Gapping both rails and using terminal tracks sounds like a good way to simplifying the wiring.  But if you gap both rails, you will have to use both the wires that come off the terminal track and connect them to a DPST (Double Pole Single Throw) switch, also available at Radio Shack.  Then you would run two more wires to your DCC command unit.  If you find that a short occurs when your locomotive crosses the gaps going onto a siding, just reverse the plug in that sidings terminal track.

Did they happen to mention if/when they will be upgrading their software to work with Windows XP or Vista?  Running only on Windows 95, 98 or ME makes it rather obsolescent.

Using the exact scale for H0, namely 3.5 millimetres per foot, 60 miles per hour works out to 12.126 inches per second.  But 12" per second is close enough.  I like to look at this as two 40 foot box cars per second, and if two go by every time I count one thousand and one, one thousand and two, etc., then that is close enough to 60 miles per hour for me.  If only one goes by per second, then it is 30 mph, a good speed for a drag freight.  A yard speed limit might be 10 mph.  Since this is 1/3 of 30 mph, it must take three times as long for a car to go by.  So that is one 40 foot car every 3 seconds.  Coupling speed in real life is around 3 mph, or 1/10 of 30 mph.  So the train should be moving only one box car length every 10 seconds.  (in scale, it is hard to couple at this speed, even with Kadees, if picking up a single car.)

Incidentally, using box car lengths makes this speed check independent of scale.  And if the cars are 50 foot, you will still get closer that just guessing.  Next time you get caught at a level crossing, you can pass the time figuring the speed of the passing train.   

Is this a plug-in installation?  Or is it a hardwire job?  If it is a plug-in, then you could have a short between one or more connections under the board.  Or a short in the wiring.  Or you could have the plug not properly lined up with the socket.  If it is a hardwire, then I would look for a misconnection or a short.

If you have a multimeter handy, you could measure the resistance between the motor connections and the pickup connections on the locomotive.  Then between the light connections and the pickup connections.  On this locomotive, the pickup connections are the halves of the frame.  It is best to do this with the decoder disconnected.  If it is the plug-in installation, you can either use a spare wiring harness to connect to the socket or pieces of solid wire the same size as the pins on the decoder plug.  Smaller wire may not contact properly; larger wire may spread the socket pins so that they can no longer contact the decoder plug.

Unfortunately, it is not a single part that is required for the current limiting circuit.  You would probably be better off turning off the power to the sidings as suggested by Gene.  What you would need is a gap in one rail at the begining of each siding (right next to the turnouts.)  You can remove one rail joiner and install an insulated joiner (available from your local hobby shop) at each gap.  Then, to turn your sidings on and off, you can use toggle switches from Radio Shack.  Simple on/off switches have two terminals.  For each siding, connect two wires to a toggle switch, one to each terminal.  Then connect the wires to a rail with a gap in it,  one wire on the turnout side of the gap, the other wire to the siding side of the gap.  Don't solder them too close to the gap or you might melt the insulated joiner, allowing the gap to close back up.  You will need as many sidings and as many toggle switches as you have dc locomotives on the layout.

I believe you are headed in the right direction for what you want to do.  The purpose of the light is to limit the current if you accidentally bridge the two systems.  To do this, it must limit the current to a value that your command station can handle.  For the 1.5 amp E-Z Command, you probably should not use a lamp that draws more than 1.0 amp at 12-14 volts.  This will limit you to one or maybe two active switchers in the yard area.  Using a power pack rated at 1 amp is NOT a solution as the 1 amp limit is normally set by a circuit breaker.  In case of a short or bridging, the current can go higher, much higher, until the circuit breaker kicks out.

If you are into electronics, it is relatively simple to install a circuit in your power pack to limit the current electronically.  This sets a maximum current that cannot be exceeded.  The problem with lamps is that they still allow an excess current for a short time, although this is not usually a problem.

I see there is still a great deal of confusion when it comes to matter electrical.  But Stewart has one thing absolutely right - It is better to over engineer than under engineer.  If you don't know the numbers or don't want to do the calculations, then err on the side of caution.  It is a lot easier and cheaper to put in a larger than necessary wire the first time than it is to tear out and replace too small a wire.

