bagman Please excuse my ignorance, but when you talk about installing a resistor, is that for each of the LED's ?
Please excuse my ignorance, but when you talk about installing a resistor, is that for each of the LED's ?
Yes.
bagman...I should add that there will be 4 LED's, each one attached to 4 x IRDOT (InfraRed Detection Of Trains) units...
bagman When you say connect the LED's in series, does that mean "+" terminal is connected to "-" terminal ? cheers and thank you all again
When you say connect the LED's in series, does that mean "+" terminal is connected to "-" terminal ?
cheers and thank you all again
Yes. Technically, it's anode to cathode, but the same idea. I have done this many times, with both incandescent lamps and LED's. True, if one goes out, the rest of the string (2 or 3 lamps or LED's) will go out, which will happen with LED's or lamps in series, but since they're drawing less current and a slightly lower voltage, the chances of this happening are reduced. There's just as much a chance that the resistor tied in series with the LED's will blow. The main advantage with LED's is they draw less current and hence, produce less heat. A major advantage with plastic buildings. A resistor does produce heat. Where you have a large building with lots of lights, the LED's will be less likely to damage the building with the heat, like a lamp or resistor will.
Right after Christmas I picked up a 60 bulb tree light set of "clear" LED's on sale (they look more blue to me).Almost the entire strand is used in my 9 stall roundhouse, wired 3 in series, then each group connected in parallel with each other, powered by an old wall wart that was a battery charger for an old cordless drill. The resistors would be another disadvantage in a large project like this, you have to calculate the power draw of the resistors as well as the LED's to determine how large a power supply you'll need. I use a 400 mA power supply on the roundhouse, If I had used resistors, I would need 3 times that, or 1200 mA (1.2 Amps) With a 12 volt power supply, I'd need 4 times that size (1.6 Amps).
Concur - it is generally NOT a good idea to put LEDs in series. And not just from the fact that if one blows, they all go out. Tolerances are such in manufacturering LEDs that even if they come fromt he same batch, they may have differences in forward voltage and current. Putting them in series would result in some getting more current than others for a given resistor. This may be OK for cheap Christmas lights (which is what they do - the LEDs are in series and there's a diode to filter the 120VAC) but the idea here is to not have to repalce the model railroad equipment after a year or two.
When LEDs are placed in parallel, you use a resistor for each LED, calculated based on each LED's rating. Each LED will then get the proper voltage and current. You can even mix and match in parallel, say one LED that needs 3.6 volts and another that uses 2.1, so long as you use the correct resistor for each one, they both will work perfectly well.
Rule of thumb to keep it simple - one LED, one resistor. If you have a lot to hook up, you can buy resistors in quantity from a place like Mouser - about 2 cents each. 2 cents vs possibly blowing the LED trying to get fancy - a bargain for sure.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Driline bagman Hi guys Thank you all for your replies. When I said the output was 40mA that is for each IRDOT not the LED. I'm not sure what the LED's are rated at. When you say connect the LED's in series, does that mean "+" terminal is connected to "-" terminal ? cheers and thank you all again Your typical LED is 20 to 40ma or so, so thats fine. Since your power source is 12v, use a 470 ohm resistor. DONT put them in series. If one blows, then all the lights will go out. You are going to install them in parallel. So yes, hook up + to the positive side of each LED, and then - to the negative side. On most LED's you can tell the positive side by the LONGER of the two leads coming off the LED. The resistor will be soldered in series.So solder one lead of the resistor to the positive side of the LED, and then solder the OTHER side of the resisitor to the + side of the 12v power source. Somebody needs to post a diagram of this. If I had time I would.
bagman Hi guys Thank you all for your replies. When I said the output was 40mA that is for each IRDOT not the LED. I'm not sure what the LED's are rated at. When you say connect the LED's in series, does that mean "+" terminal is connected to "-" terminal ? cheers and thank you all again
Hi guys
Thank you all for your replies.
When I said the output was 40mA that is for each IRDOT not the LED.
I'm not sure what the LED's are rated at.
Your typical LED is 20 to 40ma or so, so thats fine. Since your power source is 12v, use a 470 ohm resistor. DONT put them in series. If one blows, then all the lights will go out. You are going to install them in parallel. So yes, hook up + to the positive side of each LED, and then - to the negative side. On most LED's you can tell the positive side by the LONGER of the two leads coming off the LED.
The resistor will be soldered in series.So solder one lead of the resistor to the positive side of the LED, and then solder the OTHER side of the resisitor to the + side of the 12v power source. Somebody needs to post a diagram of this. If I had time I would.
Hi Driline
Many thanks again to you and the other members of this forum who have replied to my post.
You guys are very generous in sharing your knowledge and time in answering questions put to the forum.
Greatly appreciated.
Many thanks again
cheers
Bagman
rrinker I would go with a 470 ohm or 560 ohm resistor, if the MAX current for the LEDs is 40ma. Running right at the limit is usually not the best thing, and running somewhat less current through it will still generate plenty of 'light' (IR), extend the life of the LED, and prevent a slight surge in the DC output from exceeding the current limit. --Randy
I would go with a 470 ohm or 560 ohm resistor, if the MAX current for the LEDs is 40ma. Running right at the limit is usually not the best thing, and running somewhat less current through it will still generate plenty of 'light' (IR), extend the life of the LED, and prevent a slight surge in the DC output from exceeding the current limit.
Yep...thats good advice.
Phone wire, or network cable, works fine for this, because of the low current. If you have more than one, the 8 conductors of network cable will allow up to 7 LEDs (7 LEDs, plus a common for all of them) in just a single cable - keeps things neater. Or if you have a lot - 50 pair phone cable! But beware the same colors repeat in 50-pair cable - there's two of every color!
R=E/I......... Resistance = Voltage divided by current.
You'll probably need a 300 ohm resistor in series with the positive lead on the LED so you don't blow it.
12/.040 = 300
Due to the low power requirements of LED's you can use a small gauge wire (higher AWG number is smaller wire) like #20 or even telephone wire if you can find it. Much smaller than that gets hard to strip and solder
You need to know the voltage rating of your LED's to tell if this is the right power supply. LED's can be connected in series to be able to connect them to a higher voltage power supply. For example, a 12 volt power supply can power 3 volt LED's if you connect 4 of them in series. It's just like incadescent light bulbs, but you have to observe polarity with LED's.
Hi there
I want to hook up some Led's to a control panel that will be approx 10 feet from the power source.
What size (AWG) wire should I use ?
The power source is 12vDC @ 1.5amp.
Hopefully I have provided enough info. If not please let me know.
As stated above, thank you in advance for your replies and assistance.
I should add that there will be 4 LED's, each one attached to 4 x IRDOT (InfraRed Detection Of Trains) units, each one of which has a maximum curremt comsumption of 40 mA.