My father, and I bought the Western Freight Candian Pacific set, and I’m thinking of adding alot of track to the layout. I’m wondering how much power can the CW80 handle until it runs out of power from laying to much track down on the layout? I would love to hear from you O scale fanatics. Thank you.
Sincerely
Ryan LaPlaca
Its more like: it puts out 80 watts. Run power wires to 1 of every 6 or so sections of track so the engine does not slow down on the far end of the track.
It is not just the amount of track you have, but each additional power drain. Every passenger car with lights, length of trains (power needed to engine), any accessories, etc. The CW80 does not have a tremendous amount of power. If you are building a permanent layout of any size, you will probably want to get more power. In most cases, it works out to $1.00/watt, so a 400 watt transformer will cost about $400. You can pick up some good deals on postwar transformers like the ZW with either 250 or 275 watts for around $150., but be carefull that the power cord is not cracked.
Another thing to consider is that if you go for one of the command control units for your layout, you can purchase 180 watt power bricks which are cheaper. If you think you may want to go that route, you might want to hold off on a new transformer to see what you’ll need. The folks here are very helpfull. Also, you should go to the search section at the top of this page and search for topics as most times they have been discussed at length and you will have the answers right away.
Welcome,
Dennis
I’ll go farther than Dennis: The power rating of the transformer has nothing to do with the amount of track and everything to do with the current that the trains and accessories draw.
The length of the track affects your train by reducing the voltage. You can compensate for the voltage drop through a long stretch of track between the transformer and the train by turning the voltage up higher at the transformer. But it will go only so high, about 18 volts for the CW-80. Higher-powered transformers tend also to be higher-voltage transformers. So, you could put in a postwar ZW, which puts out about 21 volts, and be able to tolerate 3 more volts of voltage drop, or a 24-volt Z and get 6 more volts. However, the much greater power ratings of these transformers would have nothing to do with being able to run on a longer track. A lower-powered transformer with a greater voltage range would do just as well.
The better way to handle voltage drop is to eliminate it. Then you don’t need any higher voltage from your transformer. This is what Boyd’s advice about feeding the track in multiple places is about. Voltage drop is produced by the electrical resistance of the track, which is very roughly comparable to 16 AWG copper wire. So, to get any improvement from feeders, they should be larger than that, at least 14 AWG.
However, if your track is in a complete loop and has joints that are in good condition (or soldered), you can operate a surprisingly large loop without feeders. For example, suppose that your track loop is about 80 feet long, about 20 feet square. The resistance of 16 AWG is about .0025 ohms per foot; so the total resistance is .2 ohm. But the farthest that the train can get from the transformer is only 40 feet, or .1 ohm. And, since that location is fed from both directions around the loop, the resistance is really ha
I’ve done as Bob suggests with great success - one lockon from the transformer for a loop running 19 ft by 13 ft. I also run short trains with small engines.