Perhaps this would be a good time to clear up some of the other confusion.  Just so that we are all on the same page, there are some definitions at the bottom of the page.

Electrical Power is the product of voltage and current.  That is,

Watts = Volts x Amps

But we also know from Ohm's Law that

Volts = Amps x Ohms

So we can calculate electrical power some other ways, for example:

Watts = Volts x Volts / Ohms

Watts = Ohms x Amps x Amps

The latter explains why a piece of wire gets hotter when you push more current through it.  The piece of wire has a constant resistance, in Ohms.  But to figure how much heat that wire produces, you must multiply its resistance by the current through it (Amps) two times .  Lets do an example.  Suppose a piece of wire has a resistance of .1 ohms and we push 1 amp through it.  The heat generated is .1 ohms x 1 amp x 1 amp = .1 Watts.  Now suppose we push 2 amps through the same piece of wire. The heat generated is .1 ohms x 2 amps x2 amps = .4 watts.  As we push more current through the wire, the wire is heated more and all else being equal, gets hotter.  But note that doubling the current increased the heating by four times, not 2 times.  Note also that without the resistance of the wire, no heating would take place no matter how high the current is.

So what about current (amps) as a measure of heating effect?  It is an exact measure of heating if the voltage is known.   When we look at one space heater versus another, we can compare them on the basis of watts or amps, it does not matter, as long as the voltage is the same.  That is a direct result of Watts = volts x amps.  But do not try to compare a 10 amps heater that is rated at 12 volts to a 10 amp heater rated at 120 volts or a 10 amp heater rated at 240 volts without also taking the voltage into consideration.  They are NOT all the same.  In fact, their heat outputs are 120, 1200 and 2400 watts respectively.

Motor rating is a farce.  Current ratings, and horsepower ratings calculated from current ratings, are based on stall current, not operating current.  You go out and buy a router.  Chance are you buy the one with the most 'power'.  I have one rated 3 horse power.  Wow!  But let's do a little figuring.  3 horse power at 746 watts per horse power is 2238 watts.  Plugged into a 120 volt outlet, 2238 watts is 18.65 amps.  Way too much for a 15 amp circuit.  What it means is if I jam the bit and bring the router to a grinding halt, it will draw 18.65 amps and produce 2238 watts of HEAT.  With no motion, there is no mechanical horse power at all.  If my router really produced 3 horse power, it would probably rip my arms off and instead of saying WOW, I would be on my head screaming MOM.

Another fallacy to put to rest is that all motors draw more current as the voltage drops.  I hear this over and over, but unless you have some fancy constant speed control circuit driving the motor, this just does not happen.  This fallacy probably arises from the fact that furnace, air conditioner, freezer and similar motors do burn out during brownouts.  But these motors all have starting windings that draw power, and lots of it, when you first turn the motor on.  As soon as the motor is up to speed, the starting winding is shut off and the motor runs on a much lighter running winding.  You may have noticed the lights dim for a second when one of these motors kicks in.  That is when the starting winding is drawing huge amounts of power to get the motor up to speed.  Now is you lower the voltage to one of that type of motor, such as during a brown out, then the motor slows down.  If it slows down enough, the starting winding kicks back it.  And then the motor is drawing more current trying to get its speed back up.  If it cannot get up to speed, it runs with the starting winding engaged until it burns out.  Our dc motors in our model locomotives have no such starting windings.  If we turn down the voltage, the current decreases, unless you physically change something else, like the grade or the load.  There is one exception - if you run on DCC and are using back emf, which is a fancy constant speed control circuit.

One last comment about wattage ratings.  The last two microwave ovens I bought have both input and output power ratings.  The newest is 1000 watts output (to the cooking chamber,) 1200 watts input (from the power line.)  The much older one was rated 700 watts out, 1350 watts in.  Maybe this is a Canadian thing to stop false advertising - do you remember them selling "90 watt stereo amplifiers" when the only 90 watt rating on the whole thing was the amount of power it drew from the power line?

Now let's get out there, turn on our trains, and play nice!

