reducing power on yard tracks

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Yes, but the effect will be different depending on the current draw of the locomotives - old open frame motors with high current draw or multiple motored MU lashups will be slowed more for any given resistance than low current draw can motors. If you know the current draw, it's an easy Ohm's Law calculation - E = I * R where E is the amount of voltage decrease to the siding, I is the motor current, and R is the dropping resistor's value. It may be easier to decide on a value by experiment and connect a rheostat or the speed control from a junked train set power pack to the track, set it to get the speed drop you want, measure the resistance, and buy a resistor close to that value. Note that the dropping resistor will be dissipating a lot of power: P = I * I * R so you may need to look for one of the 10 watt ceramic type resistors. Radio Shack has a limited selection of these, so remember you can put two of the same value resistors in series to get twice the resistance, or in parallel to get half the resistance. Geezer
Reply to
Geezer
Yes, unless you're using DCC. If you're talking DCC, I don't know. Why do you want to reduce power on some tracks?
P = IV If Volts are 12 and I is 1 amp, you'll need a 12 watt resistor. This is a quick sloppy calculation, and you could no doubt get by with less than 12 watts, but why take the chance. I don't know if Radio Shack sells high wattage resistors. Their normal 1/4 watt ones won't take the load.
Reply to
<wkaiser
Yes, you can, but as stated your question contains insufficient information. Mostly, why do you want to do this, how much do you want them to slow down, etc.? Usually, we use the controller to reduce power. Also, what scale/gauge, etc.
Reply to
Wolf K.
Thanks Geezer and Bill.
< Why do you want to reduce power on some tracks?
Maybe I'm going about it wrong.- DC system. Locos speed up ( if throttle unchanged) when entering switching yard with 5 to 6 foot track blocks.
If there is a better way than adding resistors- I'm all ears
Reply to
tex shalter
Ho scale, I'm looking for automatic slowdowns for unattended viewing without resorting to products by circuitron etc.
Reply to
tex shalter
Sounds like you are having excessive voltage drop on the main line blocks. How long are the mainline blocks? What size feed wire are you using?
I use discarded extension cord wire (16 or 14 gauge) for main feed buss from the power pack(s) to the blocks, and short feeder wires from it to the rails. I look for polarised cords at yard sales, the kind that have one of the two wires marked by ridges, that's the one I use for the power feed, the other for the ground (return) buss. Add feeders to each block about 4-6ft apart. (Ie, feed an 8-12 ft block the middle.) You can use smaller wire (20 or 18 gauge) if the feed buss is less than about 15 ft.
Do not use individual wires in discarded phone cable (or computer cable, etc) as your main feed wires - they are too small and will cause a noticeable voltage drop. If you must use multi-core cables, twist two or more wires together at each end, thus creating in effect a larger feed wire for your main power buss. Make sure that the wires from power pack --> block selector switch --> feeder location are all feed buss size.
Also make sure that you've wired the layout so that only one power pack at a time is connected to each block.
HTH
Reply to
Wolf K.
tex shalter skriver:
You say that the locos runs faster when entering the switch yard ?
Then you have electrical problems on your layout - not on your switchyard. Perhaps you only have few connections to your power source on the somplete lauout and more connections on your switchyard ?
Yes. Diodes - they have a 0,7 volt voltage drop across them, allmost independent from the power drawn through them.
But - different engines react different to a (for example) 1,4 volt drop.
Klaus
Reply to
Klaus D. Mikkelsen
DEpendong on how much you need to slow the trains in the section, back to back diodes might be a better fix. Most diodes drop .75 volts each.
Reply to
Chuck Kimbrough
For that purpose, resistors will work, although the slowdown could be quite sudden. You might want to have two or three such slow-down blocks at each end of the slow section. If you want smo-o-o-th slow down and speed up, I'm afraid you will have to invest in more complex circuits, in which case the Circuitron products may look quite attractive. Unless you want to start into a whole new hobby... ;-)
As other posters have pointed out, you would need high wattage resistors, which Radio Shack no longer carries in its smaller stores/franchises.
HTH
Reply to
Wolf K.
That symptom, sounds like your 'track feeders' are not as large as they should be. I.E. you are getting MORE voltage drop on the longer feeders. The Yard tracks are close = shorter feeders.
Yes, if you are looking for a 'drop in voltage', the easiest way to not have problems caused by differing 'trains /locos' would be to use 'pairs' of rectifier diodes (1/2 amp for N, 1 amp for HO, 5 amps for O).
The Diodes, IN PARALLEL, with one diode in each direction, will pass the majority of the applied power, only adding approximately 1/2 to 3/4 volt drop to the circuit. Need to slow things down more, add more 'diode pairs' (these in series with the initial ones).
Chuck D.
However, be aware, this answers your question as stated. You may not have asked the proper (whole) question.
CD
Reply to
Charles Davis
Hi Chuck,
I think the term you're looking for is "reverse parallel". Two diodes in parallel will function as a single diode except that the current handling capability will almost double.
A bridge rectifier with the AC terminals in series with the track feed and a wire, a diode or a series of diodes across the DC terminals will act as multiple reverse parallel diodes.
Regards, Greg.P.
Reply to
Greg Procter
Yeah, that may be a more understandable way of 'labeling' it. The 'bridge rectifier', using the AC in as the connection points I was aware of, but didn't mention because of having a '2 diode' drop to start with. [one diode drop ~.7V , two diode drop ~1.4V]
Chuck D.
Reply to
Charles Davis
Greg, bridge rectifier will only work the way you described if its positive and negative leads are shorted. Draw yourself a diagram if you're not sure. Such a cricuit will drop about 1.4V across it. Without shorting positive and negative terminals it will act as an open cicuit.
Peteski
Reply to
Peter W.
[ Snip ]
Not true about doubling the current capacity, from what I've read. One diode will have slightly less resistance than the other, and will carry more current. When it overloads it will burn out, leaving the other to carry all the current, and it will burn out, too.
This isn't what has been discussed here, but anyone expecting to double current capacity by putting two diodes in parallel should heed this caution.
Reply to
<wkaiser
Your right Bill. For myself, I usually use diodes rated for about 10 times expected 'normal running value' --- I.E. Half AMP expected ---- use 5 Amp diodes.
At LEAST twice the current capacity of the power supply. {Gives some cushion to keep 'short circuit' conditions from killing everything.}
Chuck D.
Reply to
Charles Davis

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