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@stan2004 posted:

Do you have a multimeter and comfortable using it?

And, with absolutely no intent on being a wiseacre, do you know your 9V battery is good?  I hope you don't say you did the "9V battery tongue test"!    Google it for a good chuckle...

Hi Stan. Yes that is brown black red resistor.  I think that's 1k.  The battery was new out of the box and I checked the voltage with our meter so it was good.  And yeah I've done "tongue tests" in my more stupid younger years....  ;-)

SO I'm thinking some sort of relay connected to the switch of the turnout.....?

It's very puzzling that you get nothing to light up even dimly with 9V DC.  An LED should do "something" with 9V DC applied thru a 1K resistor in the proper polarity.  But to the matter at hand, this suggests to me that trying to drive the signal with "only" 5V DC from the yellow wire also won't work.  Yes, you could fuss with lowering the resistor value but I'm not aware of any published information on the capability of the yellow wire to drive more than what it is already tasked for in the pushbutton controller.

So, yes, a relay that applies your 12V AC to your 3-wire signal would surely work if you can get a relay that can be triggered with a low level yellow-wire current.  As GRJ suggested, this might mean adding some kind of transistor buffer.

Frankly, if your 3-wire signal is something that is widely available and you can identify it as such, I think you'll get more feedback.  I'm wondering if this might be viewed as a one-off.

@stan2004 posted:

It's very puzzling that you get nothing to light up even dimly with 9V DC.  An LED should do "something" with 9V DC applied thru a 1K resistor in the proper polarity.  But to the matter at hand, this suggests to me that trying to drive the signal with "only" 5V DC from the yellow wire also won't work.  Yes, you could fuss with lowering the resistor value but I'm not aware of any published information on the capability of the yellow wire to drive more than what it is already tasked for in the pushbutton controller.

So, yes, a relay that applies your 12V AC to your 3-wire signal would surely work if you can get a relay that can be triggered with a low level yellow-wire current.  As GRJ suggested, this might mean adding some kind of transistor buffer.

Frankly, if your 3-wire signal is something that is widely available and you can identify it as such, I think you'll get more feedback.  I'm wondering if this might be viewed as a one-off.

The signal is made by Evemodel in China. IF someone gave me a diagram of how the relay and the transistor would be connected that would be the help I need.   The instructions call for 14v ac or rather that is what it is rated for....

I'm wondering if the signal is straight LED's and the LED's are toasted by testing without current limiting.  If connecting the 9V battery through a 1K resistor doesn't light them, I'd say the LED's are toast.  I'm assuming you tried both polarities by flipping the battery terminals, right?

But the signals work fine with AC at 12 volts.  Maybe they need 12v DC?

I think there is some incredulity of a 3-wire LED signal that requires AC voltage and does not work with DC voltage!  But let's put aside the suspension-of-disbelief and re-state the application as simply driving a 3-wire 2-aspect incandescent signal from the Yellow-wire of an FT turnout. By incandescent signal, I mean something that requires a level of AC power that one would not expect the Yellow-wire to be able to directly light up.  So now we're talking a relay...as fussing with a triac (an "AC" transistor) should not be one's first experience with DIY electronics.  In my opinion of course.

12v relay driven by FT yellow wire and 12V AC

There are many alternative but here's one I cobbled together with stuff in my parts stash.  I do not have an FT switch to provide a proof-of-concept but I made some electrical measurements that convince me that it would work.  The idea is to convert an inexpensive (less than $2 on eBay with free shipping from Asia) off-the-shelf 12V DC relay module to work with 12V AC.

Untitled

But you have to surround the relay module with maybe 50 cents of "loose" components.  The problem is you generally must buy 5, 10, 50, etc. of an inexpensive component which makes them no so inexpensive after all.  I don't know the answer to the "minimum quantity" conundrum.

I did wire up the above circuit and it works over a wide AC voltage range - 10V to 18V - which I chose as what you might get from the Accessory AC transformer output up to O-gauge command-voltage.

As noted in my first diagram, this approach assumes that the black wire on the FT controller is the same as the AC-common (outer-rail).  Again, I do not have an FT turnout but this can be confirmed fairly quickly with a meter.

For the technically curious, the trigger current into the 12V DC relay module was only 1.5 mA when driven by 5V DC.  In fact, the relay module triggered with only 0.3 mA (with about 2.5V applied).  So I'm fairly confident the FT turnout's yellow wire can supply this.  Also, I confirmed this worked with a capacitor as small as only 100uF.  This 12V DC relay module also operates down to about 9V DC.

