FAQ - Electrical
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Frequently Asked Questions


What wire sizes do you use for track power?

I use 12 gauge stranded for bus wiring because I have some relatively long wiring runs and am concerned about voltage drops.

I use 20 gauge solid wires for feeders between the track and the bus wires.

I make sure that EVERY piece of track is either soldered to a feeder wire or to another piece of track which is soldered to a feeder wire. The longest piece of track it have attached to a single feeder is 6' because nickel silver doesn't conduct that well.

NEVER trust unsoldered rail joiners to pass power from one piece of track to another!

How do I figure out what size of dropping resistor to use?

When converting a model from DC to DCC sometimes it's necessary to add resistors in series with the headlight and reverse light bulbs. This is required when those bulbs are rated for 1.5V operation since the lighting functions of DCC controls typically output around 12V to 13V (or more).

To compute a resistor size you'll need to know

  • How many milli-amps the lamps need to operate
  • How many volts the lamps need to operate
  • How many volts the DCC decoder you're using outputs on the function wires
Ugh. And it get's a bit worse because the decoder light outputs are probably a square wave so just slapping a voltmeter across them may not give an accurate measurement.

You can assume a decoder output of around 13V in HO scale. If the loco had constant intensity lighting for DC operation then the voltage for the lamps is very likely to be 1.5V. Lamp current can be anything from 15mA (milli-amps - meaning 15 1/1000ths of an ampere) to 40mA depending on how bright they are designed to be. Lets assume 20mA to start with.

Once you know the volts and the current then the resistor can be computed as follows:

Ohms law (named after a Mr. Ohm who figured it out) states:

I = E / R

 I is current in amperes
 E is voltage in volts (E is short for Electromotive force)
 R is resistance in ohms
Ohm's law says the flow of electrons (current) is proportional to how much force is attempting to move those electrons (voltage) divided by how hard the electrical path doesn't want them to move (resistance). It's alot like water flowing through a pipe. The flow depends on water pressure divided by how small the pipe is. Use a bit of high school algebra to rearrange Ohm's law to solve for resistance.
R = E / I
E is the volts across the resistor which is the DCC decoder function voltage minus the volts across the lamps.
E = 13V - 1.5V = 11.5V
So solve for R.
R = 11.5V / 0.020A = 575 ohms
The next higher standard resistor value is 620 ohms. So in theory added a 620 ohm resistor in series with the light bulb should drop the decoder output to the necessary 1.5V.

But you may have noticed that the previous calculation seemed like a bit of work? And that we had to make guesses as to the decoder's voltage output and the rated current of the lamp?

I usually take out a reistance substitution box. This is a small box with a whole bunch of resistor of different values installed in it. Mine has a rotary switch on top that lets me pick the resistance I want. I just set it to 1000 ohms, hook it into the decoder output circuit with clipleads, turn on the power then start reducing the resistance a bit at a time until the lamp seems about bright enough.

Much easier! But the danger here is I might be optimistic about how bright the lamp should be. You can get an idea of the rated brightness by hooking the lamp to a 1.5V AA battery. Observe the brightness then pick a resistor that keeps the lamp a bit dimmer.

What wattage resistor do I need?

So you figured out how much resistance you need in series with that loco's headlight for it to be the correct brightness. But do you need a 1/8th watt, 1/4 watt, or 1/2 watt resistor?

Oh boy! Time for some more math!

The equation for electical power is

P = E * I
 P is power in watts
 I is current in amps
 E is voltage in volts
We know the resistance. We can estimate the voltage as being around 12V for HO. Let's mix Ohm's law into the power equation as follows. Ohms law states that
I = E / R
So let's subtitute E/R for I in the power equation yielding.
P = E * E / R
which we solve (for the previous example) as
P = 12V * 12V / 620ohms = 0.232 watts
It appears that this resistor will need to disappate just about 1/4 of a watt. If we're adventurous we'll use a 1/4W reistor (because they smaller!). If we're conservative we'd choose a 1/2W resistor. Since the DCC function output probably isn't pure DC but is pulsed on-off-on-off-on-off etc. we can probably get by quite well with the 1/4W resistor in this case.

Can I use a single ressistor for two lamps?

My loco has dual headlights that will both be wired to the same DCC function output. Can I use a single resistor to drop the voltage for both of them to save space?

The answer is yes, but...

  • You'll need a reistor sized for the current of both lamps. Instead of 0.020 amps in the previous example you'll need to compute for 0.040 amps. This may require a 1/2 watt resistor instead of a 1/4 watt to handle the extra power.
  • If one of the lamps either burns out or gets accidentally disconnected while voltage is applied, the other lamp will get double it's rated current which will most likely turn it into a 'flash bulb'.
So I would recommend against this practice with lamps.

But what about if the head lights are LEDs (light emitting diodes)?

I don't recommend operating two LEDs in parallel without resistors in series.

This page and images copyright © 2008 by Charlie Comstock