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Offline Bates  
#1 Posted : 08 October 2018 20:34:04(UTC)
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Bates

South Africa   
Joined: 05/06/2016(UTC)
Posts: 8
Location: Western Cape,
Hello forum members,

Some advice please. I have successfully fitted warm white LED's in coaches and loco's with constant current devices on my analog trains. I am now trying to help a friend with LED installation in coaches on a digital layout. I am not very familiar with Marklin digital and have a few questions: 1. Can one simply connect a LED and current limiting resistor straight to the track (via the pick up shoe) 2. what is the constant track voltage? Forgive me if these topics have been covered before. Advice would be appreciated.

Thanks
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Offline Webmaster  
#2 Posted : 08 October 2018 20:42:29(UTC)
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Webmaster


Joined: 25/07/2001(UTC)
Posts: 11,161
Basically it's the same setup as with analog, but maybe with a slighty larger resistor and preferably with 4-diode Waterstone bridge or a voltage regulator chip instead of 1-2 diodes...

The "digital voltage" varies up to between +/- 22V constantly, compared to the analog +/- 16V at full speed.
You could make it simple and calculate the brightness of the LED's in digital about as they would be when you do the reverse pulse in analog.
Juhan - "Webmaster", at your service...
He who asks a question is a fool for five minutes. He who does not ask a question remains a fool forever. [Old Chinese Proverb]
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Offline DaleSchultz  
#3 Posted : 08 October 2018 22:47:16(UTC)
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DaleSchultz

United States   
Joined: 10/02/2006(UTC)
Posts: 3,997
Nothing beats a good test.

I would do this:

Create a simple circuit with a bridge rectifier, and a smoothing capacitor, and measure the rectified DC voltage that is produced.

Then use that value as your DC power supply voltage in the LED resistor calculations.

Perhaps someone here has already done this test...?

Dale
Intellibox + own software, K-Track
My current layout: https://cabin-layout.mixmox.com
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Offline Michael4  
#4 Posted : 09 October 2018 00:03:58(UTC)
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Michael4

United Kingdom   
Joined: 02/02/2017(UTC)
Posts: 642
Location: England, South Coast
For an alternative solution see this thread.

https://www.marklin-user...rain-Tech-Coach-Lighting

I have just had some delivered and will be installing them shortly.
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Offline H0  
#5 Posted : 09 October 2018 08:16:47(UTC)
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H0


Joined: 16/02/2004(UTC)
Posts: 15,263
Location: DE-NW
Hi!
Originally Posted by: DaleSchultz Go to Quoted Post
Create a simple circuit with a bridge rectifier, and a smoothing capacitor, and measure the rectified DC voltage that is produced.
The 22 V digital track voltage will produce about 22 V with that setup.
The 16 V AC used for analogue operation will also produce about 22 V in that case.
The reversing voltage used for analogue operation will give you about 40 V DC.

An MS2 will only give you about 16 V, but it's better to design the circuits for at least 25 V (digital only) or 50 V (digital and analogue).

Originally Posted by: DaleSchultz Go to Quoted Post
Nothing beats a good test.
Make sure the test covers the extremes. MS2 is the low end of Märklin H0 controllers with respect to track voltage, the CU 6021 is most likely the high end. And analogue operation is a completely different story.
Regards
Tom
---
"In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS
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Offline ktsolias  
#6 Posted : 09 October 2018 08:44:07(UTC)
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ktsolias

Greece   
Joined: 01/05/2016(UTC)
Posts: 610
Location: Athens
The diagram of a full rectifier is like this:

AC rectifier.jpg

In the first plan a rectifier bridge is used, (better solution, fewer components) and in the second diodes.

The output voltage of this circuit is calculated as follows:

{V(in)-1,4} x 1.4 = V(out)

Ex: (18v-1,4)X1.4 = 23,24

Another solution is to use a voltage regulator.

I don't think that is necessary for LED lighting circuits. Usually we use much more higher resistors just to dim the LED's because they are to bright.

Something that must be in our mind is that if we use to many circuits with capacitors, without a resistance and a diode, when we turn on the power supply to the layout, all the resistors start to draw current and if they are many draw so much that is more than the capacity of the power supply and looks like a shortcut, and the power supply (MS2 or CSx) cuts the power!!!

Regards

Costas
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Offline David Dewar  
#7 Posted : 09 October 2018 11:35:09(UTC)
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David Dewar

Scotland   
Joined: 01/02/2004(UTC)
Posts: 7,340
Location: Scotland
Originally Posted by: Michael4 Go to Quoted Post
For an alternative solution see this thread.

https://www.marklin-user...rain-Tech-Coach-Lighting

I have just had some delivered and will be installing them shortly.


Yep. Now use these all the time. No flickering or sliders or current conducting couplers etc. Great product.
Take care I like Marklin and will defend the worlds greatest model rail manufacturer.
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Offline H0  
#8 Posted : 09 October 2018 14:09:46(UTC)
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H0


Joined: 16/02/2004(UTC)
Posts: 15,263
Location: DE-NW
Originally Posted by: ktsolias Go to Quoted Post
The output voltage of this circuit is calculated as follows:
{V(in)-1,4} x 1.4 = V(out)
Please note that this formula is just a rule of thumb and that it applies when the input voltage is sine wave AC.
It does not apply to DC voltages or digital track voltages.

Maybe it should be V(in) x 1.4 - 1.4 = V(out).
Regards
Tom
---
"In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS
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Offline Bates  
#9 Posted : 12 October 2018 20:46:06(UTC)
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Bates

South Africa   
Joined: 05/06/2016(UTC)
Posts: 8
Location: Western Cape,
Thanks everyone for the advice and comments
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