How to split DC carrier from AC circuit?

Discussion in 'The Projects Forum' started by iakobos, Aug 8, 2011.

  1. iakobos

    iakobos Thread Starter New Member

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    I am attempting to modify an O gauge whistling tender (K-Line) to work with an existing Lionel Train transformer. The transformer powers the engine with 0-18 VAC. The whistle in the tender is activated by a switch on the transformer that sends out a 2.5VDC carrier wave on the AC line.

    What I'm attempting to do is activate a relay inside the tender with the 2.5 VDC. The relay then activates the motor for the whistle using the AC power (through a bridge rectifier because the motor is DC).

    I've been mapping out a schematic and now I'm stuck on how to separate the DC carrier from the AC at the relay coil in the tender. Since this is a preliminary drawing I'll also need to add at least one capacitor and I'm also not sure the 5V regulator is correct either so help on those issues will be appreciated too.

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  2. praondevou

    praondevou Well-Known Member

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    A simple Low-pass filter with an OPAMP will do it, but since it's cutoff frequency has to be VERY low, it may take various cycles for it's output to react.

    What you need is a differentiator which has no phase shift... It's output would need to have the same AC amplitude as the input signal (without DC). Then you compare the original signal with the differentiator signal, it's difference will be the DC-offset 2.5V...

    Still trying to figure out how to do the circuit , though...
  3. iakobos

    iakobos Thread Starter New Member

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    This is my first encounter with an OPAMP circuit. I read through the wikipedia entry on it. Someone will have to hold my hand in order to figure it out. It's going to take me a while to understand how they work.
  4. praondevou

    praondevou Well-Known Member

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    Why does the relay have to be activated by 2.5V? (Or why does there have to be a relay?) Since the motor power comes from the same side as the 2.5VDC, why not simply turn on/off the AC 0-18V?

    And, the 2.5VDC voltage generation, this is a block diagram, I assume? There is no rectification from the tranformer...
  5. iakobos

    iakobos Thread Starter New Member

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    Turning the AC on and off would stop and start the train engine. Normally the engine runs continuously and the whistle blows when ever you want.

    To answer why the relay has to be activated by the 2.5 V: the Lionel Transformer outputs 2.5 V when the whistle button is depressed.

    I don't know if the 2.5 VDC is block diagram with no rectification. I think that information is available but I'd have to do some searching.
  6. praondevou

    praondevou Well-Known Member

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    Ok, got it. So it's a block diagram, you can't get DC from a transformer output. Is the train engine also powered from the 0-18VAC? If so, that means that if the train stops the whistle wouldn't work neither?

    How did you measure the voltage? Are u sure that this is a 0 to 18VAC with 2.5VDC offset (when the whistle is ON)?
    Last edited: Aug 9, 2011
  7. iakobos

    iakobos Thread Starter New Member

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    I see. It must be going through a rectifier. (I should have known that.) I'll fix it on the schematic.

    0-18 VAC is the advertised voltage. I just measured my transformer and it ranges from 1.5-15.75 VAC. The 2.5 VDC when the whistle button is activated is a loose average. The range is from 1.2 VDC to 2.5 VDC and is dependent on the amount of AC being supplied. It starts at 1.2 VDC and goes up to 2.5 VDC when 9 VAC is supplied and goes back down to 1.2 VDC when the AC is all the way up to 15.75 VAC
  8. praondevou

    praondevou Well-Known Member

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    Could you please post the model of the transformer? Or a picture?
  9. iakobos

    iakobos Thread Starter New Member

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    Yes the train is powered by the 0-18 VAC

    Here's some photos. The orange handle is the throttle that controls the variable AC.
    When the Whistle button is depressed it sends the DC out with the AC.
    It's a CW-80 Lionel transformer.

    Attached Files:

    Last edited: Aug 9, 2011
  10. praondevou

    praondevou Well-Known Member

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    Ok, for everybody who wants to contribute to this. It's a phase angle control.

    Apparently the bell and whistle button disables the phase angle control for one half wave, thus creating an DC offset.


    Wave forms are here: http://www.youtube.com/watch?v=OEAy_B5bVX8
    Last edited: Aug 9, 2011
  11. praondevou

    praondevou Well-Known Member

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    Is anyone here who has a better idea? :confused:

    A normal integrator doesn't do it (see my previous post for wave form explanation). It would be too slow.

