Almost all of us know what happens when we run two trains (or more) in the same block (unbroken section of track) on an analog (DC) layout using one powerpack (like an MRC Tech II). I.E. they each go much slower than if running alone. And they usually balk, too. Why then, if power is always full-on on a digital (DCC) layout doesn't the same thing occur after adding multiples of locos to layout ? Is it merely that main power supply is always adding more voltage (which it has in reserve,say) as each loco is added ? If so, how does this happen ? And, thus, isn't there a limit to amount of power available ?
It is a good question. It turns out that the track doesn't provide the power to the motors in DCC as it does in DC. The decoder has a say in the power distribution. Think of the decoder as a power broker. It is the first recipient of power from the tracks, and it decides how to meter out the power to the various functions you'd like running. It gets full voltage, but it doesn't parcel out any usable voltage to the motor leads until you dial in the first speed step on your throttle. If you only want to turn on the lights, you can do that and the engine will remain still. You can't do that in DC (maybe the new Blueline engines from BLI can, but I have no experience with them). If your decoder provides sound, it will run the speaker immediately that you power the tracks...which is instantly getting the full 17-20 volts. Again, no motion from the engine until you reach for the throttle function and begin to dial in speed. However, the thing to remember is that the motor in each case, DC and DCC, is only a lowly DC can motor. So the decoder has to convert/rectify the AC current it gets (a form of AC), and convert it to DC. As the voltage rises, the motor will want to do more work, and it will draw more current, or what we call amperage. All modern can motors run with just a few tens of milliamps until you begin to impose a load on them. Then they'll draw what they need until they max out their capability. For a sound decodered engine with a single can motor, that is likely to be in the order of 0.5-0.7 amps. As you add engines with lights, passenger cars with lights, sound systems, and longer heavier trains behind them, the amperage draw for the whole system will rise, just as it will on a DC layout. At some point, the base command station will sense an overload and trip its circuit breaker. It will do this on DC controllers as well. The only way your DC engines should slow when you add others is if there is a significant voltage drop. If you have sufficient power and good connections and feeders from place to place, voltage should remain high and the engines should continue to run apace, except that they will begin to near the max for the circuit breaker in terms of heat produced. The circuit breaker is a slightly resistive device that is designed to react to increases in temperature. Those increases in temperature come about as more amperage is being drawn through the narrow wires and rather poorly conducting nickel silver rails. At some point, it lets go and your system shuts down. So, remember this: it's still DC to the motor in either case, and the motor gets what you dial in for DC and what you tell the decoder to give it in DCC. It is just that in DCC the decoder is designed to function like a radio does, happy with the voltage it is assigned by design. If you dial in more volume, you are asking the system to do more "work", and that is what you do to the decoder, and it in turn to the motor, when you dial in more speed on your throttle. The voltage to the tracks stay constant, but amperage (working current) will be "zero" until you begin to parcel out voltage to the motor via the decoder. Does that make sense?
This really boils down to Ohm's law - again . The speed of a motor relates to the voltage supplied. The voltage that reaches the motor depends on the voltage at the supply point, the current being drawn and the resistance between the supply and motor. Volts are lost in the resistance. V=IR, which in this case is volt drop = current x resistance. The wiring and track provide an amount of resistance so adding an extra motor doubles the current and doubles the voltdrop. The change in volts is seen as a drop in motor speed. This is true of DC and DCC. If the resistance is significant the drop will be too. The controller (either a DC powerpack or DCC decoder) introduces a controlled 'resistance' to throttle the voltage and hence regulate the motor speed. Cruder controllers will tend to act like wire/track resistance and the voltage will drop more with increasing current. More sophisticated types of controller don't suffer the same loss, or in some cases (eg. back emf types) actively work to negate the loss. I don't know about the guts of the MRC Tech II controller, but to be fair to it the effect you see could be due to wire and track resistance and not the controller itself.
