Do you need help figuring out which resistor to use to keep from burning out your
1.5V lights on your DCC equipped locomotives? This tutorial will walk you through
the basics and give you the information you need to prevent them from burning out.
The logic can be applied to light bulbs of any voltage as long as you know the voltage
and the current, which is usually expressed in mA. For instance, typical 1.5V lights
consume 15mA, but we need to convert it to amps, which will give us 0.015A.
Let’s start off defining the variables we’ll be using:
E – This is the difference in electrical potential between two points in a circuit.
The unit of measure for the potential is volts (V).
I – This is the current that flows on the wires. The unit of measure for current
is amps (A).
R – This represents the resistance of components and determines the flow of current
through a circuit. The unit of measure for resistance is ohms ( ).
P – This represents the power. In our case we are interested in the amount of power
dissipated by the resistor. The unit of measure for power is watts (W).
Next up, the relationship and formulas for E, I and R. By placing these variables
in a triangle it should help you to ‘see’ and remember the formulas:
As you can see, once you set up the triangle it’s very easy to derive the formula
to solve for the unknown. For example, if you need to solve for R, then cross off
R and look at the relationship between the remaining variables. In this case we
have E over I. In other words the resistance (R) is equal to the potential (E) divided
by the current (I).
The last formulas we need are to calculate the power dissipated by the resistor
once all the other variables have been calculated:
As a matter of fact we only need the first formula where the power (P) is equal
to the potential (E) times the current (I).
Next we’ll draw the circuit to help with the calculations.
For a resistor we’ll use this symbol:
For the light we’ll use this symbol:
This is the full circuit for the class lamps on function 6. We’ll break the circuit
down and redraw it between the x’s, which will represent one lamp.
Here is the simplified version that makes the calculation much easier:
We need to start plugging in some numbers so we can figure out what resistor we
need. What do we know? Well, we can figure out what the potential (E) is across
the common (blue) wire and the function (green) wire by measuring it with a voltmeter.
Typically this is around 13-14V (track voltage), but you can run into problems if
you take your loco to a friends layout and the voltage is higher, as I’ll demonstrate
later. For right now I’ll do all the calculations based on 14V and show you what
could happen if the voltage is higher.
The next numbers we can plug in are for the light. We know the lamp is rated at
1.5V (E) – 15mA (I). This means that at 1.5V the lamp will consume 15mA. At a lower
voltage the current (I) will be lower and a higher voltage will result in higher
current. Go to high with the voltage (E) and the lamp will burn up. Now first convert
the 15mA to amps (A) by dividing by 1000, which gives us 0.015A. We now have two
(E and I) of the three variables for the lamp which means we can use one of the
formulas to figure out the resistance of the lamp. For the purpose of the calculation
we will assume the resistance (R) remains constant over the typical (useful) voltage
of the lamp. Most likely the resistance will change some as the voltage changes,
but that is beyond the scope of this tutorial.
With E and I known we can use the first formula to find R:
Now we’ll fill in the variables and see what else we can figure out:
There are now two more variables we can figure. First, the current (I) through the
resistor must be equal to the current through the lamp, which is 0.015A. Next, the
potential (E) through the entire circuit is 14V. The potential through E23 is 1.5V,
therefore the potential (E) through E12 must equal 12.5V. Now we have:
Now we can solve R for the resistor using the same formula as before:
Lastly we can solve P for the resistor:
The resistor will dissipate 0.188 watts. This will require a 1/4 (0.25) watt resistor.
Typical resistor sizes for our application would be 1/4, 1/2 and 1 watt. Now we
just need to find an 833 ohm 1/4 watt resistor, but of course you won’t. The closest
value is 820 ohm. Other resistor values in the neighborhood are 750 and 910 ohm.
Now let’s figure out the new values of E and I for the lamp when we plug in the
820 ohm resistor. Add the resistance of the lamp (100 ohm) to the resistor (820
ohm) for a total of 920 ohms for the entire circuit. We already know the potential
(E) for the circuit is 14V. Now we can solve for the current (I) of the entire circuit:
As you can see there is a slight increase in the current (I) through the circuit
due to the smaller value of the resistor being used when compared to our original
calculation. If we finish off all the calculations we come up with the following
values:
If you made it this far you should be able to figure out the necessary resistor
for any lamp that may require one. But just in case here is table showing the voltage
and current for the lamp and the power dissipated by the resistor for selected circuit
voltage (usually track voltage):
Just in case you plan to take your loco to another layout, take note that if you
use a 750 Ohm resistor for your 13.0V circuit and then take your loco to a friends
layout which runs at 15.0V you may have some problems. At this point you are pushing
the 1.5V lamp at 1.76V, which could cause it to burn out. In addition, the resistor
will need to dissipate 0.23W of energy which is pushing the limits of a 1/4 watt
resistor.
At the other end of the spectrum, if you set up your lights with a 910 Ohm resistor
to run on 15.0V you could have dim lights when you run the loco on a layout with
13.0V on the tracks. The moral of the story is it’s better to be on the safe side
and have dim lights rather then burn them out!
I’ll cover the use of one resistor for multiple lights in Part 2. Be forewarned
there are some new concepts and pitfalls to learn about.
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