ENERGY - the capacity to do work.  It can be potential energy (mass at a height, the energy stored in your neighbour's car parked up on his driveway,) kinetic energy (mass in motion, kinetic energy it what crumples your fenders when your neighbour's car rolls down his driveway and coasts into your car,) chemical energy (released by chemical reactions,) and thermal energy (heat.) Measured in joules, kilowatt hours, calories, foot-pounds per second, pound-feet, etc.

WORK - expenditure of energy, such as combining fuel and oxygen (chemical energy) to blast a rocket into space (potential energy as the rocket is raised off the ground plus kinetic energy of its motion.  As we do work, we convert one form of energy into another and ultimately into heat.

POWER - work or energy expended per unit of time.  Measured in joules/second, watts, calories per minute, horse power, foot-pounds per second², pound-feet/second, etc.

VOLTAGE - electrical pressure, measured in volts

CURRENT - electrical flow rate, measured in amperes (amps)

RESISTANCE -  impeding electrical flow, measured in ohms.

Rich, may I suggest doing the calculations.

Typical DCC booster produces 5 amps at 16 volts.  Required voltage at locomotive is 14 volts.  So allowable voltage loss is 2 volts. 
Question: what is the maximum run using #20 wire to keep within the 2 volt loss?  For #12 wire?

Answer: Maximum allowable wire resistance is 2 volts/5 amps = .4 ohms.
#20 copper wire has a resistance of 10 ohms per foot.  So the maximum length of wire = (.4/10) x 1000 = 40 feet, out plus back.  Maximum run = 40/2 = 20 feet.  And this assumes that all of the 5 amps is being drawn through the entire length of that one run of wire.  On a small layout (4' x 8'), you would have to have a very convoluted wiring layout to exceed 20 feet.

The resistance of #12 copper wire is 1.6 ohms/1000 feet.  So the maximum length of wire = (.4/1.6) x 1000 = 250 feet, out plus back.   Maximum run is 125 feet. 

Why would you want to use wire with that kind of capability on a small layout?

In fact, with DCC, you could not use that length even on a large layout because of the parallel capacitance and series inductance of the pair of wires.

Just for fun, lets look at #24 wire and Tortoise switch machines.  This is a low power application where the switch machine draws only 50 milliamps at 12 volts dc.  A Tortoise machine in good condition will throw a turnout at as little as 8 volts.  So what length of #24 wire could you use?  Well, a 4 volt drop at 50 milliamps is 4/.05 = 80 ohms.  That is, we can have up to 80 ohms of wire in the circuit and still have the Tortoise machine work.  The resistance of #24 copper wire is 26 ohms per 1000'.  So we could use an astounding (80/26) x 1000 = 3000 feet, out plus back, or a run of 1500 feet.  However, take note of the proviso "in good condition."  Suppose a Tortoise on the point of wearing out needs 11 volts to run (this number is a guess, I have never measured current consuption on an almost worn out Tortoise.)  Then the length would drop to about 400 feet, more or less.

These are just applications of Ohm's Law, which says the voltage across a resistance equals the current through the resistance multiplied by the resistance.  That is, Volts = Amps x Ohms.  This can be stated two other ways (if your algebra is a little rusty.)  Ohms = Volts / Amp  and Amps = Volts / Ohms. 

The numbers for the resistance of copper wires can be found at any number of websites - just Google "resistance of copper wire."  These calculations ignore the increase in resistance of copper wire when heated.  This is minor at temperatures below the melting point of the insulation.  Listed ampacity of #20 wire is around 3 amps, which would be the limit for bell wire if installed in an insulated wall.  In free air, the limit is about double that.  And as mentioned before, #20 appliance wire, which has higher temperature insulation, has a capacity in chassis work of around 11 amps.

Dummy (no motor) B-units and sometimes powered B-unitis are run just for looks.  At other times, powered B-units are added to trains because the number of cars and/or the grades are too great for just an A-unit to pull them.  This is true independent of power source.

The Ma and Pa may have been limited to 20 mph by the trackage, but some 4-4-0's were capable of speeds up to 95 mph. 

