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Mixing RP/NCE power supplies

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G8B4Life:

--- Quote from: Alan on July 26, 2019, 12:44:58 PM ---Nope. No DCC control signal available.

--- End quote ---

Hmm, I think it might be prudent of us to explain this more thoroughly for those that aren't too familiar with DCC and how it differs from RailPro's straight DC.

Now, I'm no artist (that Gene went to another family member) and I know it so don't anyone laugh at my poor sketch.



In the sketch you can see the difference between RailPro and a typical DCC system. In particular to this thread note that RailPro has straight DC on the rails while DCC has what is called a bi-polar DC waveform on the rails. One rail swings between 0 to +15v and the other rail swings between 0 to -15v. While Rail "A" is at -15v Rail "B" is at 0v and when Rail "A" is at 0v Rail "B" is at +15v. This swinging back and forth and the time between swings is what encodes the commands that DCC decoders understand.

This is why you would not be able to drive a train from a power district powered by a RailPro power supply straight into a power district powered by a DCC system, you would get a short. To do so you would need as Alan mentioned a fancy transition section between them that is longer than all the locomotives (and cars/wagons with pickups) on a train that could switch between the two power methods.

If you have any DCC loco's and your not planning on changing them to RailPro it'd be easier to power the whole layout using the DCC system, RailPro LM's can use the DCC waveform for power but DCC decoders can not use straight DC.

Edit: This bit is apparently not correct, read Alan's reply below, then my reply to his reply explaining my justification for what I wrote.
Just an aside for everyone, in regards to the DCC waveform, most of you would probably have heard of DCC being AC, well it's not, it's bi-polar DC as shown in the sketch but most if not all scope traces you'll see of a DCC waveform are misleading and perpetuate that DCC is AC. This is because when people run a DCC waveform through an oscilloscope they hook it up to both rail outputs. In this configuration what the scope sees and displays on it's screen is a singular AC waveform. Not knowing anything about oscilloscopes (and I still know very little) this had me stumped for a long time. Hopefully you've learned something too.

- Tim

Alan:
I think your sketch is great. Very accurate and understandable. In 3D no less!

The only thing I take issue with is your statement about bipolar DC and AC. The output of a modulated bipolar DC supply, when used in two terminal configuration (train track), is an AC square wave. Therefore, is is valid to say DCC is AC on the rails. If the bipolar supply were used in three terminal configuration then we could no longer say its output is AC.

Now, what is going on inside the DCC booster to get from DC to two terminal AC is an entirely different matter and is, as you point out, a bipolar circuit. But it is best we not delve into DCC booster innards.

G8B4Life:

--- Quote from: Alan on July 27, 2019, 08:59:49 AM ---The only thing I take issue with is your statement about bipolar DC and AC.

--- End quote ---

Ok, I'll admit to being wrong when I am (and I've edited my post accordingly) however let me explain why I wrote what I did and hopefully you can explain why my reasoning is wrong without confusing me.

I base my conclusion on DCC not being AC on two things, the only other AC that I know, vis mains power and a snippet I read on MRH.

Rail A does not swing below 0v.
Rail B does not swing above 0v.
Each rail therefore does not alternate to the opposite potential. To me to be truly AC as I know it one rail would have to alternate above and below 0v (+15 to -15v) and the other rail would have to be 0v. I think I understand what your saying about 2 and 3 terminal mode and how that would differentiate between being true AC and bi-polar DC (2 terminal being 1 probe on rail A and the other on rail B, 3 terminal being one probe on rail A, one probe on a ground (0v) and a third probe on rail B. A scope would then show two traces showing each does not swing above/below 0v respectively).

So to my understanding of AC as I see it everyday, the DCC waveform is bi-polar DC and not AC.

- Tim

Alan:
The specific voltages are irrelevant. Difference in potential is the key. The alternating direction of current flow from high potential to low potential is what makes the rails AC. Current flows from A to B one half the time and from B to A the other half. The current direction alternates thus AC.

Don't get hung up on zero. it is not a magic number. Zero volts just happens to be the reference relative to natural earth ground. Since neither rail is connected to a copper rod driven into planet earth zero becomes irrelevant. One half the difference in potential between A and B becomes the zero for an alternating circuit.

Bipolar power is a 4 quadrant system meaning voltage and current can cross a reference point independently. It can source or sink ie. it can power a load or be a load.

G8B4Life:

--- Quote ---...The alternating direction of current flow from high potential to low potential is what makes the rails AC. Current flows from A to B one half the time and from B to A the other half. The current direction alternates thus AC.

--- End quote ---

Ok, That makes sense. There's more than one way to have AC than the traditional mains power way that everyone is familiar with where one conductor alternates and the other conductor is zero/neutral. So if my electricity company decided tomorrow to deliver my electricity the DCC way and not the traditional way everything in my house should work exactly the same as before?


--- Quote ---Don't get hung up on zero. it is not a magic number. Zero volts just happens to be the reference relative to natural earth ground. Since neither rail is connected to a copper rod driven into planet earth zero becomes irrelevant. One half the difference in potential between A and B becomes the zero for an alternating circuit.

--- End quote ---

I try not to, I've read words to that effect when reading while trying to understand how you can have negative voltage (and why you'd need it) however until I can find an article that can explain it visually so it "clicks" I keep on coming unstuck; remember the only actual electronics learning I've ever had was high school and that was a long time ago.

- Tim

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