Separately, while shown triggered by the FT turnout's yellow wire, this "design" can be used for the more generic insulated-rail trigger where wheel axles straddle the outer rails of 3-rail O gauge track.  There are some minor changes but the cost is the about the same; if there is interest start a different thread so not to lose focus on the original problem.

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  • 12v relay driven by FT yellow wire and 12V AC
Last edited by stan2004
@stan2004 posted:

I think there is some incredulity of a 3-wire LED signal that requires AC voltage and does not work with DC voltage!  But let's put aside the suspension-of-disbelief and re-state the application as simply driving a 3-wire 2-aspect incandescent signal from the Yellow-wire of an FT turnout. By incandescent signal, I mean something that requires a level of AC power that one would not expect the Yellow-wire to be able to directly light up.  So now we're talking a relay...as fussing with a triac (an "AC" transistor) should not be one's first experience with DIY electronics.  In my opinion of course.

12v relay driven by FT yellow wire and 12V AC

There are many alternative but here's one I cobbled together with stuff in my parts stash.  I do not have an FT switch to provide a proof-of-concept but I made some electrical measurements that convince me that it would work.  The idea is to convert an inexpensive (less than $2 on eBay with free shipping from Asia) off-the-shelf 12V DC relay module to work with 12V AC.

Untitled

But you have to surround the relay module with maybe 50 cents of "loose" components.  The problem is you generally must buy 5, 10, 50, etc. of an inexpensive component which makes them no so inexpensive after all.  I don't know the answer to the "minimum quantity" conundrum.

I did wire up the above circuit and it works over a wide AC voltage range - 10V to 18V - which I chose as what you might get from the Accessory AC transformer output up to O-gauge command-voltage.

As noted in my first diagram, this approach assumes that the black wire on the FT controller is the same as the AC-common (outer-rail).  Again, I do not have an FT turnout but this can be confirmed fairly quickly with a meter.

For the technically curious, the trigger current into the 12V DC relay module was only 1.5 mA when driven by 5V DC.  In fact, the relay module triggered with only 0.3 mA (with about 2.5V applied).  So I'm fairly confident the FT turnout's yellow wire can supply this.  Also, I confirmed this worked with a capacitor as small as only 100uF.  This 12V DC relay module also operates down to about 9V DC.

Separately, while shown triggered by the FT turnout's yellow wire, this "design" can be used for the more generic insulated-rail trigger where wheel axles straddle the outer rails of 3-rail O gauge track.  There are some minor changes but the cost is the about the same; if there is interest start a different thread so not to lose focus on the original problem.

That's a big help as I was wondering if I would have to wire up something besides a relay to the lights and switch.  Thanks for the help!

Roger

@stan2004 posted:

I think there is some incredulity of a 3-wire LED signal that requires AC voltage and does not work with DC voltage!  But let's put aside the suspension-of-disbelief and re-state the application as simply driving a 3-wire 2-aspect incandescent signal from the Yellow-wire of an FT turnout. By incandescent signal, I mean something that requires a level of AC power that one would not expect the Yellow-wire to be able to directly light up.  So now we're talking a relay...as fussing with a triac (an "AC" transistor) should not be one's first experience with DIY electronics.  In my opinion of course.

12v relay driven by FT yellow wire and 12V AC

There are many alternative but here's one I cobbled together with stuff in my parts stash.  I do not have an FT switch to provide a proof-of-concept but I made some electrical measurements that convince me that it would work.  The idea is to convert an inexpensive (less than $2 on eBay with free shipping from Asia) off-the-shelf 12V DC relay module to work with 12V AC.

Untitled

But you have to surround the relay module with maybe 50 cents of "loose" components.  The problem is you generally must buy 5, 10, 50, etc. of an inexpensive component which makes them no so inexpensive after all.  I don't know the answer to the "minimum quantity" conundrum.

I did wire up the above circuit and it works over a wide AC voltage range - 10V to 18V - which I chose as what you might get from the Accessory AC transformer output up to O-gauge command-voltage.

As noted in my first diagram, this approach assumes that the black wire on the FT controller is the same as the AC-common (outer-rail).  Again, I do not have an FT turnout but this can be confirmed fairly quickly with a meter.

For the technically curious, the trigger current into the 12V DC relay module was only 1.5 mA when driven by 5V DC.  In fact, the relay module triggered with only 0.3 mA (with about 2.5V applied).  So I'm fairly confident the FT turnout's yellow wire can supply this.  Also, I confirmed this worked with a capacitor as small as only 100uF.  This 12V DC relay module also operates down to about 9V DC.

Separately, while shown triggered by the FT turnout's yellow wire, this "design" can be used for the more generic insulated-rail trigger where wheel axles straddle the outer rails of 3-rail O gauge track.  There are some minor changes but the cost is the about the same; if there is interest start a different thread so not to lose focus on the original problem.

