Re: Multiple LEDs on FX-DO


Hello Joo,

  Sorry about your name spelling.  I don't know how to type accents on my keyboard.

  I have a new in box Titan FX-DO and just had a quick look.   There is a error on the drawing for the bottom board.  Perhaps Kelly could have a look at this.   Connector P2 is mislabeled "TRUCKS";  when, in fact, only pins 1 and 2 are for the trucks (as are the schematically parallel P1 pins 1 and 2).   Connector P2, pins 3 and 4 are to be connected to the motor (as are the schematically parallel P1, pins 3 and 4).

  That being said, the wires from P1 are labeled correctly.

  If you look at the bottom of the bottom board, you will notice a large black box component with a hole in the middle.   That is a heavy duty, full wave bridge rectifier.   If you understand how a bridge rectifier works you will know that the voltage across the + and - side (output) is track voltage minus 1.4 volts.   In your case you will have 24 V - 1.4 V = 22.6 V total.

  Kelly recommends using an opto-isolator between the lighting and the top board.   This is the safest route as the expensive decoder board is protected.  The input side of an opto- isolator is internally similar to an LED.  You connect one side to plus +5 V and the other is connected to the decoder light port via a current limiting resistor.  The output side of the opto-isolator can then be used as a simple switch.  N.B. the isolated circuit, controlled by the output of the opto-isolator, must have the GND connected to the GND of the controlling circuit (the "-" side of the full wave bridge rectifier - same as the top board P1 - 8 and P2 - 8).

  A note about LEDs in series.   While red LEDs run around 2 volts,  white LED takes about 3 volts to turn on (some actually take 3.1 or 3.2 V to turn on).  Current controls the brightness.   That's why it's absolutely mandatory that a current limiting resistor be included in any LED circuit.   A +5 volt circuit can only drive one (3 Volt) LED.  However, a +22.6 Volt circuit can drive up to seven LEDs; 3 V X 7 = total 21 Volts (which is less than 22.6 Volts leaving room for the current limiting resistor). 

  We must choose our maximum current for the LEDs.    Most electronics engineers recommend a current maximum of 20 mA.  I use varied current depending on the LED function.   Headlights are brighter than ground lights.  For Headlights I might use 15 mA (very bright) while a ground light might only require 2 mA (considerably dimmer).

  Let's do some math.  Please excuse the simplicity, but this is a forum and our audience may not know how to calculate resistance:

e.g.    22.6 V  -  ( X  3 V)  =  1.6 V        We now use Ohm's law to figure out the resistor value for a 2mA circuit.       Volts = Amps X Ohms    or  Ohms = Volts / Amps

    1.6 V = 0.002 Amp X Ohms   or    Ohms = 1.6 V / 0.002      We get 800 Ohms.   If we don't have the exact value resistor we choose a value slightly larger.  We can use 1000 Ohms (1K Ohm, 1/8 Watt) with no problems.   

  If you decide to add more LEDs, you may add a second series circuit in parallel with the above series circuit.   Use the same formulas adjusting for quantity of LEDs in the series circuits.

  I hope this helps.   Let us know if you have any questions (or corrections).

Best regards,

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