How to repeat PL's on the facia without breaking the bank

[peteski]

1.5V?!  Are you sure?

Red LEDs have Vf of around 1.7V and that is the lowest Vf of all the visible light LEDs, Yellow and green have Vf of around 2V, and blue, white and true-green are even higher (around 3V).

With 1.5V the red LED should be just barley lit.  Do you have a schematic diagram of that controller circuit?  Or you coudl send me one for a bit of a reverse-engineering. There could be a way to easily modify the circuit to produce a higher voltage/current on the output.

OOPS!  I promised earlier on not to post to this thread anymore. 

[eric220]

Unfortunately, I do not have a circuit diagram of the controller card.  I really don't want to modify it (to the extent of even adding the "override" resistors if I can get away with it) because there are going to be a LOT of these on the layout, and modification means a significant time investment.

The "override" resistor is wired into the card in provided holes.  It appears to be parallel to the existing resistor, which is a surface mount that says "202" on it.  The instructions caution against using anything less than a 100 ohm resistor, and as you can see, the single LEDs in the cab signals are pretty bright at 100 ohms.  I actually like the brightness of the LEDs when they are lit with the default resistor.  The 100 ohm is too bright, IMHO.

I've got the control wires hooked up to the cab signals, and from there the triple LEDs of the PLs are wired in parallel, thusly:



PL on top, cab signal on the bottom.  I know the PL's are wired screwy.  That was a mistake during installation (they were intended to be in series), and one that I'm not interested in going back and correcting for the purpose of this test unless I have to.

Either the cab signal or the PL will light, but not both.  If I wiggle the wiring around, the cab signal will go out and the PL will light, but only the center lamp (I'd guess the center-lamp-only thing is due to the parallel-series wiring of the PL).  If I disconnect the control wire from the cab signal, the PL lights up fine. Both continue to show 1.5V across the leads.  One of the things that I don't really understand (I'll admit to knowing just enough about electronics to be dangerous) is that the brighter aspect and the dimmer ones all show the same 1.5V. 

[mmagliaro]

Well, right off you have a problem with the way the PL signals are wired.  You have two in parallel, and then that parallel pair is in series with a 3rd LED.   The way you drew that, you will need double the forward voltage across the PL LEDs to light them up, so you will need more like 3.4v.  And Peteski is right... are you sure that the output from that signal board is 1.5 volts? 

Why are those 3 at the top wired like that?  Why aren't they just 3 LEDs in parallel?

I *think* I understand what's going on with the bypass resistor now.  They have you wire it in parallel with an existing one on the board,
so the reciprocal/parallel combination of the two will end up producing a lower overall resistance and will increase the available
current.  But I don't think that's your problem here.

What RRCir-Kit board is this that you are using?  I can look those up on line and even download their manual so I can see what's going on here.

[eric220]

Max - Keep in mind that this is only one aspect. Is it possible to wire the PL's in parallel such that the center lamp lights with all three aspects? The recommended wiring is all three in series, and the card is supposed to put out enough power to light the three in series. Indeed, when I disconnect the cab signal LED, the PLs do light up.

The card is a RR Cir-Kits 4ASD-4. http://www.rr-cirkits.com/manuals/4asd-manual.pdf

I've checked and rechecked, and the multimeter consistently shows ~1.7 volts DC across the LED leads.

[peteski]

Unfortunately, I do not have a circuit diagram of the controller card.  I really don't want to modify it (to the extent of even adding the "override" resistors if I can get away with it) because there are going to be a LOT of these on the layout, and modification means a significant time investment.

The "override" resistor is wired into the card in provided holes.  It appears to be parallel to the existing resistor, which is a surface mount that says "202" on it.  The instructions caution against using anything less than a 100 ohm resistor, and as you can see, the single LEDs in the cab signals are pretty bright at 100 ohms.  I actually like the brightness of the LEDs when they are lit with the default resistor.  The 100 ohm is too bright, IMHO.

I've got the control wires hooked up to the cab signals, and from there the triple LEDs of the PLs are wired in parallel, thusly:



PL on top, cab signal on the bottom.  I know the PL's are wired screwy.  That was a mistake during installation (they were intended to be in series), and one that I'm not interested in going back and correcting for the purpose of this test unless I have to.

You will have to re-wire your LEDs  The way you have hooked them up will not work (even if these were regular light bulbs, not LEDs).  You cannot mix serial/parallel connections like that and expect all the LEDs/bulbs to have the same brightness.

Let's number the LEDs by their vertical location, from top to bottom: 1-4.

LED 1 and 3 are hooked in parallel, then that parallel pair is hooked up in series with LED 3.  I'll cover LED 4 later.  Since LEDs need around 1.7V to pass enough current to light them up (at their standard brightness).  You would need 3.4V for LED 1 through 3 to light up.  Then, no matter what voltage/current you apply to that circuit, LED 2 will always be brighter. Why? Because LED brightness is related to the current passing through it.  Since the LED 1 and 3 are in parallel, each of those LEDs will pass half the current going through LED 2.  No ifs or buts about it.

