On the last WPF thread (Weekend Picture Fun) MisterBeasley suggested it is possible to wire LEDs in series to reduce the number of wires where multiple LEDs are being used close together. He did it with a 10 LED signal bridge and it worked fine.
My question is: how do you calculate resistor value? In my case I am currently installing headlights and tail lights in HO scale vehicles. Each has two warm white and two red LEDs. When each LED is wired separately the resulting bundle of wires coming off the bottom of the vehicle is larger than I would like it to be.
Any suggestions?
Thanks
Dave
I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!
Voltage of devices in series adds up. However, with red and white LEDs, they have different drops and wiring them in series will result in uneven current distribution. You could get away with wiring the two white headlights in series, and the two red taillights in series. Two white LEDs in series, if typical 3.5V drops, would be 7V, plug into the calculatiosn exactly the same as a single LED> 12V power source, -7V drop = 5V, 10ma current would be 500 ohm resistor. Not a standard value, and white LEDs are usually very bright so go one higher. 560 ohms.
Red LEDs are more commonly about 2.1 volts, so 12V - 4.2V drop = 7.8 volts. 10ma there would be 780 ohms, also not a standard value, but red LEDs are usually dimmer than white so you can try one lower, 680 ohms. That would be 11ma.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
The first thing I would do is put them all together in series with some convenient resistor and just see how they look. For a vehicle, I would want the white headlights brighter and the taillights a bit dimmer, so it may work out. If not, then wire up the two sets as Randy has suggested.
My inclination would be to start with a 1K resistor. Again, it's for a vehicle, and the low-level lighting you'll get with the larger resistor is more appropriate than what you'd want for a locomotive headlight or a signal bridge.
For a few dollars, you can get a set of short clip-leads, which makes this sort of bench testing a quick, solderless process. Let us know what you come up with.
It takes an iron man to play with a toy iron horse.
rrinker ...However, with red and white LEDs, they have different drops and wiring them in series will result in uneven current distribution...
...However, with red and white LEDs, they have different drops and wiring them in series will result in uneven current distribution...
That's not correct. Current is the same throughout a series circuit. However, what you may run into is that you need to power the LED's at different current levels to get the desired brightness, which you can not do when they are wired in series. As long as the total voltage drop of the LED's is less than the supply voltage, you can power any combination of LED's in series as long as they are acceptable all running at the same current level.
As an example, with 2.1 volt red LED's and 3.5 volt white LED's and a 12 volt power source, you could power two red and two white LED's in series(2 x 2.1 + 2 x 3.5 = 11.2). To power them at 10 ma would require an 80 ohm resistor( (12 - 11.2) / .01 = 80). The closest standard 5% value is 82 ohms, which would give you (12 - 11.2) / 82 = 0.00976 = 9.76 ma. As long as both the red and white LED's look acceptable at that current(or whatever current you choose), then that will work fine.
Another advantage of powering LED's in series is a reduction in the total current draw of the circuit. In the above example, the total current draw would be 9.76 ma, whereas if all of the LED's were wired in parallel, the total current draw would be closer to 40 ma, the sum of all parallel legs of the circuit.
In an idealized model that's correct, but diode junctions are not ideal resistors and the internal resistence varies, even among the same model LED from the same manufacturer.
Same thing happens in reverse when you attempt to parallel two power supplies. Theoretically, they should have the same voltage and the total current output should sum, since they are in parallel. In reality, they are not evenly matched due to component variation, and one ends up overloaded while the other hasn't yet hit maximum.
rrinker In an idealized model that's correct, but diode junctions are not ideal resistors and the internal resistence varies, even among the same model LED from the same manufacturer...
In an idealized model that's correct, but diode junctions are not ideal resistors and the internal resistence varies, even among the same model LED from the same manufacturer...
I'm not sure, but I think you are referring to my comparison of the current draw of LED' s in series or in parallel. Let me clarify that when I mentioned powering the LED's in parallel I meant with each one having a resistor to limit it's current but each LED + resistor group wired in parallel with the other ones. If you have 4 LED's, each with a resistor to limit it's current to 10 ma, and each LED + resistor wired in parallel, you will have a 40 ma current draw; however, with the LED's in series and a resistor to limit the current to 10 ma, you will have a current draw of 10 ma even though you still have each LED powered by 10 ma.
