Hello ! Help needed regarding model railway lighting ?.

[Chris Uddin]

Mark F you are an absolute star my friend thank you very much indeed, I think I already have these resistors and the only thing I don't have is the diode but funnily enough the picture I was about to show on here before you messaged me was of the one you said not to get so thanks for that !, is there still no point in putting a long strip of 18 the full length of the coach and doing what you have said ?, I know it means using more led's but it also means that I don't have to cut 3 strips of 3 and joins each of them together by means of soldered wire ??. Chris.

[MarkF]

You could use a 1N4001 for testing if that's all you have.  Or any other rectifying diode.  I just wouldn't use it as a permanent solution.

You still might want to use a strip of LEDs the full length of the coach in order to have uniform lighting.  However, some darker spots might make it look realistic (burnt out lights in the coach or selections of the coach without lights).

If you really want to get fancy, you could put a DCC decoder (similar to waht's in the locomotive) in the coach.  Then you could turn ON different areas during operation.  Or control dimming if you connected the LEDs instead of a motor.

[Chris Uddin]

So just before I go and turn in for the night because here in the UK it's 02:10 p.m. Sunday morning and our clocks went Forward by one hour last weekend so we've lost an hour's sleep here, so can I ask Mark am I right in thinking the higher the number on the diode the dimmer the light will be ??.

[MarkF]

So just before I go and turn in for the night because here in the UK it's 02:10 p.m. Sunday morning and our clocks went Forward by one hour last weekend so we've lost an hour's sleep here, so can I ask Mark am I right in thinking the higher the number on the diode the dimmer the light will be ??.

Yes if I understand your question.  The more groups of three you have the higher the current demand will be.  With the same feed resistor (i.e. the 1KΩ for example) the higher the voltage drop across said resistor.  Leaving less current for each LED group.

This is all trial-and-error since LED brightness varies greatly among different LED types.

[Brumby]

... am I right in thinking the higher the number on the diode the dimmer the light will be ??.

No.  The difference between 1N4001, 1N4004, 1N4007, etc. is the reverse voltage they can tolerate.  That is, the voltage they can block (without exceeding their design capability) when the polarity is opposite of what you need.

For example, I have a bunch of 1N4004 on hand because they can handle up to 400V - and they're cheap as chips.  A 1N4007 can handle up to 1000V.

The brightness will be affected by the forward voltage drop - which, for these diodes, will be pretty much the same.  So, there will be no brightness differences.

Usually, you select a diode that has a lot of headroom.  For your application, the 1N4001 at 50V is a bit close to the voltages you are working with and while it could probably work quite satisfactorily, it is safer to go with something higher in voltage.  That's why I have 1N4004's as part of my "jellybean" stash.

[MarkF]

So just before I go and turn in for the night because here in the UK it's 02:10 p.m. Sunday morning and our clocks went Forward by one hour last weekend so we've lost an hour's sleep here, so can I ask Mark am I right in thinking the higher the number on the diode the dimmer the light will be ??.

Yes if I understand your question.  The more groups of three you have the higher the current demand will be.  With the same feed resistor (i.e. the 1KΩ for example) the higher the voltage drop across said resistor.  Leaving less current for each LED group.

This is all trial-and-error since LED brightness varies greatly among different LED types.

Sorry.  Are you asking about multiple diodes or multiple LEDs?

1N4000 Series diode specification:  (Get the cheapest you can find that's not a 1N4001)

[Chris Uddin]

Mark F - you have been a great help and I really appreciate the time that you've taken to try and guide me through what are to me big hurdles ?, so I will endeavour to try what you and others have suggested, I will also try take pictures as I go so that you can see how I'm doing and I look forward to being in contact with you again soon once I've started that process, so once again my friend thank you very much, I've really enjoyed chatting with you and the other helpers and I look forward to chatting with you again soon and if there is anything else that you can think of or if I have problems in the meantime I'm sure I won't hesitate to ask and I'm sure you won't hesitate to contact me, have a good day till we speak again, kind regards, Chris.

P.S. Thank you for clarifying that and in relation to the last question I was talking about using one diode and one resistor per coach with one long length of LED lights which I said before was 18, the full length of the coach but it is probably better as these are close together to do 3 blocks of 3 with one diode and one resistor per coach and the diode I will probably choose would be the 4004 on your recommendation, I was thinking of buying a selection of these diodes just so that I can see what each one does and then pick the one that suits best but the brightness is not down to the the diode as you've already said but down to the resistor and I have loads of the resistors that you pictured so happy days.  I just have to learn a little bit about resistors now ?, Chris.

