I have found a plenty of documents and books that model how the voltage across a capacitor behaves within a transient RC circuit, using the following equation:
$$V_C=V_{MAX}(1-e^{-t/RC})$$
Unfortunately, I have found no resource that discusses how to mathematically model an RC circuit, were one to provide a linearly increasing voltage source as an input.
Attempting to substitute VMAX in the equation above, for a linear equation, results in an equation that converges towards the linear equation, meaning current would cease after a time (I=(VS-VC)/R). This is obviously untrue, since we should be seeing current approach a constant value with time, as given by:
$$I_C=C\frac{dV}{dt}$$
I am fully aware how the voltage across a capacitor would behave with a linearly increasing voltage source, there are plenty of simulators that display that, and I can even think of a physical explanation for the results. What I wish to know is how one could mathematically model the voltage across a capacitor with a linearly increasing voltage source, in a similar way to the equation that models voltage across a capacitor in transients.
Best Answer
This answer is all about converting the circuit to a transfer function in the frequency domain then multiplying that T.F. with the Laplace transform of the input to get the frequency domain equivalent of the output. Finally, a reverse Laplace operation is performed to obtain the time domain formula for the output.
The Laplace transform of a low pass RC filter is: -
$$\dfrac{1}{1+sRC}$$
This is the frequency domain transfer function so, if you multiply this by the frequency domain equivalent of a ramp (\$\dfrac{1}{s^2}\$) you get the frequency domain output: -
$$\dfrac{1}{s^2(1+sRC)}$$
Using a reverse laplace transfer table this has a time domain output of: -
$$t + RC\cdot e^{(\frac{-t}{RC})}-RC$$
See item 32 on the table or, if the formula didn't have an obvious table entry you can use an inverse laplace calculator that solves it numerically like this one.
The calculator allows you to build the formula and enter a numerical value for RC. I used an RC value 7 in the above example so I could see how that number propagated to the final answer. The last hurdle is substituting that propagated value of 7 with RC. In other words it's a numerical solver but nevertheless a very useful tool to have to hand: -