Randy and CSXRobert!
Here we have The Great Debate!
Seriously, thank you for your input, as well as MisterBeasley's.
I think the message I am getting here is to try a few experiments. First, I will hook all four LEDs up in series and, starting with a 1K resistor and progressing to lower resistances, I will be able to see if I can get the desired brightness out of the LEDs wired in series. I have already discovered that in the first couple of cars the tail lights are too bright unless they are placed in a location where brake lights would be applied, i.e. at an intersection. If I can't get the appropriate brightness with one resistor I will have to use two, or possibly more.
As Randy suggested I have discovered that there is a rather wide variation in resistor values necessary to reduce the brightness in LEDs from the same manufacturer. In the car I am working on right now, one red LED needs a 10K resistor and the second only needs a 2.2K resistor to achieve the same lower level of brightness appropriate for late 50's running lights. I guess that is what you get for purchasing Hong Kong el cheapo LEDs.
This is fun!
Thanks guys!
I think in the end you said the same thing as my first reply, ie, put the two reds in series with a resistor, and put the two whites in series with a resistor, and put the pairs in parallel. If the resistors fit inside the car, only 2 wires would come out, if not, then 3 wires would have to come out. Common side for the two series sets, and then one independent wire for each set for the resistor. But do not attempt to put all 4 in one parallel circuit.
Yes. Experiment.
You are working with low voltages and LEDs. The worst that could happen is that an LED will explode.
It is not like you are working with atomic fusion.
ROAR
The Route of the Broadway Lion The Largest Subway Layout in North Dakota.
Here there be cats. LIONS with CAMERAS
BroadwayLion It is not like you are working with atomic fusion.
If your layout is solar powered, you are.
But yeah, that's why clip leads are your best friend. I know people who would spend an hour doing the math, then breadboard the circuit up only to find that it works exactly as designed, but the LEDs are too bright.
Has anyone seen a LED explode? I've toasted a couple, but all I got was a brighter-than-normal moment, and then I had one of those Darkness Emitting Diodes (DED.)
It didn't physically explode, but I connected one to a 9V battery once, with no resistor, and there was an audible pop. Don't make too many DEDs, as I've heard Lucas in England is very proactive on patent enforcements.
I also once tried to explode a rather large electrolytic cap by exposing it to well above the rated voltage in the wrong polarity (after locating it on the opposite side of a concrete wall) but alas, nothing actually happened.
Back when I was a tech-rep in the stereo biz, I can remember seeing the results of big power supply capacitors really blowing up, they're really dangerous. big dents in the steel covers, sometimes with the circular imprint of the caps in the covers. They shoot off like rockets!
Jay
C-415 Build: https://imageshack.com/a/tShC/1
Other builds: https://imageshack.com/my/albums
Hence why I used a concrete wall for cover. Guess I just didn't do it right - or rather, wrong, enough to make it explode. No fun anymore, do that sort of thing and you'll probably be tagged as a terrorist.
rrinker It didn't physically explode, but I connected one to a 9V battery once, with no resistor, and there was an audible pop. Don't make too many DEDs, as I've heard Lucas in England is very proactive on patent enforcements.
Lucas Electronics named after it's founder Joseph Lucas a.k.a. "The Prince of Darkness". A reputation for their electronics parts well deserved (as anyone who ever had to maintain a British motorcycle from the 50s/60s can attest).
A POP is a miniature explosion. Been there, done that. Check the LED after with a magnifying device.
Different color LED's with have different color intensities and the same current though all. Even the same color LED might have different intensity.
A simple search of the Internet for LED series parallel with bring up a bunch of links that you can store in Favorites for more study. A picture is worth a thousand words. You will find many possibilities for circuits, more than anyone can post here.
No doubt, many links will have an on line LED calculator in the link.
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.