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Offline Kaspar  
#1 Posted : 24 November 2022 15:30:49(UTC)
Kaspar


Joined: 09/12/2010(UTC)
Posts: 17
Location: norway
I recently purchased a "Marklin box of stuff" from an auction. Among many items was 15 Marklin Hobby signals, all of which are missing the end connector. So they only have the 2 stripped blue cables coming out. I've tested them all with my trusty DC power supply and they work and behave as expected (red/green by reversing polarity)

My blocks are already operated manually with switches on the facia panel. Is there a smart/simple/cheap way to use these signals in my existing setup? I have the DC power available but need some electronic switch or something that triggers a change in polarity when I turn the blocks on or off.

thanks
Kaspar
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Offline marklinist5999  
#2 Posted : 24 November 2022 15:53:19(UTC)
marklinist5999

United States   
Joined: 10/02/2021(UTC)
Posts: 1,870
Location: Michigan, Troy
You can use either the Marklin momentary contact control boxes, or other brand momentary contact switches. If you use constant contact switches, the solenoids will burn out.
The lights stay lit using the yellow wire connected to power, and M track ground, common, or transformer ground.
Offline H0  
#3 Posted : 24 November 2022 18:04:38(UTC)
H0


Joined: 16/02/2004(UTC)
Posts: 14,706
Location: DE-NW
Originally Posted by: marklinist5999 Go to Quoted Post
You can use either the Marklin momentary contact control boxes, or other brand momentary contact switches. If you use constant contact switches, the solenoids will burn out.
The Hobby signals discussed here do not have solenoids, they only have anti-parallel LEDs inside.

Originally Posted by: Kaspar Go to Quoted Post
Is there a smart/simple/cheap way to use these signals in my existing setup?
I don't have a very simple solution at hand.
A simple circuit with one or two transistors per signal can probably be used to show green when the track has power and show red otherwise. Having two DC supplies will probably make it simpler.
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 Kaspar  
#4 Posted : 24 November 2022 19:57:02(UTC)
Kaspar


Joined: 09/12/2010(UTC)
Posts: 17
Location: norway
How about if I used a 3PDT switch?

Connect track power to 2 - block to 3

5 and 6 connect to signal

4 and 9 are DC +

6 and 7 are DC -

When the switch is in first position the block gets no power and signal gets +/- (red). When the switch is in second position the block gets power and signal gets -/+ (green)


Switch
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Offline H0  
#5 Posted : 24 November 2022 20:04:55(UTC)
H0


Joined: 16/02/2004(UTC)
Posts: 14,706
Location: DE-NW
Originally Posted by: Kaspar Go to Quoted Post
How about if I used a 3PDT switch?
Yes, that should work.

I was thinking along the line of wiring the signals without modifying the switch panel.
Getting three-pole switches for the switch panel is a simple solution.
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 Purellum  
#6 Posted : 24 November 2022 20:20:59(UTC)
Purellum

Denmark   
Joined: 08/11/2005(UTC)
Posts: 3,455
Location: Mullerup, 4200 Slagelse
Cool

Originally Posted by: Kaspar Go to Quoted Post
How about if I used a 3PDT switch?

Connect track power to 2 - block to 3

5 and 8 connect to signal

4 and 9 are DC +

6 and 7 are DC -


This would work, if you correct your little error BigGrin

I assume the signals have build in resistors to protect the LEDs, otherwise you have to add a resistor to each signal.

You could also, if you use AC and 2 diodes - and maybe 2 small capacitors - just use a 2PDT switch BigGrin

One side of the signal connects to "AC ground", the other side of signal connects to 5.

Positive half wave from AC trough a diode connects to 4, and negative half wave from AC trough a diode connects to 6.

If you get confused by my solution, just use the 3PDT switch as you suggested Cool

Per.

Cool
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I, the copyright holder of this work, hereby release it into the public domain. This applies worldwide.

In case this is not legally possible:
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Offline Kaspar  
#7 Posted : 25 November 2022 12:35:06(UTC)
Kaspar


Joined: 09/12/2010(UTC)
Posts: 17
Location: norway
Originally Posted by: Purellum Go to Quoted Post
Cool

Originally Posted by: Kaspar Go to Quoted Post
How about if I used a 3PDT switch?

Connect track power to 2 - block to 3

5 and 8 connect to signal

4 and 9 are DC +

6 and 7 are DC -


This would work, if you correct your little error BigGrin

I assume the signals have build in resistors to protect the LEDs, otherwise you have to add a resistor to each signal.

You could also, if you use AC and 2 diodes - and maybe 2 small capacitors - just use a 2PDT switch BigGrin

One side of the signal connects to "AC ground", the other side of signal connects to 5.

Positive half wave from AC trough a diode connects to 4, and negative half wave from AC trough a diode connects to 6.

If you get confused by my solution, just use the 3PDT switch as you suggested Cool

Per.

