Detect 240V with a PIC

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  1. - 23rd October 2008, 14:58 #19
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    dhouston
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    240 (E) / 10000000 (R) = 0.000024 (I)
    0.000024 * 0.000024 = 0.000000000576 (I squared)
    0.000000000576 * 10000000 = 0.00576 watts (I squared * R)
    or
    240 * 0.000024 = 0.00576 watts (I * E)

    The clamping diodes will conduct when the voltage is > Vdd (+ diode drop) and when the voltage is < Vss (- diode drop) so, in essence, you have the resistor in series with a diode with 240V applied. The above calculations have ignored the forward voltage drop of the diodes as that's insignificant in comparison to 240V.
    Last edited by dhouston; - 23rd October 2008 at 15:04.
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