[Chris Uddin]

Here is an ebay link - would these be any good to me ?, are they a better way of using a diode ??


https://www.ebay.co.uk/itm/232772146305?hash=item36324d1481:g:w6QAAOSwCRRa~Xi~

[MarkF]

Buying a selection of the 1N4000 Series is not very useful.  The 1N4004 are good up to 400V which would cover 90% of any place you might want to use a diode.  All the diodes in the series will react almost the same.

If you really want to have different samples, a better selection of general purpose diodes would be:
  - 1N4000 Series @ 1A
  - 1N5390 Series @ 1.5A
  - 1N5400 Series @ 3A

Three different current ratings.  Pick one in the middle of the voltage range (200V to 600V).  Cheapest you can find.
The only real difference in each series is its reverse voltage capability.

[MarkF]

Here is an ebay link - would these be any good to me ?, are they a better way of using a diode ??


https://www.ebay.co.uk/itm/232772146305?hash=item36324d1481:g:w6QAAOSwCRRa~Xi~

Those are surface mount.  I would stay away unless you plan on building circuit boards.
Surface mount parts are hard to structurally mount unless you're putting them on a board.

Diodes with wires (i.e. through-hole) are a better for general purpose usage.

Don't buy from eBay.  Get them from Digikey or Mouser or Farnell.
Too many counterfeit parts on eBay, Amazon and others.

[MarkF]

Model Railroading is in my blood too.
You might be interested in a recent topic of mine:
   https://www.eevblog.com/forum/projects/any-old-time-ho-model-railroaders-out-there/msg3996535/#msg3996535

Also, I've been following the Chadwick Model Railway.  He has lots of good tips.
   https://www.youtube.com/channel/UCMjSsrcmA4BNeAnh2NxYp0Q/videos

[Chris Uddin]

Yes I watch Chadwick model railway too, I'm glad to hear that you're also into model railways so that's brilliant news. Ok so I'll stay away from eBay, here is a link to information regarding the the DCC unit I am using and I think I may have to use a resistor with a higher amp output ?, Would I be correct ?, I'll let you decide Mark, cheers myfriend. Chris.
https://www.bachmanntrains.com/home-usa/dwg/dwgs/Introduction_to_Dynamis_DCC_System.pdf

[MarkF]

You have to start somewhere. Just pick a value and see what you get with the 18 LEDs if that's wbat you settled on.

If you have a DC power supply, you can set it to 14V and try resistor values with you LED strip. Noting that the diode will drop an additional 0.5V when you get one. Causing the LEDs to be slightly dimmer.  It will get you in the ballpark.

[sw2022]

Mark F - What would a bridge rectifier do then? ( small square black box with 4 pins on it ), is that something I would need or use in this case,? Is this the same as a variable resistor ??.

It straightens out the track polarity such that the LED circuit always has the correct polarity regardless of the track polarity.  Connect the AC side to the track pickups and the DC side to the LED driver.  I'm not sure this is necessary on a DDC system.  Still, it protects the driver circuitry.

You could use a bridge rectifier.  But on a DCC system a single diode will do just as well.  And be much cheaper.

Edit:
On a DCC system the frequency of the track power is much higher than 60Hz.  So, losing the negative portion doesn't have that much of an impact.  And since DCC is positive and negative, it doesn't matter how you connect to the rails.

I suspect the bridge rectifier idea, as opposed to using a cheaper diode, came about because of the need to accept different polarities from the two pickups either end of the coach here? If you always know which pickup input is +V then a single diode is fine. If the inputs can change around then you would need to feed the LEDs via a bridge rectifier connected to both pickup pairs to get correct polarity for the LEDs

Steve

[Chris Uddin]

Please refer to the pictures  -  I've just put the light strip inside the roof for the coach just to see how bright it is against the dark tinted windows, what I'm using to power it is a combi controller and it's just wired crudely but firmly, now when I put the Direction controller one way it won't light up but when I change direction the other way and turn the power up it starts to get brighter and the more I turn the dial the brighter it gets until it's turned up full and this is without using a resistor, after about 10-15 secs on full power something happens and the strip goes dim, can you tell me what's happening, is something about to burn out because I'm a little unsure.? Personally I would like it at that full brightness or if not a little less bright,  I think that having this many led's within the length of the coach looks fine, it gives an even light spray with no dark spots but at this stage I'd be happy for it to just be constant, so I wondered if there's any thought as to what might be happening, thanks, Chris.

