In the past I've seen several posts from people requesting circuits to indicate the position of turnouts, so I thought I would share what I have come up with. I originally developed these working with Kato Unitrack turnouts, but they will work with any turnout that has a live frog that switches polarity with the turnout and/or is power routing. These circuits light one of two LEDs depending on which route for which the turnout is set. The first circuit works with a live frog: (I did not include resistor values because that will depend on your track voltage, selection of LEDs, and how bright you want the LEDs.) The second circuit works with power routing turnouts(you double the circuit and put one on each leg of the turnout): Some advantages to these circuits are they work with DC and DCC and they are independent from the method of turnout control. Being independent from the turnout control is especially nice on turnouts like Kato's because the LEDs will match the turnout whether you throw it remotely or by using the slide on the side of the turnout. Some disadvantages are on DC, the circuit only lights when power is applied, and brightness varies with the amount of power(although in my experiments, if there was enough power to make an engine move, I could easily see which LED was lit). Also, the version for a live frog can be hard to wire, for example, on a Kato #6 turnout you would have to open the bottom of the turnout to wire it to the frog. Another neat feature is on Kato turnouts you could have the diodes and resistors in the roadbed and the LED's coming through to the top, made into a signal even, so that you could have the turnout and indicator in one self-contained unit. Let me know what you think about it!
How would one determine the resister value if they knew what the voltage and LED was? Is there a chart somewhere?
I'm not absolutely positive but I think it is determined by the equation V=IR. So if you were running 12V and using a 30mA LED you would need a 400 ohm resistor. Of course you would want to make sure you didn't blow it out so you wouls actually use a higher value. A 500 ohm resistor will be good to 15V.
V=IR(volts = current x resistance) is correct, but you left out the voltage drop of the LED. With a 12 volt supply and an LED with a 2 volt drop, to run it at 30ma(0.03 amps) would require (12 - 2 )/ 0.03 = 333 ohm resistor. If you can't find a resistor of the value you calculate, use the next size up.
Oh man this is like physics class all over again! Thanks for putting in the effort, i just might use these!
Hopefully this step by step helps...(By the way the formula is "Ohm's Law") 1) Find "V", the needed voltage drop for the resistor, by taking the voltage at the power source (e.g 12 volts) and subtracting the rated voltage for the LED (typically 2 or 3 volts). (If you have more than one LED in the circuit, subtract all of them. e.g., for 2 LEDs each rated at 2 volts, subtract 4 volts) 2) Find "I", the current for the circuit, by looking at the max current for the LEDs you are using. (Do not multiply the current by the number of LEDs in the circuit. Use the value for one LED.) 3) Apply the formula V/I=R. R will be the value for the resistor. 4) Pick a resistor with a resistance value equal to R or higher. Higher values will result in dimmer LED brightness. A value lower than R will probably result in your LEDs being destroyed.
Nice explanation, I admit that I was a little confused about the “voltage drop”. It’s been over 25 years since I did any electronics stuff and 40 years since I learned ohms law. I forgot that the LEDs only operate around 2 volts. Thanks.
Sorry, forgot about the voltage drop. I have mostly used 600 ohm resistors on the layout and haven't burnt any out and they seem bright enough for me. So I never really did the math. I just used what works. Eric