Quote from: glennk28 on February 09, 2007, 08:52:07 PM
A Water analogy is often used with electricity.  You can only put a certain amount of water through a hose of a given diameter. Likewise, you can only get a certain amount of electricity thru a given size wire.

I like the water analogy, but why don't we take it a bit further.  The water pressure in my house is 40 psi (pounds per square inch.)  If I connect up 100 feet of water hose, it will produce 5 gpm (gallons per minute) of water flow.  The pressure difference from one end of the hose to the other is 40 psi.  Now lets suppose I want to get more water out of that hose.  If I increase the water pressure in my house to 80 psi, the same hose will now produce 10 gpm.  At 320 psi, it will produce 40 gpm.  In other words, all I have to do to put more water through the same hose is increase the pressure difference from one end to the other.  Of course, if I increase the pressure too much - boom, my hose blows up.

It works the same with electricity.  If I take 10 feet of #20 wire (the 'hose',) I can push 1 amp (the 'water flow')  through it with a voltage drop (the 'pressure difference') of .01 volts.  If I put 10 amps through the same piece of wire, I need .1 volts to push it through.  Note that this is the voltage difference between the ends.  If there is zero voltage at the far end (the wire is shorted to a large conductor at the far end) then the input voltage at the near end must be .1 volts to drive 10 amps through.  If we have 12 volts at the near end, and drive 10 amps through the wire, we will have 11.9 volts at the far end.  The voltage difference is 12.0 - 11.9 = .1 volts.  As you see, it is the pressure difference between the ends of the hose that drives the water through the hose, just like it is the voltage difference between the ends of the wire that drives the current through the wire. 

What happens if we drive too much current through a wire?  The insulation melts off the wire, rendering it useless or the wire itself melts.  Typical insulations used in wiring chassis will fail at around 11 amps.  The wire itself will melt in 5 seconds at about 68 amps.

So what does all this mean?  It means you can use 20 gauge appliance wire to wire your model railroad if the wires are short enough and the current is low enough.  On a 4' x 8' layout, few runs of wire would exceed 12.5 feet (down a long side, plus down a short side, plus a little bit.)  So out and back would not exceed 25'.  At 10 ohms/1000 feet, the resistance would be only .25 ohms.  At 4 amps (enough for 8 locomotives) the maximum voltage drop would be 1 volt.  Not enough to worry about in either dc or DCC.

But lets up the size of the layout.  On my own layout, I have a couple of DCC power districts that are 50 feet from my power manager (electronic circuit breakers.)  Will 20 gauge wire do the job?  Well, 50 feet out and 50 feet back is 100 feet.  That is 1 ohm at 10 ohms per 1000 feet.  I run 5 amps to each of these districts.  So the drop could be as high as 5 volts.  Much too high.  So in this case, 20 gauge wire just won't do.  Based on price per foot and availability, 14 gauge house wiring is a good choice.  At .5 ohms per 1000 feet, the voltage loss is only .25 volts at 5 amps, and it works very well.

Want to super size the layout?  How about 1400 feet of wire carrying 10 amps on an outdoor layout?  Well, it turns out even 10 gauge wire is inadequate, if it is the only thing carrying the current. 

So sometimes the experts are right, and you do need big wires.   But in usher-42's case, 10 gauge or anything near it would be ridiculous.  Common bell wire would be just fine.

No.  For that you would need a more complicated, full function DCC command station.

Most of Bachmann's capacitors are sort of like yellow M&M's with legs.  But they may also be square.  Sometimes they are blue instead of yellow.

If they have copper wire around them, they are inductors.
If the are cylindrical with four coloured stripes, they are resistors.
If they are black with a single white stripe, they are diodes.

This photo shows all four:

Any ultra sonic (silent) decoder, with or without sound, NMRA compliant or not, can have problems with the capacitors.  It is only the non-silent decoders that pulse the motors in the audible range that can deal with the capacitors.

TOC is correct about removing them - it may help but never hurts, unless you live in certain parts of Europe.

People heavily into DCC often remove the capacitors, the socket, the adapter board, the whole works.  This helps free up a little more room for the speaker and its enclosure.

As Bachmann decoders are a Lenz design, the usual trick of using a 1000 ohm resistor across the tracks should work.