## Rising Voltage Charge across the Capacitor:

To begin with, taking advantage of "a picture is worth a thousand words", let's see just how easy it is to make a basic chart that will show us how this all works. Sure, I could give you a chart all made up, but you will be able to learn and understand all this a lot easier if you do it as we go - believe me! Pull up and print TC_Grid (54Kb PDF file), and then we'll see just how easy it is to plot and use important information for those Time-Constants. As we start to put information points on this chart, here is what they represent. We have a fully discharged capacitor (Ec=0V) that is placed in series with a resistance and a voltage source. What we will discover is that the capacitor charges at a non-uniform rate, and that more specifically, there are points in time (called Time Constants) where the rate of voltage charge across the capacitor will diminish as time goes on. Theoretically, the capacitor really never gets fully charged! What we will find is that at the , the1st Time Constantvoltage across the capacitorwill charge to 63.2% of the maximum voltage provided by the source. Let's assume a source of 100VDC, which would give us a capacitor charge of63.2V. Note that the remaining voltage, which is the difference between that across the capacitor and the source voltage is 36.8V (100V-63.2V).

Special Note:The actual figure is a little above 63.2%, as actually63.21206%, which accounts for the apparent number discrepancies you will find below. To plot the correct values use the RED values shown.Next we will find that at the , the2nd Time Constantvoltage across the capacitorwill charge to 63.2% of the remaining voltage (that 36.8V), and above the existing charged voltage(63.2V). We find that this is therefore 63.2% of that 36.8V, which will give us 23.26V to add on to the currently existing 63.2V to give us a new voltage across the capacitor of86.7V. We would calculate our remaining voltage as 13.5VNext we will find that at the , the3d Time Constantvoltage across the capacitorwill charge to 63.2% of the remaining voltage (that 13.5V), and above the existing charged voltage(86.7V). We find that this is therefore 63.2% of that 13.5V, which will give us 8.5V to add on to the currently existing 86.7V to give us a new voltage across the capacitor of95.0V. We would calculate our remaining voltage as 4.98VNext we will find that at the , the4th Time Constantvoltage across the capacitorwill charge to 63.2% of the remaining voltage (that 4.98V), and above the existing charged voltage(95.0V). We find that this is therefore 63.2% of that 4.98V, which will give us 3.14V to add on to the currently existing 95.0V to give us a new voltage across the capacitor of98.14V. We would calculate our remaining voltage as 1.83VFinally we will find that at the , the5th Time Constantvoltage across the capacitorwill charge to 63.2% of the remaining voltage (that 1.83V), and above the existing charged voltage(98.14V). We find that this is therefore 63.2% of that 1.83V, which will give us 1.16V to add on to the currently existing 98.14V to give us a new voltage across the capacitor of99.33V. We would calculate our remaining voltage as 0.67VNow, if you happen to have a "French Curve" or such variable curve drawing tool (actually, you could just about free hand it), you can draw in an approximation of this curve, at least close enough to show the ideas that we want to present.

- The charge across the Capacitor will get closer and closer to the 100% mark, but as you can see: 63.2% of what's left, and 63.2% of what's left .....never really gets there. Folks simply tell you that after 10 of these Time-Constant time frames, who will actually know just how close we might be. It's like being told to go half way to the wall, and then half way to the wall, and half-way to the wall ... when will your nose touch the wall??
- The growing charge across the capacitor is definitely not linear with time.
- Note especially that the voltage increase at the very beginning is
, and then gradually slows down as the voltage begins to reach its maximum value.very rapidNotice that nothing has been said about how much time is involved!That is because, regardless of the actual time involved.this principle is consistent- Now is when it's proper to say that the element of time is related to the value of the series resistance and the value of capacitance as a simple
product of the two. I.e. a resistance of 100Kohms and a capacitance of 10microfarads will give us a product of 1.0, which represents a time factor of simply 1 second per Time-Constant.- With these values in mind and looking at the chart we just drew out, that simply means that in 1 second the charge across the capacitance will reach 63.2% of the source voltage, and that in about 3 seconds it will reach 95% of the source voltage. Simple after all, isn't it?

## Falling Charging Current of the Capacitor:

An equally important consideration is that as the voltage across the capacitor rises, the current demand will diminish in the same logarithmic fashion, following the same curve, except that it is inverted, in that it starts out very high, rapidly diminishing at first, and then diminishing less and less as time progresses. To help understand a couple of key issues here, we need to grasp an often overlooked feature. Where it's easy to see that the voltage across the capacitor is initially zero when we begin the charge, most of us fail to realize that . This dead short will of course demand the maximum amount of current that the source can supply through the series resistance for this very first instant.this same capacitor is initially a dead shortAs time progresses, and the voltage charge across the capacitor rises toward the 100% mark, the current demand diminishes toward 0% from the 100% point (intentionally redundant with an earlier statement).

