leds and reversing the polarity for use

[keitaro]

ok so i want to use my remaining strips in my models i was looking at either the acw or accw for flicker free.

 

just wanted to confirm this will reverse the polarity to make the leds work when reversing?

 

it looks as though it does but wanted to clarify with some one who knows electronics.

 

http://www.trainaidsa.com/shop-leds.shtml

[KenS]

They're described as having a rectifier, which is essentially a set of diodes that takes AC (or DC of either polarity) and turns it into DC with a single polarity (they typically have four pins, two marked for AC with a "~" and one each "+" and "-").  Even without the rest of the text I'd expect either to do exactly what you want. 

 

That they also describe them as "for polarity control" just confirms that.

 

The ACW won't eliminate flicker (it appears to just be the rectifier).  The ACCW adds a capacitor to store power (a small amount) to keep the LEDs lit over short interruptions.  This will work (I've made similar circuits myself) but the size of the capacitor determines how long a dropout (we're talking milliseconds) can be bridged.  With really poor contact you might still get some flicker.

[rankodd]

The real question is, how big is the ACCW? There was someone looking into this before, and making it small enough to be unobtrusive was going to be quite difficult.

[KenS]

I don't know how big the ACCW is, but it's not going to be inconspicuous in a heavily-windowed car, the components just don't come that small. It looks like they've bundled them on both sides of the wire, so it's probably about 8mm in cross-section and 15-20mm long judging by the size of the wire and what I know of the likely components.

 

A typicial circuit needs just a rectifier (abt 4mm high and 6x8mm) and a capacitor (5+mm high and about the same width and depth as the rectifier) plus a small resistor if you want to limit inrush current. You could tuck these in the end of a car where they'd be hard to see if you custom-built something, but they'd still be visible.  And that assumes use of a tantalum capacitor, which is a bit pricey. An electrolytic one is about 8mm in diameter and 10-15mm long.

[rankodd]

Ah - I think it was your work I was thinking of, Ken.

 

I'm pretty sure the ACCW isn't intended for use in n scale trains.

[KenS]

Maybe not, but it looks like you could use it, it just wouldn't be inconspicuous.

 

My project is still incomplete.  I've built a couple of prototypes that work well on my layout, but don't have it down to the point I'm happy with the effort required to assemble one (I'm going to need HUNDREDS of these things, I want them easy to make).

 

But the components are pretty small, as you can see in this photo.  That yellow box is the capacitor, which costs more than the entire ACCW (it's $1.73 in quantity 100+) so I don't think they're using one this compact.

[keitaro]

i emailed for dimensions he replied away till 2nd august and will tell me then.

[Guest Closed Account 1]

Couldn't the LED leads be soldered to the pick up strips and that'll fix the flicker?  My heavier cars don't flicker.

[KenS]

It's unclear where the problem is.  Soldering the LEDs to the L-shaped pickups doesn't help (I did that on one car).  My suspicion is that the tabs on the trucks are making intermittant contact with the brass strips that run the length of the car.  They can't be soldered, since the truck needs to turn.  And simply adding weight to the car didn't seem to help much either.

 

I've noted that the motor car has much less of a problem ("much less" is not "none" however), but I think there's more than weight at work there.

[Guest Closed Account 1]

Yes, of course.

 

Too much bend in the strips prevents the trucks from articulating and they could bind when the pickups miss the pickup strips.

[keitaro]

i think it's a combo of things.

 

I put a lump of solder on the pins to make better contact but still same flickering i can see they are hitting fine.

 

I was planning on soldering another pin to make them longer and hit the bottom pickup rather than using the strip they provide like the Micro ace headlights do for example but havn't gotten around to it.

[Guest Closed Account 1]

Is a decoder required to illuminate a LED on DCC?

 

Otherwise, how would a 3 leg bi-polar LED be wired to track power?  Want to put an LED in a search light car.

[KenS]

Well, a three-leg bi-polar LED is an odd choice for a searchlight as they're usually designed to show two colors (like red/green), why not use a white or yellow two-lead LED?

 

If you want to be able to show two colors, then you'll need a decoder with a "common anode" 3-lead LED (center lead is longer than the other two).  Hook the long leg to the blue (common) wire, and the other two each to its own function output.  Don't forget the required resistor (it can go between the blue wire and the long lead, but you're safer using one on each short lead instead, just in case you accidentally turn both functions on, which could blow the LED with only the one ressitor). Regardless of where you put it use the same size.

 

Calculate resistor size as usual (use a calculator) based on the LED forward voltage and amperage (2V and 20 mA if you don't have better numbers from a data sheet) based on the maximum track voltage you want to survive. To be safest, assume 23V, but if you know you're using something specific you can use a lower track voltage.

 

With a normal LED you don't need the decoder (unless you want to turn the LED off), but there are a couple of things.

