Hi,

I apologize straight off if I am asking some really dumb questions here. I recently bought some Marklin 89391 block signals and a Viessmann 4811 block signal (I believe that they are effectively the same product). Looking at the documentation that came with the signals, the Marklin signals need a 16V power source and the Viessmann needs 14-16V. The accessory output on my 67013 controller is marked as 11V; so am I correct in thinking that I am going to need an alternative power source to power these signals? If so, can anybody recommend a solution (I only have four signals)?

Secondly, I was going to control the signals manually via a Viessman 5550 universal switch. This comes with a series of plugs that I guess you use to attach the relevant wires to the switch. Are you just meant to stick the wire through the plug and then ram it into the relevant hole on the switch? It doesn't seem to give a very secure connection. Also how much bare wire should protrude from the tip of the plug?

Finally, when I was playing around with one of the 89391 signals, a resistor fell off one of the wires. When I examined the wire, it looked like the insulation had never been stripped i.e. it looks like the wire was just stuck onto the resistor with a bit of insulation tape and there may never have been a proper, soldered, metal to metal contact. Has anyone else come across this problem? Obviously, on this example I am going to have to remove the tape still stuck to the resistor and resolder the connection, but I am not going to be too impressed if I have to do this for all of them.

Thanks,
Carim

Edited by user 14 December 2015 17:12:40(UTC)  | Reason: Changed status from "Question" to "Solved"

Hi Carim

I suspect that what you are looking at is a solder joint that has been covered with heat shrink tube after it has been soldered (what you describe as tape).

I suspect what has happened is the wire has been nicked when the insulation was stripped off, and the solder hasn't wicked up the wire and under the insulation. The wire has then broken off at the end of the insulation leaving the rest of the wire in the solder joint on the resistor lead.

The best way to remove the heat shrink is to use a sharp knife (a scalpel type of knife is ideal) and cut it length wise, and then split it open to peel it off the lead. If you have a Maplin store near you, they may stock heat shrink tube of a suitable size (you will probably need the smallest size they have) otherwise you can get it from Farnell or RS or Rapid. The new heat shrink will shrink to half (or possibly a third) of its unshrunk diameter.

On re-reading this before posting I realise I am assuming the heatshrink is only over the resistor leads. If they have used a single piece that goes right over the resistor body then it probably has not shrunk small enough to go tight on the flexible wire, which it needs to do to provide the mechanical protection to the wire to help prevent exactly what has happened.

If you were nearby I would say bring it over and i could sort it out.

Alan

Hi Alan,

Thank you for the offer and those excellent instructions. Fingers crossed, I should be able to do the repair it on my own (I think it would be a four hour round-trip to you). If I mess it up with the soldering iron, I'll come running back for help!

Kind regards,
Carim

Did you check the polarity of the voltage?
The signals need DC - and if the DC from your 67013 has the wrong polarity then you won't see lights.

Hi Tom,

If you look at the second Viessman document in post #6, it suggests either AC or DC. Prior to seeing this, I thought that LEDs needed DC and that I would have to use a bridge rectifier in the circuit if my power source was AC (as it is). I thought that there may be something in the 5550 switch acting in conjuntion with the diode on the signal that effectively converts the polarity - or am I totally wrong?

If I added a bridge rectifier to the 11V AC power source, I would get a DC current but it still looks like insufficient voltage to operate the 5550 switch. (Also what voltage on the bridge rectifier - is higher better than lower?).

I am fast coming round to the conclusion that I need another transformer or I have to be happy with non-working signals on my layout!

Carim

Are you sure your power source is AC and not DC?
The 67013 comes with a DC wallwart and I don't think it will generate an AC output when you feed DC into it.

Solenoids work with either AC or DC.
With a DC voltmeter you can check if the output is DC (if you have one).
You can try the signal with reversed voltage from the controller.

The Viessmann 5550 is simply a switch and does not have a minimum voltage (although it does have a 24V max rating). As long as your wiring from the LED signal and power pack are correct, it will simply apply power to either the red or the green LED (if we're still discussing the two color block signal).

The red and green LED wires from the signal should connect to the 5550 along with the grey (?) accessory output from the transformer. The other wire (with the heat shrink diode) is attached to the yellow accessory output.

Now that I say that, did you re-attach the diode backwards? They are direction sensitive.

Cheers!

NOTE: In the 5550 wiring diagram, it shows how to wire the signal and box if you also wish to control track power with it. If you simply want to control the signal, then the red and green wires go into the appropriate holes for the one switch. Does that make sense?

Gene,

Thanks for this - this is the sort of level that I need - all makes sense now. I haven't touched the wire with the diode on it, just one "green" resistor fell off. I had got confused as I thought that with LEDs, the resistors had to be placed ahead of the LED on the +ve wire. But according to DCC Concepts' literature, whilst that is best practice, it doesn't really matter. I will repair the broken signal and make a test set-up and let everybody know how I get on.

Carim

In the end, the circuit works as long as you don't exceed the maximums for the LED (that would be bad). As long as the resistor (correctly sized for your power supply) and LED are in series, the voltage drops across each no matter the order in the series. Obviously the LED (it is a diode, after all) has to be connected correctly.

I'm so glad you have it working! I really like my 8939x series signals.

Cheers!