Somebody twist my arm. Please!

Discussion in 'DCC & Electronics' started by Pitchwife, Aug 17, 2005.

  1. Pitchwife

    Pitchwife Dreamer

    Recently I've been giving more thought about going with DCC on my new N scale layout. It's still in the planning stage so all options are still open. My original thought was to just go with DC because of the initial cost of DCC. I've worked in electronics so the wiring isn't be a problem. Also I already have two MRC Tech4 280 dual controllers plus an older MRC controller. With the layout designed as it is 5 engines would be about the max that would be running at one time anyway. So I figured I'd be set. Then when I got to visualizing the layout in the room I came to a realization. If I were the only one running trains it wouldn't be a problem. Even two would fit OK. However, if I had enough people over so that everyone was running an engine consist or switch yard it would get very crowded. That started me thinking more about DCC.

    The idea I've been kicking around lately is to wire it for DC and start buying and installing decoders in my engines without wiring them into the motor circuit.

    This brings up my first question. Can a DCC decoder equipped engine run on a DC layout? If they can, I could just do the complete install when I mounted them.

    Then as I get enough of them ready, I can go ahead and get the DCC controller and accessories and convert the track wiring to DCC.

    So the second question is, is this a practical way to go about it? How difficult would it be to change the wiring from one to the other? Would it be possable for me to set up the wiring so that all it would take to change would be connecting/removing jumper wires.

    The third question then would be, if it is practical, what system would best lend itself to this purpose. I figure that if I'm going to do it, I'll do it right. Sound, light, the works.

    The only thing I would probably not use it for would be throwing turnout switches. I plan on building switch machines (my own design) for all of the turnouts, but everything about the trains themselves would be controlled by DCC. However even that is open for debate.

    So, given these parameters, what would everyone advise? I realize that just about everyone will have different suggestions, but I would like to hear from anyone who has an opinion.

  2. Gary Pfeil

    Gary Pfeil Active Member

    Most DCCdecoders will operate on analog, but not all. Digitrax will, as far as I know soundtrax will not. I can't speak for others, you can check before buying. A properly wired DC layout needs no mods to run DCC. What is proper? Wiring to handle heavier current, basically. While there are differing opinions on what is "right" I use 14 gauge for bus wire and 22 gauge for track drops. Regardess of DC or DCC, I like drops on every piece of rail. Plenty of people say that's overkill. If you put a drop on every other piece of rail and solder every other rail joiner, that would be ok. As far as which system goes, I've used both Digitrax and NCE and both are great. I can't speak on the others as I've never seen them.

  3. VunderBob

    VunderBob Member

    The Atlas decoders have a jumper set that will absolutely convert a decoder to DC operation.

    Others have a CV that when set to 1 will let the decoder detect the presence of DC and run on it, but I don't remember which ones do that. Probably Digitrax, and maybe Lenz...
  4. tillsbury

    tillsbury Member

    Some people have created a DC system (or modified their existing one) such that one cab is DCC, and the other is DC. This allows you to run either (a) entirely as DC, since you just set the DCC throttle to loco 0, or (b) some on DCC and some on DC, or (c) flick every switch over to the DCC throttle, and run an entirely DCC system. It sounds to me like a very neat solution if you've already got most of a layout, or if you have a lot of analogue locos... All the DCC locos I've tried work on DC without alteration except an old Life-Like switcher, and I can't remember what kind of decoder that has. Sorry, that wasn't very useful. Atlas VO-1000's, for example, work fine (I think they're Lenz?). Most Digitrax drop-ins ditto. It's no good if you have to set a CV to make them work DC -- if you could do that you wouldn't need them on DC!!! Easy to test though -- take a 9v battery into the shop and put it across the wheels...

    On the other hand, since you said "...if I'm going to do it, I'll do it right..."... that's exactly what I thought, so I bought a Radio Super Chief set and just got on with it...
  5. Pitchwife

    Pitchwife Dreamer

    So all I would have to do is connect all of the operating blocks together to convert the wiring, right? If I ran 14 gauge wire under all the track and used remote switching to turn a block on/off I should just have to bypass the switch and go from the buss to the track.

    The only reason to run heavier wire is because all of the locos are running on the one circuit rather than just what happens to be on a particular block, right? I just want to verify what I think I know. :confused: Another thing, I seem to remember reading on another thread that DCC was AC current. If that is so, couldn't you put step-up transformers on the line to remote areas, like they do with pole mounted power transformers?

