Yoohoo Sean!!! Question for you!!

Hehe

 

btw, the loco is a Kato GE ES44AC with a TCS K1D4 decoder

 

Yes,to be operated independently of the headlight,the ditch lights have to be wired separately and the decoder needs a separate function to control them.

 

I do believe that that unit has light pipes that connect to the headlight to light up the ditch lights and not independent Bulbs or LED's.
It can be done with this decoder,

But you will have to install some very tiny LED's in the place of the light pipes.
I enjoy a challenge give me a shout if you need help getting it done. I work with LED installs that are as small as a pen point, I would also swap out the stock orange white LED's for some true warm whites, they would look so much better!

 

Bwoogie,

Sean is cxorrect, as usual. I have two of the Kato N scale ES44ACs that I just put Digitrax decoders in.

 

Oh those are light pipes. That makes sense. Well i just received my smds. Do i need to put resistors on them?

 

Yes resistors are needed. My K1D4 put out 12.8 volts from the function pads, with a Zephyr. I asked Sean a bit back and he uses 14 volts as the supply voltage to determine resistor size. It is probably a safe thing to do, in the event of a surge. This is the easiest site to use to determine the correct resistor. http://led.linear1.org/1led.wiz

 

Note that frequently you'll also want even larger resistors, since running an LED at maximum current frequently makes it way too bright for the intended use. I don't think any of my LED's are actually running at anything close to 20ma, and my headlights are very bright, and markers are visible and then some.

 

I'm just clarifying John.
We have an led rated for 3v, 20mA, with a minimum draw of 1.5 volts.
Using 12 volts we are told to use 470 ohm 1/2 watt resistor, determined from ohms law and rounded to the highest available resistor. This is not running the led at it's max output correct? This would be running it at the manufactures recommended voltage? So using a higher value resistor will only do so much, once you reduce the 12 volts below 1.5 volts it won't work anyway. If we move up to a 560 ohm 1/2 watt resistor our led will still work but there is very little visible change in the luminance output. Jumping up to a 680 ohm 1/2 watt will prevent the led from lighting do to lack of voltage.
My question is, what does using a larger resistor do for you? It should let the led have a longer lite life from using less amps. But a led using manufacture specs will last a long time anyway. The lighting output will not change much since we are dealing with 1/2 watt increments. It will not affect the temps of the led. It would cause more heat from the resistor using a larger one. Which could be a bad thing crapped into a n scale shell.
I just asking because I in that dangerous stage were I know stuff but not all of it hwell:

 

I'm just clarifying John.
We have an led rated for 3v, 20mA, with a minimum draw of 1.5 volts.
Using 12 volts we are told to use 470 ohm 1/2 watt resistor, determined from ohms law and rounded to the highest available resistor. This is not running the led at it's max output correct? This would be running it at the manufactures recommended voltage? So using a higher value resistor will only do so much, once you reduce the 12 volts below 1.5 volts it won't work anyway. If we move up to a 560 ohm 1/2 watt resistor our led will still work but there is very little visible change in the luminance output.

If you have a white LED that has an operating voltage of 3 volts, a 470 ohm resistor will be running it at 19ma, or basically full power.

I = E / R

9.0 volts / 470 ohms = .01915 amps (19.15 ma)

I typically would run a headlight (the brightest thing I normally do) with a 1k resistor in that situation and be running about 10ma. They're plenty bright with that current.

Jumping up to a 680 ohm 1/2 watt will prevent the led from lighting do to lack of voltage.

Nope, that isn't the way it works. You can use a pretty large resistor and the LED will just get dimmer and dimmer. There is a cutoff point when it won't be visible, but that's down to less than 1 ma through the LED.

 

hmmm.. all i know is i have SMD SMT 0402's

 

also how exactly do you mount the leds? glue?

 

If you link the page you ordered the leds from we can help get the info you need to determine resistor size.