P.S. The resistors I've used have been pictured in an earlier post above and I found these to be too dim, in fact they're way too dim so if I need to use a resistor I need something that is going to make the light strip just as bright as shown in the picture. ?

[Chris Uddin]

pictures

[MarkF]

I was assuming the windows were clear.  A very dark tint will require a much brighter light.

I looked up the Combi controller.  It's controlled output is 0V to 12V up to 0.75A.  It should be enough to drive the 18 LED strip.

• Placing the direction switch in reverse may destroy the LEDs and the reason for the diode when using DCC.
  
  
• The DCC power after the diode will be at least 14V.  Two volts higher than the 12V LED strip.  Hence, the need for the resistor.
  
  
• I'm not sure why the LEDs dim after a few seconds.  Do you have a meter?  Can you measure the output of the Combi while lighting the LEDs?  Is the voltage dropping?  It could be the Combi is overheating which would mean the LEDs are pulling much more power than I suspect.
  
  
• You could try just driving one 3 LED group to see if it's an overheating issue.
  

[MarkF]

Mark F - What would a bridge rectifier do then? ( small square black box with 4 pins on it ), is that something I would need or use in this case,? Is this the same as a variable resistor ??.

It straightens out the track polarity such that the LED circuit always has the correct polarity regardless of the track polarity.  Connect the AC side to the track pickups and the DC side to the LED driver.  I'm not sure this is necessary on a DDC system.  Still, it protects the driver circuitry.

You could use a bridge rectifier.  But on a DCC system a single diode will do just as well.  And be much cheaper.

Edit:
On a DCC system the frequency of the track power is much higher than 60Hz.  So, losing the negative portion doesn't have that much of an impact.  And since DCC is positive and negative, it doesn't matter how you connect to the rails.

I suspect the bridge rectifier idea, as opposed to using a cheaper diode, came about because of the need to accept different polarities from the two pickups either end of the coach here? If you always know which pickup input is +V then a single diode is fine. If the inputs can change around then you would need to feed the LEDs via a bridge rectifier connected to both pickup pairs to get correct polarity for the LEDs

Steve

No.  It's to accept different polarities of the track.  Not the pickups. 

The bridge rectifier idea comes from the old days when the track was powered with DC.  The track polarity is swapped in order for the train to go in reverse.  The bridge provides the same polarity no matter which direction the train is running.

[RJSV]

   That controller has no 'deliberate' explicit current limit setting, right?  Because one of the mistakes (I've ) made, on the test bench, is to apply voltage, straight to my opto-isolater device (LED input).
   I had hard time understanding, that LED voltage is a kind of 'reaction' measurable if you've got a multimeter across one LED while you do the supply increase.
You need to apply an adjustable CURRENT, that's what your main resistor is limiting. By adjustable, I mean by swapping that main resistor (approx 1 K to 3 K or more).
   It's a hard 'rule' to respect, as I'm thinking, why not just 'turn up' the voltage, until light starts glowing.
I was...actually, waiting for your narration to describe a
 "a horrific snap sound, with a TON of black smoke..."!
   
   Also, have heard about same applied ' current Vs voltage' issue with charging little batteries; must use 'current', not voltage.
To do that, just include your 'main' voltage dropping resistor (2k to 4 k), and THEN play with turning up the controller.
Sorry, if you getting maybe too much advice, from all directions, but that's the process. I'd stick with markf giving the best advice, probably because has model RR experience.
   

[rstofer]

The common LED datasheet has two numbers you care about:  V_f (forward voltage drop) at some nominal (recommended) operating current and the associated I_f (forward current that will produce that voltage drop).

To get maximum recommended output, both numbers must be met.  More often than not, I_f should be scaled back to limit brightness.  The common red LED is rated for 20 mA but is entirely adequate at 10 mA and perhaps even 5 mA.

So, E=I*R (Ohm's Law) and R=E/I but E, in this case, is the voltage that the resistor must drop at the rated current for a given Vcc

R = (V_cc - V_f) / I_f

It's just that simple.  Note how V_cc - V_f is exactly the voltage across the resistor and when added to V_f gives V_cc.

