You don’t need #8 or even #12 wire for feeders. #12 wire is rated for 20 AMPS @ 120 volts. enough for 20 100 watt light bulbs. We are using ,at the most, 18 volts. Do the math. I have a 200’ " G " outside RR. When I drove the golden spike I hooked up a set of feeders from an old HO power pack with #24 wire & ran 2 locos around for some time without the wires even getting warm or the engines slowing down as they got further from the power pack. Since then I’ve installed a more sophisticated control system.
True. 15amp x 120 V = 1800 watts. 1800 watts divided by 24 volts (DCC system) comes out to 75 amps
Power = Amps * volts
TRUE, BUT you are wroking with VIOLTAGE DROP not Current carrying capacity. Remember that in DCC your sending PACKETS of information to the decoder. IF you have to much resistance your packet will not be readable.
With that said, you use #12 for Bus up to 30 ft (45 ft max) your feeders can be as small as 24 AWG we use 18 awg since the run is only 18" max.
The myth based on Current Capacity is false BUT based on resistance it is true.
Well, that’s all true, but we’re not talking about finding the minimum wire size to carry the amperage here. The whole idea of using big, thick wires is to reduce the line loss, or voltage drop, from one end to the other. If you start with that 18 volts at your booster, it could easily drop down to 16, 14 or even 12 volts by the time it gets way out to the end of a large layout. The smaller wires will have more loss, and will make this problem worse, while the larger wires will have less loss.
In theory, you only need one pair of wires to run from the booster to the rails. However, if you have a loose rail joiner, you could lose power completely at any track joint, and even if you solder all your rails you will still experience voltage drop down the line. (The G-gauge track mentioned will have less of a problem, because it is much larger diameter and being outside I’d assume all connections are soldered. In fact, I’d imagine that G-gauge rail is probably a better conductor than bus wire!)
Also, when you’re talking about #12 wire, I assume that you mean the track power bus, not the actual “feeders,” which is usually the term used for the short wires from the bus up to the rails.
The responses are very correct. Current carrying capacity is NOT the issue here. There are tables all over the place which show the resistance per foot of various size wire. With this value, you can use Ohm’s law to determine voltage drop for a given current load.
One thing to keep in mind is that if you are running through circuit breakers downstream of your boosters, the maximum current that can be pulled through that bus wire is whatever the breaker is set for. Say you have a 10 amp booster, but your power district is fed through a circuit breaker set to trip at 2.5 amps. Well then you can;t have 10 amps going to that power district, so the MAXIMUM voltage drop would be 2.5 x (length of wire x resistance per foot), NOT 10x.
It doesn’t take much voltage drop to see a change in loco performance. Under 1 volt is probably best.
If you are building a barn-size layout, rather than running #00 bus wire to prevent voltage drop over a huge length, consider remotely locating boosters closer to the track they will be powering.
–Randy
where you been Randy?
Yes, Randy, but what about degradation in the command signal being sent to all those remote boosters?
(How’s that for a setup as to why Digitrax is a cool DCC system … [swg] )
Well besides the fact that unless you were 100% wireless you’d already have the wiring in place to plug in a booster at the far reaches, compared to all the other systems which run a separate cable for the throttle bus vs the booster bus…
The big difference is the low current draw on the ‘booster bus’ portion of the setup which naturally has less voltage drop, plus the signal in at least Digitrax’s case is differential relative to a common wire in the bus wiring, and I suspect this is the case with other systems as well. Differential signals are well protected from outside interference. And a volt or two drop in the signal feeding the booster won’t mean anything to the output of said booster
–Randy
Boy, this one has been going round and round for a long time. There are even many tables on the web - some are great and some not so great.
Here are a few numbers.
#12 untinned copper at 77 degrees fahrenheit (25C) has a specific resistance of 1.62 ohms per thousand feet.
The resistance of 100 feet (to use a round number) of wire is 0.162 ohms
Remember that the length of wire is out and back so this will be a 50 foot run.
If we draw a full 10 amps (more than many boosters can supply and a hell of a lot more than any five modern locos will draw), and we draw this all of this current at the furthest end of the buss, then the wire will have a voltage drop E=I*R of 10 *0.162 or 1.62 volts.
16 volts in at the booster and 14.38V out → worst case <–.
Lesser loads will lower the “lost” voltage proportionately. Moving the loads in closer to the booster will lower the voltage drop as will distributing the loads.
One needs to choose wire size based upon bus length, (maybe you can run two busses in parallel to two different areas and shorten the length) load current (how many locos and how hungry are they) and load distribution along the layout.