 

First, you'll need a rectifier, or at least a power diode, to block reverse current.  It's possible you don't need that, since half the time the current will be going in the right direction,  And some (very limited) experimentation I did showed that reverse currents didn't seem to damage LEDs that had adequate resistor protection. But personally I wouldn't take the chance since LEDs are easily damaged by high reverse voltage.  Just because there's no immediate failure doesn't mean it won't damage the LED over time.

 

And then you need a resistor, as above.

 

If the reason you want to use a three-lead LED is because it will light yellow with both leads active it might be possible (I've never tried anything like this, so it could just result in smoke).  You'll need a full-wave rectifier and one resistor.  Connect the rectifier AC inputs to the two track pickups as usual.  Connect the different-length LED leg to the appropriate output of the rectifier (+ for a long leg, - for a short one) through the resistor and leave the other output unconnected.  Then connect the two other LED legs to the two track pickups.

 

This should work, as one LED leg will always be identical to the rectifier output (and thus do nothing) while the other will be opposite (and thus light its half of the LED).  The rapid switching of the DCC signal will make it look like both are lit simultaneously, and you'll get the "both legs powered" color.

 

But since you don't get any less circuitry this way (in fact it requires a full-wave rectifier, whereas an ordinary LED will work with a half-wave rectifier) there's no reason for doing it this way.

[The_Ghan]

KenS,

 

I've been using 1/4W resistors, based on calculations from this link: http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator

 

I got worried when I used your link: http://led.linear1.org/1led.wiz

 

So I checked a a couple more calculators: http://ledcalc.com/ and http://ledcalculator.net/

 

I'm using 14v, 1.8v and 20mA as the three values.  Your link recommends 1/2W but the other links recommend 1/4W.  I'm using the Digitrax 2012 supply, which is always set to 14v and, apparently actually supplies 13.8v.  I don't want to be splitting hairs but should we all be using 1/2W resistors instead of 1/4W?

 

Also, I've not done any LED for in-car lighting yet.  Can you provide a link to an appropriate rectifier or power diode please?

 

Cheers

 

The_Ghan

[keitaro]

Just on this and led strips you can get a 5 led strip 10cm 5.6ma a led though so too much power being eaten.

[KenS]

The wattage of the resistor depends on the voltage being dropped in it.  With 14-1.8=12.2V @ 20 mA, that's 244 mW, or just a hair shy of 1/4 W.

 

But, if you actually use real resistors, they only come in certain sizes.  With the recommended 680 ohm resistor, you're dropping 0.02 x 680 = 13.6V, and 13.6 x 20 = 272 mA, or 10% over 1/4 W, hence the recommendation.

 

The first calculator you linked actually adds a safety margin, and notes that the power rating for the resistor should be 0.407W (or just shy of a half-watt).  Confusingly it also gives a "calculated resistor wattage" that's below 1/4 watt, but I think it's using the ideal resistor for that.

 

Use 1 1/2W resistor to be safe.  A 1/4 may work under most uses, but why take chances?

 

I'll post more later on rectifiers and diodes.

[KenS]

LEDs are diodes, which means they only work on one polarity of DC. Further the have a low tolerance for reverse voltage.  A typical 5mm white LED with  20mA current and a 3.2V forward voltage can tolerate no more than 5V in the reverse direction.  Since a resistor only lowers voltage when current is flowing, a LED connected through a resistor to a 12V source will experience 12V in the reverse direction.  So you need to convert DCC (which is an AC waveform) into DC in the right direction.  The circuit component for this is a rectifier (or a diode that can tolerate high reverse voltages).

 

A rectifier is simply an assembly of diodes that converts AC to DC.  Another name for the full-wave version is "bridge rectifier". For model train lighting these can be had in relatively small sizes that can support up to half an amp of current (500 mA) and are resistant to 50V of reverse voltage.

 

A full-wave rectifier uses four diodes and will create a DC output that varies from zero to full voltage as the AC cycles, but is only at zero for an instant.  A half wave rectifier is a single diode, and does the same thing half of the time, and outputs zero volts the other half.  With DCC, which is a square wave, the full-wave version produces a near-constant fixed DC voltage.  However, there's typically 1 - 1.2 volts lost, so on a 13.8V Zephyr, you should get about 12.8 - 12.8 V.

 

For my car-lighting projects, I use a small surface-mount model made by Fairchild that has a 1V drop and is limited to 500 mA, which is far more mA than any car lit with LEDs will use.

 

The only problem with this is that the metal tabs on the case are tiny, and very difficult to solder to (it can be done, with heat-sink clamps, something to hold the rectifier motionless while you solder, a good iron, and patience).  A slightly larger one designed for through-hole mounting can be used that has longer legs and is easier to work with.

 

Both are typically around 4-7 mm on a side, and 3-5mm thick. But be careful, as most rectifiers are much larger.

