richhotrainHow about now, Bob, after these most recent four posts? Rich
Don't get me wrong, I admire these guys with the electrical know-how to figure circuits out, and understand why they work the way they do. Kudo's to all of you.
-Bob
Life is what happens while you are making other plans!
farrellaa rrinker They even out from rail A to rail B because each half is identical, but with half wave lighting you are taking a reference from the center of the waveform to one side. It may not be much, but if say nothing but preamble packets were being sent, 0 bits have a longer effective 'on' time than 1 bits. The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice). I knew there was a reason I wanted to get a scope. This is getting rather esoteric and of not interest and of no real use to the average modeler. I agree it's going to be unnoticeable unless something is very wrong with your DCC system, or you actually do have zero stretching in use. I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration. --Randy Randy, Could you explain this in English! LOL!
rrinker They even out from rail A to rail B because each half is identical, but with half wave lighting you are taking a reference from the center of the waveform to one side. It may not be much, but if say nothing but preamble packets were being sent, 0 bits have a longer effective 'on' time than 1 bits. The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice). I knew there was a reason I wanted to get a scope. This is getting rather esoteric and of not interest and of no real use to the average modeler. I agree it's going to be unnoticeable unless something is very wrong with your DCC system, or you actually do have zero stretching in use. I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration. --Randy
Randy,
Could you explain this in English! LOL!
Rich
Alton Junction
After some investigation, I'm going to back-pedal on my response ....
Having not looked at a scope view of the DCC sine wave in some time, I was thinking the frquency between 0 and 1 packets were the same. Now that I took a refresher, I now see again that the frequency between the bits is symetrical. For some reason, I had in my mind there was equal time gaps between each packets in the duty cycle.
Sorry Randy .... I'm switching camps ....
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
Mark R.Randy's theory does make sense...
How? Again, symetrical square wave, half wave power = power on half the time, off half the time, regardless of frequency.
CSX Robert rrinker ...The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice)... rrinker ......I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration... I don't see how you are coming to this conclusion. If you are using half-wave power, and the waveform is symetrical, then the power will be on 50% of the time, regardless of frequency (within the working limits of the diodes).
rrinker ...The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice)...
rrinker ......I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration...
I don't see how you are coming to this conclusion. If you are using half-wave power, and the waveform is symetrical, then the power will be on 50% of the time, regardless of frequency (within the working limits of the diodes).
Randy's theory does make sense, BUT - the time factory involved almost makes it negligatable. The command station can send up to 180 3-byte packets per second. That's 4320 bits per second. Even if - for some reason - there is an unusually high number of 0 bits compared to the number of 1 bits, the time factor alone would negate any of this visually in the LED itself.
rrinker...The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice)...
rrinker......I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration...
I wish I knew electronics like you do, life would be so much easier for me since I switched to DCC. It's like when Microsoft introduced the Windows GUI and we didn't have to know DOS commands in order to do anything with a pc. I rely on DecoderPro so much that I forget what CV's do what. And, without guys like you on the forum, I most likely would still be using DC and a bunch of toggles!
CSX Robert Actually, if you are not using zero bit stretching (controlling a DC loco), then the packets being sent will not affect the voltage - the two halves of the wavform will always be equal, so half wave power will always be approximately half the DCC voltage. If you are using zero bit stretching, then both the DC loco speed setting and the packets begin sent will affect the voltage.
Actually, if you are not using zero bit stretching (controlling a DC loco), then the packets being sent will not affect the voltage - the two halves of the wavform will always be equal, so half wave power will always be approximately half the DCC voltage. If you are using zero bit stretching, then both the DC loco speed setting and the packets begin sent will affect the voltage.
