DCC Expert Help requested regarding voltage drops through long bus wiring

DCC is not AC. It doesn't measure as AC. I suspect that you would measure about the same with no load in the set up.
Jerry Zeman wrote:

: Dear All; : : The following thread is going on at the NCE-DCC board. This is my : latest posting, and I and others have posted about 10 threads on this : subject. : : : Here is a brief summary. If I take 12 gauge solid wire, 84 feet long : on both legs, and hook it up to a 2.5 ohm, 40 watt load, I get a 3.75 : VAC drop at the load. The readings were taken with a digital Beckmann : meter. 10 gauge stranded wire gives the same results. : : Why such a large loss occurs with DCC I cannot explain. I did not see : voltage drops like this when I ran Dynatrol command control, which is : a 14 VDC system. : : Are there any electrical engineers out there that can explain this?
Jerry, I see that Mark has given you great advice and you have followed through so the mystery gets deep. I wish I could make more of a contribution... but here's Alan Gartner's site that discusses wire resistance and has resistance tables for 100 Hz, 10 kHz (IIRC, DCC is 8 kHz) and 100 kHz that might be helpful.

JZ>Dear All;
JZ>The following thread is going on at the NCE-DCC board. This is my JZ>latest posting, and I and others have posted about 10 threads on this JZ>subject.
JZ> JZ>Here is a brief summary. If I take 12 gauge solid wire, 84 feet long JZ>on both legs, and hook it up to a 2.5 ohm, 40 watt load, I get a 3.75 JZ>VAC drop at the load. The readings were taken with a digital JZ>Beckmann meter. 10 gauge stranded wire gives the same results.
JZ>Why such a large loss occurs with DCC I cannot explain. I did not JZ>see voltage drops like this when I ran Dynatrol command control, JZ>which is a 14 VDC system.
JZ>Are there any electrical engineers out there that can explain this?
AC and DC have much different duty cycles. A better test would be to use an oscilloscope and see what the peak voltage is across the load. That will help you see the internal AC impedance of the source. Your meter is attempting to calculate an RMS value of the square wave DCC signal. This is different than pure DC.
Jeff

Jerry, The following from the NCE manual!
POWER REQUIREMENTS The Command Station and Power Station can be powered by the same transformer as long as the transformer is sized large enough. Recommended transformer: 16v at 6 amps or more.
The key wording here is "sized large enough". 16V at 6 amps means the PS should be able to deliver 16V loaded at 6 amps. If you are just powering a booster a supply of 5 amps will do!

Gentlemen, I am no E.E. but a would-be M.E. so I will step up to make a fool of myself (that always seems to Draw them out of the Woodwork). I wonder what the internal resistance of the power supply is. If your load is a fairly pure resistance (not a motor) then you are drawing about 4.8 amps at 12 v. If your power supply has an internal resistance around 0.8 ohms, then your voltage drop at 4.8 amps would be 0.8 * 4.8 or 3.75 v.
On the other hand, if your load is a motor, whose resistance is 2.5 ohms stopped, its resistance will go up considerably when it is turning, because it will be generating a back emf. It would then draw less current. Still, perhaps it still is the internal resistance of the power supply.
On the other other hand, these newfangled power supplies like to have some kind of IC regulators in them, which should keep the voltage from dropping so much with applied load.
I wonder if there is not some reactance going on somewhere, capacitance or inductance or something. Is your 12 gauge wire perhaps still on its spool... A DCC signal is most certainly AC, because it sure isn't DC. WILL SOME ELECTRICAL ENGINEER PLEASE SHUT ME UP? Cordially yours, Gerard P.

GP>Gentlemen, GP> I am no E.E. but a would-be M.E. so I will step up to make a GP>fool of myself (that always seems to Draw them out of the Woodwork). GP>I wonder what the internal resistance of the power supply is. If GP>your load is a fairly pure resistance (not a motor) then you are GP>drawing about 4.8 amps at 12 v. If your power supply has an internal GP>resistance around 0.8 ohms, then your voltage drop at 4.8 amps would GP>be 0.8 * 4.8 or 3.75 v.
GP> On the other hand, if your load is a motor, whose resistance is GP>2.5 ohms stopped, its resistance will go up considerably when it is GP>turning, because it will be generating a back emf. It would then GP>draw less current. Still, perhaps it still is the internal GP>resistance of the power supply.
GP> On the other other hand, these newfangled power supplies like to GP>have some kind of IC regulators in them, which should keep the GP>voltage from dropping so much with applied load.
GP> I wonder if there is not some reactance going on somewhere, GP>capacitance or inductance or something. Is your 12 gauge wire GP>perhaps still on its spool... GP>A DCC signal is most certainly AC, because it sure isn't DC.
GP> WILL SOME ELECTRICAL ENGINEER PLEASE SHUT ME UP? GP> Cordially yours, GP> Gerard P.
Nahh. You are doing fine for a would-be ME ...
Jeff

Since when does an electrical signal have to be AC when it is not DC? What is lightening? AC in the traditional sense implies a sine wave. People would be much better off to forget the AC/DC thing when talking about DCC, it does not make a for good analogies.
Think of it like a high current computer network protocol and things make more sense.

>AC in the traditional sense implies a sine wave.

DCC "alternates" between a positive and a negative voltage. It wouldn't have to be "traditional" to cause an effect in a winding, as suggested.

S>> A DCC signal is most certainly AC, because it sure isn't DC.
S>Since when does an electrical signal have to be AC when it is not DC? S>What is lightening? AC in the traditional sense implies a sine wave. S>People would be much better off to forget the AC/DC thing when talking S>about DCC, it does not make a for good analogies.
S>Think of it like a high current computer network protocol and things S>make more sense.
One could easily argue that it is a non-sinusoidal AC. The fact that is has a zero crossing point and changes polarity puts it "squarely" in the AC camp. Or it could be argued that it is a bipolar PWM (pulse width modulation) technology, which it really is.
Jeff

Yes, all of which is really the main point. One cannot treat it like the AC they are used to with Lionel trains and expect the same results. Just calling it AC confuses people who understand traditional analog electricity.