Extra switch for cobalt

[zedcell]

Hi all can anyone help me out with a bit of a problem.

Im using cobalt point motors and am using both the switches one for point operation and the other for led signals.

Would now like to add a lenz bm1 module linked to the signals for spad protection at certain points around the layout.

From the research i have done its become clear there is not much information on this, but from what i can make out is it possible to add another switch to the cobalt in the form of a 12 volt DPDT relay is this the right path for achieving this or are there any other options.

Any help would be much appreciated.

 

[Brian]

Hi
I think you have a couple of options...

1) Use the LEDs power supply voltage to do two jobs... One to feed the LED and the other to operate a relay for one direction of the motor. The relays coil working voltage matching that which is applied to the Cobalt contacts to feed the LEDs (Quite often this is 12 volts DC?). Then via the relays contact you switch whatever is needed. So in one position of the Cobalt/point the relay is de-energised and in the opposite position the relay is energised. While the LEDs still show their correct indications too. Connect the relays other coil tag to the LED negative supply.

2) Use a small micro switch with lever, The lever is operated by the Cobalts moving drive pin so as when the Cobalt is over one way the micro switches lever is pressed in and when the other way its released.

Option 2 wiring...

 

 

Edit to correct text.

Edited January 11, 2016 by Brian

[zedcell]

Hi Brian thanks for the reply on this very helpful, have visited your web site on many occasions.

Option 1 seems the route i would favor, as electronics is a bit of a minefield for me could you confirm if this dpdt 12 volt relay would be suitable for the job http://www.squirestools.com/files/e22.pdf

catalogue # 22-0002.

 

Regards Richie

[Brian]

Hi Richie

 

The relay will be fine if your LED Supply is 12volts DC.

 

Note the use of some strip board might make wiring the relay easier?

[zedcell]

Hi Brian, thanks for the tip on the  strip board ive now ordered the strip board and relays.

Sorry if im making hard work of this, i hope im slowly getting my head around this.

Would i be right, looking at the 8 connections on the relay i would only have to use the first 2 as per drawing for use as a switch.

Not sure where to connect on relay to activate the bm1 module on relay. 

Edited January 12, 2016 by zedcell

[Brian]

Hi

The relay you linked to on the Squires web site - 22-0002 is a Double Pole Double Throw (DPDT) contact arrangement. That is to say it has two electrically separate change-over sets of contacts that both move together as the relay coil is powered (energised) and de energised (Power removed).  

Looking at the Squires catalogue page there is no actual contact arrangement shown for the 2Amp contact relay, but opposite is a similar relay with a lower amperage contacts - 1Amp, so I'll assume the two are the same (Please check when you have the relay).  

The two off set pins at one end are the coil connections.  These are pins 1 and 16.   The two rows of pins (three per row) further along are the contacts on each side of the relay.  Using the drawing for the lower current version relay, the two end two pins 8 & 9 are the Common moving contact.  On each side 8 or 9 make contact when the relay is de energised to 6 or 11 (8 to 6) and (9 to 11). When the relay is energised the contacts change over and 8 makes to 4 and 9 makes to 13.

The relay uses what is called DIL (Dual In Line) configuration and pin numbering.  Looking down on the relay pin 1 is top left and then numbering runs numerically down to pin 8 at the bottom.  Crosses over to the righthand side and runs back up from pin 9 up to pin 16 at the top right. In a sort of U shape.. Note some pins are missing. But by using the spacing between pins and all being the same distance apart, you can see which one have been omitted -  These are 2 & 3 plus 14 & 15.   The drawing is shown looking at the underneath of relay.  So for simple switching On when the relay is energised use pin 8 and pin 4 or pin 9 and 13.

