dtabor Based on my last question, some mentioned terminal strips for wiring a DCC system. Most of what Ive read/watched shows a continuous buss line and the feeders attached every 3-6' along that, and the best practice is to keep the feeders as short as possible for less power loss. Wondering what the advantage of a terminal strip is and how that works with feeders. It seems if I have a terminal strip in a certain place, that the feeders from that strip to the track would get long. Do you use a small terminal block at each feeder point? Seems that would be a ton of terminal blocks if you had a long main line. There are some youtube videos showing terminal blocks/strips but I cant find one that shows the whole process of wiring the actual feeder points.
Based on my last question, some mentioned terminal strips for wiring a DCC system. Most of what Ive read/watched shows a continuous buss line and the feeders attached every 3-6' along that, and the best practice is to keep the feeders as short as possible for less power loss. Wondering what the advantage of a terminal strip is and how that works with feeders. It seems if I have a terminal strip in a certain place, that the feeders from that strip to the track would get long. Do you use a small terminal block at each feeder point? Seems that would be a ton of terminal blocks if you had a long main line. There are some youtube videos showing terminal blocks/strips but I cant find one that shows the whole process of wiring the actual feeder points.
I have a similar question regarding wiring for DCC using terminal strips. I was informed in another model railroad forum that I could easily connect all my feeder wires (for the main oval, turntable service tracks, turntable bridge track, and DCC controller) to a terminal strip without any worries, though it was also recommended that I use an IDC (Insulation Displacement Connector) and bus wires. The type of track I'll be using is sectional, with predetermined lengths of track molded to a plastic roadbed, so I'm wondering if I would need to add a bus wire/additional feeders or if the terminal strip would work just as well.
Run a main bus harness, tap it with t-taps, and run tap to a terminal strip. From terminal strip run feeders (if you need to)Advantage: Makes maintenance and modifications a lot easier.
Don - Specializing in layout DC->DCC conversions
Modeling C&O transition era and steel industries There's Nothing Like Big Steam!
dtaborThank you sir, this makes total sense. Thank you!
Mike
dtabor Wondering what the advantage of a terminal strip is and how that works with feeders.
I'm not sure this ever really got answered in the mess that your thread unfortunately turned into. Ideally, you have the shortest bus runs possible with the shortest feeders possible (within reason). Hard to do placing your DCC system at one "end" of the layout. If I recall, your layout has figure 8 shaped benchwork, yes? To place your DCC booster at a location and run bus wire around the room and across the center of the "8" would likely mean a fairly long bus. Not necessarily unworkable, but could be better. Place your booster along one side of the "8", lets say the left side, right at the center. Connect your booster to a terminal strip with all the screws on one side connected as has been shown in the thread. Then run a bus from the terminal strip along the top loop of the "8". Run another bus along the bottom loop of the 8, also connecting it at the terminal strip. Finally, run a third bus across the center of the "8". Makes for an easy way to connect up a wiring plan like this that minimizes wire run. Obviously you drop your feeders from each section down to the corresponding bus. While you could simply solder all the busses together and eliminate the terminal strips, should a problem arise, and you've electrically isolated the various sections of your layout fed by each respective bus, it's a simple matter of unscrewing and disconnecting a single post on the terminal strip to isolate a part of the layout for troubleshooting purposes.
Thank you sir, this makes total sense. Thank you!
SeeYou190He is another new participant here that had his thread SPIKED! and reduced to silliness and garbage.
Really? Is this neccessary? (No). Yes, there were some misinformed comments, that were corrected, followed by the usual personal attacks (even though I saw none of the "I'm smarter than you" attitude of the past) but we've moved on from that.
This was the OP's question:
dtabor Wondering what the advantage of a terminal strip is and how that works with feeders. It seems if I have a terminal strip in a certain place, that the feeders from that strip to the track would get long. Do you use a small terminal block at each feeder point? Seems that would be a ton of terminal blocks if you had a long main line.
He is another new participant here that had his thread SPIKED! and reduced to silliness and garbage.
This does not surprise me at all.
-Kevin
Living the dream.
LastspikemikeSo possibly damaging the booster. The train won't be affected.
CSX RobertThose deocders are not expecting to have a voltage difference between the same side front and rear pickups and can be damaged if they do.
more concisely ...
any loco/car with multiple pickups can "bridge" a gap between boosters when wheels are on opposite sides of a gap and the electrical path joining those pickups is susceptible to damage due to the current that flows during the time the gap is bridged. however, it is still unclear how "excessive" the "bridged" current is
and some decoder circuit boards may include the path between pickups. that current does not flow thru the decoder electronic components (e.g. bridge rectifier).