They even out from rail A to rail B because each half is identical, but with half wave lighting you are taking a reference from the center of the waveform to one side. It may not be much, but if say nothing but preamble packets were being sent, 0 bits have a longer effective 'on' time than 1 bits. The effective 'area under the curve' is greater for a string of 0 bits than for a string of 1 bits (possibly - if the exact pulse widths are just right, it could be that they are equal, it just takes more 1 bits per a given time slice. The DCC frequency is not constant, the 1 bits have a shorter wavelength and higher frequency, but if in a 100ms time sample, you have enough 1 bits, the mean voltage could be the same as that from the fewer 0 bits that would fit in that same time slice). I knew there was a reason I wanted to get a scope. This is getting rather esoteric and of not interest and of no real use to the average modeler. I agree it's going to be unnoticeable unless something is very wrong with your DCC system, or you actually do have zero stretching in use. I do think however that a string of say 5 zero bits will have a higher mean voltage (common to rail, not rail to rail) than a string of say 20 one bits that fits in the same time. More math is needed to compare the typical zero duration with the typical one duration.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
farrellaa richhotrain farrellaa Just want to share the MRC decoder diagram I made (for connections only). -Bob PS: The tabs on the far right were not used (had no trace to them??) I was re-reading this entire thread this morning and noticed Bob's question on the diagram about the two unused tabs on the far right of the decoder. I assume that those are the rear headlight function outputs. Rich Rich, The reason I noted the two far rear tabs; I tested them with a VOM and got no reading? I tried connecting an LED using test leads and nothing worked. Since this is an A unit and doesn't have rear lights I thought maybe the mfg didn't use them (this really didn't make sense but ????). -B*ob
richhotrain farrellaa Just want to share the MRC decoder diagram I made (for connections only). -Bob PS: The tabs on the far right were not used (had no trace to them??) I was re-reading this entire thread this morning and noticed Bob's question on the diagram about the two unused tabs on the far right of the decoder. I assume that those are the rear headlight function outputs. Rich
farrellaa Just want to share the MRC decoder diagram I made (for connections only). -Bob PS: The tabs on the far right were not used (had no trace to them??)
Just want to share the MRC decoder diagram I made (for connections only).
PS: The tabs on the far right were not used (had no trace to them??)
I was re-reading this entire thread this morning and noticed Bob's question on the diagram about the two unused tabs on the far right of the decoder. I assume that those are the rear headlight function outputs.
Rich,
The reason I noted the two far rear tabs; I tested them with a VOM and got no reading? I tried connecting an LED using test leads and nothing worked. Since this is an A unit and doesn't have rear lights I thought maybe the mfg didn't use them (this really didn't make sense but ????).
-B*ob
That would tell me there were no copper traces to those tabs.
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.
farrellaa hon30critter I believe the major difference you will find between Bob's situation with an MRC decoder and your situation with a Soundtraxx decoder is that the lighting outputs on the Soundtraxx are operating at 12 volts already as opposed to the 1.5 volts on the MRC. You can use the existing lighting contacts without having to find another source for 12 volts on the board as Bob had to. This is precisely what I was concerned with, as I didn't know how to get 12 volts to run the LED, being this decoder didn't have the normal NMRA wire colors or even solder tabs. All my other Genesis F series locos were easy to upgrade to LED's. -Bob
hon30critter I believe the major difference you will find between Bob's situation with an MRC decoder and your situation with a Soundtraxx decoder is that the lighting outputs on the Soundtraxx are operating at 12 volts already as opposed to the 1.5 volts on the MRC. You can use the existing lighting contacts without having to find another source for 12 volts on the board as Bob had to.
This is precisely what I was concerned with, as I didn't know how to get 12 volts to run the LED, being this decoder didn't have the normal NMRA wire colors or even solder tabs. All my other Genesis F series locos were easy to upgrade to LED's.
Rich:
Thanks for pointing that out. I should have read dumbasapost's information more carefully. I was referring to decoders meant for general useage and not those supplied as factory original equipment.
Dave
I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!
hon30critterI believe the major difference you will find between Bob's situation with an MRC decoder and your situation with a Soundtraxx decoder is that the lighting outputs on the Soundtraxx are operating at 12 volts already as opposed to the 1.5 volts on the MRC. You can use the existing lighting contacts without having to find another source for 12 volts on the board as Bob had to.
I nominate this thread as the Best Thread on the Electronics and DCC forum this year. Extremely informative. Yes, as Randy pointed out, we have discussed wiring the function outputs and common tabs before, but I cannot recall ever discussing using the track power for LED voltage where insufficient power is provided on the decoder itself. You learn something new every day.
