How much track voltage for DCC?

[modelmaker51]

Paul3:

IF you read further in that NMRA bulletin : "To keep the DC component of the total signal at zero as with the "1" bits, the first and last part of the "0" bit are normally equal to one another."

They don't mention AC anywhere in that bulletin.

And it's not internet rumor, if all the manufacturers say it's half-wave DC (you'd think they would know what they're building) and you're the only one that disagrees, who are we to believe.

[grayfox1119]

Paul3........do you remember ELI the ICE man from electronics school? In an Inductive circuit, the voltage leads the current, in a Capacitive circuit, the current leads the voltage. You don't have that with DC and don't have that with DCC.
This is a good discussion, it is bringing back cobwebs from the past.

[daveb]

Paul3 is correct.
DCC is an AC signal, it starts from zero, rises to its maximum then drops passing through zero to reach the maximum again but with the opposite polarity. AC can have many waveforms not only sinusoidal.
The motor sees a "halfwave DC signal" as the decoder selects either the positive or negative going part of the DCC signal to determine its direction of rotation. This is why the direction of a DCC equipped loco is not dependent on track polarity.

Dave.

[daveb]

A good description of the DCC signal is to be found in Chapter 3 of the book Digital Command Control by Ames, Friberg & Loizeaux,
where they refer to it as bipolar (which to me is another term for AC, but possibly now used to differentiate from the normal sine wave household supply).

According to my knowledge a DCC signal is " A pulse width modulated AC square wave".

Dave

[train18393]

Wow, what a discussion!!! ELI the ICE man is good especially when making adult beverages. Just thought I would lighten up the conversation. My answer remains the same as my earlier post, call the DCC signal what you will. I would call it a Bi-polar, pulse modulated signal,but what I would call it has no bearing on anything.  What the manufactures call it is usually their lingo and may mean nothing as well. It does go positive and negative, usually by the same amount but not when you have stretched pulses. The reading of the inexpensive voltmeter will vary depending on what the signal is doing (pulsewidth.) I believe when adjusting the voltage you are adjusting the peek voltage. I do use an inexpensive Radio Shack $9.99 voltmeter to measure the track voltage. I know it is not accurate. I have True RMS voltmeters and an O Scope, and lots of other expensive working tools, but know what?  I don't usually use them for measuring DCC signals. I use the Cheepo RS meter. I use it to take comparitive readings, and to see if a signal is present. I do not use it for a reference. If all your decoders are working and none are overheating your voltages are fine. 

Now here is what I will really get flamed for. When you reduce the track voltage output of many DCC systems you are loosing that voltage by electronically converting it into heat in the DCC command station (or whatever you want to call it.) That is not necessarily a good thing, but it is another discussion and does not answer the original question. Someone could start a new thread on what to call the DCC signal and/or where does the voltage go when adjusting your command stations output.

 

Paul

Dayton and Mad River RR. 

[rrinker]

That's why you shouldn;t use a 25 volt toy train transformer to power your command station - especially if you set the output to 14.5V or less.

The power supplies designed for DCC system, at least the ones Digitrax and NCE sell, put out 15 volts, which is just perfect for an HO 14.5v output or an n scale 12v output. This is the issue with some of the others - their DCC supplies put out 19v or more.

                                        --Randy

[jbinkley60]

There is some mangling of definitions going on here.  DCC is a digital communications protocol made up of pulses that are coded in a format that decoders can read and write to/from.   This is true digital technology.  Thus digital logic circuits, processors and similar technology can be utilized to manipulate the pulses.  If it were AC (i.e. analog) then there would be no definite 1 or 0 bit definitions because true AC has no point along the sinewave where two consecuvite time intervals on the waveform have the exact same amplitude.  Thus it is continually changing.  You may find a repeating set of events in subsequent wave intervales but not within the same 360 degree rotation of the sine wave.  That would be a violation of the sinewave function. 

