LEDs so that the light comes out the sides as well as the end.

Bob Nelson

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Posted by Anonymous on Thursday, June 10, 2004 10:17 AM
Chief,
when you wrote passanger cars I thought you meant real cars (autos). When I read lionelsoni's post I realized you meant train passanger cars.

Sorry,
Angelo
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Posted by Anonymous on Thursday, June 10, 2004 10:13 AM
Lionelsoni,

I woke up yesterday and hadn't the foggyest idea how to do anything with a led. All I knew was that you could test batteries with them. You took me through and now I feel ready to do some wiring. I am waiting for the leds, when they arrive I will test them with the resistors (I picked up 100 for 24.00 yesterday). Thanks again for all the info I will post the outcome.


Angelo
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Posted by lionelsoni on Thursday, June 10, 2004 9:40 AM
Angelo, we're getting close.

Don't decide on the resistance value until you try it with the actual LEDs; and start with 1000 ohms, as I've said before. It might be bright enough for you. You probably don't have a stock of resistors to play with; so you should probably calculate the power rating for several resistances in advance and buy one of each, to save trips to Radio Shack or whereever. Then, after you settle on the resistor that gives you the right brightness, go back and buy the other 13.

Your power calculations look right to me.

Of the voltages that you list, I would use 10 volts for 2 pairs. You want the transformer voltage to be comfortably above the 7 volts of the LEDs. The current and therefore the brightness depends on the difference between the two; so you don't want a situation where a slight change in one or the other makes a big change in the difference between them. However, you would just be wasting power to make the transformer voltage any higher than that.

As for putting LEDs in passenger cars, I would not count too much on saving power, unless you run one of the newer schemes with a constant, high track voltage. The reason is that you would not want to have more than one LED pair in each circuit, so that the lights would still be on at low track voltage and would not vary too much in brightness as the voltage changes. This would involve dissipating much of the power in the series ("ballast") resistor, rather than turning it to light in the LED.

LEDs' light output actually varies less than incandescent lamps'. They have the additional advantage that the light color does not change with brightness, which seems to me to make the varying brightness less noticeable.

Bob Nelson

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Posted by Anonymous on Thursday, June 10, 2004 8:35 AM
ChiefEagls:

I know that people are converting their passenger car lights to LEDs. I haven't done it myself.

Daylight Don almost certainly can tell you how to do it. Lionelsoni almost certainly can, too.

Tony
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Posted by Anonymous on Thursday, June 10, 2004 12:34 AM
Hey chief they use them in caddys, crown vics, and some traffic lights (red). The new trend is putting blue ones on the hoods of civics. In the windshield wiper spray nozzle housing.

Angelo
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Posted by Anonymous on Thursday, June 10, 2004 12:25 AM
I think I got it.
I just got some leds, they are 2.6-3.8v the specs say they are 3.5.
They also say 11000mcd, when I was buying there were numbers from 3000 to 20000. Does that make a diffence in what I am using them for?

Tell me If I got it down pat. I figure I will use 330ohm for example.
-I am going to have two pairs per resistor. So that’s 2x 3.5 which is 7
-Lionel RW transformer voltage 9/10/15/19 (whichever pins connected)
(stuck at full throttle, I use only for scenery, got it like that, and left it)

Out comes: .8/ .7/.5/.4 k1
Outcomes: .12/.18/.33/.43 k2

9X9=81/330=2454x.12= 1.9x .025x2=.5
10x10=100/330=.3030x.7=.21x2=.5
15x15=225/330=.681x.33=.225x2=.5
19x19=361/330=1.09x.4=.437x2=1
so according to the answers I would use a 330ohm with .5 for any one of the first 3 and a 330ohm with1for the last one... ?

What do you think lionelsoni are my calculations correct?

I want them to be a bit dim but bright enough to see the detail in the walls and ceilings of the station. Which would you use?

Thanks again for all the info,
Angelo

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Posted by ChiefEagles on Wednesday, June 9, 2004 11:19 PM
My quetion has been, can you wire LEDs to be used in passenger cars? Passenger cars pull a lot of voltage and wattage. LEDs do not and give off a lot brighter light. We are using them in boat trailer lights now [marker, turn and and tail lights] as well as boat interior lighting.

 God bless TCA 05-58541   Benefactor Member of the NRA,  Member of the American Legion,   Retired Boss Hog of Roseyville Laugh,   KC&D QualifiedCowboy       

              

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Posted by lionelsoni on Wednesday, June 9, 2004 3:44 PM
The 100 ohms was just an example resistance for which you might want to compute the power rating.

Let me modify your diagram a little:

(t1)---ledpair---ledpair---ledpair---Resistor---(t2)

Notice that I have replaced (-) and (+) with (t1) and (t2), to represent the terminals of the transformer. (I wanted to avoid - and + because they imply a DC power supply.) I have changed "led" to "ledpair", since each "ledpair" is two LEDs wired back-to-back, for mutual protection.

