Converting AC/DC for Rail Logic Signal System

The system senses current pulling through a block,not really necessary in 3 rail. I would just use the outside insulated rail method and relays. Much more reliable as no lighted cars are needed. 

System detector is for DC in and passes track power,3 amp or 6 amp depending on model. A bit inadequate for O gauge.  If you rectify the current you would have DC to the track and the whistle would blow and damage to some boards could result. . 

Dale H

I'm not sure it would be the cheapest, but you could use outside rail detection to drive an AC optocoupler.  The transistor side of the optocoupler would switch a separate DC power supply running through your logic board.

But that would make sense only if you really needed those particular boards to drive your signals or whatever.

Professor Chaos, could you point me in the direction of an optocoupler that could run off 24VAC pass 12VDC on the transistor side?

I use an AC source to drive 12-volt DC relays on the club layout. The way I do it is to feed one AC lead on a bridge rectifier from the 10VAC accessory tap. The other AC lead goes to the insulated rail. The DC leads feed the relay coil. When a freight car or locomotive rolls through the insulated section there's a DC feed sufficient to trip the 12-volt DC relay. You could run that output through a light bulb to generate the current draw to trip the sensors.

Just a thought.

I've used the PS2505.  (The link is to a 4-channel;  there are 1- and 2- as well).

The specs claim up to 80v and 50ma on the output.  However, I've only used it for 5V logic so I'm not sure if the output will pass 30ma @ 12V.  If not you may need another transistor.  Of course, you would need a resistor on the input to limit LED current to 80ma or less.

And, obviously, because it's AC the output will turn off 60 times per second (not to mention what happens with intermittent track contact).

If you need a consistently high signal you can use a capacitor on the output;  I use a capacitor + Schmitt trigger to get a clean 5V logic signal from the intermittent AC third rail voltage.

Thank you very much Dale, Matt, and Professor.  Plan on using at least two of the directions between my normal blocks and some complex turnouts.

Hi Marty

If you decide to use relays with the insulated rail method,it is described here.

http://www.jcstudiosinc.com/BlogShowThread?id=410&categoryId=

Dale H

Gentlemen, I am not sure you even need the relay in this case. If you use the insulated outside rail as your presence detector then the wheels will act as a simple switch between the insulated outside rail and the other outside rail. If the other outside rail is your layout common then all you need is to connect your block detector between a small 12 volt power supply and the insulated rail. Connect the other side of the 12volt DC supply to your layout common. Even a small wallwart should be suficient as these detectors generally do not require much current to operate. When a set of wheels makes the connection the detector will detect the current between the 12 volt supply and the common rail. Oh yes, I almost forgot, to prevent a short connect a small light bulb, 12 volt or larger, between the 12 volt supply and the detector or between the detector and the insulated rail. A relay or other smoothing circuit should not be necessary as the detector should have the necessary circuitry built in.

Al

Al

Maybe I am missing something. If you look at the second schematic,the track circuit of the hot goes in and out of the detector and requires DC current. If you rectify it then DC will go to the trains. I do not see anyway around it,other than building another system to fit in with whatever is already there.  You can make a system with a proper resistor and get an appropriate signal 15 ma or whatever to some device.

But if you go to the trouble of making insulated outside rails,a simple relay will give binary language for computer language and power the accessory directly. 

Dale H

Dale, normally this circuit works by sensing the current drawn by the powered unit or by the current drawn by some resistance placed accross the wheels. If the outside rail is insulated you have two circuits. One AC from the center rail through the train load to the neutral common outside rail, the other DC from the insulated outside rail through the wheel sets to the Common outside rail. Since these circuits are normally designed to filter out the dropouts one normally encounters in this circuit it should not be necessary to use the relays to filter the action. Relays can always be used of course but in this case they should not be necessary.

Al

Note that the two power supplys, the track and the accessory DC should be isolated from each other or you might run into problems.