LED Bulb Replacements dying one by one.....

Just curious regarding how this will make any difference in the life of the bulbs - LEDs should last a long time unless they receive voltage in excess of what they're rated for.

The bulb (and switch machine) can be powered independently by using a plug in the side port that is connected to a constant voltage source.

BTW, I replaced mine with LEDs because the incandescent bulbs were too hot when I set the voltage high enough to throw the switch.  They've been there for a few years, but usage has been infrequent.

Well, if you run TMCC then 18vac is high. You can reduce the voltage for the switch guts and bulb. It takes a plug into tha side of the switch.

Here is a thread where it was suggested that the issue could be attributed to the high voltage transients introduced by the switch machine solenoids and that the typical TVS diode may be a good solution (and doesn't require modifying the switch machines!). I would place a 1.5KE33CA TVS in parallel with the bulb as close to it as you can. You could connect one end of it to the screw terminal on the AUX plug (the other end goes to your common ground/outside rail/center post on switch machine)

https://ogrforum.com/topic/leds-in-o22s

Hey @Mallard4468...

I use the 18V rated LED bulbs and the constant voltage port and run the switches at 18V already (my layout has 22 switches).

I saved all the original bulbs (and have tons more from passenger cars I converted to LED light strips) and plan to put the incandescent bulbs back in and add simple DC circuits to power the bulbs.

@bmoran4

Thanks....I missed that whole discussion. I'm aware of TV issues. I will definitively check that discussion out.

IMO, reverse bias is more likely to kill LED's that the transients.  The typical white LED has a reverse voltage specification of around 5V.  18 volts AC has a peak reverse voltage on the LED of over 25 volts.  Even considering the a dropping resistor, that continued over-spec reverse voltage takes it toll on the LED device.

I used to install tons of LED's with a resistor for locomotive headlights.  After a year or so, some of them would start failing.  Since we all know the LED's should last longer than most of us, I started looking into the issue.  Adding a blocking diode to the headlight installation has solved that problem, haven't lost one of those LED's since.

@gunrunnerjohn, what you say makes sense for discrete LEDs some install. Are you implying that the vendors out there marketing direct replacement LEDs with E10, BA9s and similar for use in AC environments at various AC voltages, including 18VAC, do not account for the reverse bias as they should? Are there (non-destructive) measurements an equipped and curious individual could make to determine this?

@bmoran4 posted:

@gunrunnerjohn, what you say makes sense for discrete LEDs some install. Are you implying that the vendors out there marketing direct replacement LEDs with E10, BA9s and similar for use in AC environments at various AC voltages, including 18VAC, do not account for the reverse bias as they should? Are there (non-destructive) measurements an equipped and curious individual could make to determine this?

I know that some of these bulbs have been disassembled to find only a resistor, others have had a diode as well.  Since both types have been observed in the wild, the answer is yes.  There was a discussion of this not long ago here.

I don't know offhand if you could non-destructively test for the presence of a diode.  I suppose you might be able to measure reverse leakage before the LED croaks, but that would be a science project.  Given the cost of a single bulb isn't that significant, it's easy to just rip one apart and see what they put in side.  If there's no reverse voltage protection diode, don't buy any more of that brand.

I may have found the thread that could be the one you mention: https://ogrforum.com/...cement-led-longevity

Very disappointing to see the one dissected product in that thread.

It would be nice to know which vendors have appropriate LEDs and which ones mismarket.

@bmoran4 posted:

It would be nice to know which vendors have appropriate LEDs and which ones mismarket.

I wouldn't hold my breath for the vendors to tell you that their product is a sub-standard design!

John: Did you not once post a link about some acceptable LED,s for the Post War Extruded Passengers cars and would they not also work in the 022 switches?

@RJT posted:

John: Did you not once post a link about some acceptable LED,s for the Post War Extruded Passengers cars and would they not also work in the 022 switches?

I may have, but I've posted a lot of stuff.   Finding it is another matter...

@Junior posted:

Hey @Mallard4468...

I use the 18V rated LED bulbs and the constant voltage port and run the switches at 18V already (my layout has 22 switches).

I saved all the original bulbs (and have tons more from passenger cars I converted to LED light strips) and plan to put the incandescent bulbs back in and add simple DC circuits to power the bulbs.

I'm a bit confused as to your reasoning.  Presumably you initially chose to pony up the cash for LED bulbs.  This was because you wanted to REDUCE POWER CONSUMPTION.  But the LED bulbs had a high failure rate.  So now you're going back to the incandescent bulbs?  IMO the high failure rate has been explained (correctly) by the inadequate design of the specific LED bulb you purchased.  But this can be remedied by a 5-cent diode.  So if you are re-wiring the switch anyway to allow an external DC source to power JUST the incandescent bulbs...

