I'm in the process of replacing the incandescent bulbs in an Athearn locomotive with LED bulbs. I know I will need to add resistors (and I already have them) to keep the LED happy with 12 volts coming from the decoder, but what about going ahead and placing heat shrink tubing completely over the resistors after soldering them to the wires? Does this cause any issues with the resistors overheating and failing, or is this not a concern in this application?
Good Luck, Morpar
Morpar but what about going ahead and placing heat shrink tubing completely over the resistors after soldering them to the wires?
after? typically put the tubing on before
but there's no need unless there's a possibility of the metal leads contacting another electrical component
greg - Philadelphia & Reading / Reading
I always use heat shrink tubing on the end of the resistor that connects to the LED to minimize the possibility of shorts.
Rich
Alton Junction
gregc Morpar but what about going ahead and placing heat shrink tubing completely over the resistors after soldering them to the wires? after? typically put the tubing on before but there's no need unless there's a possibility of the metal leads contacting another electrical component
Really? How do you make the solder joint (wires to resistor) with the heat shrink in place? I always put it in the final position and shrink it AFTER doing the soldering work.
Same as Rich, I sometimes cover the resistor with heat shrink. If you're worried about heat then that seems to me that you're running too close to the rating of the resistor and you should bump up to a half-watt or better.
Some of my passenger car lighting involves strings of up to ten SMD resistors and a 1/8 or 1/4 watt resistor gets too warm for my liking. It may not melt plastic but it may also fail earlier than I'd like. In a locomotive you are generally only running one or two LEDs on a resistor so shouldn't be a problem with heat dissipation.
Good Luck, Ed
MorparI always put it in the final position and shrink it AFTER doing the soldering work.
i usually cut the tubing to size and slide it down the wire (or whatever), solder the wires then slide it over where it needs to go and shrink it down.
gregc Morpar I always put it in the final position and shrink it AFTER doing the soldering work. i usually cut the tubing to size and slide it down the wire (or whatever), solder the wires then slide it over where it needs to go and shrink it down.
Morpar I always put it in the final position and shrink it AFTER doing the soldering work.
What he's asking is if there would be any issues with putting heat shrink tubing over resistors. His question has nothing to do with the order of operations or process for using heat shrink tubing.
i have an old HP oscilloscope that apparently has a resistor that has a power rating too small for the application. it gets hot and over time fails and needs to be replaced.
a 1/4W 1k resistor passing 16 ma will be near its rated power. it's not obvious how this affects it.
another option is to use stripped insulation from a heavier gauge wire that is slide over the solder joint without covering the resistor
gregca 1/4W 1k resistor passing 16 ma will be near its rated power
You cannot get a 16 volt drop if supply is 12-15 VDC.
Start with the assumed voltage, 15VDC at decoder or power source.
15/1000 = .015 amps
I squared-R gives .225 watts at resistor.
I use 1/8 watt resistors and do not worry about heat.
Heat Shrink tubing could shrink even more and tear open.
I've seen that happen in industrial settings.
So from the answers actually related to my question it sounds like going ahead and sliding a single piece of heat shrink over the resistor and soldered connetions in lieu of using 2 separate pieces (1 for each side of the resistor) won't be a problem. After many years of industrial experience I NEVER leave an electrical connection bare, as sooner or later it will be a problem. I really didn't think there would be an issue with heat, but figured it was easier to ask than to have a strange problem down the road.
Never had a problem with covering a resistor with heat shrink. Some are decades old. Heck. Look at automotive circuits. Some are completely encapsulated in epoxy or silicone and they don't fail from heat.
Pete.
Greg posted a trap.
Power is also I^2 x R. You don't need to "assume" a voltage. (0.016 A)^2 x 1000 is 0.256W.
Actually, accepted practice is to always use 2X a resistor rating. They're cheap enough and some are still small enough at 1/2W, too - because yes, you really don't want to find out what happens to the resistor.