What everyone is talking about is how does DCC get sent down the "wire" onto the tracks, be able to go over some distance without the waveform becoming "non-square" over distance and how does it also carry a current component to power the locomotives and such.  The DCC spec calls it a baseline encoding method.    What this is, is a differential voltage signal between the rails.  So think of rail A as the positive rail and rail B as the negative rail.  The voltage difference between the two rail equals the total amplitude of the DCC pulse.  Here's an oscilloscope snapshot of rail A.

 

 

You can clearly see it is a 0 to 15V positive pulses.  Rail B would be the other half of the wave form.  In this case I used ground as the reference point for the measurement.  Decoders look at the entire waveform differential between the rails but what they reallu care about is whether the pulse is a 1 or a 0.  Now comes the question of bit polarity.  If you look at the PDF posted above, you will notice that the voltage referemces between the rails doesn't change whether the bit is a 1 or a 0.  In other words whether the bit is a 1 or a 0 the A rail will still have a positive polarity pulse with regards to ground and the B rail will have the opposite.  This doesn't change.  If it did then the power that decoders pull from the pulses would constantly be changing polarity and the decoders would need full wave bridge rectifiers in them to compensate for the changing voltage references, also the concept of forward and reverse would have no meaning.  So if you look again at the PDF on page 1 it describes the bit encoding method.  It also talks about keeping the DC component at zero.  What they are saying is that the the two halves of the waveform (the Rail A half and the Rail B half) both be equal and opposite in duration and polarity.  In other words if a 1 bit was encoded simply as a long positive voltage pulse on Rail A then the longer the pulse the more is starts to look like a pure DC voltage for the duration of the pulse cycle.  This is bad for transmitting current down a transmission medium and will also cause more voltage drop because this DC component and the resistance of the rails.  This technology and encoding method is very similar to what the telcos use to send T-1, DSL and similar signals long distances down twisted pair wires but also provide power for repeaters and similar devices where local power is not available. 

I hope this helps shed some light on the topic.  It is timely because I had the scope out today as I am working on designing a decoder add-on that can utilize an auxiliary power supply but use DCC to turn off/on lights with a decoder function command, vary the brightness of lights using the decoder speed function and provide straight 12VDC for powering control panels, switch machines etc.  I needed to do some testing.  I'll post the circuit and parts list when I am done.  It looks like for around $50  (including the decoder) I can provide up to 3A of output for these functions all under DCC conrol and without using power from the rails (except a few milliamps to power the decoder).

 

 

[grayfox1119]

Thank you Jeff!!! I hope now we can put this baby to bed.....many of us have been saying this, but you have shown it graphically to clarify the confusion. Maybe you can use a differential amp and a ladder network to switch on/off bits to help your circuit design.

[jbinkley60]

Thank you Jeff!!! I hope now we can put this baby to bed.....many of us have been saying this, but you have shown it graphically to clarify the confusion. Maybe you can use a differential amp and a ladder network to switch on/off bits to help your circuit design.

I am actually looking at something more simple than that.  Basically start with a simple power supply.  Transformer, fuse, bridge rect and cap.  Then add a 12V regulator for the fixed voltage output for control panels, switch machines etc..  Then add a simple 12V minature relay off of the decoder function leads for an on/off function control.  For the variable voltage portion I plan to simply use the motor leads with a cap, regular diode, resistors, fuse and a TIP120 darlington transistor to drive a variable voltage off of the decoder.  The decoder will simply be a Digitrax DH123D or similar.  It can be expanded in a number of ways.  Add another relay and function control for multiple on/off zones.  Add a bigger transformer and rectifier for more current.

Right now I am verifying the DC isolation capabilities of the decoder since with the variable supply I will mixing two different source supplies together.  I blew up one decoder today already.  They don't like either motor lead taken to a true earth ground potential.  Fries the decoder almost instantly.  It looks like a floating ground will be fine.  I've got a little more testing to do.

