That's exactly what I do Randy,
even though my layout is wired for twin cab control (Old DC system) I only ever connect one controller to the layout at a time. There can never be DC and DCC connected to the layout at the same time. As I only use a PowerCab which has an upper limit of 2 amps for DCC, my existing wiring works fine.
The diodes are only connected to power dead ends such that DC engines can be backed out, they stop automatically on their way in due to the diode blocking the current flow once the engine has all power pickup wheels inside the dioded end section. This prevents DC engines from going off the end of the track (and crashing to the floor in some cases).
Guess I'll just have to put better stops (Piles of dirt etc), at the dangerous dead ends.
Alan Jones in Sunny Queensland (Oz)
That is 'bad' That's why this is not a good method - either the whole layout runs DC, or the whole layout runs DCC, it's the only safe method to use both on the same layout. At some point, someone will not be paying attention and cross a block boundary between DC and DCC and cause problems.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
IansaWhat happens when loco bridges the DCC & DC current " NOTHING" no damge will be done to decoder.
I confess that I don't understand any of what's being discussed here. But I'm willing to accept that there will be no damage to the decoder if the loco bridges the gap. But what effect, if any, will there be to all the DCC components connected to the track on the DCC side of the gap when they get backfed DC?
AS the op has DC and DCC the simple answer is fit a DPDT switch to each isolated section so each section can be powered by DC or DCC.
Turn DC op off in all decoders in CV 29 and when op DCC switch DPDT switch to DC power.
When the decoder see's DC power the loco will stop. To power loco by DCC again just switch to DCC power and off you go.
What happens when loco bridges the DCC & DC current " NOTHING" no damge will be done to decoder.
I once had an automatic signalling system using a variation of this method.
Loco/train would automatically stop at red signal when isolated section, actvated by train in block in front, was changed from DCC to DC.
It was done with block detection and relays but that is another story.
BTW, get rid of the diodes if using this method.
Cheers
Ian
CSX Robert betamax If you look at the diagram and remove the 4 diode block, leaving the single one, the waveform will still look the same, except one will not have a 2.8V drop, while the other will still have a 0.7V drop in amplitude. No it won't. With the assymtrical DCC module you have two paths for the two phases to take. One phase will be blocked by the four diodes and pass through the one diode, creating the 0.7 volt drop.The other phase will be blocked by the one diode and pass throught the four diodes, creating the 2.8 volt drop. Without the second path, the phase that is blocked by the one diode wlil not have any way of passing through.
betamax If you look at the diagram and remove the 4 diode block, leaving the single one, the waveform will still look the same, except one will not have a 2.8V drop, while the other will still have a 0.7V drop in amplitude.
No it won't. With the assymtrical DCC module you have two paths for the two phases to take. One phase will be blocked by the four diodes and pass through the one diode, creating the 0.7 volt drop.The other phase will be blocked by the one diode and pass throught the four diodes, creating the 2.8 volt drop. Without the second path, the phase that is blocked by the one diode wlil not have any way of passing through.
You are correct, I made a mistake. I started thinking of current flows from HV measurements I've been doing lately and accidently applied that to diodes. Diodes conduct the current, if the diode is reverse biased the current can't flow. I was thinking of capacitors with respect to current flows to measure high voltages (by which I mean 100,000V or more).
Since it is exacting work, sometimes you forget you are not doing that when looking at something else.
rrinkerOne input line to the processor connects before the bridge
That makes much more sense. It's what I would have guessed, but guessing gets me in trouble.
"A man's got to know his limitations" Dirty Harry
I have the right to remain silent. By posting here I have given up that right and accept that anything I say can and will be used as evidence to critique me.
rrinker One input line to the processor connects before the bridge, with a few other discrete components to limit the apmplitude to within the tolerances for the processor input line. Simply not shown on that schematic since the main purpose of that one is to show you how to connect a keep alive that powers the entire decoder and not just the sound part. --Randy
One input line to the processor connects before the bridge, with a few other discrete components to limit the apmplitude to within the tolerances for the processor input line. Simply not shown on that schematic since the main purpose of that one is to show you how to connect a keep alive that powers the entire decoder and not just the sound part.
Look below.
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
One thing I still don't understand from this decoder schematic is how the processor picks the digital signals after the the rectifier. It would no longer see the polarity reversal that represents the "data".