    One other comment on your post Gary. When I was first learning about electronics I was taught that even if you ran a heavy gauge wire most of the distance, if you put a smaller gauge wire at the very end you were, in effect, putting a resister in the line, Trying to funnel all the current the larger wire carried through the smaller wire. Scince a small gauge wire has less current handleing capacity it would therefore have more resistance. I would think that you would be better off runninng 18 gauge wire as the buss and the drops. Or running 18 gauge as drops to the 14 gauge buss. 22 gauge wire is awfully small. The track itself has much more current handeling capacity that that does. I suppose if you ran a lot of them though. Just some idle thoughts. :) Thanks for the input, I appreciate all of it. :wave:

    This sounds like something to look into, as I will be running everything DC with decoders installed until I'm ready to get the control system. I'll check them out for price, features, etc. Also I just want to be sure that all decoders work with all systems, or if there are some that don't, which ones they are. Please correct me if I'm wrong on any of this stuff.

    Right now the layout is still on paper. I'm one of those oddballs that needs to plan everything out in as much detail as possable before even buying lumber for the benchwork. I can be a free spirit in other aspects of my life. Having fun with MRRing doesn't include tearing up track or rebuilding benchwork because something didn't work. :curse:

    All of the decoders will need to work without any address setting because I don't plan on buying the DCC system until I am ready to switch over entirely. My reasoning is that I don't want to have most of my engines sitting around collecting dust because I can't afford to put a decoder in them yet, and can only run one DC on the DCC system at a time. First the decoders and everything running DC. Then when all or most engines have decoders (and I have been able to sock away some cash :rolleyes: ) I'll get the DCC system.

    The Radio Super Chief set, is it good and expensive or just expensive? Those two don't always go together. :p I don't want/need one that will make my coffee in the morning, just one that will do what I want it to without costing an amount equal to the GDP of a small nation. :eek:

    So, does this sound like a feasable plan? I will have to do my homework on this, but I appreciate everyone giving me pointers so that I'm not just out there fumbling around in the dark.

    Thanks :thumb: :thumb:
  6. 60103

    60103 Pooh Bah

    I worked on a layout where a couple of sections had been wired for cab control (and we had operated that way) and they were put into a DCC layout. It went smoothly for the most part (I think the problems we had were problems on DC as well).
    You should not plan to operate part DC and part DCC at the same time. Apparently, if DCC meets DC (overrunning a block, say) you get terrible problems -- to the DC power supply.
    DCC is not exactly AC (long debate on the DCC forum). I don't think transformers would boost it, especially as it's a very high frequency square shaped wave. Best bet is thick bus wires and possibly power districts if you get over 5 amps.
    The small gauge feeders shouldn't be a problem if they're short. Remember that they won't be carrying much current (not the full 5 amps, certainly) and they are in parallel (mostly) and are joined by a hunk of rail.
    There are a couple of differences between DC and DCC wiring. Pure DCC wiring tries to eliminate dead tracks in favour of having them fully powered all the time.
  7. Pitchwife

    Pitchwife Dreamer

    Thanks David

    That is the entire reason for planning this the way I'm thinking of. Putting the cart before the horse, so to speak. I understand that most people buy the DCC system and then install the decoders as they are able. I'm doing it the other way round. The process would go as follows:

    1) Strictly DC operated, adding decoders as I can afford it and stashing away $$ till I'm ready to make the plunge.

    2) All or most of my engines have decoders.

    3) Open up the stash of cash and buy the DCC system that I have determined to be the best choice for my needs.

    4) Convert the track wiring to DCC and install the system.

    5) Enjoy!

    Thanks :wave:
  8. tillsbury

    tillsbury Member

    The issue only being that you seem to be thinking of a lot of costs for 'the DCC system' (FX: "Hallelujah" in chorus...). Whereas actually you might find that you spend a whole heap of money on decoders and wire and mucking about and DPDT switches before you get anywhere. Just a thought.

    Anyway, I can report that your decodered engines will almost always work on DC, and that the Super Chief is great. Sure, not always are expensive things good, but never forget that the best things are always expensive.

    If you're concerned about the cash outlay, why not just use a smaller selection of locos to start with?

