Gang: ready to convert my DC layout to DCC. Total track length is 32 teet, HO scale. I use lamp cord as my DC bus. Can I use this for DCC or do I need to replace it with something else. Many thanks, Andy
So, is that the little skinny brown stuff that some "home"extension cords are made of?
Even if it is, I bet it’s 14ga. stranded. I would say with the length your talking, it would be OK.
Mike.
Lamp cord is 18 gauge. Still, for your short-ish run, it would probably work ok. IF it isn’t too much of a hassle, it might be worth your time and effort to bump it up to 14 gauge stranded.
Mark.
Wiring is very important with DCC. DC wiring is usually inadequate for DCC. The lamp cord used for your is probably 18 gauge, which is adequate for feeders but not for the bus.
For the run you plan, 14 or even 12 gauge is a better choice. So upgrading your wiring should be considered. Poor wiring can cause problems such as runaway decoders and inadequate short circuit protection. DCC systems put a lot more current on the track than analog power packs, so this is important.
Since a DCC system can supply more power, voltage drop becomes an issue, because you can and will run more trains… If the voltage drops the current will increase, meaning more heat produced by your decoders and booster. More heat is never good with electronics.
why? 5 Amps is required to cause a 2V drop at 30 ft (18g wire has 6.4 mOhm / ft)
why? what’s a “runaway decoder”?
why? since there’s less voltage why don’t you say there will be less heat ???
Model Railroader Georg Ohm, playing with his Marklin train set, didn’t think current increased with a voltage drop, otherwise he would have invented a different law V=I*R
Andrew:
Our club uses 16 guage automotive wire. I don’t know our size, but we have four power districts, with three subdivisions in each.
Our record is 18 people each running an engine. You are probably good.
I’ve seen some lamp-type zip cord in 12 gauge, but it is not very common. Speaker wire is also a type of zip cord, too, but most places charge a premium price for speaker wire. Some is extra flexible, too.
The drawback in my mind would be trying to differentiate “rail-A-rail-B” while under the layout when the color of each conductor is the same and all you have is a little rib to differentiate each leg.
This outfit (and others) offers a black/red zip cord down to 2 ga.
https://powerwerx.com/red-black-bonded-zip-cord
Still, there are better choices.
Regards, Ed
Some years ago our club switched from DC to DCC and we used #14 with about six to eight inch #24 feeders. Number 12 we thought was overkill and we had the NCE five amp Power Pro.
PFM turnouts. Non by DCC.
We ran as many as ten sound locos with no issues on code 100 rail.
At home I kept #22 when I switched to DCC. I only have the Power Cab. No big deal.
If your layout is wired, give it a try. That is what I did.
Rich
The speaker wire I use is more expensive because there is no oxygen trapped in the insulation with the copper. It is called “OFC” on the packaging. It is also 10 gauge made up of strands that are only 0.004" in diameter. Not sure if any of that really does any good or not.
I am not a DCC user, but this does not sound right.
-Kevin
All: many thanks. Everyone is very helpful. You are the best folks. Andy
OFC refers to the copper alloy itself, not oxygen trapped by the insulation. That’s why they call it Oxygen Free Copper instead of Oxygen Free Cable. The alleged advantages are very slightly improved conductivity and more resistance to corrosion. I understand it makes a big difference in particle accelerators.
Pretty much a gimmick at audio frequencies, but as audio gimmicks go, it’s a pretty mild thing. #10 is overkill unless you are running 1000 watt amplifiers.
Now if you said you were using directional speaker wires, or directional line cords, or those silly isolation pyramids under the speaker wire to keep vibration out of the speaker signal…
Adequate wiring is important so circuit breakers can trip. Too much resistance int he wire means the current flow won’t exceed the trip point of the breaker, so realtively high currents can flow through things not meant to handle them (pickup wipers, the thin wires used inside the loco, etc). This is why passing the quarter test is important. Set (do not push down) a quarter or equivalent size local coin across the rails and various spots all along the tracks, every foot or so. In each case, the circuit breaker should trip. If not, the wiring needs beefing up in that area.