Why not just insert ONE 5-cent diode to power ALL your switches?  That is, if you've already cut the wire to separately power all the switch bulbs, a single 1-Amp diode can power dozens of LED bulbs...even the LED bulbs of inadequate design.

I may be missing something, but it seems the original objective was to lower power big time (5x to 10x power reduction) between LED and incandescent.  Why go back to incandescent bulbs?!

Sorry I've been gone so long everyone. Today was the day to take down the outside Christmas lights....just wrapped that up!

Okay....I believe I finally found what I was looking for....sort of.

Below is a screen shot of the process. But instead of leaving the bulb wire connected to the constant voltage post, it could be removed from the post, extended some what and connected to its own AC power source with the diode. That way the bulb was not being driven at 18+ volts...I could set it for what ever voltage I wanted. Maybe I'm way off base on  this approach...but it makes sense to me.....

Hey @Stan2004....

I bought the LEDs prior to the latest round of inflation so they were cheaper than what they cost now. And maybe its an issue of my being penny wise and pound foolish.....but in my mind, it would be cheaper to replace diodes for pennies (if I had to) vs. replacing LED bulbs for dollars. And I don't feel I'm getting the bulb life that I expected.

@stan2004 .....

Just reread your last post. I guess I'm suffering from a severe case of "horse blinders". I was so focused on the approach of making changes at a component level, I totally missed the possibility of approaching this project at a more global level .

I'm a bit confused on how best to approach this change at a layout level. Would you be willing to provide a little insight? Is it literally as simple as inserting a diode in-line on the dedicated power supply to the?

@Junior posted:

...

Okay....I believe I finally found what I was looking for....sort of.

Like you say, "sort of" .  What's described is a simple technique (a 5-cent diode) to reduce the voltage applied to an incandescent bulb.  The lamp is dimmer, the bulb lasts much longer, and operates much cooler so nothing melts.  See if these pictures help.

On the left is the original situation.  An incandescent bulb is not polarity sensitive and it "lights up" on both the positive and negative halves of the AC voltage.  A diode blocks half of the AC signal so only one polarity goes through; in the example above with the diode oriented as shown ("band" on the right) only the positive voltages get to the bulb.  So the bulb is powered essentially half of the time as before so it is dimmer and cooler.  BTW, in the pictures I only show the "hot" side of the circuit; the power supply common is assumed to be connected to the common of the bulb.

Your situation is slightly different.

You have an LED (instead of an incandescent bulb).  An LED is a Light-Emitting-Diode so it is a diode in itself.  Hence, unlike a bulb which lights up on both halves of the AC voltage, the LED only lights up on the positive halves of the applied voltage.  So the negative voltages do not contribute to the LED brightness.  And to add insult to injury the negative voltages do not play well with the LED and eventually destroy it.  So in your case, the external diode is providing the function of blocking the negative voltage from reaching the LED.  And since the LED only lights up on the positive pulses anyway, the lack of negative pulses does not affect the brightness.

The two cases are similar in many ways in that the external diode is extending the life of the light...but via slightly different electrical "mechanisms."

Let's take your situation at "face value."  That is, everyone agrees that there are LED lights out there that have the diode built-in ... but that is not the matter at hand.  But if you have extracted the "hot" wires to your "unprotected" LEDs then you could indeed bus them and apply a DC power supply so that the LEDs only see positive voltages.

I don't know how difficult it is to perform surgery on the switches to extract the "hot" wire to your LEDs.  That is not what I'm commenting on.  What I was saying is a single 5-cent diode (e.g., 1N4001, 1N4002, 1N4003, etc.) would allow you to use whatever AC voltage you had been using.

Hope this helps...

@stan2004 .....

Your explanation is absolutely awesome! As my Wife likes to say, "I always get lost in the corn flakes!" meaning I'm buried in the details (I was a programmer, she was a business analyst. I did the detail coding, she design business systems to replace/enhance business processes).

The last piece of the picture I never quite understood (especially in this case) is what about the negative side of the circuit?

An incandescent bulb gets its negative volts from the switch frame; which gets its negative volts from the outside rails (I detailed this for my own clarity).

When substituting the incandescent bulb with a LED, the LED now gets its positive from the half wave AC coming from the diode (in your example, the positive half). Would there be a "clash" between the half wave AC coming into the positive on the LED and the negative coming from the switch frame? Or does this become a non-issue because the positive side of the LED is only getting the positive half wave AC from the diode?

Stan, this is one of the most straight forward explanations of the need for a diode with AC that I’ve seen.  I’ve seen writeups but not with simplified diagrams that assist in understanding.  Thanks!