 

When using the strip board for mounting the relay it is essential to cut the copper tracks in a row downwards and roughly in the middle of the relays position, cutting the tracks between each sides row of pins or you'll have a short between each side! Use a sharp 3mm drill bit or  a special spot face cutter tool.  Perhaps a little OTT would be to use one of the ready made DIL PCBs?  While it is larger than needed it makes mounting the relay simpler and wiring to it easier too... http://www.ebay.co.uk/itm/2-SETS-OF-PC-PCB-UNIVERSAL-STRIP-BOARD-IC-PIC-91x45-Electronic-Components/151722770399?_trksid=p2047675.c100009.m1982&_trkparms=aid%3D222007%26algo%3DSIC.MBE%26ao%3D1%26asc%3D34996%26meid%3D483841a4a4ec417ba29efdd22787db24%26pid%3D100009%26rk%3D1%26rkt%3D10%26sd%3D161428545597

Edited January 13, 2016 by Brian

[zedcell]

Thanks brian for the invaluable information on this much appreciated.

[zedcell]

Hi Brian

I know this post is now over 5 years old but i am now due to lockdown ready to tackle this project .

All the parts have been purchased could you please draw my a wiring diagram to make this as easy as possible .

Connecting a cobalt classic powered via 9 volt dc to the relay for berko led signal and bm1.

                                          Regards Richie

[John ks]

I can see 2 ways to control a relay from the cobalt

1---- Use a set of contacts in the cobalt to control the relay

2----- Provided that that the 12V relay will work with 9V then control the relay from the point switch ( relay coil is in parallel with the cobalt motor) 

If you decide to go for the 2nd method it would be wise to test the 12V relays on 9V to see if they work reliably

 

A disadvantage of method 2 is that the relay is controlled by the point switch 

If the point fails to change the relay still operates & your signals will give a false indication

On a real railway i wouldn't consider this circuit to be fail safe but for a toy train model railway it should be OK

The advantage is you still got the 2 sets of C/O contacts on the cobalt

 

The diode D2 (flyback diode)supresses back EMF from the relay coil

Diode D1 only allows the relay to energise when Orange is positive & Grey is negative 

With the point switch in the drawn position the relay will be off

Any 1A 100V or higher diode would be suitable eg !N40041, 1N4004 or similar

 

 

John

[Nigelcliffe]

If wanting a relay to separate things John's is fine.    

 

But, there is a MUCH simpler way, no relays, no other stuff.  One switch on Cobalt for the frog, other switch for the BM1 brake section isolator.    

 

Connect the common wire from the LED signal to the frog wire going to the track.  Take the other wires from the signal (each via their resistor), and connect one to each of the DCC bus wires.   As the frog changes (which is when the blades are moved), the signal will change. 

 

Doing it really well would mean a small increase in the resistor values from those rated for 12v (but probably doesn't matter).   And, would add a single diode in the common line to the frog to resist reverse bias over the LEDs (theoretically reverse bias will cause a LED to break down over time, in practise most modern LEDs seem remarkably robust against it).  

 

 

- Nigel

 

 

 

 

 

Edited April 6, 2021 by Nigelcliffe
thought of even simpler approach than previous one..

[zedcell]

Found this very helpful thanks.

[zedcell]

10 hours ago, Nigelcliffe said:

If wanting a relay to separate things John's is fine.    

 

But, there is a MUCH simpler way, no relays, no other stuff.  One switch on Cobalt for the frog, other switch for the BM1 brake section isolator.    

 

Connect the common wire from the LED signal to the frog wire going to the track.  Take the other wires from the signal (each via their resistor), and connect one to each of the DCC bus wires.   As the frog changes (which is when the blades are moved), the signal will change. 

 

Doing it really well would mean a small increase in the resistor values from those rated for 12v (but probably doesn't matter).   And, would add a single diode in the common line to the frog to resist reverse bias over the LEDs (theoretically reverse bias will cause a LED to break down over time, in practise most modern LEDs seem remarkably robust against it).  

 

 

- Nigel

 

 

 

 

Thanks for the reply on this what size resistors would you recommend using, the track bus is 5amp.

                                            Regards Richie 

 

[Nigelcliffe]

6 hours ago, zedcell said:

Thanks for the reply on this what size resistors would you recommend using, the track bus is 5amp.

                                            Regards Richie 

 

The theoretical answer requires information about the signals, or the resistor recommended/supplied for 12v use.   The current available at the track bus is not relevant, though the track voltage is relevant. 

