Talk: Wire Sizes and Spacing
Rationale behind the tables
One thing that is hard to find on the internet tis a table indicating recommended wire gauges for DCC based on scale and distance. Scale determines the current (mainly) and the voltages used. Layout size may also be a factor with larger scales.
Many electrons have been inconvenienced over the years regarding the size of wire needed, with no read conclusion.
A quick search with google will offer a variety of opinions. The first opinion always is related to cost: They want to save as much as possible on wire and justify their choices there.
The sound opinion is based on resistance. This is where the argument goes off the rails really quickly. For example, they will quote a number of ohms/1000 feet and work from there.
Example: 10AWG is a good reference, as its resistance is ≈1Ω/1000 feet. As referenced on the page, every three steps where the gauge number is increasing the resistance of the wire is doubling.
The problem: This is DC resistance. DCC is not DC. It is a pulse train, turning on and off thousands of times a second, and it is loaded with harmonics. A wire when presented with such a signal doesn't behave according to the simple rules of DC, because it now has additional properties in the form of reactive components. Where a coil of wire may be a short at DC, at 60HZ (AC) it isn't.
Many of the opinions are based entirely on the DC resistance of the wire and completely ignore the reactive properties of the wiring. Also ignores that inductive properties are also related to the gauge of the wire.
Good practice is to keep your power bus wires close together to reduce the reactive component, and there are tables which demonstrates this on the page.
I'm having a hard time buying into that table for recommended wire gauge based on voltage drop. I agree that large wire gauges (like 12 and 14) are good for a whole host of other reasons...
- Mechanical strength
- Ease of stripping to attach feeders
- Local availablity at hardware stores
- Low price
but for low voltage drop? That doesn't stand up to scrutiny. It seems to be misleading for the average modeller.
Let's work out the wiring gauge needed for a medium size 2'x40' N scale end-to-end layout. The table would imply that needs 12AWG buses.
First, a few assumptions...
- 20' average bus length (assuming centered booster)
- 10' average distance to loco (assuming even distribution)
- 20' round trip for current
- Solid core wire
- Inductance is negligible, (esp. if bus is twisted)
Let's work out how many trains we can run with 12AWG with only 2V drop on such a 40' layout.
12AWG is 1.8mΩ/ft @20kHz (including skin effect at 20kHz, see http://chemandy.com/calculators/round-wire-ac-resistance-calculator.htm) so 20' is 36mΩ Max acceptable Voltage drop 2V I=V/R = 55A
That's over 200 trains running with a typical load of 250mA each!!
So 12AWG definitely seems like overkill.
OK, so lets work out what we really need for a more realistic worst case of a 5A booster maxed out...
Max currrent 5A Resistance = V/I = 400mΩ 22AWG = 13mΩ/ft @20kHz (including skin effect) so 20' is 260mΩ
So 22AWG is comfortably within the limit of 400mΩ
Obviously, the gauge requirement goes up if you put the booster one end and bizarely managed to run all you locomotives at the max 5A at the far end of the layout. So now the average round trip is 80'.
Now to stay below 400mΩ you need 16AWG. That's still a long way off 12AWG.
I can see why the situtation might be different with NTrak modules, where a module may end up in a very long string of modules with the additional resistance of connectors every 4'.
--Train depot (talk) 17:32, 20 May 2013 (EDT)
What the above calculations fail to take into account is that the bus lines have a lot of reactance, and that is directly related to frequency. DCC is not a simple sine wave, it is a complex waveform.
At 10 kHz the wiring's impedance is predominately reactive due to inductance, and it adds a lot to the total impedance.
12AWG will have three times the impedance than that shown in the example above. A 36' run is 0.5Ω which at 5A = 2.5Volts. At the end of the run, a 1.25V reduction will appear (2.5/2), which is almost 10% of 15V. Maximum VDROP should be 0.75V, which will be achieved at 1.5A. A couple of locomotives could do that.
For reliability, these have to be considered as the worst case scenario. Building too close to the edge often has hazards.