Discussion in 'General Electronics Chat' started by Jotto, Apr 20, 2011.

  1. Jotto

    Thread Starter Member

    Apr 1, 2011
    Think I have found one that I can play with a bit.

    Current limiting resistor

    5-1.15/.01=385 ohm, 390 ohm standard.

    Load resistor

    5-(0.7+2.0)/.01=230 ohm, 240 ohm standard.

    Basically the same as you posted previously, but now I can change it up a bit.

    Current limiting resistor

    24-1.15/.01=2285 ohm, 2.4k, e24

    Load resistor

    24-(0.7+2.0)/.01=2130 ohm 2.2k, e24

    Still need to learn a bit more on using datasheets, but I think I have this correct using your numbers for the LED.
  2. SgtWookie


    Jul 17, 2007
    It is not a good idea to use my username in the title, as it may discourage others from attempting to answer your question; and I may be offline for awhile very soon.

    It would have been better if you had simply replied to the original thread, here:
    Hopefully a Moderator will append these messages to that thread.

    This looks OK.

    I guess you're talking about LED1 and R2 in this schematic:

    If so, then you're OK.

    If you're talking about the IR emitter side being supplied with 24v and a current of 10mA, then you're fine.

    Yes, that works too.

    Don't forget to calculate the power requirements for the current limiting resistors.
    For example, the 24v supply on the IR emitter side:
    P= 22.85 * (22.85/2400)
    P = 22.85 * 9.52mA
    P = 218mW
    Now, for reliability's sake, we multiply the result by 1.6
    2.18*1.6 = 348mW
    1/4 Watts is not enough, you'd need to use 1/2 Watt rated resistors. If you tried it with 1/4 Watt resistors, they would get quite warm, and the reliability would suffer.
    Last edited: Apr 20, 2011
  3. Jotto

    Thread Starter Member

    Apr 1, 2011

    thanks for the help.