OK I try to make simple
Formula
I(t) = C dv /dt
C is capcitance, dv/dt rate of change of voltage over time
Question : what should I know to find out dv/dt?
I think you have the wrong idea about dv/dt here.
dv/dt describes the change of voltage with time. The voltage in question is what you are applying to the capacitor: you have control over it. So you know how it changes over time. C is a scalar value, Capacitance, which is defined as the
ratio of the charge impressed upon a capacitor to the change in potential caused by that change. This can be restated as the quantity of charge (usually expressed in a count of electrons) which must be added to raise the voltage by one unit of electrical potential.
For our system of electronics, we use the Farad which is defined as one
coulomb, which simultaneously is considered a dimensionless number (being the charge that passes any point in a circuit in 1 second when 1 amp of current is flowing) and a measure of electrons (6.24... x 10^18 e-), and is the charge to add to a capacitor of 1F capacitance to raise its potential by 1 Volt. (This is the fundamental basis of electronics units, incidentally: 1A for 1s moves 1 C of charges. 1C added to 1F raises potential 1V. 1A through 1Ω causes a 1V drop in potential, etc.)
So, if you have a capacitor with 1F capacitance and 1V potential, you know it has 1 coulomb of charge on it. If we add another coulomb, the charge rises by 1V to 2V. Similarly, if you apply 1A of current for 1s to the capacitor, charging the already-charged plate, and stop, the potential across the capacitor will rise by 1v, but if you apply the current to charge the other plate for one second, the potential will become 0V, regardless of the absolute number of electrons on either plate.
dv/dt doesn't describe how much voltage is applied, but rather the rate of change. In fact, this is a rather simple thing to determine, because it is saying that, if the change of applied voltage is constant (i.e., constantly rising or falling) the current created will be constant. This is where most people's heads explode. So let's look at it differently.
We know that a constant current will provide the same amount of charge per unit time. The voltage, according to I = C dv/dt (where I is average current, and i would be instantaneous current) that C is a constant and I is (because we're providing a constant current) a constant, and therefore the only thing that changes is voltage across the capacitor. Therefore, we know, if dv/dt applied to C is a constant, I will remain constant.
Similarly, we can tell that, if there is no change in applied voltage, dv/dt = 0 and I =0. Without changing the voltage on the capacitor, we won't cause current to flow. At all.
If dv/dt is a very high rate of change, a high current will be involved. If dv/dt is very small, a very low current will be involved.
If dv/dt is varying, the current rate will vary as well as the first derivative of the voltage! This is important because if v=A sin(t), then current will vary as cos(t), and the amplitude will be dependent on A and C.
These are all qualitative features of i = C dv/dt. The fact of the identity means you can calculate values when you know values, but this is really secondary to understanding what the equation says.
The circuit that implements i = C dv/dt is simply a capacitor with a controllable current source, ideally without source resistance, etc. In the Real World, you won't see these things work out exactly, mostly because the charge rate will be controlled by the source internal resistance, the resistance of the wires, and the leakage of the capacitor. Because of these things, which we can theoretically lump into one resistance, we can picture it (as a realistic world view with ideal parts) as a controlled current source with no internal resistance, in series with all of the circuit resistances, in series with an ideal capacitor with all of the circuit's capacitance lumped together. The result, then, is to modify the instantaneous i by taking into account the time constants (which are R X C in seconds).
When the capacitor is in a series circuit with a resistance (which it _always_ is in the real world), the effect is the same as an RC circuit which has a signal source to the left, a series capacitor, a resistor from the capacitor's right lead to ground (which we declare to be the terminal of the source which is not connected to the capacitor, and our reference 0v point for measurements) and the output taken across the resistor. Why do we take the output across the resistor?
- the resistor is a passive component whose voltage is independent of active-component effects, so we can be sure that its voltage is an accurate representation of the current flowing through it.
- the current in a series circuit is the same at every point in the circuit
- so measuring the voltage across the resistor gives us a scaled readout of the current, I=E/R. For a 1Ω resistor, I will be the same value as E, in amps (because E/R is V/Ω = A).
Would something bad happen if we measured across the capacitor? Possibly. We might provide an increase of leakage current effect by paralleling the measuring device (which requires some current to be able to make a reading). The capacitance of the capacitor might resonate with or otherwise affect the measuring device. So we go for the safe component.
This means that the capacitor voltage can be
deduced (calculated when we have numbers) from the resistance voltage. Now, we can control the circuit current by controlling the voltage across the entire string, we know the resistor voltage, and we can subtract that from the capacitor to know the capacitor voltage. And what we find is:
- with an applied pulse of voltage V for a given time t, the resistor's voltage instantly at start is 100% of the applied voltage. This is because the capacitor acts like a short, i.e. just a wire.
- as the capacitor starts to charge, it develops a voltage, tiny at first, but increasing quickly, and this voltage is subtracted from the resistor (because the entire circuit is provided V for the period t, and the capacitor and resistor act as a simple voltage divider at any instantaneous time.)
- After one time constant, if t is longer than one time constant, C will have acquired 63% of the charge it is going to be able to hold (based on its capacity) and that time constant's length is R x C.
- After 4 times constants, it will be up to about 98.12%, and after five, it will have risen to 99.3%. At this time, the resistor will be dropping a very low voltage, about .7% V. This will decrease at a decreasing rate until, some time far in the future, the resistor will drop 0v, the capacitor will drop 100% of the voltage (because of the charge stored in it) and no current will flow.
If, then, the source is disconnected, no current will continue to flow because of the open circuit, but if it is left connected and we reach time t, it will drop to 0v. At this point, we have (through the source) the equivalent of a short between the input end of the capacitor and ground. Now, the capacitor will discharge. The Capacitor will become, effectively, the source, and its entire output will be dropped across the Resistor, which will allow maximum current. Not infinite current, of course, because the resistor is not a short. Instead, the current that flows is the Capacitor charged voltage divided by the Resistor's resistance: i=e/R. (Remember, small i and e are instantaneous values.) As the capacitor discharges through the resistor, its voltage falls, according to the i = C dv/dt equation, and that reduces the voltage across the resistor. As e -> 0, I -> 0. The discharge of the capacitor is governed by the same physics as the charging, so after one time constant, 63% of the charge will have left, and the capacitor's voltage will have dropped to 63%.
What if the input from the controlled source is sin(t)? -cos(t)? A triangle-wave? A square wave? Work out each of these qualitatively first, then play with the math and see if your results match your qualitative expectations.
It is this working out of qualitative effects that is most useful in electronics: the numbers can always be done afterwards to determine if what ought to work really will. But it is unlikely (enough to be called serendipitous) that sitting around manipulating math is going to result in a working, useful circuit.