If you're getting that behavior with DCC then either the wiring isn't properly sized or the power supply isn't of sufficient capacity. If you run DCC you should have at least one RRAmpMeter from Tony's Train Exchange. Use it to test the output of your booster, the track power with no locos, with one, with two, etc. You will see a voltage drop over distance with either, so the fact that there is some drop is not the issue. However, if the drop from your booster output to where you measure on the track is more than 1-1.5v with DCC then your block is too long, or your wiring is not heavy enough. The drop should be the same with locomotives running unless you're pushing the limits of your power supply or booster.
Mike and Leo covered it pretty well. Many DC controllers do not offer regulated output, the idea being that you don't need it as you are (presumably) running only one loco (or one consist) from a power pack at a time and if things are too slow, just crank 'er up! If you were to put, say, a regulated 4 volts on the rails, you would not see the speed drop as you add locos. DCC decoders in effect regulate the voltage to each motor (assuming more-or-less regulated DCC voltage to the rails, which is typical with today's DCC systems), so you don't see speed drop as you add locos.
My bold No it won't be the same (unless it is too small to measure). This idea has had members using voltmeters to fault find track where trains slow or stall but find 'no problem'. If there are no locos or other current drawing devices in a section then no current is drawn. Therefore (from the equation above and assuming there is not a complete open circuit) V=IR has I=0; therefore the volt drop V=0. So checking track voltage with nothing running only checks for open circuits. If you are running the power supply or booster hard the output voltage may indeed reduce, but this loss is additional to that in the track/wiring. To check voltages under load is tricky and will usually need some extra equipment to provide a load. Using locos has the obvious problem that they will keep moving . Ideally a load that pulls a couple of Amps (R = about 7 Ohms) should be connected across your voltmeter probes to give a known load. The meter can then be used as normal by putting the probes on opposite rails which will both supply the load and measure the voltage. You can then quickly see where the volts are lower, and hence where your feeders or track are not up to standard. The RRampMeter instructions describe how to do this using an auto lamp as the load, but you could use a power resistor if you want (and this has the advantage of not causing a heavy surge when first connected).
I reread the OP and realised this bit hasn't really been answered. The answer is sort of yes. A good quality power supply, DCC booster or DC power pack will (once set at a voltage) maintain the same output voltage as the load changes (= more or less current is drawn; more or less locos are present). In practice the voltage will change a bit because it gets very expensive to hold it rock steady - the amount of variation is referred to as the pack's 'regulation'. The regulation has limits which may be quoted somewhere in the documentation; usually from 0 to the maximum output current rating of the pack, but sometimes a smaller range. Which brings us to the last question - yes, there is a limit. It is usually given in Amps for power packs, etc. When you draw more current than that, the voltage will likely start to drop off quite noticeably and eventually either the unit will trip or start smoking ... or on a bad day, do both and then trip your whole house out Transformers (as opposed to power packs) are often rated in Watts (or VA). To get the current rating divide the Watts (or VA) by the output Volts. So a 25 VA, 16V transformer is rated for about 1.5 Amps.
My bold: The tail light on the RRAmpMeter is a good method. My point is, DCC boosters are (supposed to be) regulated power supplies. The idea of a good regulated supply is that it maintains a constant voltage. Indeed, my Astron 20 Amp 13.8v DC supply (a nice quality ham supply) puts out 13.8 volts on my digital volt meter at all loads up to 16 Amps (it's rated continuous max), then it drops very slowly to about 13.5 volts at 20 Amps (peak max). In my experience testing NTrak Layouts with an RRAmpMeter (a true RMS, DCC aware meter; your experience with a standard volt meter will be different) is that you may get 12.5 volts output from the booster, 12.2 volts at the furthest point with no load, and 12.1 volts at the furthest point with 4-5 locos running. 0.0-0.2v max difference. This makes sense, most boosters are 5 Amp regulated supplies. At a half amp a loco running even 5 at once is only half load. They should maintain very tight regulation at that power. However, you can easily make the drop be larger. If rather than using 12ga bus wire, you use 18 ga zip cord then the drop over 100 feet will be 2 volts, rather than 0.2 volts. Hence my comment that if you're seeing drops with "a couple of locos" it's probably a wiring problem. So to be more specific, if you're drawing under 60% of your boosters capacity, and you have a quality power supply feeding a quality booster, and you have good wiring and power districts with reasonable sizes, the drop between no-load and the 60% load should be so small that it is nearly indetectable. 0.0-0.2 volts, maybe up to 0.4. Larger drops indicate (mostly) poor wiring, but less likely poor boosters, poor power supplies, or poor grounding.