Hookup should be simple.  115VAC for the power supply and two leads to the track.  Outputs will be 12VDC fixed, 12VDC on/off decoder controlled and 0-14VDC variable decoder controlled.

 

 

[daveb]

If the DCC signal is not AC then why are bridge rectifiers needed in order to measure current with a DC ammeter and as voltage dropping devices ?

Dave

[DALCruiser]

You might find the following link useful for constructing a simple circuit for using a DC meter to measure the DCC track voltage.

http://jdb.psu.edu/nmra/dcc-voltmeter.html

I have not tried it but it looks like it should do the job.

 

Dave

 

[jbinkley60]

If the DCC signal is not AC then why are bridge rectifiers needed in order to measure current with a DC ammeter and as voltage dropping devices ? Dave

Because DC ammeters are based upon straight DC and the easy way to take a pulse signal and convert the partail duty cycles of the waveform into pure DC is via a bridge rectifier.  The problem is that you don't end up getting a true RMS value of the current because the duty cycle of the pulses is not the same as the standard .707 RMS value for traditional AC sinewave, when going through a bridge.  This is why standard multimeters only give an approximate value of the steady state current.  The PDF above shows a better way to convert the DCC pulses into their steady state DC components and then they mention using a small resisitor for measuring the current component.  This is  a traditional method for measuring both voltage and current in a pulse technology.  The thing to keep in mind that the current draw and actual voltage on the rails is an instantaneous value based upon exact spot on the pulse cycle you choose to look at. During a pulse it should remain steady state through the end of the rise time and to the beginning of the falling time.  The current component will follow the same cycle.  In AC the current and voltage are often out of phase with each other due to the types of loads (inductive or capactive in nature) the AC voltage must deal with.  Voltage leads current with inductive loads but current leads voltage with capacitive loads.  In pulse technology capacitive loads look like a rounding  of the corners on the pulse.  Inductive loads may look like an overshoot on the rising edge.  With what we ae dealing with, I doubt we will see much of this. 

 

 

 

[daveb]

AC or DC on tracks.

Please read DCC for Beginners PDF on Tony's Trains web site under Basic Principles.

Dave

[MRRSparky]

It looks like you are measuring one of the funtion output voltages.  Isn't there some voltage consumed by the decoder before getting to the F(0) output?  I think it depends on the mfgr. of the decoder, but I seem to remember that the funtion output voltage can be 1.5V to 2.5V less than what is actually on the rails.  To measure DCC track voltage, couldn't you measure one rail to ground and double it?

Scott Groff

Adirondack & Saratoga Rwy.

Sumner, WA

[train18393]

Engineer Jeff,

You said it well. I was trying to be brief so as to not confuse some, but guess I should have gone into more detail. What you have said concerning this is 100% correct. One thing you didn't address is the streched bits some DCC systems use to operate non DCC equipped locomotives with DCC equipped engines. From what I understand you can even MU them with the Digitrax(and probably others,) which I have never tried. The streched bit should further confuse some who are trying to combine analog definitions and technology with digital. The analog part ends with the power input to the DCC command station.  As long as we have fun with our model railroad that is all that matters.  

Paul

NYC,  Dayton to Springfield fan    

[jbinkley60]

I completed my design for a decoder adapter that can be used to turn on/off layout lighting (or anything else which requires 12vDC), that also has a constant 12vDC output (for control panels, switch machines etc..) and has a 0-14v DC variable output to adjust layout lighting brightness.  All are controlled via DCC and utilize a Digitrax mobile decoder to drive the adapter.  No power is used from the rails, expect to power the decoder itself.   There are two versions.  Here is the schematic for one version:

If you want to read more about it and see the unit I assembled go to:

http://www.thebinks.com/trains/decoder%20adapter.html

 

 

[jfugate]

You might find the following link useful for constructing a simple circuit for using a DC meter to measure the DCC track voltage.

http://jdb.psu.edu/nmra/dcc-voltmeter.html

I have not tried it but it looks like it should do the job.