For the leds double sided tape might be the easiest for this install, in my opinion. Determining where to cut the light tubes is going to be the harder part. You will need some glue for this part, either a gel CA glue or Arlene's Tacky glue. Arlene's is a user friendly way to go.

 

If you link the page you ordered the leds from we can help get the info you need to determine resistor size.

I bought them on ebay. here's the specs

*edit* check the next post for a better sized pic.

 

You could mount them with a drop of Cyanoacrylate glue (Super glue) or even a drop of hot glue, or epoxy.

 

Read the instruction but I do believe that decoder already has the resistors on the decoder board for the function outputs.

 

Their web site causes confusion with Sean. I thought that too but it goes into detail in the help section. I had 12.8 volts on mine.
http://tcsdcc.com/public_html/Customer_Content/Technical_Info/Tech_Info/Tech_Index.html

 

12V and a 470 Ohm series resistor on an LED with a Vf of 3V will draw 19mA.
(12V-3V)/470 = 0.019 Amp, near but under the MAX current if the Power supply never goes above 12V (did you measure it?) Many inexpensive power supplies are not well regulated and can be off their nominal value be 5 to 10%. So a power supply marked as being 12V may be 13V. Then with a 470 Ohm the LED current is 21mA which is over the MAXIMUM rating.

LEDs are current controlled devices but have a minimum voltage to start conducting (the 1.5 you listed) and a typical voltage called Vf or Forward Voltage. Use the Vf as the voltage drop across the LED in resistor calculations.

Jumping up to a 680 ohm 1/2 watt will prevent the led from lighting do to lack of voltage.

False. The LED will draw 13mA and still be pretty bright.
(12V - 3V)/ 680 Ohm = 0.013 A

My question is, what does using a larger resistor do for you?

The resistor is what limits (or sets) the current through the LED.

As to heat for the resistor or how to determine the resistor power rating.
Power (Watts) = Voltage across times Current through. So for the 470 Ohm resistor the voltage across it is 12V -3V = 9V. an the current is 19.1mA (calculated above) so it will dissipate 13 V * 0.0191A = 0.172 Watts.

I just asking because I in that dangerous stage were I know stuff but not all of it

Then buy a handful of cheapish red LEDs and a bunch of resistor values and try them. Do use your DVM to measure Vf of the LED and the voltage drop across the resistor. Once a few let out their magic smoke you will be smarter.

The 14V Sean uses for calculations is a good idea for the reason I mention above and will not make much difference in the LED brightness. I typically run LEDs rated for 20mA max at 5mA or less.

By the way, the resistor calculator you linked to above is not completely correct. However it is conservative and chose a larger value resistor than the pure calculation does. Try it and then calculate with the equations I used above.

Have fun.

 

Stan, You are exactly right! My bad!

waltr, Good advice! :thumbsup::thumbsup:

 

lets try that again...

 

I would use a 1k resistor or larger.
Your never going to get into trouble using too large a resistor for value and like everyone has said.
#1 Running an LED anywhere near the max amps is just a bad idea!
#2 You don't want to let the magic smoke out of the LED!

What size wire are you planning on using for hooking up of the LED's?
I use 1/8watt SMD or wire film resistors for these, attach the resistors in a manor as not to rest againt the body as they will get warm or hot and can warp or distort the body.

 

bwoogie - first off, sorry about siderailing your thread hwell:

Walt, John and Sean thank you for the explanations:thumbsup: I see where I was going a stray. The manufacture's data is the max rating and not the recommended. It is more important to be concerned about the amperage draw then voltage supply.
Just to follow up "So for the 470 Ohm resistor the voltage across it is 12V -3V = 9V. an the current is 19.1mA (calculated above) so it will dissipate 13 V * 0.0191A = 0.172 Watts" this should be (9v*0.0191A =0.1719 or .172 Watts)
I really need to go back and redo the engines I have done

 

Ohms law is absolute, it doesn't lie.