    I just don't want to see you wasting so much time when you know where you're going, otherwise you're going to be so cross :mad: that you wasted so much time getting there -- once you're there. :thumb:
  9. MasonJar

    MasonJar It's not rocket surgery

    I agree with tillsbury - I think that you should go directly to DCC.

    In your initial post, you mention running up to 5 engines at once. That implies at least 10 blocks in the layout. By the time you fiddle with all the wiring and (electrical) switches that you need for that, you can probably get yourself into a entry or mid-range DCC system.

    I have the Digitrax Zephyr, which can now be had for well under US$200. The Zephyr will even let you run one non-decoder-equipped loco (although it makes a funny noise! ;)). You can also hook up up to two "smooth DC" (i.e. non-pulse) power packs for use as "jump throttles". I assume that one dual controller could take the place of two separate power packs.

    I hope that helps.

  10. baldwinjl

    baldwinjl Member

    I'll jump in on a couple of the technical questions you raised that were not completely addressed.
    DCC is NOT AC, nor is it really DC. I guess you'd call it modulated DC. Anyway, the transformer idea isn't totally off, in DCC they are called boosters, and handle the DCC waveform properly. The thing that is a bit different is that the input to the booster isn't the end of the line going to the track, but a separate signal line from the controller.
    As someone said, not all of the current that does through the bus wire goes through the feeder, unless there are only locos in being fed by that feeder. While the resistance per length in the feeder is clearly greater than the bus, the length is short, and the current is low, so the impact is small. Ideally, you'd use the biggest bus wire you could, no matter what the feeder size. But working with (and buying, for that matter) anything bigger that 12 gauge is not worth the trouble or expense. As far as feeders, big is good, but you need to be able to solder to the rails, and make it invisible. 18 is certainly big enough, I expect in N scale 22 probably is, but I haven't looked up the numbers. I think you'd like to be able to handle 2 or 3 amps, but it won't often be that high, and won't be that high for long, so you can probably even cheat a bit, as long as they are protected, and short.
    There is an excellent DCC clinic in the making at It has a lot of good info, and explains things really well.

  11. Pitchwife

    Pitchwife Dreamer

    I think that personally it would be more frustrating to have spent the money on the system and have it sitting there and only be able to run a couple engines at a time (and it sounds like a hybrid system would be more troble than it is worth) and then add one engine at a time till I'm up to full capacity. If I do it the other way, I can be running all of my engines, and then when I'm ready for it, I install the system and convert the wiring over and continue to go at full capacity without any slowdowns and without any hybrid headaches. Oh, and by the way, I've got lots more time than money. :D :D :D
    Wireing isn't a problem. I've done enough on other projects that it will be the least of my worries. As far as converting the wiring from DC to DCC, a few well placed, hidden switches or jumper points is all that I need. When I'm ready I just hook the DCC boxes up (I'll need at least one reverse loop unit and maybe two) and flip the switches or connect/disconnect the jumpers. Since it sounds like the decoders can be fully wired in beforehand I'll just have to program them (probably the most difficult part of the whole operation), and I'm off and running, still at full capacity. I doubt that I can buy even an entry level system for the price of 10 or 12 switches. I will be able to get a full-featured system without messing with a starter system that I will outgrow.

    I do appreciate all of your input, even if I don't take your advice. :thumb: :D :D
  12. Gary Pfeil

    Gary Pfeil Active Member

    Clark, no need to add additional switches and/or jumpers. When you add DCC just connect its track outout to one of the block switches inputs, replacing a power pack. As was stated above, then you just put all the block switches to the DCC position. A benefit to doing it the way you want to is that you have block switches you can use to kill selected blocks when you need to find a short. There are other, high tech DCC ways to do this (power shields, additional boosters, etc), but for a road the size of yours they really aren't needed. A short anywhere on the layout will shut down the whole layout, thats where the above items are handy.
  13. Pitchwife

    Pitchwife Dreamer

    Great Idea Gary. :thumb: I would have only thought to jumper blocks together to make one large circuit, but your idea makes a lot of sense. Lots easier to do too. Thanks for the tip. :thumb: :thumb:
  14. seanm

    seanm Member

    One quick comment from me that I didn't see addressed.

