Bi-color 2-wire do I connect it?

Discussion in 'The Projects Forum' started by cazksboy, Apr 23, 2012.

  1. cazksboy

    Thread Starter Active Member

    Nov 9, 2009
    I'm afraid I have a question that stumps me but most of you could do in your sleep - so I'm hoping you'll indulge an occasional hobbyist with minimal experience and no theory training!

    I'm modifying a "universal" motor (this one) so that it will reverse direction, for use driving my watchmaker's lathe (I'm a clockmaker/watchmaker). It's a DC motor with an internally-mounted bridge diode allowing it to run from the wall AC outlet. I am basically following the attached diagram for wiring. I removed & replaced the bridge diode externally in a separate enclosure alongside a DPDT center-off toggle switch so I can reverse polarity to the motor's brushes. I managed to burn out the original bridge diode, and I replaced it with this one.

    MY QUESTION: I want to add a bi-color 2-wire LED (this one) to the circuit just to glow different colors depending on the motor's spin direction.
    So how do I hook up the LED in the circuit? I know I need a resistor to achieve a voltage drop so I don't burn the LED (the bridge diode puts out 110 VDC and the LED only needs 2 VDC) but that's as much as I know.

    I need to know the following:

    1. What value for the resistor (ohms and watt-rating)?
    2. Should the resistor be wired in series or parallel with the LED?
    3. Should the resulting LED/resistor pair be placed in series or parallel with the motor's brushes?

    Maybe the LED/resistor pair is hooked up some other way, I don't really know. But if someone can just explain in clear descriptive English, I know I can follow directions and accomplish it!

    Thanks in advance....
  2. panic mode

    Senior Member

    Oct 10, 2011
    i guess you are still confused


    but are you sure you get 110VDC? did you measure it?
    when AC is rectified you may get up to 170VDC.
    it will be somewhere between the two (110 and 170VDC)

    If supposedly we have
    Iled=0.010A (10mA)


    R= 123/0.01=12300 Ohm

    Actually nearest standard value is 12k so current will be a bit higher than 10mA.

    Power is calculated as P=V*I = 123*0.01=1.23W

    but this is actual heat dissipated by resistor, to survive that you need resistor that is rated little bit higher such as 1.5W, or 2W ... or even 5W (larger device has larger surface area so it will feel less hot even though same amount of heat is dissipated).


    in worst case (motor not connected but LED and resistor still running, peak voltage is 170VDC)
    you get P=2.4W. This is assuming that you are pressing and holding that switch continuously (not sure if it is momentary action or not).


    using V=170 and R=12000 you get 2.4W.

    So to cover for this, resistor should be 3W (or larger).

    the resistor and LED must be placed in series. this circuit is then placed in parallel with motor.
    if you connect it in series with motor, then LED resistor will limit current to motor to that same 10mA and motor will not work.

    so how about this?
    Last edited: Apr 23, 2012
  3. cazksboy

    Thread Starter Active Member

    Nov 9, 2009
    panic mode, thanks! That makes it absolutely clear. I appreciate your patience and effort...

    By the way, I installed the new Radio Shack bridge diode and got the motor working just fine. But I confirmed the output DC voltage, and it measured between 109 VDC and 109.4 VDC.

    Thanks for the math formulas, they will help me in the future.
  4. Audioguru


    Dec 20, 2007
    I don't know if your AC is 110VAC, 115VAC, 117VAC or 120VAC so I assumed 120VAC.
    I don't know if your meter is cheap junk that is wrong or is a good one.

    A cheap meter measures the peak of the sine-wave then attenuates its reading x 0.707 times. Then it shows 120VDC when the 120VAC sine-wave is full-wave rectified which is too high since the real average is 108VDC.