Bob

@Junior posted:

...

The last piece of the picture I never quite understood (especially in this case) is what about the negative side of the circuit?

An incandescent bulb gets its negative volts from the switch frame; which gets its negative volts from the outside rails (I detailed this for my own clarity).

When substituting the incandescent bulb with a LED, the LED now gets its positive from the half wave AC coming from the diode (in your example, the positive half). Would there be a "clash" between the half wave AC coming into the positive on the LED and the negative coming from the switch frame? .

...

I think I understand what you're asking but maybe not.  By "negative side" of the circuit I believe you're referring to the lamp connection to the switch frame which is hooked up to the outer-rail or train transformer common.

I like to think of this common connection as the reference level or in measurement terms the "zero" of the circuit. In the graphical representation of the AC signal, the common or zero is represented by the horizontal center line.

Perhaps you're thinking that if you look at a battery (a DC device), its two terminals will be labeled "+" and "-".  Or if you look at an LED  (a DC device), its two terminals might be labeled "+" and "-".  Or many polarized capacitors (a DC device) will have one of its terminals labeled "-" with the implication that the other terminal is "+".  In other words, lots of "-" of negative symbols with DC devices rather than "common" or "0". 

One way to think of it is for DC devices, during proper operation the "+" terminal should be more positive than the other terminal...or equivalently the "-" terminal should be more negative than the other terminal.  This cannot be said for AC devices since the polarity alternates so you can't really call one terminal +/positive and the other -/negative.

So getting back to the matter at hand, the external diode only passes positive voltage pulses to the LED
"+" terminal, so this terminal is always more positive than the LED "-" terminal (case). 

Hopefully this does not add to the confusion!

@gunrunnerjohn posted:

IMO, reverse bias is more likely to kill LED's that the transients.  The typical white LED has a reverse voltage specification of around 5V.  18 volts AC has a peak reverse voltage on the LED of over 25 volts.  Even considering the a dropping resistor, that continued over-spec reverse voltage takes it toll on the LED device.

I used to install tons of LED's with a resistor for locomotive headlights.  After a year or so, some of them would start failing.  Since we all know the LED's should last longer than most of us, I started looking into the issue.  Adding a blocking diode to the headlight installation has solved that problem, haven't lost one of those LED's since.

John,

I wish I had known about this issue sooner, l had my street lights die one after another. I even had some of my switch indicators pop and shatter. I replaced the LEDs in the street lights and have now powered them through a buck converter. Will need to look at replacing the LEDs at the turnout and use a blocking diode there.

Thanks now for that information.

Ray

@Rayin"S" posted:

I wish I had known about this issue sooner,

It's what happens when you read the spec sheets for the components.   Truthfully, I never used to give it a thought until I started losing LED's, then I decided to find out why.

To me it is a no-brainer and I just figured all were engineered to account for that because otherwise is just ludicrous. I guess I'm going to have to secret shop a few vendors and perform some dissections to prove which vendors have the goods.

@gunrunnerjohn posted:

It's what happens when you read the spec sheets for the components.   Truthfully, I never used to give it a thought until I started losing LED's, then I decided to find out why.

I suspect/expect that many of the products sold to O gaugers are really designed for the automotive and amusement markets, so DC operation and correct polarity can be assumed due to the prevalence of negative ground and bayonet sockets with grounded shells. But if you look at listings for "reversible" bulbs, like the 194 and its variants, the suppliers talk about "polarity bias." If you insert the bulb the wrong way it won't work.

Like everything else, O gaugers are one of the very few user of low voltage AC, so there is very little that is "purpose built," Just like digital panel meters suitable for our use, to give a single example.

Yep, one of the reasons why I've done a lot of little modules that are purpose-built for O-scale AC voltage.  Add in the issues of DCS signal attenuation, and you'll find no "purpose built" stuff, at least I haven't found any.

@stan2004

Were you a teacher at some point in your life? Your explanations and diagrams are simply the best!

Sorry about confusing negative with ground. It's easy to forget that AC and DC are two different animals.

"By "negative side" of the circuit I believe you're referring to the lamp connection to the switch frame which is hooked up to the outer-rail or train transformer common." Yes.

"So getting back to the matter at hand, the external diode only passes positive voltage pulses to the LED
"+" terminal, so this terminal is always more positive than the LED "-" terminal (case)." This is where I'm not getting it. I understand what the diode is doing; blocking the negative pulses from the AC coming in. But in this case, what completes the circuit to the LED; allowing it to light up? Is it the connection to the switches Common? Does this "work" because the diode is blocking the negative pulses on the positive side of the LED; which allows only negative pulses from Common on the negative side of the LED?

