# 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
2
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

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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
790
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