Kudo to woodone !
rrinker...this is half wave power because it is only one half of the DCC wave form...
True, and reminds me of something I meant to mention in my previous reply but forgot - even though track voltage is usually ~1.5 volts greater than the function common voltage, since it is half-wave power the average voltage applied to the function is about half that. This can be a good thing if you are adding a decoder to an engine that has a 12 volt incadescent bulb because it will give you more realistic lighting (a full 12 volts is usally to bright), cooler temps, and longer bulb life.
rrinker...The only downside of this is that the effective voltage will vary based on the DCC packets being sent, and also if used on a layout that allows address 00 to control a non-decoder loco, the effective voltage of the half wave connecion will vary as address 00 is used.
I believe the major difference you will find between Bob's situation with an MRC decoder and your situation with a Soundtraxx decoder is that the lighting outputs on the Soundtraxx are operating at 12 volts already as opposed to the 1.5 volts on the MRC. You can use the existing lighting contacts without having to find another source for 12 volts on the board as Bob had to.
dumbasapost:
However, you must use a resistor in series with the LED in the circuit. It doesn't matter which lead you attach the resistor to, but many of us try to always attach the resistor to the same polarity lead in each installation just for consistancy. It makes it easier to understand what you have done if you have to go back and re-examine the circuit down the road. A 1K ohm (1000 ohms) is the most common value used. If you are modelling older steam where you want the headlight to be dimmer then a higher value resistor can be used.
Not sure if you knew that so I thought I would mention it.
Regards
Thanks for posting your results Bob. I was planning on installing SMDs myself so I found this very helpful. I'm dealing with Soundtraxx boards so the solution may be slightly different, but this thread helped point out some of the potential issues and solutions I may run into. I found it very interesting that DCC control is on the negative end of the LED, rather than the positive. Thanks again!
richg1998 Very nice. A good example of a picture being worth a thousand words. Nice work. Hope you find a solution. Let us know when you do. I have seen many complaints about poor 1.5 volt lighting in Athearn locos. Rich
Very nice. A good example of a picture being worth a thousand words. Nice work.
Hope you find a solution. Let us know when you do. I have seen many complaints about poor 1.5 volt lighting in Athearn locos.
Thanks, I did find a solution as shown in the drawing and as posted earlier. It was a little confusing since MRC didn't follow NMRA standards in their decoder wiring or layout. With the help of the forum members it was resolved with satisfactory results. I will have to post a photo of the engine with the lights on.
-*Bob
richhotrain richg1998 Issues like this can be confusing for those with limited electrical DCC experience. We shall see. I would certainly try the suggestion. Rich I, for one, fall into that category. Once the OP solves his problem, and assuming that he reports back that the rail power suggestion worked, I hope that someone more knowledgeable than me will take the time to explain this in more detail. I always thought that the wiring of loco lamps, headlight, rear light, ditch lights, were always dependent upon their assigned function outputs. For one thing, the function output voltage is typically lower than the track voltage, or so I thought. And, I always thought that using the correct function output tab was critical to the proper application of CV values for advanced performance of the lamps. So, do you not even need to use the function output tabs as long as the other leg of the LED or incandescent is wired to the accompanying common tab for the designated function output tab? Anxious to learn more about this whole topic, and anxious to hear if the woodone solution indeed works. Rich
richg1998 Issues like this can be confusing for those with limited electrical DCC experience. We shall see. I would certainly try the suggestion. Rich
Issues like this can be confusing for those with limited electrical DCC experience. We shall see.
I would certainly try the suggestion.
I, for one, fall into that category. Once the OP solves his problem, and assuming that he reports back that the rail power suggestion worked, I hope that someone more knowledgeable than me will take the time to explain this in more detail.
I always thought that the wiring of loco lamps, headlight, rear light, ditch lights, were always dependent upon their assigned function outputs. For one thing, the function output voltage is typically lower than the track voltage, or so I thought. And, I always thought that using the correct function output tab was critical to the proper application of CV values for advanced performance of the lamps.
