I have three very small micro LEDs:http://www.ebay.com/itm/201116774969?_trksid=p2060778.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AITSpecs are V(forward) = 3.4V, 0.015A (15mA) each current draw.I have a Digitrax TL1 decoder, which provides 0.125A of current, and I attached a multimeter to the function output of the Digitrax TL1 when the function is turned out, and I got around 11.8V (let's call it 12V).So I want to find a way to attach one single resistor to the three micro LEDs in parallel.Questions: 1. Do I use the voltage that I measured from the multimeter of 12V or do I use the DCC voltage at the track, which is about 13.4V? 2. As for the voltage drop for Ohm's Law, I assume I use the 12V instead of the 13.4V from the track voltage. Is this correct? 3. If each LED is in parallel with eachother, the voltage across each LED is 3.4V as taken from the link above. So the voltage drop across the resistor would be 12V - 3.4V = 8.6V, correct? 4. If each LED is drawing 0.015A of current, and they are in parallel, then the total current draw is 0.015A + 0.015A + 0.015A = 0.045A, correct? 5. So I have V = 8.6V, I = 0.045A, and R is calculated by R = V/I = 8.6V/0.045A = 191.1 ohms for the single resistor.Problem is that I don't trust this value because the Digitrax documentation for the TL1 recommends a 470 ohm resistor.Are my assumptions and math sound?I'm trying to get these hooked up to the end observation car platform on my N scale Polar Express for my son.
1. Use the output voltage from the decoder
2. see #1
3. correct, though why're you wiring them in parallel? (you say there are marker lights -- might be easier to wire them in series)
4. no, unless all three are on at the same time (else, it's just 0.015) correct, although why're you running them in parallel?
5. No. I'll go into more detail below.
5.1 470 ohm resistor @ 12v probably means they're calculating for a single 25 mA LED. (12/470 = 0.025)
5.2 we have a 12V (nominal) input voltage, and need to drop it to 3.3v for the LED (yeah, I know you said 3.4 ... but whatever). 12 - 3.3 = 8.7 / 0.015 = 580 Ohms . 580 is not a standard resistor size, but 620 is ... so we'll use a 620 ohm, 1/4 watt resistor and be golden. (granted, this is for ONE LED)
5.3 Assuming we're running all three in parallel (as in your original post), then we'll need a 220 Ohm 1W resistor (well, that's the nearest standard size. Your 190 Ohm result sounds about right on the raw math - but it's not a standard size).
5.4 Assuming we wire them in series (because we can), then we only need a 150 Ohm 1/8 watt resistor, since the three LEDs are dropping 9.9V themselves. (2.1v / 0.015 = 140 ... which again isn't a standard size).
-Dan
Builder of Bowser steam! Railimages Site
Just a slight modification - most of those little LEDs are VERY bright. The 15ma rating on the LED is a MAXIMUM, so you want to calculate using a much lower current value, both to protect the LED in case the voltage gets slightly higher than the measured 12V, and also to keep them from looking like miniature suns. As little as 5ma per LED might be plenty. In parallel that's about 573 ohms, not standard, but 560 ohms is, which would be a little more than 5ma each. Series would be 360 ohms, also non stantdard, but 330 ohms would work and be about 5.5ma.
For the 560 ohm resistor for parallel, a 1/4 watt resistor would be enough. For the series 330 ohm, the 1/8 watt would be fine.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Yeah, definitely agree with Randy on the "drop it farther than you think you need". It's a miracle I got things (close to) right with having barely half a cuppa coffee right now.
However, I'd still look to get a 560 Ohm 1/2 watt resistor. My math shows it putting out just over 1/4 watt when running (0.26 watt), which would probably still be OK -- but hey it's only like 50 cents more on the 5-pack for the 1/2 watt vs. the 1/4 watt; and with my luck, it'd fail when I needed it most (e.g. Christmas, trainshow, etc.)