LastspikemikeIt occurs to me that the voltage differential is only between two rails with the same polarity (phase) so that can't affect the decoder adversely.
no. its the difference in voltage on the same rail across the gaps between boosters
greg - Philadelphia & Reading / Reading
gregchas nothing to do with the decoder. the decoder sees the resulting track voltage the question is how much excessive current is flowing between the boosters when there's a difference in their output voltages and the gap between them is bridged. (i've posted measurements indicating there is little excessive current)
The decoder can be damaged if it has separate same side front and rear pickups. The loco or car can be damaged if it's current collecting scheme can't handle the current flow.
Lastspikemike CSX Robert Lastspikemike I don't see how a 1v differential could make a difference. As the resistance approaches zero, the maximum current approaches infinity. Theoretically, any voltage difference coupled with a low enough resistance can produce the maximum current that can be supplied by the booster. A 1 volt difference through a .2Ω resistance will produce 5 amps (1/.2=5). Sure. I'm having trouble envisioning a decoder that can handle 16v no problem bring affected by an additional volt or even two. Mr Puckett at least implies that the decoder can be damaged by the additional current. Not seeing how the decoder resistance path changes as a result of this issue. I had thought decoders were designed for 20 plus volts and some, such as ESU, will include over voltage protection also.
CSX Robert Lastspikemike I don't see how a 1v differential could make a difference. As the resistance approaches zero, the maximum current approaches infinity. Theoretically, any voltage difference coupled with a low enough resistance can produce the maximum current that can be supplied by the booster. A 1 volt difference through a .2Ω resistance will produce 5 amps (1/.2=5).
Lastspikemike I don't see how a 1v differential could make a difference.
As the resistance approaches zero, the maximum current approaches infinity. Theoretically, any voltage difference coupled with a low enough resistance can produce the maximum current that can be supplied by the booster. A 1 volt difference through a .2Ω resistance will produce 5 amps (1/.2=5).
Sure. I'm having trouble envisioning a decoder that can handle 16v no problem bring affected by an additional volt or even two. Mr Puckett at least implies that the decoder can be damaged by the additional current. Not seeing how the decoder resistance path changes as a result of this issue. I had thought decoders were designed for 20 plus volts and some, such as ESU, will include over voltage protection also.
It's not an additional volt or two, you're not going from a decoder being powered by 16v to one being powered by 17 or 18v. The voltage difference is occuring across a circuit that is not supposed to have a voltage difference difference. The 16 volts is from left side to right side rails. The difference between boosters is from front truck to back truck, on the same side.
I'm not as familiar with HO-scale decoders, but most N-scale decoders actually won't ever see that voltage difference, because most have only one left rail and one right rail pick up, or if they have more than one they are going to the frame which has so little resistance the decoder won't see any significant voltage difference. In these cases you can still have damage to the loco somewhere such as burned wires or melted plastic from too much heat. Some decoders do, however, have four separate inputs, front left, front right, rear left and rear right. Those deocders are not expecting to have a voltage difference between the same side front and rear pickups and can be damaged if they do.
Lastspikemike I'm having trouble envisioning a decoder that can handle 16v no problem bring affected by an additional volt or even two. Mr Puckett at least implies that the decoder can be damaged by the additional current. Not seeing how the decoder resistance path changes as a result of this issue. I had thought decoders were designed for 20 plus volts and some, such as ESU, will include over voltage protection also.
I'm having trouble envisioning a decoder that can handle 16v no problem bring affected by an additional volt or even two. Mr Puckett at least implies that the decoder can be damaged by the additional current. Not seeing how the decoder resistance path changes as a result of this issue. I had thought decoders were designed for 20 plus volts and some, such as ESU, will include over voltage protection also.
Alton Junction
LastspikemikeMr Puckett at least implies that the decoder can be damaged by the additional current.
has nothing to do with the decoder. the decoder sees the resulting track voltage
the question is how much excessive current is flowing between the boosters when there's a difference in their output voltages and the gap between them is bridged. (i've posted measurements indicating there is little excessive current)
gregcPuckett said the one booster tries to raise the voltage of the other booster to it's level, drawing an extra amount (?) of current from the higher voltage booster. this makes sense from a linear circuits perspective and Kirchhoff's law. current would flow from the higher voltage source into the lower voltage source. but boosters aren't linear circuits, they have regulators and other non-linear devices that act like circuit breakers.
this makes sense from a linear circuits perspective and Kirchhoff's law. current would flow from the higher voltage source into the lower voltage source. but boosters aren't linear circuits, they have regulators and other non-linear devices that act like circuit breakers.