So is the answer simply that DC power packs don't have good voltage regulation? There is no reason they COULDN'T, it is just that most aren't offered with the electronics to increase current to maintain voltage. (Right?) Otherwise I don't understand why there should be problems in DC.
Sounds about right Yes and no. As said above, DC packs have a knob that allows you to set the output voltage to control your locos, so there isn't any great need for regulation unless you need trains to run unattended at constant speed. For that you can get BEMF controllers that will adjust the volts to maintain constant speed under varying conditions, but that is not really voltage regulation as it involves getting speed feedback from the load and adjusting the output voltage to correct it (the speed). Sorry to go on about it but power supplies don't "increase current to maintain voltage". Current flows as a result of voltage (the water analogy is very useful). Regulated supplies adjust their internal workings (pumps) to maintain the output volts constant across a range of current draw. It's the load(s) that determines the current that actually flows.
To continue to illustrate the relationship between amperage (current) and voltage as I did at the outset, I will relate a story. My father was a mining engineer for many years and gained a rather substantial reputation as a consultant due to his successes where he was employed. IOW, he would get farmed out for a week or two once or twice a year by his corporation for a fee. In one instance, after having been on the ground of his host's concentrator and crushing operation for about three days, he happened upon a group huddled around a crusher and its control panel. My dad was trying to sort out recovery issues (often related to a mix of variables in concert, such as tonnage of ore pushed through the mill, the fineness of the grind from the crushers right through to the smallest ball mills, and so on..) My father asked the group what was going on and they said they felt the 200 hp GE motor driving the jaw crusher was running hotter than the specs indicated were optimal for long life. My dad felt the motor's outer casing and agreed that it was hot. Looking inquiringly at the shift boss and the electrician who were pondering this development, they seemed to have no good idea what to do. My dad asked what the voltage was that was being provided for the motor service. The elecrician glanced over and responded, "360". My dad said, "Raise it to 400." They raised it, and within a couple of minutes the outer casing began to cool noticeably. The motor was going to draw all the amperage it needed for the work forced upon it...by the density and hardness of the rock, but also by its volume per unit time. At a regulated voltage that was too little for the "force" required, the motor heated up into the problem zone. When my dad had them raise the voltage, it eased the flow of the amperage, for want of a better term, and made it more readily available to the starving motor. Think of sucking a golf ball through a garden hose....few of us would be happy doing it, but you don't play golf until you have the golf ball. In our context, the thin wires have a small cross section which tends to restrict the passage of the current-carrying voltage. They will also heat up when you begin to force "work", or amperage, down those wires by running hard-working can motors through that powered section. Does that help?
Sites like this are quite useful too: Voltage Drop Calculator One difference you need to understand is that as the wiring distance increases, DC Voltage drops much faster than AC voltage (hence by AC is used for long distance transmission lines). 12v DC system, 5 Amps, 12ga cable, 100 feet of cable, 16.7% potential drop. 12v DC system, 5 Amps, 14ga cable, 100 feet of cable, 26.2% potential drop. There's a reason why the power pole spec for NTrak used 12ga wire. The previous 18ga zip cord recommendation was not good for DC, much less DCC. The loss over NTrak distances is too high, you see it in your DC trains speeding up and slowing down, and in the DCC trains not working when the voltage drops too low. This is also why almost always is the drop a wiring issue. Usually the bus gauge is not heavy enough, but sometimes it's not enough feeders or other issues. So check the distances. Even in an 8'x4' layout you may have runs of 20 feet time you consider the actual wire distance. Paying an extra few cents for 12ga (or 10ga, if you have a large layout) will make your layout perform much better down the road.