Thank you gentlemen! My thought about wiring them in parallel was that if one failed, I would not have to replace the entire bank of LEDs. Series might be the way to go.
If you click the link on the top of this thread (the ebay one), there is a link in the auction to the spec sheet. Can't really tell from the spec sheet where the negative and positive wires (blue or white decoder wires) should be hooked up. I assume the resistor goes on the blue wire? And if I wire these in series, do I need a single 330 ohm resistor?
Edit:I tried one resistor. Success! All three LEDs light up in series. The end platform for the observation car is going to thrill my son. I also plan on adding two kids and a conductor to the observation car for that scene in the movie where they discover the North Pole. Awesome.
gatrhumpyMy thought about wiring them in parallel was that if one failed, I would not have to replace the entire bank of LEDs.
This is a common concern, but if just one LED fails, why replace all of them? If a single LED fails, it can quickly be identified using an ohm-meter. Since an LED is diode, it will conduct in one direction (polarity).
In fact, once you wire them up, you can check that they are all wires correctly and verify their polarity using an ohm meter. Wiring in series will require less current/power and fewer resistors.
greg - Philadelphia & Reading / Reading
Problem with parallel is that if one LED goes, the current the remaining 2 get goes way up. The key is that LEDs re current devices, not voltage devices like an incandescent bulb. LEDs drop a certain voltage, incandescent bulbs need a certain voltage to light. Exceed the LEDs current, it blows. Exceed the bulb's voltage and it blows.
Parallel is best for bulbs, for the very reason that if one blows, the other two stay lit. Series is best for LEDs, at least if they are all close in ratings - current in series is shared equally, so if you put a 25ma LED in series with a couple of 10ma LEDs, you have to add a resistor so that no more than 10ma flows. The 10ma LEDs will be full brightness, the one that can handle 25ma will be somewhat less bright (though probably not a lot in this contrived example - but if we set it up so that those 10ma LEDs got no more than 5ma, the 25ma LED would not be super bright by a long shot).
Series LEDs, so long as you don;t add up more voltage drop than the power supply voltage, also allows a smalelr resistor - both because there is less voltage to drop int he first place, and also because the current in series is shared, not added like it is in parallel, so you have just the defined current flowing through the resistor, whereas with parallel LEDs you have 3x current for 3 in parallel. This determines the wattage of resistor required.
Since you got it working, the rest here is pointless - blue is positive in the decoder, resistor can go anywhere in the series circuit, blue wire before any LEDs, between the second and third LEDs, on the white (or whatever function you want) wire before the LEDs - it doesn't matter. For headlights, I usually put the resistor on the white or yellow wire and runt he blue common to each LED. 'Tradition' says you don't put the resistor in the middle of the LED string, but electrically it makes no difference. If doing so would make things fit better mechanically, don't hesistate.
A breadboard is handy for building and measuring circuits without soldering.
Jim
For 3 identical LEDs, you should be able to wire in parallel, but if you have dissimilar LEDs, series is the only way to go. I was wiring a signal bridge with red, yellow and green LEDs, and my parallel circuit just didn't work. Only one of the LEDs would light, as their current demands were different. Once I went to series they all worked fine.
It takes an iron man to play with a toy iron horse.
MisterBeasleyFor 3 identical LEDs, you should be able to wire in parallel, but if you have dissimilar LEDs, series is the only way to go. I was wiring a signal bridge with red, yellow and green LEDs, and my parallel circuit just didn't work. Only one of the LEDs would light, as their current demands were different. Once I went to series they all worked fine.
i assume you tried wiring the 3 LEDs in parallel thru a single resistor.
the current drawn by a diode (LED) varies exponentially with voltage -- a slight increase in voltage results in a large increase in current. The LED that draws the most current at the least voltage will "turn-on" first and the voltage across the other LEDs will never be high enough for the others to draw enough current to be visible.
while this is most likely true for LEDs of different types, especially different colors, it may also be true of LEDs of the same color and type -- they may not all be the same brightness.
gregc gatrhumpy My thought about wiring them in parallel was that if one failed, I would not have to replace the entire bank of LEDs. This is a common concern, but if just one LED fails, why replace all of them? If a single LED fails, it can quickly be identified using an ohm-meter. Since an LED is diode, it will conduct in one direction (polarity). In fact, once you wire them up, you can check that they are all wires correctly and verify their polarity using an ohm meter. Wiring in series will require less current/power and fewer resistors.
gatrhumpy My thought about wiring them in parallel was that if one failed, I would not have to replace the entire bank of LEDs.