–Randy
Greg, line resistance takes place over distance at a given voltage. If the voltage drops enough, the short detection circuitry in the base unit of DCC systems won’t detect the fault, it won’t trip, and it’ll let whatever has the least tolerance for current burn up. Among the first things to go are the components in a decoder.
Runaway decoders are decoders that continue to behave per their last received packet instruction. If the DCC signal is so weak that the decoder can’t read it over the noise, the decoder will continue to meter the same voltage to the can motor as it did with the last instruction. It won’t obey anything except a cessation of rail power, which the emergency button ‘should’ effect for the operator. If that doesn’t work, there’s always pulling the plug…literally.
Finally, if you continue to draw the same current, but use less voltage to ‘carry it’, you’ll get hot-running components, especially a motor doing work with the current, or the amperage. If an industrial motor is running hot, but you need it to perform the work it’s meant to do, and with limited current, you’ll have to change something, and that is done by raising the voltage. I learned this from my father, a mining engineer, who instructed an electrician to pot up the voltage running to a crusher motor that was almost too hot to touch. When they raised the voltage by 40 volts, the motor began to cool. Same amps, higher voltage, cooler motor.
Then it must have been Hornby modeler Joule who invented the correct law; you maroons forgot something when you stopped using E for voltage and I for current for plain old DC and got your little letters mixed up.
V (or E) is simply IR. (That’s little Georgie’s law)
P = EI. (That’s the thing that poor Betamax is being blasted for.)
Combining, P = I squared R (with an exponent, not an asterisk which connotes a form of multiplication).
One of you soi-disant electrical experts tell me what happens if you have a constant load in watts and the voltage starts sagging. Now how does that differ if you reduce voltage some other way?
(Seriously, now that I’ve had my little Manhattanite chop-busting fun: if all you’re concerned about is the voltage drop from distributed resistance in the wire, then the available voltage goes lower following Ohm’s law. It’s when you actually try extracting power down near the end of the wiring run that the ‘other’ thing starts to become an issue … see resistance heating and Kirchhoff’s laws.)
And with a power factor of 1, ie a pure resistive load with AC, or with DC, P=VI if you solve P-I squared R by replacing R with V/I and simplifying. So if P remains constant, but V is going dowm, I has to increase to maintain the same power.
–Randy
sorry, but things need to make sense.
don’t understand what voltage has to do with it. The wire has a resistance regardless of the current flowing thru it or the voltage across is.
how much of a voltage drop is “enough”? is 2V at 5A too much
which part of the DCC system needs to detect the short? how can there not be enough voltage at the booster or a circuit breaker connected to the booster for it to not detect a short on its output
if there’s insufficient voltage reaching the decoder why should it be damaged?
how can a voltage drop introduce noise? the signal is the polarity reversals?
how much of a voltage drop is required for the decoder to be unable to see the polarity reversals carrying the signal?
how low can the track voltage go before the decoder stops operating (<~5V)?
if it’s not using BEMF, it’s applying the same PWM signal to the motor and there is effectively less voltage across the motor.
If it is using BEMF to maintain speed, an equivalent amount of current (> PWM duty cycle) is flowing thru the motor (it’s not running any faster or slower unless it was
I used to smoke cigarettes and the pack was labeled LS/MFT!
2V is definitely too much drop, that’s very noticeable as a speed change in the train. .5V is about a smuch as you really ever want. It was different with most DC systems, since you never ran wide open anyway, or at least anyone operating realistically did, if the train slowed at the far away point due to loss in the wire, you could just gradually creep up the speed control.
–Randy
of course it is. but that’s at 5A? it would only be 0.5V at 1.25A and even that’s a lot a current.
But we’re talking about the DCC track voltage, not necessarily the motor voltage. if you’re operating at less that max speed step, the PWM duty cycle can be increased to maintain the motor voltage. If the decoder uses BEMF to maintain speed, it adjusts the PWM duty cycle automatically.