 

Assuming a track voltage of approx 16v, then resistor should be approx 35-40% larger than that for 12v use.    I don't know what is supplied with the signal, but a quick online search suggested its 1K.  Therefore a 1.3K to 1.5K resistor might be better for use on the track bus (or, just use 2K - ie. two 1K in series). 

 

 

 

  

 

[ikcdab]

I have found that resistor values for a 12v feed are very tolerant. It depends on the designed brightness and your desired brightness... Which will be two different things. I normally start with 1k and move up. I have some DCC concepts led ground signals and they recommend 10k ohms. 

[zedcell]

On 06/04/2021 at 12:17, John ks said:

I can see 2 ways to control a relay from the cobalt

1---- Use a set of contacts in the cobalt to control the relay

2----- Provided that that the 12V relay will work with 9V then control the relay from the point switch ( relay coil is in parallel with the cobalt motor) 

If you decide to go for the 2nd method it would be wise to test the 12V relays on 9V to see if they work reliably

 

A disadvantage of method 2 is that the relay is controlled by the point switch 

If the point fails to change the relay still operates & your signals will give a false indication

On a real railway i wouldn't consider this circuit to be fail safe but for a toy train model railway it should be OK

The advantage is you still got the 2 sets of C/O contacts on the cobalt

 

The diode D2 (flyback diode)supresses back EMF from the relay coil

Diode D1 only allows the relay to energise when Orange is positive & Grey is negative 

With the point switch in the drawn position the relay will be off

Any 1A 100V or higher diode would be suitable eg !N40041, 1N4004 or similar

 

 

John

Great information thanks .

If I decided on option 1 where would I connect the 3 wires from the signal.

[John ks]

If i understand your requirements correctly then this drawing is what you want

 

The connections in the drawing are for the Cobalt Classic Omega Point Motor.

The original Omega had 8 connections & terminals 1 & 8 are connected to the Point control switch & the internal switches were terminals 2  thru 7

 

Assuming that your signal uses LEDs 

The polarity of the 12V dc*  for the LEDs will depend weather the signal is common Positive or common Negative

* (the signals could probably be supplied from the same 9V you use for the cobalts)

 

Common Positive will require the 12V DC on the left (R1 End) to be Positive

Common Negative will require the 12V DC on the left (R1 End) to be Negative

 

Depending on how your signal is supplied it may have 1 (R1 , R2& R3 are not required) or 2 (R2 & R3, R1 is not required) resistors 

 

If the signal shows the wrong aspect for the route selected then connect the relay coil to term 4 instead of term 5 on the cobalt 

OR 

Swap the wires from the Red aspect to the NC relay connection & the Green aspect to the NO relay connection

As I don't know what relay you are using i cannot show where the physical for the relay

John

 

 

 

 

[zedcell]

Hi John

Thanks for walking me through this much appreciated.

The cobalts are the originals with the 8 terminals and are set up and working for point operation via a 9v dc supply.

The signals i want to add are the berko 2 aspect and are common negative with just the 1 resistor.

Not sure on the best option for power supply to the leds would be the 9V DC from the cobalt or from the 12V DC from a  dedicated signal bus that i originally installed for this.        

The relay would be the one in the link on page 4 12v DPDT 2A on page 4 at the top code 22-002                          http://www.squirestools.com/files/e22.pdf   

What would you recommend for the above and would you have a diagram.

Thanks again .

                                          

Edited April 8, 2021 by zedcell

[John ks]

5 hours ago, zedcell said:

Not sure on the best option for power supply to the leds would be the 9V DC from the cobalt or from the 12V DC from a  dedicated signal bus that i originally installed for this. 