You cannot check an LED with an ohm meter. I have used LED's since 1972. Tried that many years ago.
Use a 9 volt battery and a 1k resistor. Some super bright LED's you might have to experiment with different resistors or use a 5 k pot in series to find the brightness you like and then get the closest standard value.
With three LED's, I have used a single 1k resistor. It is rare that an LED will burn out. Enough current to burn one will usually burn all.
I will add, to date all my analog and digital meters show max resistance on all scales and the diode test scale. Both directions and good batteries in the meters.
Maybe there is a multimeter that will check LED's now but I have yet to see one.
I belong to a SoundTraxx DCC group and a couple years ago this was discussed. Quite a few bought a LED tester on ebay which ships from China. I think they did cost about $6.00 at the time with free shipping.
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
Ok, to follow up, Google multimeter to check led's. Some are doing it.
To date I have not been able to with my meters. Read in different DCC forums not too long ago that others could not.
it depends on the battery voltage in the ohm meter. Older meters with a 1.5V battery cannot measure the resistance of an LED requiring 3+Volts. Modern digital meters may have 9V battery which should be sufficient.
So I have an update. I want to add another micro LED to the top of the observation car (just like in the movie), actually, this one:
http://www.miniatronics.com/Merchant2/merchant.mvc?Screen=PROD&Store_Code=M&Product_Code=12-001-12&Category_Code=6_1&Product_Count=1
The specs are not there, so I assume it's a 3V LED. How would I wire that in series with the three 3.3V LEDs I have in there now? What resistor value would I need? Maybe 100 ohms?
Those are red LEDs, probably 2.1 volts Vf and likely a different brightness rating than the ones you have for the other 3. I'd run this one seperately in parallel with the rest, not add it to a series chain, because it does have different characteristics.
So if my calculations are correct (amperage of 0.005A going to parallel red LED, voltage drop of 12-2.1=9.9V needed), and your assumptions are correct (2.1V), I have a second resistor value of R2 = 9.9V/0.005A = 1,980 ohms. I could use either a 1.8kohm or a 2.0 kohm resistor, correct?
This is assuming the other branch of the parallel circuit contains the three LEDs and the other 330 ohm resistor.
I got it to work! Used a 470 ohm resistor for the parallel branch for the red LED out of the back. Looks great.
Well done! N scale even!!
Dave
I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!
Not sure what the current limit is on those red LEDs, since it doesn;t state in the link (are they marked ont he package, by any chance?), but with a 470 ohm resistor and the known 12V function output, you are putting 21ma into the LED. That's likely close, but shouldn't be over, the limit. Vf on the red LED could be as high as 2.7V, as well, instead of just 2.1V, which puts the current under 20mA.
Since the red LED + resistor are in parallel to the white LED string and its resistor, there's no need to factor that in when calculating the resistor for the red LED. However, you need to add up the current flow through the white LED string plus the current through the red LED and make sure it doesn;t exceed the limit for the function output. In this case, it doesn't, even running a set of parallel LEDs at a full 20ma, it would take 6 of them to get near the 125ma function limit.
Should be getting easy now, you've done it a few times. The formula was always easy for me to remember: E=IR. E is used for voltage - Electromotive force. I is current, in amps, and R is resistence in Ohms. My Dad's name started with E, and he worked at IR (Ingersoll Rand). Know any 2 and basic algebra shows how to determine the missing value.