I understand the operation for capacitors, which make a lot of intuitive sense to me, and seems deceptively simple. I know that the derivative of the voltage across a capacitor (the slew rate, if you will) is proportional to the current, for a given capacitance. So when the voltage is not changing much, like when it is connected across power supply rails, and the capacitor has little-to-no current flowing through it, and it appears as a high impedance. Fair enough. And when I am winding up the wick on the constant-voltage power supply, so that the slew rate is constant, and the voltage is changing at a constant rate, the current will be relative to how fast that voltage rising/falling. Again, that makes sense. I have proven that to myself time and time again. The technical math way of showing all that is this: dV/dT * C = I or with units plugged in: d(volts)/d(seconds) * Farads = Amps (or something like that) Likewise, I know that I can do things in reverse, and that forcing a controlled current through the capacitor, that the slew rate of the voltage across the capacitor will be proportional to that. In other words; the voltage will be integrated over time (as it steadily rises or falls). I do not like integrals in math (esp. when they require by parts or partial fractions!), but the concepts do come in handy in practical design! This is also my 2nd favorite way of imagining what an integral are! (My favorite is actually the water cup or well analogy, where a water faucet or hose is a function, and a the level of water in the cup, pool, well, etc. is the integrated result. That makes the function of integrals really clear and deceptively easy LOL!) So, I basicly have been trying to figure out how to take these simple, easy-to-understand relationships, and take ohms law, and have a super basic RC circuit, with 5V, 5 ohms, and 0.1F. I have so far figured out how to take ohms law, substitute I in the capacitor formula, and get a function. HOWEVER, this is where I get stuck. -------------------------------------------------------------------------------------------------------- Here is the circuit: +5V---(V2)-----/\/\/\/\/\/----(V1)-----| |-------0v----(grd) So after having a bit of a think, I have figured out that I really need to account for 2 different V's. The 5V power supply, V2, and the voltage across the capacitor, V2. I know that the current flowing through everything in a series circuit is the same, so then I can easily figure out current by calculating the voltage drop across the resistor which is this: I = (V2 - V1) / R So now, lets plug that into the mysterious capacitor derivative thingy: (All I did was substitute the I in the capacitor formula with the that ohms law formula above.) dV(1)/dT * C = (V(2) - V(1) ) / R. -------------------------------------------------- Now, I just plug in the values, and simplify as much as I can, to make it more friendly to my eyes. C = 0.1F, and R = 5 in my case, since I am using a 5 ohm resistor, 100mF capacitor, and I know V2 = 5V, since that is the output of a 7805 voltage reg. d(V(1))/dT * 0.1F = (5V - V(1) ) / 5ohms. That dV/dT is a bit hard to look at, so I will use V prime, or V' to represent that instead: V(1)' * 0.1 = (5 - V(1)) / 5 I do not know how to simplify that further though, to end up with that weird inverse exponential curve that is the time constant thingy, with that decaying charging curve. I have V on one side, and V' on the other side. :( So this is how far I got: 5 + (1/2 * V(1)') = V(1) EDIT: I HAVE ADDED THE PROOF HERE IF YOU WANT IT