Charging time for C2 capacitor in attached figure

Thread Starter

aamirali

Joined Feb 2, 2012
412
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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Mike33

Joined Feb 4, 2005
349
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!
 

YokoTsuno

Joined Jan 1, 2013
43
Are you familiar with the Laplace transform method?
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
 
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