Well, there you go - it was actually tried and the result was as expected - the speed dropped by about half. Because the decoder wa only getting half the DCC waveform. And also witnessed proof that DCC is AC - half the time the voltage flows in one direction, and half the time it flows the other way.
Where's the disconnect on diode functionality here? A diode only passes current in one direction, less the loss in the junction, which for a silicon diode is .7V. It blocks all current flow in the opposite direction, until you exceed the peak inverse voltage and fry the whole thing.
If you take that Asymmetrical DCC illustration - one diode faces one way, four diodes face the other. So on one half of the waveform, there will be a one diode (.7V) drop in the amplitude. In the other direction, there will be 4 x .7 = 2.8V drop in amplitude. Result is an asymmetrical signal (relative to the common point). One side will be .7V less than the peak, the other side will be 2.8V less than the peak.
Remove those four diodes, and in one direction you will get a .7V drop, in the other you will get 100% loss of voltage as the diode blocks it. Your waveform will now show half of the signal, going to .7V less than the peak, and the other half will no longer exist.
Or even look at the decoder circuit (less the keep alive) - there's a bridge rectifier to get DC for the rest of the circuit. If DCC wasn;t alternating, the bridge rectifier wouldn't be needed. If it only went from ground to plus 30V you wouldn't need a bridge rectfier to get DC out of it, it would already BE DC, just pulsed DC, and all you'd need would be to filter it to get smooth DC. But it's not, it's true AC, and you need a rectifier to get DC out of it.
edit: ok, there are FOUR diodes in the example circuit, I can count...
gregc carl425 gregc isn't it possible that half the waveform is sufficient to provide enough power for the locomotive to continue at the last speed it was commanded. I guess it would be possible. What scares me though is that the decoder would see something like old school pulse-power. Who knows what that will do? Decoders are too expensive and too annoying to replace to risk killing one. here's at least one decoder design that has a capacitor to filter the supply to the decoder circuit. As I said, only half the DCC signal would still provide enough power to for the circuit to work fine with the decoder set to a slow speed.
carl425 gregc isn't it possible that half the waveform is sufficient to provide enough power for the locomotive to continue at the last speed it was commanded. I guess it would be possible. What scares me though is that the decoder would see something like old school pulse-power. Who knows what that will do? Decoders are too expensive and too annoying to replace to risk killing one.
gregc isn't it possible that half the waveform is sufficient to provide enough power for the locomotive to continue at the last speed it was commanded.
I guess it would be possible. What scares me though is that the decoder would see something like old school pulse-power. Who knows what that will do? Decoders are too expensive and too annoying to replace to risk killing one.
here's at least one decoder design that has a capacitor to filter the supply to the decoder circuit. As I said, only half the DCC signal would still provide enough power to for the circuit to work fine with the decoder set to a slow speed.
Sorry to drag this one back up, but in that diagram, the capacitor / diode / resistor at the bottom left is not used as a filter, that is a home-brewed stay-alive circuit. Note that it is separate from the outline of the decoder itself.
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
betamaxIf you look at the diagram and remove the 4 diode block, leaving the single one, the waveform will still look the same, except one will not have a 2.8V drop, while the other will still have a 0.7V drop in amplitude.
What you have with a single diode is a half wave rectifier, and that will clip off one phase of the signal. Google some images of a half-wave rectifier waveform, like this one:
It's the same concept.
CSX Robert betamax ...When current flows through the four diodes from the track and then returns via the grounded phase, the signal is clipped by 2.8V. When the phase flips, the current only flows through one diode, reducing its amplitude by 0.7V... That is exactly how the Asymetrical DCC module works, but the question was what will ONE diode, by itself, do. It will drop the signal 0.7 volts in one direction but, there is no alternate path for the current to take when the polarity is reversed so it will completely block the signal in the other direction.
betamax ...When current flows through the four diodes from the track and then returns via the grounded phase, the signal is clipped by 2.8V. When the phase flips, the current only flows through one diode, reducing its amplitude by 0.7V...
That is exactly how the Asymetrical DCC module works, but the question was what will ONE diode, by itself, do. It will drop the signal 0.7 volts in one direction but, there is no alternate path for the current to take when the polarity is reversed so it will completely block the signal in the other direction.
If you look at the diagram and remove the 4 diode block, leaving the single one, the waveform will still look the same, except one will not have a 2.8V drop, while the other will still have a 0.7V drop in amplitude.