Another factor is the track itself. If you use multiple feeds (and you should) the track will carry some of the total current and thus lower the current through the bus and the voltage drop.
Sorry, I don’t have a resistivity table for nickle silver rail at hand. A wild guess would be code 100 is similar in CSA to #16 wire and has a specific resistance twice that of copper. Maybe 4 ohms per 100 feet, 0.4 ohms per 100 feet. I
#14 copper has a specific resistance of 2.57 ohms / 1000 feet
#10 has 1.018
#8 has 0.6404
Personally I use 12 gauge and 14 gauge for busses. My layout is 29 x 12-1/2’ walk in G shape.
Hopefully this will help someone and my four years at RPI will not be wasted.
If you can’t
IWhich one was the prof at RPI, Halliday or Resnick? Was the guy as really disliked as folks used to tell us? Or are we talking before your time?
Don’t forget resistive losses at each termination/connection.
According to the electronics engineers I work with, the DCC waveform can be considered to be high-frequency alternating current. The voltage drops being bandied about are based on DC current - AC doesn’t see those voltage drops (at least not to the same magnitude as DC), which is why power transmission is AC and not DC, by the way. So using the DC voltage drop calculations for a DCC layout gives wrong answers. That’s what the avionics guys, who deal with high-frequency AC waveforms of various types all the time, tell me. I’m not a subject matter expert, though, so if anyone who is an expert knows this is NOT true, please explain it to me.
(But I use 12-gauge buss wires just to be safe) [:D][:D][:D]
It’s true that AC experiences less linear loss than DC, which is why we use AC to power our homes. Edison was way wrong about that (he wanted DC), but probably had an ulterior motive - less distance capability would mean many more generating stations and more money for him.
However - there are other factors to consider, although at the low voltages used by DCC they may have minimal effect. Take a look at those tall high tension towers next time you pass one. See how the wires are arranged in a pattern? That’s not accidental or random - the pattern actually affects the line loss. However, at a fraction of the voltage (15 volts as opposed to 150KV and more) these effects may not really matter, even when the bus wires are right next to each other. Signal degredation seems to be more of an issue over a very long straight bus run, which is why it is mentioned that twisting the bus wires instead of running them straight and parallel to each other. DCC is also much higher frequency than AC power line transmissions, which reduces the loss effects. But the capacitance of the parallel wires will server to start rounding off the edges of the square waves. This is where twisting helps, just like in network cables (which also carry square wave signals)
–Randy
Why make things complicated? Just stick to what works!!
Hi Dave,
Halliday must have been before my time. I had Resnick for physics. Has to buy his books too. Huge lecture hall with hundreds of students, some of the demos were fun but the recitation sections were with grad students who didn’t care too much or couldn’t teach.
Resnick wasn’t hated too much, at least not by my compatriots. Now as for professor slumlord or was that slemrod …
Are you an alumni?
Stories grow with time I guess.
Larak (Karl spelled backwards- sort of)
Actually AC DOES see those voltage drops. All electricity sees DC voltage drops. In addition, AC will see other losses caused by reactance, coupling blah blah blah.
In theory AC will have a LARGER loss than DC. I practice the difference is probably negligable for our purposes. I have been unable to find the bandwidth figure (or even upper frequency limit) for DCC but my guess is just above audio.
In any event, the highest frequencies are just for control at extremely low current. The power is created from the carrier. Sure it’s a square wave but that’s just for efficiency in the booster transtors. The nice square corners are totally unneccessary to operation. Otherwise we would be using litz wire or wave guides (giant pipes) to carry the signal.
This would be a fascinating topic to study in the lab. I’ll put it on my some day list. Meanwhile I like your theory of using #12 “just to be safe”.
We use AC to power our homes simply because you can transform the voltage level up and down very easily and thus the current levels inversely. Nothing to do with losses at comparable current levels.
Power = current x current x resistance. Power is lost in wires as heat. Double the voltage = halve the current = divide lost energy by FOUR. Hey we can use that to our advantage.
Edison must have hated Tesla!
Ain’t physics wonderful [:D]
He hated George Westinghouse more.
#### I HAVE SEEN THE LIGHT!!!
AC doesn’t suffer the losses of DC because power loss is Current squared times resistance, so with AC you can reduce the current, and thus the power loss, by upping the voltage on the line. Since watts = Volts times amps, if we double the voltage we have the current for the same power transmission, which means the losses will be one quarter what they are at the lower voltage.
Great!
So for a given voltage, like 16 volts AC or DC, line losses are the same. Okay. So the high frequencey DCC waveform suffers the same losses as an equal-voltage DC wire.
Rats! I thought I was onto something there for a minute.[sigh]