i measured the current thru a small resistor (~0.8 Ohm) between a 3V battery and 4V regulated output and between a 5V (7805) regulator and 10V regulated output
my Fluke measures ~132 ma thru the resistor between the battery which measured 3.3V at that time and the regulated output that measured 3.9V. according to Ohms law, the current should be 750 ma, which suggests additional resistance in the path, possibly the shunt in the meter.
i measured 33 ma thru the 0.8 Ohm resistor between the 5V regulator and 10V output, which per Ohms law should be 4.2 A. The current should be significantly higher with a 5V vs 1V across the resistor. but as i stated, the 5V regulator is not a linear circuit and won't behave like a simple battery, the transistors in the regulator block current flowing into the regulator.
the experiment demonstrates that Ohm's and Kitchoff's Laws are questionalble when non-linear electronics are involved.
it's not that simple
LastspikemikeI don't see how a 1v differential could make a difference.
CSX RobertI was trying to keep the example simple
DigitalGriffinI kept it simple
Everything should be made as simple as possible, but not simpler. -- Einstein
i am probably over simplifying as well.
what is the voltage difference when the gap is bridged?
LastspikemikeMr Puckett would have been better advised to lay a coin across the A rail insulated joint to short it out directly. How much current could the mismatched outputs push through the coin?
yes, some real data and understanding would be helpful
if it is truly a 1V difference, the current thru the bridge can be excessive, 2A = 1V / 0.5 Ohm, as mentioned, while much less (~0.5A) is drawn by the loco?
gregc DigitalGriffin As a general rule of thumb I try to keep the difference < 0.7V and 3 amps. That's 2.1 Watts of power, but enough most diodes and circuit mosfets/transistors can handle it. why should there be any excessively high current flow across the "bridged" gap since the booster for that district is already supplying sufficient current? what diodes or transistors are passing any more current than the booster is already limited to? DigitalGriffin Interestingly enough, Larry posted a video about this topic this very morning. Puckett said the one booster tries to raise the voltage of the other booster to it's level, drawing an extra amount (?) of current from the higher voltage booster. this makes sense from a linear circuits perspective and Kirchhoff's law. current would flow from the higher voltage source into the lower voltage source. but boosters aren't linear circuits, they have regulators and other non-linear devices that act like circuit breakers. and presumably there is a higher voltage on the unregulated side, the input side of the booster allowing the output voltage to be adjusted. if a slightly higher voltage (< 1V) is present on the regulated output, i don't see how it can flow into the higher votlage unregulated side of the booster. in other words the booster simply stops passing any current even a simple common emmitter BJT regulator has the emmitter terminal connected to the output which will look like a reverse biased diode if the voltage at the emmitter is > the base voltage of the transistor. even if you do adjust the booster voltages to be the same, one will inevitably be lower when current is being drawn, especially at the booster boundary with presumably the maximum length of wire to the booster. of course you should try to balance them, but boosters need to handle this situation (it would have be intesting to demonstate the currents from each booster when the gap is bridge and the voltages are not the same) 1835
DigitalGriffin As a general rule of thumb I try to keep the difference < 0.7V and 3 amps. That's 2.1 Watts of power, but enough most diodes and circuit mosfets/transistors can handle it.
why should there be any excessively high current flow across the "bridged" gap since the booster for that district is already supplying sufficient current?
what diodes or transistors are passing any more current than the booster is already limited to?
DigitalGriffin Interestingly enough, Larry posted a video about this topic this very morning.
Puckett said the one booster tries to raise the voltage of the other booster to it's level, drawing an extra amount (?) of current from the higher voltage booster.
and presumably there is a higher voltage on the unregulated side, the input side of the booster allowing the output voltage to be adjusted. if a slightly higher voltage (< 1V) is present on the regulated output, i don't see how it can flow into the higher votlage unregulated side of the booster. in other words the booster simply stops passing any current
even a simple common emmitter BJT regulator has the emmitter terminal connected to the output which will look like a reverse biased diode if the voltage at the emmitter is > the base voltage of the transistor.
even if you do adjust the booster voltages to be the same, one will inevitably be lower when current is being drawn, especially at the booster boundary with presumably the maximum length of wire to the booster. of course you should try to balance them, but boosters need to handle this situation
(it would have be intesting to demonstate the currents from each booster when the gap is bridge and the voltages are not the same)
1835
At the risk of sounding like I'm making excuses, I kept it simple for those who don't know how the internals of Bipolar junction transistors work.
And yes I'm assuming a lot of things like near zero resistance for 5 amps @ 1V. But it is possible with certain pickups.
The point being, the closer you keep it to 0V difference the better. Adding good feeders to the end of each block is the best way to help settle this.
So in short, I'm trying to keep it simple. But you do raise valid points.
gregc CSX Robert Of course, if that resistance drops to 1Ω, the current max jumps to 1 amp, and with a 1.5 volt difference and 1Ω you get 1.5 amps. does this account for all the other resistance between the 2 boosters: the resistance of the wiring as well as the effective impedance within each booster?