Hi all,
I tried a variety of decoders. The digitrax one I have did not respond to the direction change, the engine kept going slowly in the same direction, it would NOT stop or change direction. I have two non-sound Atlas diesels, one responded, the other didn't. My sound equipped Atlas loco responded and the sound continued to play. I didn't leave it in the diode section for very long so I don't know if the sound would play for very long or just a short time. All my Bachmann diesels responded.
So the diodes will definately be shorted out for DCC operation.
cheers and thanks everyone for your ideas and suggestions, I would not have tried it deliberately without your feedback.
mfm37 If it were me, I would just run an engine over the diode section and see what happens. It is not going to fry the decoder. Martin Myers
If it were me, I would just run an engine over the diode section and see what happens. It is not going to fry the decoder.
Martin Myers
Martin, I tried that with a DCC Bachmann 44 Ton switcher with original decoder (I didn't mind if I had to replace it), and it just slowed down as if running on half voltage. I am using an NCE Powercab. I was able to change direction and the engine responded and reversed out of the diode section. I'll try a couple more non-sound decoders and see what they do.
Oh, and Randy, I am old that's why I use very old tricks ;).
cheers
betamax carl425 betamax No, it will not eliminate the negative signal, as there isn't one. It will just insert a 700mV drop when the rail is held to ground and the other rail's signal passes through it on its way back to the booster. Sorry dude, but you seem to be misunderstanding either the basic function of a diode or the nature of the DCC signal. The diode will only pass current in one direction. That is it's purpose in life. The voltage drop is a side effect that they are exploiting to create the asymetry. The DCC signal goes from +16v to -16v at intervals that vary to indicate either a binary 0 or 1. If you put one diode in the path, depending on orientation, it takes out either the + or - phase of the signal. Look at the picture in the wiki you referenced. They have 4 diodes in one direction and one in the opposite direction. This creates a 4X drop in one phase and a 1X drop in the other. What the OP has is ONE diode at the end of a DC powered spur that is used to stop the loco as it heads in, but when you reverse the polarity you can go out. This is a VERY OLD trick that has been used for years. The reason it works is that the diode only passes current in one direction. Also from the wiki you referenced: DCC is a bipolar signal. One rail (A) is held to ground, the other (B) is connected to the V+ source, then this flips through the switching circuit in the booster, where A is now positive and B is held to ground. Viewed through a scope you will see what looks like ~30V p-p signal. When current flows through the four diodes from the track and then returns via the grounded phase, the signal is clipped by 2.8V. When the phase flips, the current only flows through one diode, reducing its amplitude by 0.7V. Since there is no common point, there is no way to create a voltage below the common point.
carl425 betamax No, it will not eliminate the negative signal, as there isn't one. It will just insert a 700mV drop when the rail is held to ground and the other rail's signal passes through it on its way back to the booster. Sorry dude, but you seem to be misunderstanding either the basic function of a diode or the nature of the DCC signal. The diode will only pass current in one direction. That is it's purpose in life. The voltage drop is a side effect that they are exploiting to create the asymetry. The DCC signal goes from +16v to -16v at intervals that vary to indicate either a binary 0 or 1. If you put one diode in the path, depending on orientation, it takes out either the + or - phase of the signal. Look at the picture in the wiki you referenced. They have 4 diodes in one direction and one in the opposite direction. This creates a 4X drop in one phase and a 1X drop in the other. What the OP has is ONE diode at the end of a DC powered spur that is used to stop the loco as it heads in, but when you reverse the polarity you can go out. This is a VERY OLD trick that has been used for years. The reason it works is that the diode only passes current in one direction. Also from the wiki you referenced:
betamax No, it will not eliminate the negative signal, as there isn't one. It will just insert a 700mV drop when the rail is held to ground and the other rail's signal passes through it on its way back to the booster.
Sorry dude, but you seem to be misunderstanding either the basic function of a diode or the nature of the DCC signal.
The diode will only pass current in one direction. That is it's purpose in life. The voltage drop is a side effect that they are exploiting to create the asymetry.
The DCC signal goes from +16v to -16v at intervals that vary to indicate either a binary 0 or 1. If you put one diode in the path, depending on orientation, it takes out either the + or - phase of the signal.
Look at the picture in the wiki you referenced. They have 4 diodes in one direction and one in the opposite direction. This creates a 4X drop in one phase and a 1X drop in the other.
What the OP has is ONE diode at the end of a DC powered spur that is used to stop the loco as it heads in, but when you reverse the polarity you can go out. This is a VERY OLD trick that has been used for years. The reason it works is that the diode only passes current in one direction.
