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LEDs on AC Accessories circuit

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Posted by rrinker on Thursday, October 8, 2020 8:49 AM

Overmod

 

 
rrinker
 Rail to rail isn't 30V - it's typically around 15V. One way to measure track voltage without a special meter is to measure Rail A to common, and Rail B to common, then add them.

 

I thought DCC spec was at least 14V either plus or minus (measured from the respective rail to common) with the modulation arranged as on old modems to keep the net electron charge transfer equalized.  That by definition would give you not less than 28V rail-to-rail of the power supply, were you able to get inside and measure it.  If the actual modulation of the 'square wave' signal is only ~7V, which as I recall is around where some sound chips become active, I'd be MUCH less surprised why even small booboos in track lead to functional dropouts...

 

 

 But as one rail goes to +7 relative to the common, the other rail goes to -7 relative to ground - so rail to rail is 14V, not 7. Yes, there is a short zero crossing, but it's a fast rise/fall square wave, so apart from the slope not being perfectly vertical since no transistor responds that fast and there is always capacitive and inductive factors in the wiring and that giant capacitor that is the track, you pretty much have a cosntant voltage near or at the peak, not some in between that drops below the operating voltage of the decoder. The decoder can't 'see' the common - it's only connected to the rail terminals. Like in typical US house wiring, L1 to neutral is 120V, L2 to neutral is 120V, L1-L2 is 240V. The track outputs on a DCC booster are like L1 and L2, neutral is typically the case, or in the case of Digitrax, a specific terminally incorrectly labeled GND. You need this common linked between boosters to handle the current flow that occurs when a split pickup loco crossed the gaps between two booster districts.

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Modeling the Reading Railroad in the 1950's

 

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Posted by Overmod on Thursday, October 8, 2020 9:48 AM
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Posted by MisterBeasley on Thursday, October 8, 2020 10:59 AM

Some years ago, I built a signal bridge for two tracks, and wired the LEDs to indicate how the adjacent turnout was set, and also the next one down the line.  I have red, yellow and green LEDs in the circuit, and wired the bridge portion with magnet wire to reduce the wire size.

What I discovered was that the different colored LEDs had sufficiently different resistances, and some would not light when wired in parallel with others.  For this circuit, it wasn't too difficult to re-design the wiring for series rather than parallel, but if you're using different LEDs on this bus, it might be something to think about if you get weird behavior.

It takes an iron man to play with a toy iron horse. 

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Posted by gregc on Thursday, October 8, 2020 11:43 AM

rrinker
But as one rail goes to +7 relative to the common, the other rail goes to -7 relative to ground - so rail to rail is 14V, not 7.

that's not correct

if there's a 14V DC supply, the H-bridge is alternately connecting 14V and ground to each rail.  the rails are +/- 14V relative to each other.   

if you attached a scope across the rails, you'd see a squarish wave between -14 and +14V.   if the scope reference is the B rail, it shows 14V on the A rail while 0V is on the B rail and then -14V on the A rail when the h-bridge connects the A rail to 0V and the B rail to 14V.

if you looked at the rail voltage with a scope relative to common/ground of the DC supply, each rail would look like a sqarish wave between 0 and 14V.

 

rrinker
Like in typical US house wiring, L1 to neutral is 120V, L2 to neutral is 120V, L1-L2 is 240V. The track outputs on a DCC booster are like L1 and L2

a DCC booster is not like L1 and L2.

L1 and L2 are both 120V AC but 180 degrees out of phase. they share a common ground (literally ground, a water pipe in your house.

unlike DCC where one rail is 0V from the DC supply when the other is 14V, when L1 +171Vpk, L2 is -171 Vpk

greg - Philadelphia & Reading / Reading

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Posted by SpaceMouse on Thursday, October 8, 2020 6:11 PM

gregc
what are you planning to use these LEDs for in what looks like a sheet of balsa?

RR_Mel
I guess I missed the point.  The board in your picture has what appears to be 60 1KΩ resistors fed by a bus and 60 holes for LEDs.  A simple 12 volt DC power source should work fine once the LEDs are installed, the resistors look ready to go for 12 volts DC.

That's mounted just under the benchwork on the edge of the layout. Any time I install a structure, I can light it without crawling under the layout. 

Chip

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Posted by RR_Mel on Thursday, October 8, 2020 6:28 PM

SpaceMouse

 

That's mounted just under the benchwork on the edge of the layout. Any time I install a structure, I can light it without crawling under the layout. 

 

???????????  Explain please


Mel


 
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Posted by Mark R. on Thursday, October 8, 2020 6:41 PM

Looks self explanitory to me .... negative bus across the top and positive bus across the bottom, broken into individual taps each having their own resistor.

Mark.

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Posted by rrinker on Thursday, October 8, 2020 7:14 PM

 Rail to rail may measure double with a scope, but the AC volts is half value - decoders don't see 30V, neither does an LED connected across the rails. It sees 15V.

AC meter with no RMS capability will give a true value, since Vpeak and Vrms for a square wave are the same. Both my bench RMS meters measure 1V lower than the peak shown on my scope. (meters both show 13.86V AC, scope shows Vp 14.86

                                  --Randy

 


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Posted by gregc on Thursday, October 8, 2020 8:18 PM

what does the scope show peak-to-peak with a 14VAC waveform?

how do you generate that waveform from a DC supply?