Cool


Yes, oops a little typo there, signal should be in 5 and 8.

But as far as resistors go I have sent 16v directly to the signal and it still works so can assume there is adequate resistance in place (?)

I am curious about your solution with 2PDT switch- but which diode would you use and how (direction) would you wire them? What would be the use of the capacitors?
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Offline Purellum  
#8 Posted : 25 November 2022 13:30:58(UTC)
Purellum

Denmark   
Joined: 08/11/2005(UTC)
Posts: 3,455
Location: Mullerup, 4200 Slagelse
Cool

Originally Posted by: Kaspar Go to Quoted Post
but which diode would you use and how (direction) would you wire them?What would be the use of the capacitors?


Almost any diode would work, like 1N4001 or something else you can get cheap.

One diode should point towards the switch, the other should point away from the switch,
this way you'll get one half of the AC on one of the switch's "input" and the other AC half wave on the other "input".

( If you run digital, and don't mind using "digital power" to supply your signals, everything can actually be done with a 1PDT switch )

The capacitors is to avoid flicker from the signals.

Per.

Cool
If you can dream it, you can do it!

I, the copyright holder of this work, hereby release it into the public domain. This applies worldwide.

In case this is not legally possible:
I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.

UserPostedImage
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Offline marklinist5999  
#9 Posted : 25 November 2022 13:33:40(UTC)
marklinist5999

United States   
Joined: 10/02/2021(UTC)
Posts: 1,870
Location: Michigan, Troy
I was thinking he had non led signals.
Offline H0  
#10 Posted : 25 November 2022 14:23:45(UTC)
H0


Joined: 16/02/2004(UTC)
Posts: 14,706
Location: DE-NW
Originally Posted by: marklinist5999 Go to Quoted Post
I was thinking he had non led signals.
The only signals that work with the "72751 box" are the LED Hobby signals, so that piece of information made it clear for me.
The Hobby signals are very simple and the circuitry for the light change is in the switch box.
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 Kaspar  
#11 Posted : 25 November 2022 15:21:37(UTC)
Kaspar


Joined: 09/12/2010(UTC)
Posts: 17
Location: norway
Originally Posted by: Purellum Go to Quoted Post
Cool

Originally Posted by: Kaspar Go to Quoted Post
but which diode would you use and how (direction) would you wire them?What would be the use of the capacitors?


Almost any diode would work, like 1N4001 or something else you can get cheap.

One diode should point towards the switch, the other should point away from the switch,
this way you'll get one half of the AC on one of the switch's "input" and the other AC half wave on the other "input".

( If you run digital, and don't mind using "digital power" to supply your signals, everything can actually be done with a 1PDT switch )

The capacitors is to avoid flicker from the signals.

Per.

Cool

Per,

I do run digital indeed (CS2) but keeping my blocks and turnouts analog because it works fine with my style of running trains (and is lot less expensive BigGrin. So I am curious now how you would do this with 1PDT (unless of course you mean adding decoders). And what is the disadvantage to using “digital power” ?

Best
Kaspar
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Offline Purellum  
#12 Posted : 25 November 2022 19:21:47(UTC)
Purellum

Denmark   
Joined: 08/11/2005(UTC)
Posts: 3,455
Location: Mullerup, 4200 Slagelse
Cool

Originally Posted by: Kaspar Go to Quoted Post
how you would do this with 1PDT


I just made a drawing, and realized that my idea will not work - unless you use Schottky diodes and call it a brake section instead of a stop section Blushing

This is what I wanted to do:

On your 1PDT switch:

Connect track power to 2 and block directly to 3

Also to 3 a diode pointing away from 3

To 1 a diode pointing towards 1

The other two ends of the two diodes connects together and this connection goes to the signal - other side of signal goes to ground from track power.

This part will make the signal work as wanted BigGrin

What I missed in my head was that you will get two diodes in series, pointing from 3 to 2 on the switch; which means that you will have half of the digital signal going to the tracks when your switch is off - but if you use the right type of diodes = Schottky, this will become a breaking section.

In a breaking section the loco will stop; but lights and sound will stay on BigGrin

Braking section with Schottky diodes - a little into the thread: https://www.marklin-user...thout-capacitor---solved

In this situation I would not use capacitors, if so they have to be relatively small ( and bipolar ) and would most likely create a small interference of the digital signal every time you change position on the switch........... Blink

Originally Posted by: Kaspar Go to Quoted Post
what is the disadvantage to using “digital power” ?


The disadvantage of using digital power is that it is limited and expensive, it has to come from your controller or a booster Cool

Of course your LED signals doesn't consume a lot of power; but I still mention it as a general rule BigGrin

Per.

Cool
If you can dream it, you can do it!

I, the copyright holder of this work, hereby release it into the public domain. This applies worldwide.

In case this is not legally possible:
I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.

UserPostedImage
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