Hi Stan,

SO If found a 12v DC power source and the signals lit up just fine....  I wonder now how that might change the circuit you created?

As long as the 12V DC source has its own wall-outlet connection like a wall-wart or an HO transformer output, then:

12v signal revised

The key connection is between the DC- (or DC common) and AC common.

As shown, you can separately decide to power the signal with your original 12V AC or with the new 12V DC.  If using DC, this assumes the signal's common (black wire) is the minus DC side...typically referred to as a common-cathode LED signal.

Unfortunately, with the relay module shown you still need the 5 cent diode to work with the yellow wire from the FT turnout.  While I show a 1N4003 diode, it can be virtually any diode/rectifier.  (Note to the technorati;  the diode is needed because this relay module type uses an AC-input opto-isolator which would interpret the +5V and -5V signal from the yellow wire as the same!).

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  • 12v signal revised
Last edited by stan2004
@stan2004 posted:

As long as the 12V DC source has its own wall-outlet connection like a wall-wart or an HO transformer output, then:

12v signal revised

The key connection is between the DC- (or DC common) and AC common.

As shown, you can separately decide to power the signal with your original 12V AC or with the new 12V DC.  If using DC, this assumes the signal's common (black wire) is the minus DC side...typically referred to as a common-cathode LED signal.

Unfortunately, with the relay module shown you still need the 5 cent diode to work with the yellow wire from the FT turnout.  While I show a 1N4003 diode, it can be virtually any diode/rectifier.  (Note to the technorati;  the diode is needed because this relay module type uses an AC-input opto-isolator which would interpret the +5V and -5V signal from the yellow wire as the same!).

Hi Stan I ordered one of the relays you used and as soon as I get it, I'll let you know how it works.

Thanks again

Roger Elliott

I'm getting notifications for this thread so post here when you get your goodies.  I don't use FT turnouts but am curious to see what happens.  I figure these relay modules with the built-in buffer transistors to drive the relay coil will have the eternal gratitude of the "yellow wire" electronics that surely wasn't designed to directly drive a relay coil.

@stan2004 posted:

As long as the 12V DC source has its own wall-outlet connection like a wall-wart or an HO transformer output, then:

12v signal revised

The key connection is between the DC- (or DC common) and AC common.

As shown, you can separately decide to power the signal with your original 12V AC or with the new 12V DC.  If using DC, this assumes the signal's common (black wire) is the minus DC side...typically referred to as a common-cathode LED signal.

Unfortunately, with the relay module shown you still need the 5 cent diode to work with the yellow wire from the FT turnout.  While I show a 1N4003 diode, it can be virtually any diode/rectifier.  (Note to the technorati;  the diode is needed because this relay module type uses an AC-input opto-isolator which would interpret the +5V and -5V signal from the yellow wire as the same!).

Hi Stan,

Well, I got the relay chip and wired everything up as you show in your photo but it didn't work.  One signal light remains lit but it doesn't switch when I throw the  turnout switch.  I tried moving wires around and got the same results.  Any idea as to what I did wrong?

1, When you apply the 12V DC to the module, does the LED in the center of the module light up indicating power is applied?

2. Is the plug-on/press-on jumper on the left 2 (of 3) pins as indicated in my photo "jumper set to H"?

3. Does the relay ever "click" on at all?  When the relay turns on, the 2nd LED near the lower left of module should turn on.

4. Is the "-" wire of the 12V DC supply tied to the black wire of the FT turnout (i.e., the outer-rail of the layout)?

@stan2004 posted:

1, When you apply the 12V DC to the module, does the LED in the center of the module light up indicating power is applied?

2. Is the plug-on/press-on jumper on the left 2 (of 3) pins as indicated in my photo "jumper set to H"?

3. Does the relay ever "click" on at all?  When the relay turns on, the 2nd LED near the lower left of module should turn on.

4. Is the "-" wire of the 12V DC supply tied to the black wire of the FT turnout (i.e., the outer-rail of the layout)?

Hi Stan

Happy Thanksgiving!

1. There is an LED that says PWR and it goes on when I plug in the 12V.

2.Yes, I set the jumper to high

3. No I have not been able to get the relay to click.  There is another LED that is more in the middle of the board but I have not seen it light up.

4. I did run a wire from DC "-" to the AC GND of the AC power supply

I have attached a pic of how I wired it up.  The diode is in place on the yellow wire and connects to the purple jumper tied into the "IN" connection.  The Orange jumper goes to the AC GND from DC "-"  The black jumper goes from DC "-" to COM on the relay

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