    If the feeders are short, the small size does not matter. Up above someone said putting a 22 gauge feeder on an 14 gauge buss was like putting in a resistor. ONLY if the 22 gauge wire is really long. Think of it this way... how big are the wires to a DCC decoder?? They are (I believe) 30 gauge wire!! Of course, they are very short, so...... I think you get my point.
  15. Pitchwife

    Pitchwife Dreamer

    I understand what you are talking about seanm, but even a short length will make a considerable voltagee drop. If you don't believe me take a piece of stranded 18 gauge wire and put it across the terminals of a 9v battery. It will short the battery out and drain it. Now take one strand of that wire and do the same thing, with a fresh 9v of course. It will melt the wire in half. Even a short run with fine wire will make a difference.
  16. seanm

    seanm Member


    Then how do they use the 30 gauge on the decoders with out them dropping the voltage.

    On a practical note, I am using 22 gauge feeders that are all less then 12" long going to 14 gauge buss. I put a feeder to both rails about every 3-6 feet. My track passes the short test over the entire layout and I notice no speed drops.... so even if it causes some minor drop in voltage, it is not hurting my system at all. Even with several lashed up unites in a block, the 22 gauge wire never gets warm... so how is it wrong and what would I gain by by going to larger feeders?
  17. Gary Pfeil

    Gary Pfeil Active Member

    Clark, the reason the small drops aren't a problem is that there are many many drops, not one. I have minimal electronics knowledge, but voltage drop is not the real potential problem with small feeders, handling current would be. In your example, it is not voltage drop which causes the wire to melt. Most DCC units provide 5 amps, some more. However, it is the bus that sees the accumulated currents of the locos and lights being run. Individual sets of feeders see only some small portion of that. As someone stated above, the rails themselves will carry most of that load to other feeders.

  18. baldwinjl

    baldwinjl Member

    One more stab at explaining why feeder wires and the little wires to the decoder do not burn up, and do not cause excessive voltage drop.

    Any wire in a circuit is effectively a resistance, or a load in the circuit. The value of the resitance is related to the size of the wire, and its length. There are tables that show the resistance of a particular gauge of wire as x Ohms per foot. So for any gauge and length of wire, we can figure an approximate resistance. The other load in the circuit is/are the decoders. We don't really have a resistance value for them, but for example lets say that a decoder draws 0.5 Amps on a 12 Volt circuit. If it was a pure resistive load, its resistance would be 12/.5 = 24 Ohms. That's just a wild guess, but it is good enough for the purposes of this discussion.
    So, starting simple, with only one decoder, we have the bus, the feeder, the decoder, the return feeder, and the return bus all providing resistance to the circuit.

    So the total resistance in the circuit is 2B + 2F + D. The voltage drop in each element is the current times the resistance. Assuming 0.5 Amps again, the total drop through the wires is 0.5x2x(B+F), and the drop through the decoder is what's left (the voltage doesn't increase to make up for the loss through the wire, so it just gets what's left). For things to work right we want the voltage loss through the wires to be less than .5V (another number out of thin air, but not unreasonable). So the resistance through the wire need to be less than 0.5 Ohms.

    There could be other numbers, but here are the first I found:
    12 gauge 0.0016 Ohms per foot
    18 gauge 0.0064 Ohms per foot
    30 gauge 0.103 Ohms per foot

    So, let's say the feeders are a total of 2 feet long, that gives 0.0128 Ohms. Just for fun let's add 6 inches of 30 gauge in the loco, for 0.05 Ohms. That leave about 0.43 Ohms for the bus, or 268 feet total.
    To allow for 5 Amp draw at the end of the bus, which is probably never going to happen, everything gets divided/multiplied by 10, so the bus would only be 27 feet total. But you can see that the feeder only contributes a small amount to the total, maybe 2 or 3 percent at worst, so the key is in the bus wire, and not the feeder.

    In the case with the 9 volt battery, there was no load other than the wire, and the bigger the wire the more current would flow. The larger wire was able to handle more current, so it did not melt (as quickly). The small wire actually had a larger resistance and drew less current, but plenty to melt it.

    Anyway, it was the long way to get there, but the point is that short feeders don't cause excessive voltage drop, and can easily handle the current involved.
  19. Pitchwife

    Pitchwife Dreamer

    baldwinjl, you sound like an electronics professor I had. :D What I did not take into account is the number of feeders, overlooking the equasion 1/Rt=1/R1+1/R2+1/R3... etc. With lots of feeders the overall loss is minimal. I have seen the light! :D :D :D :D

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