Now, adding LED 4 to the circuit. This LED is wired in parallel with the LED circuit of LED 1-3 I described above.  Since LED 4 needs 1.7V to conduct current (to light it up), and the LED 1-3 circuit needs 3.4V, LED 4 will be brightly lit and 1-3 will not light up.

If you were to place LED 4 in parallel with LED 2 (like LED 1 and 3 are connected) then the cirut woudl work (still needing 3.4V). Assuming that all the LEDs are identical parts, the current would be evenly distributed between those 4 LEDs. It is not the best way to power LEDs, but it would work.

The 202 resistor you mention is a 2000 ohms resistor.  I suspect that the output circuit supplies more than 1.5V, but since you are measuring it with an LED attached, you are simply measuring the Vf of the LED.  Try to measure the voltage of an output which is turned on with no load attached.  That will give you a good idea of the maximum voltage the circuit can supply.

[eric220]

Peteski - LED 2 is in fact brighter when lit, and I knew that would be the case . I'm willing to live with that in a testing/proof of concept environment. The math regarding the PLs (LEDs 1-3) makes sense, but it falls apart with the fact that when the cab signal is disconnected, they light up. What I don't get is why the PLs light up when disconnected from the cab signals (LED 4) but not when hooked up in parallel.

I'll check the voltage of an unused output later tonight.

[peteski]

The math regarding the PLs (LEDs 1-3) makes sense, but it falls apart with the fact that when the cab signal is disconnected, they light up. What I don't get is why the PLs light up when disconnected from the cab signals (LED 4) but not when hooked up in parallel.

I'll check the voltage of an unused output later tonight.

It makes perfect sense. LED 4 basically shunts down the available voltage to 1.7V (or whatever the Vf of that LED is.)  That is because it is in parallel with the PL signal circuit (which has LED 1-3 hooked up in a combination series/parallel way). The output circuit supplies a voltage through a resistor (wither the 2000 ohm or little less than whatever resistor you add in parallel with the 2000 ohm one).  I suspect that the open circuit output voltage is much greater than 1.7V.   If you were to measure the voltage across the output when only the PL signal is attached, it will most likely be close to 3.7V

EDIT:
If that circuit can supply voltage greater than 5.1V then you should be able to hook up LED 4 in series with your PL signal and they will all light up.  How brightly will depend on the maximum available voltage and on the value of the resistor you install in the circuit (paralleling the 2000 ohm resistor).

2nd EDIT:  According to the manual: This means that the signal LEDs have up to 9V to allow for
reliable series operation of up to 3 LEDs of any color.

So, there you the answer - open output voltage should be close to 9V.  Assuming that the Vf of all the LEDs you use is around 1.7V (and the current of 10mA (0.01A) will make them bright enough for you, then (after hooking up the repeater LED in series with the PL LEDs, we can figure out the value of the resistor you need.

3 x 1.7V=5.1V  That is the voltage needed for the LEDs to operate.  The 9V is the voltage supplied by the circuit (ahead of the 2000 ohm resistor).  So 9V - 5.1V = 3.9V  That will be a voltage across the current limiting resistor you will install.  So, 3.9V / 0.01A = 390 ohm. Actually, with the 2000 ohm resistor still in the circuit (in parallel with the 390 ohm resistor you would add), the LED current will be slightly over 10mA (more like 12mA), but none of this is exact (since I don't have the specs for your LEDs).  But that gives you a rough idea what to expect.

[mmagliaro]

Peteski has pretty much explained all the issues here.
I would just clarify, that in general, you need to remember that the LEDs are diodes.  They do not behave like resistance
loads.  There will always be 1.7v across an LED (well, a typical red one, anyway), no matter what the supply output voltage is.
By definition, an LED drops exactly its forward voltage, no matter what (until you overload it and it fries). 

That's why it will fry if you don't limit the current with a resistor.   

In other words, there will always be 1.7v across a single LED, and always 3.4 across two in series, no matter what else is going on
in the circuit.  (There are boundary cases, such as when the supply voltage is too low, or the current limiting is too severe to fully turn on the diode, but in general, this fixed limit can be thought of as always true).  So if you have two in series, and then
one in parallel with that pair, there will be 1.7 volts across the whole thing (because the single LED will force the voltage to be 1.7), and that will not be enough to light the two in series.

[eric220]

So quick fix, I should put LED 4 in series with the PLs?

[peteski]

So quick fix, I should put LED 4 in series with the PLs?

Yes, plus add the resistor I calculated in parallel with the on-board 2000 ohm one to increase the overall brightness. Unless you'll be happy with the brightness using the 2000 ohm resistor.