I would put about 3 pairs of LEDs with each resistor. However, since you will have only 2 pairs in each chandelier, you might want to put only 2 with each resistor, so that each chandelier is an independent circuit, or try 4 pairs (2 complete chandeliers).

In any case, wire up the LEDs and try it out with 1000 ohms first, then reduce the resistance in steps until it is bright enough to suit you. You can probably go down by factors of 3, that is, from 1000 ohms to 330 to 100 to 33, etc., then fine tune it if none of those is quite what you want. At each step, be sure to use a resistor whose power rating is high enough, according to my instructions above.

Here is another example: Suppose your transformer voltage is 16, you decide to put 2 LED pairs per resistor, and the LED forward voltage is 3.5 volts. You will start with 1000 ohms, but you need to calculate the power rating for that resistor. k1 is 3.5 x 2 / 16, or .4375, as above, so we use .43 for k2. The power that the resistor will dissipate is (16 x 16 / 1000) x .43, or .11 watt; so you can use a 1/4-watt resistor. Suppose this is not bright enough, so you go to a 330-ohm resistor. Its power is (16 x 16 / 330) x .43, or .33 watt, so you should probably move up to a 1/2-watt or 1-watt resistor. If you take the next step, to 100 ohms, the power will be 1.1 watt and you will need a 2-watt resistor. Suppose that is too bright, but 330 ohms was too dim. You might next try a 220-ohm resistor, whose power dissipation will be (16 x 16 / 220) x .43, or .5 watt; so you use a 1-watt, 200-ohm resistor--and the lighting is perfect!

You're not driving me crazy. I like this stuff! Keep on asking questions until I explain it well enough that you understand.

Bob Nelson

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Posted by lionelsoni on Wednesday, June 9, 2004 3:14 PM
I have a way that's not too complicated (I hope) to compute the actual resistor power rating. This might be worthwhile because, when there are lots of LEDs in the string, the resistor power goes down quite a bit. Here's how it works:

Multiply the LEDs forward voltage by the number of LED pairs, then divide this by the transformer voltage. Call this "k1". Using k1, look up "k2" in the following table:

k1 k2
0 1
.1 .83
.2 .68
.3 .55
.4 .43
.5 .33
.6 .25
.7 .18
.8 .12
.9 .08
1 .05

Finally, calculate the power as I described above, by squaring the transformer voltage and dividing by the resistance, then multiply the result by k2. To continue my example that resulted in 2.5 watts, suppose that there are two pairs of 3.5-volt white LEDs. So k1 is 3.5 x 2 / 16, or .4375. To be safe, use .4 for k1 and get .43 for k2. Multiply 2.5 watts by .43 to get a resistor power value of about 1 watt.

Bob Nelson

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Posted by ben10ben on Wednesday, June 9, 2004 2:48 PM
One end of the resistor goes in series with one of the leads of the LED. This is all that there is to hooking it up.
Ben TCA 09-63474
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Posted by Anonymous on Wednesday, June 9, 2004 2:43 PM
OK I think I have this all down except for a few things.

Is this correct:
(-)-------led------Resistor-----(+)

I want to run a string of 14 chandaliers each with four leds, how would I plug this figure into the rating formula.
lastly where does the 100ohm come from? Is that the standard rating for the leds or a single type?

Thanks for the info
Hope I'm not driving you crazy .
Angelo
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Posted by TexasEd on Wednesday, June 9, 2004 2:35 PM
Bob,

How are you doing? Good to see you posting on this forum. Let me know if you want to run trains some time. I may modify my layout to accomodate your double stacks by lifting the upper level 1/4"

Ed W.
http://www.trainweb.org/ttat
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Posted by lionelsoni on Wednesday, June 9, 2004 1:28 PM
You can have as few pairs as you want, even just one pair. The resistor takes up the excess voltage. Wire the LED pairs in series and put the resistor in series with that. Then connect the ends of the entire series string to the transformer.

The resistance of the resistor depends on the number of pairs, the transformer voltage, and the brightness that you want. I suggested starting with 1000 ohms since that is almost certainly dimmer than you would want and very safe. The calculation of the power rating of the resistor is complicated; but, if you calculate it as if there were no LEDs at all, you will get a conservative value that is sure to be greater than the actual power dissipation.

To calculate the power rating that way, square the transformer voltage, then divide by the resistance to get the power in watts. For example, 16 volts squared is about 250, which divided by 100 ohms is 2.5 watts. I would probably double that to 5 watts to be on the safe side and not have a very hot resistor.

The modern white LEDs are so bright that I would not worry about exceeding their current rating. They will be brighter than you would want long before you reach their full output.

Bob Nelson

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