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Offline Mjn2ts  
#1 Posted : 31 March 2025 23:57:28(UTC)
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Mjn2ts

United States   
Joined: 04/10/2021(UTC)
Posts: 56
Location: Maine, Lewiston
I'm having a hard time trying to understand how to determine what resistor I need to use in a locomotive that I'm converting to led lights. In the documentation for the decoder there is no "supply voltage" listed, only "Current load at the light outputs" and that says less than 250 milliamps then there is "maximum voltage" and that says less than 40 volts. I'm not sure what to use for the "supply Voltage". Its a Marklin 60977 decoder if that helps. The info for the led was pretty straight forward:

Luminous Intensity: 1560/2180mcd
Forward Voltage: 2.1/3.1V
Viewing Angle: 30°
Minimum Operating Temp: -30°C
Forward Current: 30mA
Size: 3mm
Reverse Voltage: 5V
Wavelength: 624/6500nm
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Offline H0  
#2 Posted : 01 April 2025 07:59:58(UTC)
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H0


Joined: 16/02/2004(UTC)
Posts: 15,432
Location: DE-NW
Originally Posted by: Mjn2ts Go to Quoted Post
The info for the led was pretty straight forward
The slash "/" makes me think that the chart contains data for two different types of LEDs.
Do not go beyond 20 mA. Modern LEDs are typically bright enough at 2 mA or even less.

The output voltage of the decoder is input voltage minus the voltage drop inside (rectifier, output transistors). Assume 24 V
LEDs should still be very bright with a 10 kOhm resistor and can be dimmed further using decoder programming.

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 Mjn2ts  
#3 Posted : 01 April 2025 10:29:25(UTC)
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Mjn2ts

United States   
Joined: 04/10/2021(UTC)
Posts: 56
Location: Maine, Lewiston
Originally Posted by: H0 Go to Quoted Post
Originally Posted by: Mjn2ts Go to Quoted Post
The info for the led was pretty straight forward
The slash "/" makes me think that the chart contains data for two different types of LEDs.
Do not go beyond 20 mA. Modern LEDs are typically bright enough at 2 mA or even less.

The output voltage of the decoder is input voltage minus the voltage drop inside (rectifier, output transistors). Assume 24 V
LEDs should still be very bright with a 10 kOhm resistor and can be dimmed further using decoder programming.


It is 2 different leds in one, they are red/white with 3 prongs. While searching I came across an older post that seemed to indicate that the supply voltage was basically track voltage or the output of the controller, so roughly 19volts. If I assume your 20mA is accurate that would put my resistor at around 805ohms. I think...
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Offline H0  
#4 Posted : 01 April 2025 12:13:53(UTC)
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H0


Joined: 16/02/2004(UTC)
Posts: 15,432
Location: DE-NW
Originally Posted by: Mjn2ts Go to Quoted Post
While searching I came across an older post that seemed to indicate that the supply voltage was basically track voltage or the output of the controller, so roughly 19volts.
It's around 19 V for a CS3, but around 22 V for a CS2 with a transformer and around 23 V for a 6021.
When you do the maths for 24 V your pretty good on the safe side.
I made conversions where I have three white LEDs in series. There was a 1k5 resistor on the decoder board, but I added 20k and the light was still on the bright side.

You can easily test the LEDs with a bridge rectifier and different resistors (10k is a good start) to see how bright they are.

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 kiwiAlan  
#5 Posted : 01 April 2025 21:43:01(UTC)
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kiwiAlan

United Kingdom   
Joined: 23/07/2014(UTC)
Posts: 8,454
Location: ENGLAND, Didcot
Originally Posted by: Mjn2ts Go to Quoted Post
Originally Posted by: H0 Go to Quoted Post
Originally Posted by: Mjn2ts Go to Quoted Post
The info for the led was pretty straight forward
The slash "/" makes me think that the chart contains data for two different types of LEDs.
Do not go beyond 20 mA. Modern LEDs are typically bright enough at 2 mA or even less.

The output voltage of the decoder is input voltage minus the voltage drop inside (rectifier, output transistors). Assume 24 V
LEDs should still be very bright with a 10 kOhm resistor and can be dimmed further using decoder programming.


It is 2 different leds in one, they are red/white with 3 prongs. While searching I came across an older post that seemed to indicate that the supply voltage was basically track voltage or the output of the controller, so roughly 19volts. If I assume your 20mA is accurate that would put my resistor at around 805ohms. I think...


20mA is the maximum safe current for an LED, but if you run them at that you will burn your eyes out. I would use a resistor of 8.2k or 10k as suggested by H0 as a starting point, and I think you will find they are plenty bright at that current.
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Offline Mjn2ts  
#6 Posted : 01 April 2025 23:11:51(UTC)
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Mjn2ts

United States   
Joined: 04/10/2021(UTC)
Posts: 56
Location: Maine, Lewiston
Originally Posted by: kiwiAlan Go to Quoted Post
Originally Posted by: Mjn2ts Go to Quoted Post
Originally Posted by: H0 Go to Quoted Post
Originally Posted by: Mjn2ts Go to Quoted Post
The info for the led was pretty straight forward
The slash "/" makes me think that the chart contains data for two different types of LEDs.
Do not go beyond 20 mA. Modern LEDs are typically bright enough at 2 mA or even less.

The output voltage of the decoder is input voltage minus the voltage drop inside (rectifier, output transistors). Assume 24 V
LEDs should still be very bright with a 10 kOhm resistor and can be dimmed further using decoder programming.


It is 2 different leds in one, they are red/white with 3 prongs. While searching I came across an older post that seemed to indicate that the supply voltage was basically track voltage or the output of the controller, so roughly 19volts. If I assume your 20mA is accurate that would put my resistor at around 805ohms. I think...


20mA is the maximum safe current for an LED, but if you run them at that you will burn your eyes out. I would use a resistor of 8.2k or 10k as suggested by H0 as a starting point, and I think you will find they are plenty bright at that current.


Think I'm gonna end up using the 8.2k, it was bright but not too bright and I can always turn it down in the decoder settings. Still not sure how that works, the online calculator I found is saying 800-1000.. Thanks for the guidance!! Jeff
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Offline H0  
#7 Posted : 02 April 2025 08:01:55(UTC)
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H0


Joined: 16/02/2004(UTC)
Posts: 15,432
Location: DE-NW
Originally Posted by: Mjn2ts Go to Quoted Post
Still not sure how that works, the online calculator I found is saying 800-1000.
The calculator finds the value for 20 mA, but 2 mA already give very bright light.
Trust the calculator if you want to build a desktop lamp, but use bigger values for locomotives.

P.S.: Dimming the LED with the decoder will lead to high-frequency flickering that might become visible when filming, therefore it makes sense to increase the resistor value.

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
UserPostedImage
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thanks 2 users liked this useful post by H0
Offline Mjn2ts  
#8 Posted : 02 April 2025 10:18:52(UTC)
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Mjn2ts

United States   
Joined: 04/10/2021(UTC)
Posts: 56
Location: Maine, Lewiston
Originally Posted by: H0 Go to Quoted Post
Originally Posted by: Mjn2ts Go to Quoted Post
Still not sure how that works, the online calculator I found is saying 800-1000.
The calculator finds the value for 20 mA, but 2 mA already give very bright light.
Trust the calculator if you want to build a desktop lamp, but use bigger values for locomotives.

P.S.: Dimming the LED with the decoder will lead to high-frequency flickering that might become visible when filming, therefore it makes sense to increase the resistor value.


Thanks for the help!! Jeff
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