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
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.
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.
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
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
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
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
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
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.
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.
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
[ 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.
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.
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