Charging time for C2 capacitor in attached figure

Discussion in 'General Electronics Chat' started by aamirali, Apr 10, 2013.

  1. aamirali

    Thread Starter Member

    Feb 2, 2012
    415
    1
    How to calculate charging time of C2 cap in figure attached.

    If its only C1 & R1 then
    tcharge = tf + (ti - tf)(e^-R/tC).

    I had calculated it mathematically also.

    But how to do it for C2 here
     
  2. Mike33

    AAC Fanatic!

    Feb 4, 2005
    349
    25
    The trouble seems to be the fact that the voltage at the node where R1/C1 meet is rising, rather than instantaneous. Makes it very hard to predict the charge time of the R2/C2 combination using the standard formulae!

    I don't have any mathematical way for you to figure out the C2 issue, but I did sim it for you....1 time constant (where each cap charges to 63% of the supply voltage) occurs at 1.8ms for C1, and 3.2ms for C2.

    There must be calc involved to find the answer mathematically, ha ha!
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Are you familiar with the Laplace transform method?
     
  4. YokoTsuno

    Member

    Jan 1, 2013
    41
    9
    I was thinking the same:

    1) Laplace transform of two identical cascaded low pass filters.
    F(s) = 1/(C²R²s² + 3RCs + 1)
    F(s) = 1/(R²C²(s² + (3/RC)s + 1/R²C²))

    The left side of denominator has the form:
    D(s) = (s+a)(s+b)
    Hence:
    F(s) = (1/R²C²)(1/(s+a)(s+b))
    F(s) = (1/R²C²)(1/(s+a))(1/(s+b))

    2) Reverse Laplace
    F(t) = (1/R²C²)(e^(-at).e^(-bt)
    F(t) = (1/R²C²)(e^((-a-b)t))

    You just need to resolve a and b.
    D(s) = (s+a)(s+b)
    D(s) = s²+ (a+b)s + ab
    ab = 1/(R²C²) => a =1/(bR²C²)
    a+b = 3/(RC) => b = 3/(RC) - a

    Either via a matrix or by substituting a in b and b in a, you'll get a quadratic equation which results in 2 constants which have the form k/RC and l/RC.

    Good luck
     
Loading...