 

Since DCC uses high-frequency AC, you can get away with using a diode in place of a rectifier.  Just connect the diode, resistor and LED in a row between the two track pick-ups and the LED will be lit half the time (on DC the LED will only be lit in one direction, which can be useful for making head and tail lights).  Since with DCC the interval where it's off is so short, it will appear continuously lit. The benefit of the diode over the LED (which is also a diode) is that you can use one with a high reverse breakdown voltage, typically 30 or more volts for DCC.  LEDs have a very low breakdown voltage. The critical numbers are "V_R" (maximum reverse voltage) and "I_F" (forward amperage) and in general what you want is called a "Schottky" diode (although "standard" semiconductor diodes can be used at the voltages and currents involved in N-scale train lighting with LEDs).

 

The downside to diodes is that the ones you can get are often larger than the rectifier would have been.  For example, this "Schottky" diode, which supports up to 9 Amps and withstands up to 30V, is 6.3mm in diameter and 9.5mm long.  Or this "standard" one, which supports up to 700 mA and withstands up to 200V, and is 2.5mm in diameter and 5mm long. It's hard to find smaller ones.

[The_Ghan]

Hi KenS,

 

Will a 1A Schottky such as this do the job? http://www.ebay.com.au/itm/1N5819-1A-40V-Schottky-Barrier-Rectifier-Diode-x-50-/260572570654?pt=LH_DefaultDomain_0&hash=item3cab55e41e

 

Cheers

 

The_Ghan

[KenS]

Yes, V_R=40 and I_F=1.0A should be fine.

[The_Ghan]

Thanks KenS,

 

will order some Schottkys.

 

Cheers

 

The_Ghan

[Guest Closed Account 1]

Will these Schottkys allow bicolor LEDs to work with DCC?

 

Looking to simply have an indicator LED (3 leg 3mm) at each switch so regardless of view you can see that all the switches are open or closed.

 

Or can I use the Kato 11-210 Lighting decoders to power a 2-leg bicolor LED? 3-leg?

[KenS]

A LED to show switch position needs to be aware of the state of the switch somehow.  With a slow-motion switch machine you can wire them into the DC controlling the motor.  But with a solenoid system like the Kato and Tomix switches, you need something else.  The simplest approach might be to connect some kind of microswitch to the mechanical part of the turnout, so when the turnout was set, the switch controlled power to the LED.  This wouldn't care about DC or DCC at all.

 

On DC, if you have a "power routing" switch, i.e., one where the rails past the frog are either unpowered (or powered to the "wrong" polarity") when the turnout is thrown away from them, and correctly powered when it is thrown towards them, then you could light a normal LED based on the difference in phase (with some care to resistor size and reverse-current protection).  It may be possible to make this work on DCC with a 3-leg LED, although it's not quite as general a solution.

 

If you connect the common leg (the one a different length from the others) to a single piece of rail that is always powered but changes phase when the switch is thrown (the point rails are often wired this way, but not always) with a resistor and a diode to block reverse voltage, and the other two legs to the two outside rails that are always powered.  Then at any given time one LED element will have both leads at the same phase, and not be lit, and the other will have both leads in opposite phase, and be lit 50% of the time (apparently lit continuously).  This obviously depends on the switch being constructed in such a way that there is a changing piece of rail and it's possible to wire your LED to it.

 

And because you have to connect the common lead like this, you can't work with a bit of track that's powered one way and unpowered the other, which rules out use with certain kinds of turnouts.  Using two separate LEDs would be much easier then, since all you need is a power-routing switch and it can work just like a DC system with two LEDs would. Just wire each LED, with resistor and reverse current protection, between a of rails on the frog end of the switch (one on the through line, one on the departing line).  When the turnout is thrown away from a pair of rails, that LED will either have one end disconnected, or both ends at the same phase, and either way will be unlit.  The other LED will have its ends at different phases, and be lit half the time (and appear continuously lit).

 

I'll note that I haven't tried this, and don't have experience with 3-leg LEDs, so there's a chance it won't work this way (at worst, you might short through the LED, which should trip the DCC circuit breaker in the command station, but not before frying the LED and potentially melting/igniting whatever it will touching at the time.  Experiment with caution.

[Guest Closed Account 1]

Here is the youtube article

 

 

"Uploaded by dont08901 on Mar 23, 2010 -

Using a 3 wire bi-color LED and 680 ohm resistor, you can detect the position of the SHS switch. The common is attached to the blue frog wire. The other leads go to the outer and inner rail current. Adjust the wire leads so the red LED comes on when the switch is thrown to the curved point rail"

 

My switches are manually controlled.

 

This site has LEDs pretty reasonable, at least the shipping is http://www.unique-leds.com/

[keitaro]

maybe be interesting a snippet from a japanese blog

 

Kato developing a new interior lights! 「LED室内灯クリアー」 "Clear Light Indoor LED"

 

•より均等に全面が光ります More evenly illuminate the entire surface

•プリズムが1㎜薄くなります The 1 mm thin prism

•新室内灯と互換あり With compatible with the new interior light

詳細・発売時期は発表があり次第ご報告いたします Details will be reported as soon as the release date announcement has

[quinntopia]

Thanks for all the great information on this thread!  I learned a few things about LED's that I wasn't aware of! Always impressed and enlightened by these sort of threads!