I think I may have discovered a simpler way to operate the Lionel 153 Block Signal using an insulated rail and external power source and would like to get opinions and/or verification from others. ( I am not trying to control blocks with the signal but only make use of the light change feature.) My wiring is this: Connect the common of the external power source to the layout common, i.e. the outside rails generally. Connect the supply (hot) of the external power source to the red lamp's terminal. Connect the green lamp's terminal to the layout common. Connect the signal's common (center terminal) to the control rail (insulated outside rail). As I run a train around the loop, the block signal has green on and red off when the train is not on the insulated rail and red on and green off when the train is on the insulated rail, which is just what I was hoping to achieve. I just happened to tumble on to this after working with lionelsoni's wiring directions found in my previous post ( Lionel 153 Block Signal). Again, I would appreciate comments as to what is happening electrically and perhaps verification that the wiring is sound. Thanks.
Almost.
Both bulbs will be "on" with the brightness of half the supply voltage while no train is occupying the control rail.
With the train on the rail, the green light will be shorted & go out, the red will go to the full brightness of the voltage of the power supply.
Rob
Rob,
Thanks for the explanation. I am using a 1033 as an external power supply and using the U and A terminals so I can vary the voltage. I don't know if it's just a peculiarity of my particular block signal, but with the voltage at 13.25v and the train not on the insulated track, the green light is quite sufficiently bright and the red light not visible at all (even when I covered myself and it with a blanket-I'll investigate more tonight when it's dark). At 16v the green light is very bright and the red just barely showing some illumination. With the train on the insulated track, the red is as bright as the green was previously. I'm guessing that the extra lamp in parallel that lionelsoni includes in the wiring is to deal with the red light being on when the train is not on the insulated track. Without his original help, I would not have stumbled upon what I did. And thanks very much to you for helping me understand a bit more about the electrical aspects of the hobby.
The bulbs are mis-matched, then, with the green bulb having higher resistance than the red. If you can do this with consistency, by identifying the bulb types, now you are on to something.
With identical type bulbs, you will get the results I described.
One more, and perhaps last, question.
I tried another red bulb and found the situation as Rob described above. In fact it required two 14v bulbs, as indicated by lionelsoni that it might, in order to accomplish the desired result. My question is this: What would be the rating of a single resistor which would accomplish the same result as these two bulbs?
Thanks again.
As I mentioned earlier, Albert Kalmbach's 1943 version of this circuit used an ordinary resistor in place of the extra lamp. Lamps work much better in this circuit because the current that the lamp draws varies as the .55 power of voltage, not proportional to it as with a resistor (Ohm's law). So the voltage across the lamp increases almost as the square of the current. The number-53 lamp draws 120 milliamperes at 14 volts; so putting a number 57, which draws twice that, in parallel with the red lamp means that the red lamp gets only 1/3 the current that the green one does when they are in series. Because of the near-square-law behavior of the incandescent lamps, that means that the red lamp gets only about 1/7 the voltage of the green lamp, or about 1/8 of the total voltage. Furthermore, since an incandescent lamp puts out light proportional to the 3.5 power of voltage, the red lamp in that circumstance is only about 1/1000 as bright as the green one.
We could replace that number 57 lamp with a resistor having the same voltage/current ratio (which is its resistance) when the red lamp is off and get exactly the same performance. That would be about (14/8) /.08, or about 22 ohms. The problem comes when the train turns on the red lamp, in which case the resistor is dissipating 14^2/15, or 9 watts. The number 57 lamp, on the other hand, dissipates only about 3 watts when lit. (All of this assumes a 14-volt supply.)
So Kalmbach's resistor version certainly works. It just consumes a lot more power.
Bob Nelson
Thanks, Bob, for the explanation. I am a numbers and formulas sort of person, so the actual data help me understand.
I'd use the dissimilar bulbs, it works and doesn't require extra components.
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