Either supply should be OK for the LED Signal

I would tend to use the 9V supply Providing the signal was bright enough

 

The upgraded circuit with the cobalt connections changed to better indicate the original Cobalts with 8 connections

& the pin layout of the relay

 

 

 

The spec for the relay shows "Coil Operating Range 70-150% of coil voltage" which works out to 8.4V to 18V so the relay should work OK from your 9V supply

You could run The Cobalts, relays*, & Signals from the 9V supply **

 

John

 

 

* Provided the relay works reliably on 9V, as i said earlier you may need to test it on 9V 

** Provided that your power supply can handle the extra load which wont be much Maybe 30mA for the relay & 10mA for the LEDs

[zedcell]

John am i right in thinking the relay pins 8 & 6 connect to the BM1 & PINS 9 11 & 13 return via the resistor to the 12 v dc supply on the above diagram.

If using the cobalt 9v supply what number connections would need to be made between the cobalt  terminals and the relay pins. 

Example connect cobalt terminal 5 to relay pin 8 with reference to any diodes or resistors in line.

 

[John ks]

I have added pin numbers to the relay (looking from the top of the relay)

& should be clearer which wire goes where

2 hours ago, zedcell said:

If using the cobalt 9v supply what number connections would need to be made between the cobalt  terminals and the relay pins. 

Example connect cobalt terminal 5 to relay pin 8 with reference to any diodes or resistors in line.

If you decide to run the relay from 9V then the positive for the relay connects to the positive 9V (follow the dotted red line), same for the Negative (follow the dotted black line)

If you use 12V for the relay then ignore the dotted lines

 

Relay pins 9,11 & 13 are the second set of contacts  C= Common, NC = Normally Closed & NO = Normally Open 

 

I have show the relay connected to Cobalt terminals  4 to Positive & 3 to relay coil (pin1)

Connecting Cobalt terminals 5 to Positive & 6 to Relay coil would also work 

John

 

 

 

[Nigelcliffe]

The above will work, but is excessively complex unless it is really important to wait for the point motor to throw half-way before changing the signal.  

 

Simplified version, where signal will change as turnout motor switch is changed: 

Completely Remove the relay and external 12v supply from the diagram.  

Take wire 6(relay) from signal, connect to wire 1 on Cobalt

Take wire 8(relay) from signal, connect to wire 8 on Cobalt. 

  

And there are still two un-used change-over contacts on the Cobalt. 

 

I've shown two resistors, one to each signal line. That's probably an unnecessary detail compared to one resistor, but my arrangement does ensure the reverse voltage over the un-lit LEDs is zero.  

 

- Nigel

Edited April 10, 2021 by Nigelcliffe

[John ks]

Nigel

Some times you get the idea of using a relay stuck in your head

Then someone comes along and shows you a simpler way

 

 

Thinking about it, if you needed a relay then connecting it between the orange wire & the negative should work

 

John

 

 

 

 

[zedcell]

20 hours ago, Nigelcliffe said:

The above will work, but is excessively complex unless it is really important to wait for the point motor to throw half-way before changing the signal.  

 

Simplified version, where signal will change as turnout motor switch is changed: 

Completely Remove the relay and external 12v supply from the diagram.  

Take wire 6(relay) from signal, connect to wire 1 on Cobalt

Take wire 8(relay) from signal, connect to wire 8 on Cobalt. 

  

And there are still two un-used change-over contacts on the Cobalt. 

 

I've shown two resistors, one to each signal line. That's probably an unnecessary detail compared to one resistor, but my arrangement does ensure the reverse voltage over the un-lit LEDs is zero.  

 

- Nigel

That certainly does simplify things just one question am i right in thinking terminals 6 &7 can be the feed for the BM1.

[John ks]

On 11/04/2021 at 15:21, zedcell said:

am i right in thinking terminals 6 &7 can be the feed for the BM1

Not sure how the BM1 is wired but with a 8 terminal cobalt i would think you would need terminals 5 & 6 or 5 & 7 depending on which route you are controlling

John

[Nigelcliffe]

29 minutes ago, John ks said:

Not sure how the BM1 is wired but with a 8 terminal cobalt i would think you would need terminals 5 & 6 or 5 & 7 depending on which route you are controlling

John

 

BM1 needs an "on" contact when the controlled section is "proceed at normal speed", and an "open" connection when the controlled section is "brake to a stop".   

 

(BM1 = Lenz Asymmetric DCC braking module  ).  

 

 

- Nigel