So, do you not even need to use the function output tabs as long as the other leg of the LED or incandescent is wired to the accompanying common tab for the designated function output tab?
Anxious to learn more about this whole topic, and anxious to hear if the woodone solution indeed works.
It's like this (I've posted explanations on this several times - one more of those things I really need to just type up and post on my web site so I stop repeating myself): The function 'outputs' on a decoder are not 'outputs' - they are connections to ground. The various color wires, white, yellow, green, violet, etc. connect to switches that either are open (no current flows) or are closed (they connect to ground). The blue wire is really the 'output' on a decoder, it is the positive lead. So, to make a complete circuit and light a lamp or LED, you need source of power, the load (lamp or LED and resistor) and a connection to ground. You can get a source of power from either of the rail pickups instead of the blue wire. Relative to the decoder ground, this is half wave power because it is only one half of the DCC wave form - the ground is in the middle, relative to the track power, because the track power is first passed into a bridge rectifier on the decoder. The only downside of this is that the effective voltage will vary based on the DCC packets being sent, and also if used on a layout that allows address 00 to control a non-decoder loco, the effective voltage of the half wave connecion will vary as address 00 is used. In the case of a decoder that limits the voltage on the blue wire, using half wave power can be a way to get a higher voltage, like the example of 1.5V bulbs being repalced by white LEDs, which usually need at elast 3.5V. If the limiting resistors or whatever are on the actual function wires, this won;t help. Most decoders that have individual limiting resistors often have an alternate function pad or else a pair of pads you can short to bypass the built in resistor.
Thanks for that clarification, CSX.
Incidentally, not being from Europe, it was a surprise.
richhotrain...the function output voltage is typically lower than the track voltage, or so I thought...
It is, but only by about 1.5 volts, the voltage drop of the diode bridge rectifier.
richhotrain...I always thought that using the correct function output tab was critical to the proper application of CV values for advanced performance of the lamps...
They are. You don't replace the function output lead (white, yellow, etc.. on a wired decoder) with track power, you replace the common (blue) lead.
richhotrain So, do you not even need to use the function output tabs as long as the other leg of the LED or incandescent is wired to the accompanying common tab for the designated function output tab?..
So, do you not even need to use the function output tabs as long as the other leg of the LED or incandescent is wired to the accompanying common tab for the designated function output tab?..
No you do need the function output, you don't need the common.
It should not be a surprise that this works, it is actually a common way of wiring lights in Europen locos. Ever seen a 6 pin NEM 651 plug decoder or socket (this is what is used on Fox Valley Models diesels)? They only have 6 pins because they use the rail pickup for the function common.
woodoneIf you look the way they show how to hook up the lights you will not that both lights show a common, THAT IS THE POWER TERMINAL. The other wires ( one to each bulb) is the control wire. The LED does not care where it get its power, but it must be enough ( about 3 volts) to make the LED light. With getting power from the rail you will be getting 1/2 wave form from the rail. If you read the voltage across both rails you should see about 14 volts. so 1/2 that is about 6 volts. YOU MUST USE A RESISTOR !!! 1K will work. you can use more if you want to dim it a bit.
This solution has been in the Digitrax literature for some time. See the latest Digitrax decoder manual: http://www.digitrax.com/static/apps/cms/media/documents/documentation/Decoder_Manual_V2-01_2014.pdf See page 22 for an illustration of options. Even though this is a Digitrax manual it holds for other brands of decoders as well as Woodone has shown.
Joe
OK guys, I followed woodone's advice and he was correct! The front headlight now functions as it should and goes off when in reverse and also using F0. I installed a SMD cemented to a short length of 1.5mm fiber optic rod and put it in the hole where the incandescent bulb was. It looks great and works great! Thanks to all who commented on this issue and especially to woodone. I was a little skeptical of this solution at first too, but now we all know. I did have to try a couple of the solder points at the light socket to finally get the right one. One of them kept the light on all the time.
BTW the schematic of the MRC decoder board that was posted doesn't look like mine. I made a sketch of mine and marked all the outputs that I could identify.