Almost correct <G>. I believe you are referring to "skin effect" (where AC current migrates toward the surface of the conductor, thus reducing it's effective cross-section. The higher the frequency, the greater the migration). Skin effect at DCC frequencies is negligible (a few percent at most) and for all practical purposes can be ignored, so AC and DC voltage drop would be effectively the same. The primary reason that AC is used for long distance power transmission is that AC can be stepped up to higher voltage (or down to a lower voltage) via a transformer (DC cannot be stepped up, it must be converted to AC first). At a higher voltage, less current is required to transmit a given amount of KW, so smaller cable is needed. Back to our miniature ("reduced") world though, the larger (12 gauge) wire recommendation for DCC was a result of higher expected current to each power district/block due to the need to accomodate multiple trains per district, whereas DC running only can operate one train per block (block overruns excepted). For a given voltage loss budget (a max of 1 volt or 10% was assumed in the NTRAK RP evaluations), the higher current draw of multiple DCC trains requires the larger wire.
A little clarification: The motor uses however much wattage it needs, and since watts = volts x amps, by increasing the voltage it lowered the amperage draw of the motor causing it to run cooler.
Hey MarkinLA, are you following all this? I'm just a lay person here, sorry if I'm not understanding. I'm familiar with the formulas but that doesn't help if you don't know which one to apply. Going back to the original question and cutting it in half... Why is there a voltage drop when you add locos to a DC circuit? (Please explain in terms of a formula, if you can.)
Well, this is actually a bit more complicated, and depends a bit on the type of throttle that you are using. At a high level, there are two factors at work: 1) The factor we've been discussing, cable loss. With DC this is easy to see with a volt meter. Take all of your locos off the track and spin your throttle up half way. Measure right at the output of the throttle, then the furthest point away on your layout. You'll see some drop in the voltage, probably in the range of 0.1-2 volts depending on distance and your wiring. The problem is this loss (for a given size cable) is dependent on two factors, distance, which was just measured, and current. The higher the current, the higher the loss. Going back to the calculator I posted before we see that: 12v DC, 0.5 Amp, 12ga wire 20 feet of cable: 0.3% (think 1 loco) 12v DC, 2 Amp, 12ga wire 20 feet of cable: 1.3% (think 4 locos) (Note, the calculator doesn't have 16 and 18 ga wire common in model railroads, so this next example isn't realistic for N scale, but is given to put some scale to what 12ga wire can handle; it's rated for 20 Amps): 12v DC, 10 Amps, 12ga wire 20 feet of cable: 6.7% More motors == more amps == greater voltage drop. And remember, the drop is dependent on distance. At 0 feet of cable it is 0%. You can again see this with your volt meter. Set your throttle at half speed with 4 locos running. Check the throttle output voltage, let's say it's 7 volts. Clip the throttle on the furthest leads to the track, and watch as the train goes by. It will be like 6.8v when the train is near the power pack (no amps going down the long cable run) and 6v when the train goes by. Since the motor spins based on voltage you see the train speed up and slow down. You can make the effect more dramatic by: - Increasing the load (more locos). - Increasing the distance (splice in a 100' spool of cable). - Lower the voltage (which makes it easier to see, as the motors perform poorly at low voltage). 2) Is a factor of how power supplies work. I spoke of a regulated supply. This is no different than an air tank regulator, or a water pressure regulator. It's an electronic circuit inside the power supply that tries to keep the voltage constant. It is more expensive to make a regulated supply. It's easiest on a constant output voltage. So my ham radio supply which is 13.8 volts has a very good regulator, and it always puts out 13.8 volts, no matter what the draw. An unregulated supply, BTW, might be 13.8 volts @ 0 Amps, 13.5 volts @ 5 Amps, 13.2 volts at 10 Amps. The problem is regulating a variable output. It can be done, but to do it properly becomes quite expensive as you get to lower voltages. If you buy components that can produce a regulated 13.8v @ 15 Amps, making them do 1v @ 15 Amps is extremely expensive. So most power packs use a hybrid approach. Each have their own methods, further messed up by "pulse power" and the like, but basically consider they produce a regulated internal voltage, and