Dave

Quite true, but you can also get the same result by just reading the voltage across the blue and green wires on a loco decoder and pressing F1 on your DCC throttle. Just use a regular voltage meter -- it only takes a few moments and you don't have to build anything special.

[rrinker]

I completed my design for a decoder adapter that can be used to turn on/off layout lighting (or anything else which requires 12vDC), that also has a constant 12vDC output (for control panels, switch machines etc..) and has a 0-14v DC variable output to adjust layout lighting brightness.  All are controlled via DCC and utilize a Digitrax mobile decoder to drive the adapter.  Now power is used from the rails, expect to power the decoder itself.   There are two versions.  Here is the schematic for one version:

If you want to read more about it and see the unit I assembled go to:

http://www.thebinks.com/trains/decoder%20adapter.html

 

 

 I STRONGLY recommend you put a snubber diode across the relay coil so you don't damage the function output from the spike when the relay drops. Also make sure you maintain isolation between power sources since you have the gray decoder wire connected to the fixed 12v common. Obviously it works but a robust design would probably add an optoisolator to completely disconnect the decoder from the power supply.

                                                 --Randy

[jbinkley60]

Both are valid points and folks are free to modify this design as needed.  My goal was to make it as simple as possible and yes, I may have left out some protections.  A snubber diode or capacitor would be easy to add.  The relay I am using is pretty small and I doubt the inductance of the coil is more than 1-2 mH but it isn't a bad idea.  I considered the optoisolator option for controlling the variable voltage but Radio Shack no longer carries anything which would work easily and I wanted to use parts from there, since they are everywhere.  Of course in quantities of 1-2, their prices are outrageous.  Another thing I've noticed is that the value of C3 can be dropped quite a bit.  The goal was to provide a smooth voltage to the Darlington so that the lights didn't flicker as track voltage changed under load.  When I was powering lights directly from the decoder motor outputs, I could see flickering of the lights.  At 220uf there is some delay between when you turn the speed down and when the lights dim.  More of a delayed effect than a major issue.  I also added 3 diodes in series of the common lead on Q3 to get the output voltage up to 14V instead of 12V.  The Minatroics yard lights I am powering use 16V bulbs that look better at 14V instead of 12V. 

Thanks for your feedback.  I hope somebody finds this useful.

[rrinker]

 That's quite a nifty way to filter the PWM output from the decoder. With that size cap - how long do the lights stay on if you do an e-stop on the decoder?


                          --Randy

[larak]

If the DCC signal is not AC then why ... Dave

Wow. This topic really opened a can of worms.

DCC is AC. Paul was correct. So is Dave. Paul's pdf clearly shows the centered zero line in the illustrations.

Here is something else to consider:

DCC is designed to have an average DC voltage value of exactly zero. This is done by having equal areas under the cure for both positive AND negative portions of the signals. This is done so that a DC loco will not run when placed on the track. A pulsed system would have both a symmetrical component AND a DC component, the DC loco would run.

This is similar to how pulse stretching works except that we don't directly add a DC component. Instead we fiddle with the time domain of one half of the signal, thus making it asymmetrical.

Try a spectrum analyzer or fourier analysis on a non stretched DCC signal. You will see a series of sine waves added together.

Better yet, don't - too much trouble and it takes time away from the trains. Simply place an old analog (slow) DC meter across the rails. It should read zero volts because that is the average value. Just like the average value (not RMS, just average) of house voltage is zero. zero DC component, non-zero AC component = AC.
 
"Half wave DC" is a marketing term not an engineering term, and probably refers to what the OUTPUT of the decoder feeds to the motor. The term itself is misleading at best.

In the end ... DOES IT REALLY MATTER?

[rrinker]

 ACK! No more time-domain modelling and Fourier transforms!

[selector]

What does shoeing horses have to do with MRR?

[rrinker]

 Well they DO call locomotives "Iron Horses"......