CSX Robert Of course, if that resistance drops to 1Ω, the current max jumps to 1 amp, and with a 1.5 volt difference and 1Ω you get 1.5 amps.
does this account for all the other resistance between the 2 boosters: the resistance of the wiring as well as the effective impedance within each booster?
No it does not. I was trying to keep the example simple while still showing that the current passing through the engine or train car is going to be dependent on the voltage difference and resistance of the circuit. Any additional rsistance in the circuit is, naturally, going to reduice the current draw even more.
CSX RobertOf course, if that resistance drops to 1Ω, the current max jumps to 1 amp, and with a 1.5 volt difference and 1Ω you get 1.5 amps.
BATMANNow if the know it all would come forth with some numbers.
I hope you are not holding your breath.
Maybe we will get pictures to go with those numbers.
Lastspikemike I'm not following Mr Puckett's claim that a 1v differential between boosters at the gap can push 5 amps through the locomotive or lighted car. The current is surely limited by the voltage.
I'm not following Mr Puckett's claim that a 1v differential between boosters at the gap can push 5 amps through the locomotive or lighted car. The current is surely limited by the voltage.
Well, he didn't explain it well. The boosters aren't going to "push" 5 amps through the locomotive or car, the current is still going to be limited to what the circuit draws. As an example, I measured the resistance between the front truck and rear truck of one of my locomotives at 1.5Ω. With a 1 volt difference that would draw 1/1.5 = 0.667 amps, the booster would not "push" more than that through it. Of course, if that resistance drops to 1Ω, the current max jumps to 1 amp, and with a 1.5 volt difference and 1Ω you get 1.5 amps.
DigitalGriffinAs a general rule of thumb I try to keep the difference < 0.7V and 3 amps. That's 2.1 Watts of power, but enough most diodes and circuit mosfets/transistors can handle it.
DigitalGriffinInterestingly enough, Larry posted a video about this topic this very morning.
If the end of one block is 14V and the end of the other block is 13V, that's a 1V difference. Electricity will always go from the high voltage source to the low one. If you loco or passenger car bridges the gaps with one set of trucks on one booster, and another set of trucks on another, that could be asking for trouble. (Assuming pickups on both sides like walthers passenger cars, or Athearn trucks)Now as a general rule of thumb 3Amps is what you should limit any given block to. But many boosters go from 5 to even 10 amps.
10A * 1V = 10 Watts. Trust me that's enough to melt a lot of thin circuits/plastic.
As a general rule of thumb I try to keep the difference < 0.7V and 3 amps. That's 2.1 Watts of power, but enough most diodes and circuit mosfets/transistors can handle it.
richhotrain BATMAN richhotrain I took it as a satirical question because it was asked in the midst of all that nonsense about not needing a bus but only using a single pair of feeders from the command station to power an entire DCC layout.
BATMAN richhotrain I took it as a satirical question because it was asked in the midst of all that nonsense about not needing a bus but only using a single pair of feeders from the command station to power an entire DCC layout.
richhotrain I took it as a satirical question because it was asked in the midst of all that nonsense about not needing a bus but only using a single pair of feeders from the command station to power an entire DCC layout.
A know it all, will be a know it all until you ask him for the hard facts or numbers. I asked the question not knowing myself and that inspired those with the knowledge to come forward. Now if the know it all would come forth with some numbers we could have a really good rumble conversation.
Brent
"All of the world's problems are the result of the difference between how we think and how the world works."
BATMAN What gauge wire is equal to code 83 rail?
What gauge wire is equal to code 83 rail?
The C83 Nickel Silver equivilent would be AWG 26 (copper, stranded), more or less. Remember that both nickel silver and brass are copper alloys, but have different resistances because they are not pure copper.
It also depends on the alloy itself, no guarantee that rail samples from different sources will be the same alloy, nor will the profile be identical, which could add or subtract resistance.
SeeYou190BATMAN What gauge wire is equal to code 83 rail? None. Wire is generally made from copper which has different (better) characteristics for carrying electrical current.
Sorry Kevin, next time I'll try to be more specific.
The question was posted as more of a tease to bring out those smarter than I to present some hard numbers rather than the vague chit-chat we sometimes get.
Oh ya, don't pee on an electric fence on the ranch, my poor dog just wouldn't listen.
We should go back and ask Brent what he meant by that question. At the time that he posted it, I took it as a satirical question because it was asked in the midst of all that nonsense about not needing a bus but only using a single pair of feeders from the command station to power an entire DCC layout.
Rich
SeeYou190 BATMAN What gauge wire is equal to code 83 rail?