then run it through a second circuit (much like a simple rheostat, but better) to produce the right voltage. The back end of that circuit is not really regulated, and thus also has the problem that the voltage drop is dependent on load. The way to measure this is to put the volt meter on the power pack itself, and measure with 1, and then 4 locomotives. You'll see most power packs put out a slightly lower voltage under the higher load. So there are two factors that conspire together to make DC trains "speed up and slow down" over the layout. High quality power packs with better regulation combined with over sized wire make the problem disapear, non regulated supplies (like cheap trainset supplies) with thin wire (ever see a layout with 22ga wire?) and it happens really bad. How does DCC get around all of this? Well, first, it uses a constant (high, relatively speaking) voltage. This removes the low voltage regulation problem on the power supply end, and in fact removes the variable voltage problem. It becomes easy to produce a good regulated supply cheap. Second, although there is a drop across the system to the decoder, the decoder in essense re-regulates the voltage at the motor, by taking the constant voltage and running it through a regulation and then voltage adjustment circuit. If the computer chip in the decoder says put out 5v to the motor, it can do that over an input voltage from the rails of ~10-15v. The computer controled regulator keeps that motor voltage constant. Since the wire loss (decoder to the motor) is short and doesn't change, there is no variability. Edit: I almost forgot. For all the electronic wizardry don't forget that trains slow down when they go up grades or around tight curves. The added drag from the train causes the engines to draw more amps. More engines, more amps, more voltage drop, and also harder for the motor to do it's job at low speeds. DCC decoders use "Back EMF" (which is a whole different topic) to compensate; but at the end of the day it's just the computer noticing the added load and bumping up the throttle automatically for you. If you have grades, tight curves, and long trains there may be no electronic problem, it may just be simple physics.
More specically Vloco = Vpack-Vdrop, where Vdrop = I * R. Resistance (R) is the sum of wiring and track resistance between the output of the power pack (Vpack) and the loco (and some inside the power pack as well, especially the cheapies). As you add locos, more current (I) is drawn, resulting in greater voltage drop (Vdrop) and lower voltage at the loco (Vloco). Mr. Ohm will gladly accept your donations.
Thank you. To the person who has not studied this, it is COMPLETELY unclear that the 'V' in V=IR refers to voltage drop and not voltage supplied by the pack. (Mike, I realize now you said that earlier, but I missed it.) Seems like there should even be a different symbol besides V, I must say. So in essence, the answer to the original question is simply this... In DC, the voltage that affects the speed of the loco is between the pack and the loco(s). Since there is only one pack in a block, adding a loco (or consist) to the same track affects the voltage of all the locos that pack is powering. And by ohms law, the result is that they slow down. In DCC, the voltage that affects the speed of each loco is between its decoder and the motor. Adding a loco to the same track does not affect the motor voltage for any other loco, since that voltage is controlled separately for each loco by its own decoder. Each loco is (literally) carrying it's own DC power pack with it, so it is as if it were in its own DC block. Which is how DCC works in the first place.
Forgive me if this has been mentioned, but in skimming the two pages of posts I didn't notice this being covered. DCC is a pulsed AC signal. The digital control information is encoded by width of pulse instead of phase of pulse, so that the decoder gets the same data regardless of direction of travel. In addition to reading data off of the DCC signal, the decoder rectifies the signal into DC to power the locomotive. But as I understand it decoders do not perform a digital to analog conversion to set a DC level. Instead they control speed by varying the widht of DC pulses to the motor. So for slow speeds the motor gets narrow DC pulses interleved between wide zero volt periods. As the width of the DC pulse widens, the zero volt periods become narrower and the speed of the locomotive increases. But the fact is that even DC power packs are pulsed. Every DC power pack I've ever put on an oscilloscope had pulsed at 60 cycles per second (house current frequency). In DC power packs the amplitude of the pulses varies, as opposed to the pulse width in DCC. Regards,
