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.
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
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.
OvermodThink 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,
gregcwhat does the scope show peak-to-peak with a 14VAC waveform?
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!
gregchow do you generate that waveform from a DC supply?
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.
what does the scope show peak-to-peak with a 14VAC waveform?
how do you generate that waveform from a DC supply?
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
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.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
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.
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 pleaseMel My Model Railroad http://melvineperry.blogspot.com/ Bakersfield, California I'm beginning to realize that aging is not for wimps.
gregcwhat are you planning to use these LEDs for in what looks like a sheet of balsa?
RR_MelI 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.
Chip
Building the Rock Ridge Railroad with the slowest construction crew west of the Pecos.
rrinkerBut 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.
rrinkerLike 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
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.
How does his meter lie?
http://www.sumidacrossing.org/ModelTrains/ModelTrainDCC/DCCDecoders/DCCVoltages/
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...
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.
SpaceMouse 1) Okay, I get that it will blink on and off 60 times a second. But can you tell? Will it send an autistic person into seizures? 2) Can you run LEDs and grain of rice bulbs in parallel off the same circuit. (Without the resistors.) Here's the LED part.
1) Okay, I get that it will blink on and off 60 times a second. But can you tell? Will it send an autistic person into seizures?
2) Can you run LEDs and grain of rice bulbs in parallel off the same circuit. (Without the resistors.) Here's the LED part.
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.60 LEDs running at max 20ma each would be 1.2 amps. If you want to power them with AC simply use a 2 amp bridge rectifier. I stock 1amp, 2amp and 4amp bridges that I buy off eBay.I rarely run my LEDs more then 10ma but even at 15ma 60 would be under 1 amp. I don't think the LEDs will draw even close to 20ma with 1KΩ resistors at 12 volts so the DB107 rectifier could work fine.DB107 1amp eBay cost 50 for $42W10 2amp eBay cost 10 for $1.75KBP310 4amp eBay cost 10 for $1
You could parallel the LED source with 12 volt GOR bulbs, lower voltage bulbs would need resistors to operate off 12 volts.
Mel My Model Railroad http://melvineperry.blogspot.com/ Bakersfield, California I'm beginning to realize that aging is not for wimps.
SpaceMouseI ordered a bunch of LEDs from China--you know the packs that offer 500 LEDs of various colors, etc. for like $4. They they each came comped with 500 1K resistors. I only use the clear bulbs. Is there a reason why I shouldn't run them in parallel each with their own resistor off a dedicated 12V 2 amp power supply?
Is there a reason why I shouldn't run them in parallel each with their own resistor off a dedicated 12V 2 amp power supply?
have you connected an LED and 1k resistor to your supply to check the intensity of the light? you may want to use a different size resistor
if the LED operates at ~3V, they will draw ~9ma ((12 - 3) / 1k). (max is typically 20ma) 30 will require 270 ma. so you could get by with a smaller power supply.
another approach is to wire 3 LEDs in series and another 3 with reverse polarity wired in parrallel to a single 330 Ohm resistor ((12V - (3*3V)) / 9ma). 30 LEDs would need just 5 resistors and they would only draw 45 ma. an even smaller supply
5 resistors would be easier to change if you wanted different intensity.
what are you planning to use these LEDs for in what looks like a sheet of balsa?
SpaceMouseLOL! I don't know what a CCD imager is but I gotta get me one of those.
I am quite sure that any reports of 'blinking' coming from YouTube are artifacts of a digital camera set to a high shutter speed, perhaps with a rolling shutter (which also produces the sort of image distortion early focal-plane shutters capturing fast-moving objects did!) that is imaging lines or areas 'away' from the LED the whole time that it is on. There are similar problems shooting the LED headlights on ACS-64s in many early videos of them I've seen.
I'm looking for good normal-language discussions of electronic shutter operation but so far the only ones that actually tell you much use cinematographer-speak ... or worse!
OvermodYou are almost certainly not reading any form of "AC" with that LED, you are likely reading very quickly-modulated unipolar DC. I would suspect there is some physical "blinking" going on, at a somewhat irregular rate corresponding to the modulation, and there is likely an easy way to visualize some of it by using a typical "high-speed" CCD or other staring imager set to 60Hz with very short acquisition and then slow down the 'playback' of the resulting file and see if in some frames the LED appears to be off. Of course, casually observing the LED, or even moving it in a dark room, shouldn't show flicker at all when the modulation is at data rates...
Of course, casually observing the LED, or even moving it in a dark room, shouldn't show flicker at all when the modulation is at data rates...
LOL! I don't know what a CCD imager is but I gotta get me one of those.
