Charging a capacitor in parallel with a resistor from a constant current source.

WBahn

Joined Mar 31, 2012
28,185
TS simply wants to tweak R and C for a given over-current limit.

Without R, V(t) will be reached in time t = C x V / I .
Add R in parallel will make t longer to reach V(t).

When I say "compare" I mean what is the value of t compared with RC?
Okay, so say t is 200 ms and RC is 100 ms. I compare them. t is 100 ms greater than RC. Now what? Or what if t is 50 ms? Or what if both t and RC are 100 ms?

What is the TS supposed to do after making this comparison?

MrChips

Joined Oct 2, 2009
28,151
Here is what the charge curve looks like for different values of R.

BobTPH

Joined Jun 5, 2013
6,310
Take the capacitor out and convert the Norton equivalent to a Thevenin equivalent, and you have the well known charge through a series resistor with a voltage source of V= IR.

crutschow

Joined Mar 14, 2008
31,543
Here's the simulation of the circuit and its Thevenin equivalent, showing their identical response:
In either case the equation I showed in post #16 describes the circuit behavior.