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My electrical experience is limited to ordinary soldering of resistors‎, pots, and relays to control my model railroad layout, so please excuse my lack of more advanced logic circuit and semiconductor knowledge. 

What I'd like to do is reduce 7 volts to 4 volts on a particular track when a photocell trips a DC polarity reversal ‎(½ amp max current). ‎The idea is that a train on this track would shuttle back and forth in automated mode. I've installed the photocell device, which senses the train and reverses track polarity‎, causing the train to stop and run in the opposite direction.

‎The problem is that the track has a steep downgrade and  voltage needs to be reduced to 4 volts to avoid excessive speed going downhill. Of course, I can manually vary the track voltage at the transformer, but that defeats the idea of having the train continuously shuttle back and forth in automated, hands-off mode.

Any suggestions for a circuit that would block the 7 volts and direct the current flow to a resistor or pot limiting track voltage to 4 volts, when the DC current changes from positive to negative?‎

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  • \$\begingroup\$ How about a resistor and a diode in parallel? The resistor will drop voltage (you decide how much), and the diode will mostly short out the resistor in one polarity. \$\endgroup\$
    – user253751
    Feb 5, 2018 at 3:02
  • \$\begingroup\$ It would help a lot to know exactly what "photocell device" you are using and exactly HOW it reverses the track polarity. \$\endgroup\$
    – jonk
    Feb 5, 2018 at 3:11
  • \$\begingroup\$ It's a Miniatronics RU2-1 Automatic Reversing Unit commercial product that uses photo resistors between the rails; when the train goes over the photo resistor, the resistance goes up and triggers a reverse. The train slows to a stop, waits the preset amount of time for that circuit, then reverses DC track polarity and starts the train up again in the opposite direction. \$\endgroup\$
    – JGH
    Feb 5, 2018 at 3:41
  • \$\begingroup\$ I'd need more information than a picture and a price. I'd need the schematic diagram of the reversal process it employs. Your better bet, in my opinion, is to examine it closely and work out how you can add a different voltage supply that it uses when reversing. That might also be done by adapting another circuit to its existing technique. But again, schematic needed. I know it sounds "easy" to you to just drop the voltage down. But this is also POWER, not just signal. So that's where I'm coming from here. But if no schematic, I'd rather not guess about a good approach to try. \$\endgroup\$
    – jonk
    Feb 5, 2018 at 10:38
  • \$\begingroup\$ No schematic came with the device. All I know is, it has 6 terminals for making connections: #1= required 12v DC input (+); #2= 12v DC input (-); #3/4= photocell resistors; #5/6= 7v DC output to track. The device internally swaps polarity between #5 and #6 depending upon which photocell is triggered, ie, #5 output to Track 7v DC (+) and #6 output 7v DC (-) change to #5 negative and #6 positive. How it accomplishes this internally I don't know. The point is, I want to add a downstream circuit taking the resultant 7v DC output and reduce to 4v when polarity changes \$\endgroup\$
    – JGH
    Feb 5, 2018 at 17:50

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Thanks, immibis, for your answer. As you suggested, I attached a 5 watt diode to the sensor's 7v + ‎output and a 10 ohm‎ resistor in parallel to reduce voltage to 3v when polarity reverses to negative, and that solved the problem.

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