LCC110 Question

Discussion in 'General Electronics Chat' started by huwilerp, Jul 14, 2013.

  1. huwilerp

    Thread Starter New Member

    Jan 10, 2013
    10
    0
    I have already posted this topic before, but I am still not understanding fully. If I wanted to use 9v battery in this circuit could I? If yes or no why can or can't I..

    If I am correct the max input voltage is 5V to control the LEDs..

    R/S

    Patrick



    I am having an issue understanding the Datasheet for the THE LCC100 CLAIR IC.

    I am looking at the Input Control Current to Activate and it states a maximum of 8mA. So if I am going to power this IC with 1 AA battery and if my math is correct. I would need a 187 Ohm resistor (1.5V/.008A=187) between the input and the battery?

    When looking at this Datasheet what other information should I pay close attention to?

    Where can I look to see exactly what can go through the NC/NO contacts as in Amps and Volts?

    Thanks again for your help, as I am jumping into this new world of electronics.

    R/S

    Patrick
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,086
    3,024
    The datasheet says maximum continuous current is 50mA, and that as little as 8mA will activate it. Let's shoot for 30mA, shall we? Should be plenty safe and still get the job done with room to spare.

    The voltage drop across the LED is given as 0.9 to 1.4, typically 1.2. At the worst case of 0.9V, your resistor needs to drop battery voltage minus 0.9V. Battery is 9V assuming your battery is truly 9V - you need to confirm. So the resistor needs to drop 8.1V at 30mA. Using ohm's law: ∆V=I*R 8.1 = 0.03*R and R=270Ω. That happens to be a standard value.

    I'd give that a try and if it works, you're done. If it fails, check the actual current (by voltage across the resistor). You might need to add a bit more current by using a smaller value resistor, especially as your battery voltage sags to 7V or so.

    The resistor will dissipate power as heat: I^2•R so 0.03^2*270 = 0.24W. You should use at least a 0.5W rate resistor. More is better.

    The AA battery won't work, since the voltage will sag during use and not light the LED even though it may have juice left in it. You could use a AA if you also use a joule-thief circuit form a solar light to boost the voltage. That would pulse the LED. I'm not sure what that might do to your coupler - you might need a capacitor to smooth it out.
     
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  3. huwilerp

    Thread Starter New Member

    Jan 10, 2013
    10
    0
    Wayneh,

    Can you show me the math behind using a AA battery. I really appreciate it.
     
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