OvermodBut that is not quite what he said. Remember that is the LED that 'switches the current on', not the other way around. You have picked the resistor to get one LED (or the string of them with the same polarity) to the right point. A resistor knows no polarity; it doesn't care which way the electrons in the I are moving, it only makes heat out of I^2R. So it 'ballasts' the LED or string that conducts on a given AC half-wave, then does the same thing for the 'opposite' LED or string when it comes up out of 'dioding' and starts to light on the following half-wave.
I have neither the schema nor lexicon, but I got you.
gregc had suggested that you could wire 2 LEDs in parallel with reversed polarity. this would cut the number of resistors in half, as well address some peoples concern that you should wire a reversed biased diode to limit the reverse voltage across the LED. but if you're using 12V, you could also wire 2 or more LEDs in series, further cutting the number of resistors. this may be helpful if you need to replace the resistors.
I ordered a bunch of LEDs from China--you know the packs that offer 500 LEDs of various colors, etc. for like $4. They they each came comped with 500 1K resistors. I only use the clear bulbs.
By the way, I thought cross-wiring the LEDs to the resistors was brilliant.
richg1998My NCE Power Cab has a 3mm red LED and 1k resistor tied to the output to indicate DCC, a form of AC and does not blink.
BigDaddyLED "binking" can be seen on some of the Youtube video reviews. In real life the Terminator can see it, you and I cannot.
Interestingly enough, with respect to CRT monitors with short phosphors, 60Hz flicker can be very annoying, but 72Hz is usually better and 75Hz better still -- that small a frequency difference can 'make all the difference'
There is a similar issue with 'digital cinema' where the source is at the same 24Hz presentation as 'film'. Even with motion-vector steering of a quality 'cost-effective' for home devices, just frame-doubling (to 48Hz) will not get rid of motion effects, particularly at dark/bright edges. It takes frame-tripling to present fast enough to overcome the issue.
SpaceMouseThe part I had trouble with is that Greg said since with AC, the current switches the LED on only when it is flowing in the right direction, it is only on half the time, so it only needs half the resistance.
LED "binking" can be seen on some of the Youtube video reviews. In real life the Terminator can see it, you and I cannot.
Henry
COB Potomac & Northern
Shenandoah Valley
half the resistance means twice the current. but if it's only on half the time, the power (Watts) and intensity are halved. in other words, ~same intensity as with DC when using half the resistance and AC
SpaceMouseI already have a bus with 30 LEDs pre-wired in parallel.
that's too bad.
i had suggested that you could wire 2 LEDs in parallel with reversed polarity. this would cut the number of resistors in half, as well address some peoples concern that you should wire a reversed biased diode to limit the reverse voltage across the LED.
but if you're using 12V, you could also wire 2 or more LEDs in series, further cutting the number of resistors. this may be helpful if you need to replace the resistors.
RandyChip - it's like this. If running the LED on DC, the power is always in the direction that lights the LED. 100% of the time. So the LED is whatever brightness it is based on the amount of current - controlled by the resistor. If instead of being on 100% of the time, you feed it AC, the LED now is only lit up half the time. All else being equal, it will appear half as bright. You cna compensate by using a lower value resistor, so when the LED is on, it's a brighter 'on', raising the average brightness.
The LED on the panel is wired across Rail A and Rail B, with a resistor - so it 'sees' the track voltage. Half the time, anyway - in the opposite half of the DCC wave, it doesn;t allow any current flow.
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.
A red LED is ok, as long as it has a resistor - but LEDs in general have fairly low reverse voltage limits. Whereas many regular silicon diodes can withstand 50, 100, 200, 400, or even more volts in the reverse direction before breaking down, LEDs are typically much lower. Typical red LEDs are generally OK, but a lot of white LEDs can be damaged by a reverse voltage no more than their actual forward voltage.
Chip - it's like this. If running the LED on DC, the power is always in the direction that lights the LED. 100% of the time. So the LED is whatever brightness it is based on the amount of current - controlled by the resistor. If instead of being on 100% of the time, you feed it AC, the LED now is only lit up half the time. All else being equal, it will appear half as bright. You cna compensate by using a lower value resistor, so when the LED is on, it's a brighter 'on', raising the average brightness.
richg1998Make sure you have fuses with that much power from a computer power supply. I know what those are capable of.
I was semi-joking. There are better ways to handle it.
richg1998There are online companies that are much cheaper than Amazon for parts.
I usually get most of the stuff on eBay, and it's dirt cheap if you can wait for the Chinese junk to sail here. I tried a few online dealers, but with shipping the price was comparable to Amazon.
What sealed the deal for me is that I am in the process of getting the layout ready for track. I wanted to prewire the Atlas turntable when I cut the plywood to drop it in, even though I won't convert to a wooden bridge turntable for a while. That's when I noticed that the LEDs were hooked up to AC. I wired them for 12V DC.
I want to close up the top of the layout so I can get the roadbed down. Amazon will get me parts in one day.