Still in the planning stages of some layout/structure etc. lighting projects. As I am not familiar with electronic theory much…
How much would, say, 12 LEDs wired in series draw if wired to an old Lenz100 decoder lighting output? Would it be possible?
Would the output be too dim?
Would I need 12 resistors?
I know my ques. may be goofy, unfamiliar as I am 
First, you need to know the voltage and current rating of the LED’s you’ll be using, not all LED’s have the same ratings. Then you need to know the voltage output and current limit of your decoder. If you plan on using a resistor with each LED, the current draw for the resistors will need to be added in. We won’t get into why you’re using a decoder and not wiring them straight to, say, a wall wart or old power pack. LED’s can be wired in series like incandesent lamps (except you have to observe polarity), to cut down current draw, but you have to know the above ratings first.
If you are not familiar with electronics, I would go one resistor, one LED. For twelve volts, the standard is a 1k resistor.That allows 10 ma through the LED which for most purposes is sufficient. Most LEDs, the maximum current is 20ma.
I have been an electronic tech for many years and I never, ever go with LEDs or 1.5 volt bulbs in series. Resistors are very inexpensive.
Manufacturers sometimes put this stuff in series to save manufacturing cost.
Rich
The first thing to remember is that an LED is not a light bulb. It behaves differently. A diode will pass as much current as it can until it passes too much and burns out. That is the purpose of the resistor. It limits the amount of current passed through the LED.
Depends on the LED.
That would be the best plan.
Thanks so far.
Ok, the decoder function output is 100mA
I LEDs are 3mm Miniatronics Yello-Glo LEDs.
If I were to wire 10 Miniatronics Yello-Glo LEDs in series with one resistor on one bus and connect it to the decoder with 100mA output using 12 to 13 V on the DCC system, what would happen?
Would there be any discernable light coming from the LEDs?
Some Spectrum locos have quite dim LED headlights. One I have, only 4 ma of current passing through it. I think the resistor was around 2.7 k. Not sure got rid of it and installed a 1k. The decoder voltage is about 12.6 volts which gives me about 10ma of current. Decent brightness. It would be noticeable next to a LED drawing 20ma of current but the difference is not too much from a couple experiments.
Another Spectrum I have was about the same.
Rich
Why do you want to wire them in series? That’s just asking for trouble. If they’re not all exactly identical, some will draw more current than others, it’s much harder to calculate the resistance and if one comes loose or fails, all of them go out. Wire them in parallel and attach a 1,000 ohm resistor to each one (assuming your power source is 12 volts). Like someone said, resistors are cheap so why do it the hard way.
Assuming a DCC system with a 13 volt output, the deocder output will probably be around 11.5 volts, though you really should measure it to make sure. That gives us the voltage and current of the decoder output, but you still need to know the voltage drop and current requirements of the LED’s.
You’re best bet may be a combination of parallel and series circuits. So that we have some numbers to work with, let’s assume that the voltage drop for your LED’s is 2 volts and that you want to run them at 10 mA… If you have 12 LED’s, you won’t be able to wire them all in series because the total voltage drop would be 122=24 volts, and the total voltage drop has to be less than the supply voltage. You also would not be able to run them all in parallel because the total amperage draw would be 1210 = 120 mA, greater than the 100mA output of your decoder.
One possible answer in this case would be to run 3 groups of 4 LED’S, with LED’S in each group wired in series, and the 3 groups wired in parallel with each other. Each group has a total voltage drop of 6 volts, so the calculated resistor for 10 Ma would be (supply voltage-voltage drop of LED) / current = (11.5-6)/.01 = 550 ohms. I believe the next standard size up is 560 ohms, so using that value each group draws (11.5-6)/560 = .0098 amps = 9.8 mA. The total current draw of the 3 groups would be 3*9.8 = 29.4 mA.
4 LEDs dropping 6 volts? Not most white ones, such as the Minaitronics YeloGlo and the ones Richmond Controls sells - they are 3.1-3.6 volts EACH. Most red/yellow/green LEDs are 1.8-2.1 volts each.
–Randy
[%-)] All elements in series draw or pass an identical amount of current.(Source voltage - Sum of LED voltages) / 10 mA = resistance in k ohms
[tup]Very true. And unless you have some test equipment it might be hard to find the bad one.
Oops, that should have been 8 volts, I multiplied by 3 instead of 4. Here’s the corrected math: Ccalculated resistor for 10 mA would be (supply voltage-voltage drop of LED) / current = (11.5-8)/.01 = 350 ohms. Closest standard size = 360 ohm. Current draw of each group = (11.5-8)/360 = .0097 amps = 9.7 mA. Total current of all 3 groups combined = 3 * 9.7 = 29.1 mA. Again, this is assuming a 2 volt drop per LED as I stated earlier. If these LED’s have a voltage of 3.1 - 3.6, then the most you could have in a series would be 3.
The main point if my post is the same though. Using LED’s in parallel requires more current but using LED’s in series requires more voltage, so sometimes the best solution is a compromise approach using a combination of series and parallel circuits.
Thank you all, gentlemen. As I say this is a planning stage.
Threads like this are always interesting to me because the discussion really evolves until a best answer is more or less obvious to me…and I can see others picking up info along the way too.
I appreciate the input. My little learning is a dangerous thing.