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  1. #1
    Join Date
    Nov 2007
    Posts
    980

    Some basic questions for you...

    Ok, I'm finishing my controller box for my mini lathe CNC but I would like to add an exhaust fan and a pilot lamp for the power to it. I'm using a 48 volt power supply but I'm not sure how to connect the pilot LED (3.5v) and I also have a brand new PC case fan that's 12v.

    How would I safely connect both of these to my power supply to turn on and off with the main toggle switch that I'm using?

    Would I be better off using a standard 115v pilot lamp? If so, can I still connect the fan?

    Link to the PDF of the Keling 48V P.S.

    http://www.kelinginc.net/PowerSupplyDiagram.pdf

    Thanks for your time-
    Dave
    Dave->..

  2. #2
    Join Date
    Jan 2006
    Posts
    2985
    For the led you just need to limit the current, the voltage does not matter, as long as it is under the breakdown voltage of the led. Most LED's are happy with a current of around 10-20 mA. Ohms law tells us that we need a resistor of about 2.2 Kilo ohms.

    V = I*R
    (48-3.5) = .02*R
    44.5 = .02*R
    44.5/.02 = R

    R = 2225 ohms

    Closest standard value is 2.2 K, so we'll go with that. You could also use anything up to about 4.7 K without any problems. You probably want to stay at or above 2.2 K.

    For the fan I would recommend one of 2 things to make life easy:

    1) buy a 12v "wall wart" plug in power supply and connect it to the ac power in your box.

    2) buy a 48 volt fan and install it in your box.

    You can make/buy a voltage regulator to step down 48 to 12 but it may be more of a hassle than either of the other two options.

    Matt

  3. #3
    Join Date
    Nov 2007
    Posts
    980
    Thanks very much, Matt, I do appreciate the help. I may just go with the 48 volt fan as a 12V wall wart may be a bit large to add to the box but what a great idea. I'll have to remember that one

    Dave
    Dave->..

  4. #4
    Join Date
    May 2006
    Posts
    41
    If you're using a dropping resistor for the LED don't forget that the resistor has about 46 volts across it.

    At 20 ma, it's going to dissipate an appreciable amount of power ( 46V X .02A = .9W ).

    If you use a single 2.2K resistor, you have to use a 1 watt or larger part (and a 1 watt resistor actually dissipating 1 watt gets surprisingly hot, so watch those fingertips).

    Or, alternately, if you have some common 10K 1/4 watt resistors in your shop, you could use 4 in parallel to share the dissipation. (they'll get hot too, BTW)

  5. #5
    Join Date
    Nov 2007
    Posts
    980
    Thanks for the warning but what I've now ended up doing was using the idea above for the 12V wallwart as I'm able to fit one in the case and run both the fan and led off that.

    I found an LED calculator online and yes, it warned me of the heat that would dissipate and that wasn't sitting well with me so I simply thought since the wallwart I have can handle 2 amps, why not run them both off the 12V as that would be a a little more "safer"

    Thanks again-
    Dave
    Dave->..

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