LED “binking” can be seen on some of the Youtube video reviews. In real life the Terminator can see it, you and I cannot.
But 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.
Where you see it is if you have sensitivity to motion or change in your peripheral vision, for example if a high myope (badly nearsighted) and the LED is in your peripheral vision. Normal action of the ocular muscles moves the eyeball enough that perception picks up the jitter.
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.
You 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…
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.
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?
By the way, I thought cross-wiring the LEDs to the resistors was brilliant.
I have neither the schema nor lexicon, but I got you.
LOL! I don’t know what a CCD imager is but I gotta get me one of those.
Stripped of the highfalutin’ verbiage: a modern digital camera with CMOS imager and global shutter would likely do the job of a CCD, even one with interline transfer, for the required purpose (there is no problem with low light, only timing.)
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!
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?
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 $4
2W10 2amp eBay cost 10 for $1.75
KBP310 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.
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 link
How does his meter lie?
http://www.sumidacrossing.org/ModelTrains/ModelTrainDCC/DCCDecoders/DCCVoltages/
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.
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.
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
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 please
Mel
My Model Railroad
http://melvineperry.blogspot.com/
Bakersfield, California
I’m beginning to realize that aging is not for wimps.
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.
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
–Randy
what does the scope show peak-to-peak with a 14VAC waveform?
how do you generate that waveform from a DC supply?
I see what he means.
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!
Think of it as ‘generating the waveform’ with two DC supplies, by gating them on and off very quickly with the different on times corresponding to the binary states. One supply is driven exactly the opposite of the other; think of it as the ‘second’ supply mirroring the modulation of the first’s pulse as soon as the zero crossing is reached. (That is the action that equalizes net charge due to the logic-pulse on times).
By putting the equivalent of a high-voltage diode on the decoder, no matter which