I'm using miniatronics "Grain of Dust" lights with 163 decoders, WHAT VALUE RESISTOR should I be using? I need to use the Grain of Dust size bulbs with my steam engines. I mount them in back of an MV Lens installed in a brass headlight casting. Is 800 ohms too much? Will using a 1/2 watt resistor instead of a 1/4 watt resistor get me anything?
using miniatronics "Grain of Dust" lights with 163 decoders, WHAT VALUE RESISTOR?
- Thread starter Y3a
- Start date
You can calculate the required resistor using any of the online LED calculators like http://ledcalculator.net/ . You will need to know the voltage output of your decoder function (usually 1.4 volts less than your track voltage), the voltage of the bulb and the current rating of the bulb as well.
The answer the calculator will give you will be the resistor that will drive the bulb at its maximum potential - which you probably don't want to do. In order to save stressing the bulb and make it last longer, my general rule of thumb is to take the answer and double it, then adjust to suit my own taste for brightness. The calculator will give you the absolute minimum value you can use without damaging it.
For what you are using, a 1/4 watt rating is fine, the 1/2 watt would just be that much more ability to dissipate the absorbed voltage. The wattage rating is basically the amount of heat the resistor can safely dissipate, and you won't be dissipating that much. If you are curious about the wattage, you can use this calculator with your known values .... http://www.csgnetwork.com/ohmslaw2.html
Mark.
The answer the calculator will give you will be the resistor that will drive the bulb at its maximum potential - which you probably don't want to do. In order to save stressing the bulb and make it last longer, my general rule of thumb is to take the answer and double it, then adjust to suit my own taste for brightness. The calculator will give you the absolute minimum value you can use without damaging it.
For what you are using, a 1/4 watt rating is fine, the 1/2 watt would just be that much more ability to dissipate the absorbed voltage. The wattage rating is basically the amount of heat the resistor can safely dissipate, and you won't be dissipating that much. If you are curious about the wattage, you can use this calculator with your known values .... http://www.csgnetwork.com/ohmslaw2.html
Mark.
Iron Horseman
Well-Known Member
Ohms law E=IR.
Assume max voltage from Digitrax 16V.
16V - 1.4V for the lamp = 14.6V
14.6V = 0.015A * R so R = 14.6V / 0.015A or 974 ohms. I would use a 1K.
Watts law W = EI
so 16V * 0.015A = 0.24W
Assume max voltage from Digitrax 16V.
16V - 1.4V for the lamp = 14.6V
14.6V = 0.015A * R so R = 14.6V / 0.015A or 974 ohms. I would use a 1K.
No.
Watts law W = EI
so 16V * 0.015A = 0.24W
Last edited by a moderator:
Not to pick nits but in the interest of accuracy the wattage dissipated by the resistor would be the current through the resistor times the voltage across the resistor - in your example that would be 14.6V not 16.0V.
Iron Horseman
Well-Known Member
Nits and technical accuracy are good as long as they don't obfuscate the over all message. It is always my hope that others with similar issues will read these threads and learn from them, so this was a very good nit.
