Hi all,
I originally wired my small layout for DC two cab operation and have now hooked up an NCE Powercab as one cab (Not using the other cab) and it works very well with no modifications to the layout. On some of my dead end sidings I originally put an isolated section at the end with a diode wired so a loco would stop when it went into the section, but could be reversed out.
My question is, will this diode section damage DCC decoders or the motors they are controlling if I run an engine into the dioded section of track? I know DCC puts an alternating current to the track and I don't expect the diodes to have the same effect on a DCC loco as a DC loco, but I wonder what effect they do have.
I'm sure I will accidently run a DCC engine into one of these diode sections one day and was thinking I should short out the diodes, but I ocassionally revert to running the whole layout on just DC and the diode sections work very well then.
TIA
Alan
Alan Jones in Sunny Queensland (Oz)
A single diode will lop off half of the DCC waveform - the track beyond the diode will only see the upper half or the lower half, depending on the orientation of the diode. If things run, thay will be at half speed. Some more advanced decoders can be programmed to perform various actions when they see this half waveform, like automatically come to a stop (following the momentum setting, not just instantly halting). This is Asymmetrical DCC.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
I would think that when a locos wheel bridges the Diode gap the Loco(s) will go, all the way in until the last wheel passes the gap and then not stop. Since Diodes don't block AC, wouldn't the loco move out when the locos direction is reversed? Leds light no matter how they are connected to DCC.
rrinkerThis is Asymmetrical DCC.
Nah, this is Half DCC.
In asymmetrical DCC, there are still positive and negative sides of the waveform, they are just not equal in amplitude.
AlantrainsMy question is, will this diode section damage DCC decoders or the motors they are controlling if I run an engine into the dioded section of track?
I wouldn't try it with my decoders. I don't think they would work anyway. Follow your first instinct and short them out. Get some jumpers with alligator clips so you can take them off when you're in a DC mood.
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.
the dcc waveform provides two things: power and communication.
while the diodes are more likely to disrupt communication, 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.
greg - Philadelphia & Reading / Reading
Thanks everyone,
I think I'll take Carl's advice and short out the diodes.
gregcisn'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.
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.
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.
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
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.
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.
Well, it's not going to fry the decoder. Worst case would be it simply doesn't move. Empirical testing may be in order.
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.
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.
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.
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/
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.
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:
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.
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.
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 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.
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.
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
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
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
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.
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...
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.
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.
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.
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.
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.
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
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...
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".
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.
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
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.