    I was thinking of extracting the pulse width of each halfwave and depending on it's width difference create a signal for the relay... The extraction I did but now I need some ideas.

    The attached picture shows the original waveform from the transformer. The other two waveforms correspond to each half wave pulse width.

    Attached Files:

    Last edited: Aug 9, 2011
  12. praondevou

    praondevou Well-Known Member

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    I think you can forget the previous post. I just saw in THIS link that the Lionel whistle relay is an AC- insensitive DC-relay. So whatever AC voltage you apply it just ignores it.

    Is this what you want to built into your K-Line tender? Or do they have the same circuit? Sorry about these question, it's been around 25 years since I last had something to do with this stuff... ;)

    However, if you want to use the original K-line whistle I think you'll have to find out how it works.
    Last edited: Aug 9, 2011
  13. iakobos

    iakobos Thread Starter New Member

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    The 167 whistle controller in the link was used when older style transformers didn't have their own whistle activation circuitry.

    Either today or tomorrow, I'm going to take apart a modern Lionel whistling tender and see how it works. That will hopefully give us something more concrete to utilize with the K-Line tender. The main difference between the K-Line and Lionel tenders is the Lionel uses 2.5 VDC to activate the whistle and the K-Line uses 9 VDC. Since my transformer only puts out 2.5 VDC I'm attempting to engineer the K-Line to work with the Lionel.

    While searching the web I found how the old Lionel tenders (1940-60's) distinguished between the AC and DC.
    Original: Lionel made a relay that had a shield blocking the AC magnetic flux from reaching the armature. When DC was applied, the 1 direction flux would pass thru the shield and the armature pulled down. Things could also overheat, both the relay and the transformer's selenium rectifier.

    To counter the overheating effect, Lionel created a seamless 2 step whistle-on switch. The first position put just the rectifier in series with the center rail's line.

    The second position connected a resistor around the rectifier. This changed the 1/2 wave rectified voltage back to almost full wave, with one 1/2 of the wave being smaller that the other.

    The 1st step kicked up the relay's armature and the second step was sufficient to hold it up.

    The combination of resistive rectification and the additional load of the relay and blower motor resulted in the loco going slower, so the second step also inserted an additional in-phase winding of 4 Volts which kept the hyper Voltage sensitive AC steamer motors happy.

    The Lionel relay operated for either polarity of DC.

    Note: Whenever an electronic whistle operated by an older linear transformer doesn't respond, try pressing the whistle button only part way.

    I don't know if that's any help on a modern transformer such as mine, which happens to non-linear as opposed to the older linear transformers. However, the old style whistles do work off my modern transformer. The part where it talks about the center rail (O-gauge has 3) is where the AC line (DC +) are on the center and the AC Neutral (DC -) are on the 2 outside rails.

  14. praondevou

    praondevou Well-Known Member

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    Well, from the youtube link I posted it doesn't look like it's putting a real DC on the AC line, rather it's changing the phase cut of each halfwave in order to create the DC, so it may still work inside the tender the way they described it in the link, with just a DC relay...
    However, you will see it when you take it apart.

    I would possibly see if I can take a remote controlled toy car and use it's circuitry... (they are very cheap too)
  15. iakobos

    iakobos Thread Starter New Member

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    After watching the youtube link I see what you mean about it not putting out a true DC. That's when I realized I should learn more about how the Lionel board works instead of assuming how it works.

    I took apart my Lionel whistling tender and saw that the circuit is rather simple. I drew it up and posted it below. When I measured the amount of DC it took to operate the K-Line motor it was 9VDC. I measured the Lionel and found that it operates the whistle motor with 12.9-15.4 VDC. I thought that was interesting since it means the motor doesn't operate off the 2.5 VDC from the transformer but the circuit board is converting the AC line to DC. After thinking about it a little bit, I soldered the board from my Lionel to the K-Line whistle motor and it works perfectly.

    I'm a little unsure that I have the transistor drawn correctly but I if I build or buy (shudder) a copy of the Lionel board I'll be good to go.

    Attached Files:

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