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Posted by Overmod on Thursday, October 8, 2020 11:34 PM

gregc
what does the scope show peak-to-peak with a 14VAC waveform?

I see what he means.

Think of the DCC waveform during normal operation not as 'square-wave AC' but as two 14V modulated-DC pulse signals, one between +14V to common, and the other between -14V and common, but precisely in opposite modulation, so that when one is 'conducting' the other is off.  

If you put a 'scope across the internal power-supply output the range will of course show as 28V (in other words, between +14 and -14), and this would be the defined output voltage 'rail-to-rail' of the power supply itself (the term 'rail-to-rail' not having to do with physical model track rails, or with Rail A and Rail B in the defined output, but only to the power 'rails' in the supply) except that since there is never more than 14V in one direction or the other, even in case of a short, at any particular moment, you get the effect of an equal logic signal from either rail to zero (at the 14V potential) so there is no directional 'polarity' depending on which way the locomotive happens to be facing.

For some reason this reminded me of the graphs of sine and cosine in polar coordinates, which look like exactly the same waveform but are in fact never tangent!  

gregc
how do you generate that waveform from a DC supply?

Think of it as 'generating the waveform' with two DC supplies, by gating them on and off very quickly with the different on times corresponding to the binary states.  One supply is driven exactly the opposite of the other; think of it as the 'second' supply mirroring the modulation of the first's pulse as soon as the zero crossing is reached.  (That is the action that equalizes net charge due to the logic-pulse on times).

By putting the equivalent of a high-voltage diode on the decoder, no matter which way the locomotive is facing the decoder will only see DC from zero up to 14V with the logic pulsetrain imposed on it.  That in one direction the signal train leads or lags by the most recent logic pulsewidth makes no effective difference in perceptible timing.

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Posted by gregc on Friday, October 9, 2020 5:54 AM

Overmod
Think of the DCC waveform during normal operation not as 'square-wave AC' but as two 14V modulated-DC pulse signals, one between +14V to common, and the other between -14V and common, but precisely in opposite modulation, so that when one is 'conducting' the other is off.

that would be analagous to how 220 VAC is provided from separate 120 VAC line 180 degs out of phase.

that's not how DCC is generated

there's not necessarily a bi-polar power supply: +14V, -14V and 0V, with the 0V common connected to one rail and the other rail alternately connected to +14V and - 14V.    this could be done

more conventionally, a single 14V DC supply with two outputs, 14V and 0V, are alternately connected to opposite rails using an h-bridge.   replace motor terminal with rails

the result is the following DCC waveform with the reference on one rail, not the power supply 0V common, and the other rail alternately +/-14V.   ~14V AC, Vpk is 14V, Vpk-to-pk is 28V, 

 

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Posted by rrinker on Friday, October 9, 2020 7:50 AM

 I'm not sure of the internal topology of Digitrax boosters, as the schematics are proprietary, but they do generate a common the measures half of Vpeak (not Vp-p) between it and either rail output, despite using an H-bridge output driver like most all others.

 I notice there is no such common on the PowerCab - perhaps why to expand it, you don't add a booster, you replace the built in booster with another, which DOES have a case common to attach even more boosters.

 Actually, the mystery may be revealed by the Hans DeLoof DIY booster. 15VAC is supplied to power it, same as with most DCC boosters. V+ and V- rails to drive the H bridge are generated by a simple two diodes (and some filter caps) so that gives you DC at +15 or so and -15 or so relative to the other AC input terminal. I've not taken apart my DB150 - others have said the ealier Digitrax boosters (that could run on AC or DC input) have a full wave rectifier right behind the input terminals - maybe to drive the processor, but probably not to generate the power rails for the H bridge. 

                                           --Randy


Modeling the Reading Railroad in the 1950's

 

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Posted by gregc on Friday, October 9, 2020 8:37 AM

i think you're right.   thanks

because of the higher power (10A) output of a booster, it would be simpler to use an AC power supply and two discrete high power mosfets to alternately connect +/- V to one rail rather than use an h-bridge.   (I don't know of any single chip 10A h-bridges).   the power supply 0V common is connected to the other rail and is the common connected to all boosters

two half-wave rectifiers produce +/- V.   the DCC signal from the command station can be applied to one mosfet and its inverse to the other.

the complication is short circuit protection that requires a processor which may also disable the outputs if there is no DCC signal

greg - Philadelphia & Reading / Reading

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Posted by rrinker on Friday, October 9, 2020 11:11 AM

 The schematic for the DeLoof design is in his PDF instructions:

http://users.telenet.be/deloof/booster/LocoBoosterEN.pdf

It's a low power design, 3A max. The third schematic is the simplest one - no Loconet, just a basic booster similar in function to ones like the Tam Valley units. It does indeed use a micro to cut power on shorts or on loss of input signal.

 It uses an LM675T power amp as the driver - somewhere I have a couple of these, as I think I was going to build this or a similar one that used the same amp as the driver.

                                 --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

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