Model Forum / General / Railroads / July 2005

 "Patrick Carcirieri" <PCARCIRIERI@nc.rr.com>,
 In a message on Sun, 10 Jul 2005 14:45:25 GMT, wrote :

"C> I am getting to install resistors for the bulbs on a first generation
"C> Genesis and have some questions.
"C>
"C> 1.  Can the wires on the resistor be bent for installation?

Yes.  You need to keep the wires (leads) from touching each other or any
random metal.  Electrical tape, heat shrink tubing, discarded insulation
from wire stripping, etc. can be used.

"C> 2.   Can the wires be cut?

Yes.  You need to leave enough for soldering! :-)
Actually they generally should be trimmed as short as you can and still
reach the connections you need to reach.

"C> 3.  Why do you need a resistor for each bulb?

Current and/or voltage limiting.  Incandescent bulbs have a certain
voltage rating.  Grain-of-wheat (grain-of-rice) bulbs are often 2V
bulbs.  Putting a full 12volts across them will turn them into extremely
bright and very short-lived bulbs!  Often modern headlights bulbs are
LEDs which are all about 2Volts as well and will also 'burn up' if
given a higher voltage.  The series resistor drops the voltage by
turning the excess voltage into heat.  The resistor value is determined
by Ohm's Law:

    V = I*R
    (R = V/I)
    (I = V/R)

* == multiplication, / == division
V == Voltage in Volts
I == Current in Amps
R == Resistance in Ohms.

If the track voltage is 12 volts (typical analog H0 scale track voltage,
DCC is sometimes higher, but you are probably NOT wiring the bulb to the
track in that case, but to the DCC decoder's 'headlight' function
output, which would typically be 12 volts) and the bulb (or LED) is
rated at 2 volts, the resistor needs to drop 10 volts.  Typically a LED
uses 20miliamps (.020 Amps).  You'll need to check what your bulb
actually draws, but lets assume it draws 100miliamps (.100 Amps).  This
works out to:

20MiliAmp 2V LED:

    10 / .020 == 500

100MiliAmp 2V Grain-Of-Wheat:

    10 / .100 = 100

You also need to figure the power rating of the resistor.  This is:

P = V * I

P == Power in Watts
V == Voltage in Volts
I == Current in Amps

For the 500 Ohm resistor for the LED:

    10 * .020 = .2

    A 1/4 Watt (.250W) resistor will do.

For the 100 Ohm for the grain of wheat bulb:

    10 * .100 = 1

    You'll need a 1 Watt resistor for this.

I have a resistor calculator program as part of my Model Railroad System
(http://www.deepsoft.com/MRRSystem/)

"C>
"C> Thanks
"C> Pat
"C>
"C>
"C>                                                                          

                                    \/
Robert Heller                        ||InterNet:   heller@cs.umass.edu
http://vis-www.cs.umass.edu/~heller  ||            heller@deepsoft.com
http://www.deepsoft.com              /\FidoNet:    1:321/153

                                                                         

Yes.  Best way is to use needle nose plier and hold the wire straight right
at the body of the resistor and to bend out past the needle nose.  Less
chance for breaking the wire at the body of the resistor this way.

Yes.  I would suggest using a pair of flush cutters and have the flush side
of the cutters towards the resistor.  Less mechanical shock sent down the
wire to the body of the resistor or other component and less chance for
failure of the component.

If the lamp is a low voltage lamp, let's say 6 volts as an example.  It will
be rated at some specific current flow through the lamp at 6 volts, let's
use 0.02 amps, also called 20 milliamps.  Using Ohms law we can calculate
out that the resistance of the lamp when it is operating is 300 ohms.  The
lamp will last a good long time at 6 volts, won't get too hot and will not
be real bright.  Now hook that same 6 volt lamp up to 12 volts and with all
other factors being the same you have doubled the current going through the
lamp.  The original lamp was dissipating 0.120 watts.  Now lets see what the
same lamp on 12 volts does for dissipation.  To get power I am using the
formula of  I^2 * R where I is the current in amps and R is the resistance
in ohms of the lamp.  After doing the math the result for this lamp on 12
volts is 0.48 watts.  That is four times the dissipation that the lamp is
designed for.  It will be Real Bright for a Short amount of time.  

With the above items known it is easy to see that to hook a 6 volt lamp to
a 12 volt power supply there has to be a way to make the lamp see only 6
volts.  That is what the resistor does.  Ohms law would be used to calculate
out that you need a resistor of 300 ohms to keep the current in the circuit
down at 0.02 amps.  Put this resistor in series with the lamp and the lamp
will only see the 6 volts it was designed for.  The resistor will have 6
volts dropped across itself.  In this instance I would recommend a 1/4 watt
resistor to keep the temperature down on the resistor body.  a 1/8 watt
resistor, while still within the limits of it's ratings, is going to get
real hot.  The 1/4 watt resistor while dissipating the same amount of power
will have significantly more surface area to radiate the heat away.  It will
stay quite a bit cooler than the 1/8 watt device.

For multiple bulbs the calculations get a little bit more complicated and I
don't plan on going into them here.  Suffice it to say that as long as ALL
the lamps are the same and they are all working, yes you could use just one
resistor and put all your lamps on that one resistor.  BUT, if one uses
different rated lamps on the circuit they all have to be the same voltage.
And then one would have to calculate the equivalent resistance of the lamps
when they are in parallel.  When one of those lamps open up then the rest of
the lamps see an increase in voltage.  As that happens the power dissipation
in the remaining lamps goes up and soon you have to replace all the lamps at
the same time.  

Sorry I got so long winded.  

If interested the Ohms law formula is E = I * R  Where E is voltage measured
in Volts, I is current measured in Amps and R is resistance measured in
Ohms.  

Al

MREG's Resistance Chart might be of some use:

http://www.merg.org.uk/resistor/index.htm

Bill
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