[RJSV]

I might add to that:
   Let's say, you've got 2.67 volts and 12 mA.
Now, if you swap in a very slightly higher resistor, you might see 2.65 volts. That curve is going to be very non-linear. Might mean, that the operation of the LED itself, causes slight heating, moves the place on that curve.
Maybe the voltage goes to 2.66 v. Now what ? You adjust again? Looking at your 'current' meter reading, you adjust the voltage fine adjust knob, to get back to 10 mA, your desired place.  A couple minutes ago by...now things have drifted, again.  Maybe you accidently set a small amount too high, 2.9 volts.  Led is starting to go over spec, overheats.  Doesn't sound that extreme, you 'only' moved voltage up a little. But the current increased, as that curve is, well, non-linear.
   By setting the current (maximum in your case), you have a maximum, with that extra resistor in there.
Otherwise, you can reach a point where a little off on voltage can cause current to be way off.  Plus, this is with the DCC being pulsed, it's an average rather than straight dc.
   Confusing, yeah.

[MarkF]

The 12V LED strip that I have uses 0.120A with 18 LEDs.
The strip has 180Ω resistors in each group of three LEDs.
From your pictures, your strip has 82Ω resistors. 
Twice the current.  They are driving them very hard.

This is how a LED strip is wired:


Still within the power range of your Combi power pack.
Maybe applying a reverse polarity voltage to the strip has damaged some of the LEDs and causing an excessive current draw?

[Chris Uddin]

I think rstofer and rick have added these comments to elaborate just how confusing things can be and dare I say it too see if I cry, I now have to tell you that as I was reading the comments and it was all going over my head tears did start to fall and I started to wonder what the point was adding LED lights to model railway coaches and who needs it anyway??.

Mark F, I was thinking about why the lights may have gone dim, I was doing it from a direct power feed without no resistors or diodes in line so would that have had anything to do with it?, probably not?, but I have to ask the question, there is another variable as well the red and black wires I'm using aren't as thick as the wires that come connected to the LEDs, maybe this wire is too thin and not adequate enough for the voltage hence it'll only take so much? And this originally would have been the wire that I would have used to power the track so who knows, poor wires, too many LEDs, no diodes or resistors and at this stage the list is endless.

[MarkF]

Your LED strip is designed to run directly from 12V.  The resistors to limit the current are built into the strip.  You can see them in your picture (the black squares with 820 written on them).  The extra resistor and diode are to handle the higher 14V and +/- voltage of the DCC power.  They are not need for a plain 12V DC voltage.

My guess is that your 18 LED strip should be pulling 0.24A.  Your wires are short and shouldn't need to carry much current.  The larger wires you're talking about are probably to drive the entire strip in the package which will require much more power.  I don't think your wires are an issue.  I still think you damaged the LEDs in the little piece you cut off.

I would like to see you take some measurements to determine what your Combi output is doing and how much current your strip is drawing.  Do you have a meter and do you know how to take these measurements?

[sw2022]

Mark F - What would a bridge rectifier do then? ( small square black box with 4 pins on it ), is that something I would need or use in this case,? Is this the same as a variable resistor ??.

It straightens out the track polarity such that the LED circuit always has the correct polarity regardless of the track polarity.  Connect the AC side to the track pickups and the DC side to the LED driver.  I'm not sure this is necessary on a DDC system.  Still, it protects the driver circuitry.

You could use a bridge rectifier.  But on a DCC system a single diode will do just as well.  And be much cheaper.

Edit:
On a DCC system the frequency of the track power is much higher than 60Hz.  So, losing the negative portion doesn't have that much of an impact.  And since DCC is positive and negative, it doesn't matter how you connect to the rails.

I suspect the bridge rectifier idea, as opposed to using a cheaper diode, came about because of the need to accept different polarities from the two pickups either end of the coach here? If you always know which pickup input is +V then a single diode is fine. If the inputs can change around then you would need to feed the LEDs via a bridge rectifier connected to both pickup pairs to get correct polarity for the LEDs

Steve

No.  It's to accept different polarities of the track.  Not the pickups. 

The bridge rectifier idea comes from the old days when the track was powered with DC.  The track polarity is swapped in order for the train to go in reverse.  The bridge provides the same polarity no matter which direction the train is running.

Knowing nothing about model/real railways here as you can tell, but - can the carriages be run 180 degrees reversed, or are they always run unidirectionally? If so another reason to use a bridge in a DC system I guess,

Steve