At least that is what I see from reading the manual and looking at your signal setup.

[eric220]

Shockingly (electrical pun fully intended) when you listen to people who know what they're doing around electronics:



I also wired up the approach aspect on the 1W lower arm, but there's no extra resistor on that aspect, so its so dim that it's barely visible.  The amber LEDs in the PL there are intrinsically about 1/10 the brightness of the LEDs lighting the cab signals.  If I go with the PL version, I'm going to have to find a different LED supplier.  In the meantime, I've got a lot of work ahead of me to make this thing work.

[peteski]

Eric,
if the repeater uses white LEDs (and PLs use yellow) then in general, white or blue LEDs (there really is a blue LED chip inside those white LEDs) are much more efficient in generating light than yellow LEDs.  So per given number of milliamps, white LED will have much brighter glow. I don't think you'll be able to find a white LED which is as inefficient as a yellow LED (for matching brightness). Since all the devices in a series-connected circuit conduct the same current (by law  ), that is why your repeater glows brighter than the PL signals.  There could be workarounds, but that would increase the component count and complexity.

A simplest workaround would be to add a resistor in parallel with the repeater LED. That way, part of the current at the repeater LED would be diverted through the resistor, making the repeater LED dimmer.

A simplest solution would be to use a yellow LED for the repeaters.

If you are still considering using a PL style of repeaters, then you will need to redesign your circuit because the output will not be able to supply high enough voltage to light up 6 series-connected LEDs (3 in the PL on the layout and 3 in the PL repeater.

BTW, if you are using white LED in the repeater then my earlier calculation for the resistor is off. The value as I calculated is too high (because white LED Vf is close to 3V instead of 1.7V).  No damage done - it simply means that the LEDs will have less current passing through them.

You said that you have a lot of work ahead of you. I agree. But IMO, part of the problem is your inflexibility for redesigning parts of this circuit which aren't optimal (but would require either more components, or re-wiring).

[eric220]

Eric,
if the repeater uses white LEDs (and PLs use yellow) then in general, white or blue LEDs (there really is a blue LED chip inside those white LEDs) are much more efficient in generating light than yellow LEDs.  So per given number of milliamps, white LED will have much brighter glow. I don't think you'll be able to find a white LED which is as inefficient as a yellow LED (for matching brightness). Since all the devices in a series-connected circuit conduct the same current (by law  ), that is why your repeater glows brighter than the PL signals.  There could be workarounds, but that would increase the component count and complexity.

I'm not too worried about that. The final implementation will be one or the other. The main reason I need to find another LED supplier is the physical shape of the LEDs I got are not conducive to this application.

If you are still considering using a PL style of repeaters, then you will need to redesign your circuit because the output will not be able to supply high enough voltage to light up 6 series-connected LEDs (3 in the PL on the layout and 3 in the PL repeater.

This is definitely a concern.

You said that you have a lot of work ahead of you. I agree. But IMO, part of the problem is your inflexibility for redesigning parts of this circuit which aren't optimal (but would require either more components, or re-wiring).

Two points. One: I am going to go back and rewire at least one of the PLs so that we can compare them to the cab signals (hence having a lot of work ahead of me). Two: As I've said several times, this is a proof of concept device; I don't need it to be perfect, just functional enough to explore the concept.

[eric220]

Well, after five moths of inactivity, I finally have some progress to report.  The amount of work that is going to be required to make the PL's function correctly has pretty much kept me from going anywhere near this project.  So, rather than getting my ingenious little device working, I have begun exploring the other idea that I had in regards to facia repeaters.  $50 and about 5 minutes with JMRI later,



This is using the stock JMRI images for the signal heads, and again, this is a 5 minute throw-together to see how it works.  In practice, it works amazingly well.  The signals are clearly visible from across the room.  This has the advantage over the hardware version that for all intents and purposes, it is infinitely adjustable, and that it can be configured with inputs and outputs at will.  Obviously, to do the entire layout with tablets and software will be quite a bit more expensive than the hardware version.  That said, the opportunity cost is greatly reduced, since the software version can be brought up and running in a fraction of the time.  If I do it one tower at a time, it shouldn't be cost prohibitive.

Thoughts?

[C855B]

I dunno about the JMRI version being all that expensive. Pay attention to the surplus "clearance buys" section of retailer ad flyers. I've seen - and bought - "Brand X" Android tablets for around $25 on clearance that would be considered junk or toys if you tried to use them as web browsers or personal organizers. But as single-purpose interface panels or throttles they work fine.

One trick to know however... not all of these cheap Brand X pads or padlets are Google Play-friendly. Google will apparently license Android to these junk manufacturers doing one-shot products, but then demand another licensing ransom for Google Play access, which they often do not get. If you can download the .apk app file someplace other than Google Play, you're good... if it's available only on Google Play, then you have to cruise for workarounds.