@Junior posted:

@stan2004

Were you a teacher at some point in your life? Your explanations and diagrams are simply the best!

I wonder that myself, Stan's explanations are some of the most lucid that I've ever seen.

@Junior posted:

...

Sorry about confusing negative with ground. It's easy to forget that AC and DC are two different animals.

"By "negative side" of the circuit I believe you're referring to the lamp connection to the switch frame which is hooked up to the outer-rail or train transformer common." Yes.

"So getting back to the matter at hand, the external diode only passes positive voltage pulses to the LED
"+" terminal, so this terminal is always more positive than the LED "-" terminal (case)." This is where I'm not getting it. I understand what the diode is doing; blocking the negative pulses from the AC coming in. But in this case, what completes the circuit to the LED; allowing it to light up? Is it the connection to the switches Common? Does this "work" because the diode is blocking the negative pulses on the positive side of the LED; which allows only negative pulses from Common on the negative side of the LED?

Never a teacher...but of the school-of-thought that a picture is worth a thousand words.

Right. It is the connection from the LED case to common that completes the circuit allowing the train transformer AC to power the LED.  And the diode insures that the LED "+" terminal only sees positive pulses.   A positive pulse or voltage is "positive" with respect to, or relative to, a reference.  In other words the voltage represents the difference between two points in the circuit; the difference can be positive or negative.  In this case that reference point is the common bus or outer-rail.  It is expedient for analysis purposes to think of the common as being 0 Volts or the horizontal line on the graph. 

It is indeed confusing that the reference connection might be called "common" or "ground" or "neutral" or "return" or "black" or "zero" or "outer-rail".   

Keep asking questions!

BTW. I have been consciously avoiding "current" or Amps in the discussion.  Voltage and current are as thick-as-thieves as it gets!  But I believe the issue at hand (your dying LED bulbs) can be satisfactorily explained using voltage only. 

@stan2004 ....

I GET IT! LET THE CELEBRATIONS BEGIN!

Thanks for being so patient with me. It's  discussions like this that make this forum so valuable!

Thought an addition to Stan's picture might help, but just saw your last reply.

Dave

@Dtrainmaster

Thanks Dave....it's still a great graphic! I like where you gave credit to @stan2004 for his original graphic and noted your modification to it....nice touch.

Well....after reviewing @stan2004's great tutorial and pulling the first switch motor apart, I decided to take a different approach.

I was going to run a separate wire to the diode connect to the LED and use Common available from the outer portion of the light socket. But the base of the switch motor is pretty stout steel.🤯 Plus finding a location where I could drill a hole through the base without tearing the whole switch motor apart was going to be a challenge at best (the whole thing is assembled with rivets).

Soooooo.....I decided to add a 1K ohm resistor to the negative side of the LED and add the diode to the positive side of the LED as was suggested by everyone:

After some "fiddling", I was able to shape the leads (the final position is shown in the above pic) so that the LED/diode/resistor could be fit into the light socket itself!

Using a soldering pencil with a very fine, long tip and some flux, I soldered the resistor lead to the outside (Common) of the light socket and the clipped lead of the diode to the spring inside the socket....and.....

Viola! An LED replacement that handles 20VAC (which is what I run my switches at. These are Lionel 072 switches. The throw mechanics are pretty sloppy-loose) for literally pennies on the dollar. And the switch lantern fits great and easily swivels:

So what if I need to undo this change and go back to an incandescent bulb? The solder joints used very little solder and almost come apart themselves once heat is applied to to the solder joints.

But I couldn't have done this without everyone's help and support explaining basic electricity and how AC and DC can coexist.

I'll bet some of you guys still have old-school lighted doorbell buttons:

Curiously enough the incandescent bulbs in these run off the doorbell transformer which puts out something like 16V AC.  And ironically, like the lanterns in our switches, when you push the button, the light goes out until you release it.

Sometime in the last century I'd notice the incandescent bulb in my doorbell button would burn out every year or so and I'd dutifully go to Home Depot to buy a new lighted button for a couple bucks.  It only took a few minutes to change the button so never thought much of it.  But then I looked around my neighborhood and lo and behold all my neighbors too had burned out doorbell buttons.  Everyone just shrugged it off since the doorbell itself worked fine.

So. I disassembled the button and inserted a diode-resistor-LED combo as discussed in this thread.  It was tight-quarters soldering just like the switch lanterns...but it was the principle of the matter.  The button has worked 24/7 for decades now!  I see now that doorbell buttons at Home Depot are LED (finally!) and of course the latest thing are the video doorbells.

I never put 2 and 2 together that this is the same problem until seeing @Junior's DIY persistence to understand the problem and effect a solution. 

Thanks @stan2004. This just proves that what's old is new.....again.