Also from the wiki you referenced:
DCC is a bipolar signal. One rail (A) is held to ground, the other (B) is connected to the V+ source, then this flips through the switching circuit in the booster, where A is now positive and B is held to ground. Viewed through a scope you will see what looks like ~30V p-p signal.
When current flows through the four diodes from the track and then returns via the grounded phase, the signal is clipped by 2.8V. When the phase flips, the current only flows through one diode, reducing its amplitude by 0.7V.
Since there is no common point, there is no way to create a voltage below the common point.
That is very definitely NOT how Digitrax DCC boosters work. They swing plus and minus around a common. It's a true bipolar output. One of their suggested ways to measure the tack voltage is, instead of going from rail A ro rail B, measure Rail A to 'ground' (actually common) and Rail B to 'ground' and add them. Also, they should be the same DC value unless address 00 is in use and at a non-zero speed.
Some systems have a true bipolar driver, others do as you say, run 0-30V and arbitrarily set 15V as the middle to give a 15V swing in each direction.
betamax...When current flows through the four diodes from the track and then returns via the grounded phase, the signal is clipped by 2.8V. When the phase flips, the current only flows through one diode, reducing its amplitude by 0.7V...
betamaxNo, it will not eliminate the negative signal, as there isn't one. It will just insert a 700mV drop when the rail is held to ground and the other rail's signal passes through it on its way back to the booster.
I kept one of my DC Cabs, so I could run DC locomotives when the mood struck. This was almost 20 years ago and I find I only use the DC cab now to run my turntable swing; or, check the operation of a new locomotive previous to installing a DCC Decoder. I think you will find yourself in much the same situation: DCC will become your primary method of operating and DC only for checking out new un-DCC locos, if you even buy them that way, anymore.
So, I would remove your diodes as you likely will not need them.
NP 2626 "Northern Pacific, really terrific"
Northern Pacific Railway Historical Association: http://www.nprha.org/
carl425 betamax Actually, diodes are used to create Asymmetric DCC, which some decoders will react to. Most other decoders (without the feature) will just ignore it. One diode won't have much effect. All it does is reduce the signal amplitude by 2.8V or 0.7V depending on polarity. See https://www.dccwiki.com/Asymmetric_DCC if you look at the circuit you'll see that 4 diodes in one direction are in parallel with 1 diode in the other direction. This still allows both phases to pass, but reduces the amplitude differently on each phase. This is what creates the asymmetry. A single diode as described by the OP will eliminate one phase all together.
betamax Actually, diodes are used to create Asymmetric DCC, which some decoders will react to. Most other decoders (without the feature) will just ignore it. One diode won't have much effect. All it does is reduce the signal amplitude by 2.8V or 0.7V depending on polarity. See https://www.dccwiki.com/Asymmetric_DCC
Actually, diodes are used to create Asymmetric DCC, which some decoders will react to. Most other decoders (without the feature) will just ignore it. One diode won't have much effect.
All it does is reduce the signal amplitude by 2.8V or 0.7V depending on polarity.
See https://www.dccwiki.com/Asymmetric_DCC
if you look at the circuit you'll see that 4 diodes in one direction are in parallel with 1 diode in the other direction. This still allows both phases to pass, but reduces the amplitude differently on each phase. This is what creates the asymmetry.
A single diode as described by the OP will eliminate one phase all together.
No, it will not eliminate the negative signal, as there isn't one. It will just insert a 700mV drop when the rail is held to ground and the other rail's signal passes through it on its way back to the booster.
What would happen would probably depend on which direction the loco is facing relative to the diode. First of all, all decoders mhave some amount of capacitance on the power input, probably enough to keep the loco powered. The decoder has to monitor one rail for the DCC signal. With the blocking diode in the circuit, the decoder will be able to 'see' the DCC signal on one rail, but not the other, so if it is checking the rail that it can see the signal on it will probably keep going but if checking the other rail then it will stop.
Well, it's not going to fry the decoder. Worst case would be it simply doesn't move. Empirical testing may be in order.
gregchere's at least one decoder design that has a capacitor to filter the supply to the decoder circuit. As I said, only half the DCC signal would still provide enough power to for the circuit to work fine with the decoder set to a slow speed.
You sound convinced. Give it a try and let us know what happens.
greg - Philadelphia & Reading / Reading