## Capacitive Reactance:

Capacitors, and they "REACT" by "do not like Voltage Changesabsorbing current" when the voltage across the capacitor is increased, and "gives back" the current when the voltage across the capacitor is decreased, to attempt maintaining the voltage at a constant value.When you consider this basic feature of the capacitor to absorb or give back this current, remember that the "time element" is a key factor. We saw where the initial changes were those that had the most impact on current changes. The rapidity of changes will involve this "time element" by appearing as lots of initial changes in a short time frame, in that the faster these occur, the greater the current changes within that time frame (remember how the greatest changes occured at the beginning). Considering that the more often these events re-occur, the greater will be the average current flow of the capacitor. This translates to an apparent "Lower Ohms", as a resistor of lower ohms would draw more current from a voltage source. The formula for Capacitive Reactance X _{C}=1/(2*pi*fr*C) shows that when the frequency (fr) is increased, the X_{C}(expressed in ohms) will decrease. In the same respect, increasing the amount of capacitance for any given frequency will cause X_{C}to decrease in ohms ("pi" is ~3.14159).Note that although we are using a set of curves (one for the rising current and the other for the falling current), a facet of truth is that we can look at a small segment of a curve as a slope. This slope can be related to frequency, in that low frequencies have a low rising slope, and high frequencies have a high rising slope. Compare with

Inductive Reactance

Phase Angles:When we realize that with these two curves, where the charging voltage starts at 0V, and the charging current starts at the maximum value, this is equivalent to a voltage sine wave and a current sine wave that are 90. At the instant that the voltage sine wave is at it's minimum voltage, the current sine wave is at it's maximum value.^{o}out of phase with each otherWith that same concept, with these same two curves, where the charging voltage has peaked, the charging current has diminished, this is equivalent to a voltage sine wave and a current sine wave that are again 90. At the instant that the voltage sine wave is at it's maximum voltage, the current sine wave is at it's minimum value.^{o}out of phase with each otherIf we are dealing only with pure capacitance, then the resulting phase angles between voltage and current are at the 90 ^{o}indicated, but if any resistance is involved, the resulting phase angle (for the circuit as a whole) will be altered from the 90^{o}of pure Capacitive Reactance to something less than the 90^{o}.Compare with

Phase Anglesfor Inductance

How this all fits in with Impedance.Impedance here is the result in the Ohms of the Capacitive Reactance that may exist, and the combination with any Resistance. Remember that the Capacitive Reactance in Ohms is a result of the frequency applied to the amount of Capacitance. Since there is a phase angle of the voltage across the capacitor vs. the current though the capacitor of 90 ^{o}, and the phase angle of the voltage across the resistor and the current through the resistor is always 0^{o}, the combination of these two will result in ohms that are not simple series or parallel computations. The phase angle will always be some value between 0^{o}and 90^{o}.This is why, on a Vector Chart, that the values associated with the capacitance are at 90 ^{o}to those values associated with the resistance.This also simply means that with any given RC circuit, that the resulting Impedance and Phase Angle will vary with the applied frequency.

## Imaginary Numbers:

## This description makes you feel like someone is pulling your leg, but here's where that concept comes from:

When you either apply voltage across a resistance, or shove current though a resistance, it causes a "power loss" as lost energy. Suppose though that you press down on a spring, or compress air into a device, you are "storing energy" (neglect for the moment that there is a miniscule amount of loss in this process). This stored energy can be recovered later. Now, we need to remember that a capacitor "stores energy" in its "Electrostatic Field". Remember also that voltage and current go hand-in-hand here to accomplish this. When you try to increase the voltage charge across a capacitor, it draws current from the source to accomplish this. This "stored charge" is in the form of energy stored in the electrostatic field, and if there is no change, there is no current flow. Current flow would only exist during voltage changes. If the voltage across the capacitor is applied to a resistive load, then current would now flow into that resistive load as the electrostaic field collapses. At some time in the past, someone (or a collection of "some ones") decided that where the energy is stored and not lost, the numerical values represented for that action should be represented by "Imaginary Numbers", rather than "Real Numbers" (Real Loss of Energy). These "Imaginary Values" are plotted at 90 ^{o}from the "Real Values".

## Now, for a real trick:

## Special - How we can utilize Constant Current Sources for a special effect.

Remember our discussions about Constant Voltage Sources (Thevinin's Equivelant Circuits), and Constant Current Sources (Norton's Equivelant Circuits)? A very interesting phenomenum occurs when we apply a constant current source to a capacitor to charge it. We will find that the charging voltage across the capacitor is a " " effect, instead of a RC "Time Constant" curve, in that the voltage will rise at a Linear Rate.LinearThe " " of this voltage across the capacitor is simply based on the amount of the "Constant Current" and the value of the capacitance. Inmathematics, they call this "Linear Integration".Rate of ChargeIn fact, remember that "Capacitance" is defined by the "Charge in Coulombs" that will occur for an applied voltage of 1V. (Although the definition of a Coloumb is 6.25x10 ^{18}electrons, I usually just say to think of this like a "bucket-full" of electrons). I.e. a 1microfarad capacitor will only hold 1microcoulombs at 1V, and it would take 100V to raise that charge to 100microcoulombs.The reason for bringing this up, is because it allows us to think of what's happening across the capacitor in terms of the resulting voltage, as we dump whatever fraction of Coulombs into the capacitor as a charge (or to suddenly "dump" the charge into a load of some kind, like an electronic flash unit). The purpose of such a circuit would be where we need a "Linear Voltage Change" (often called a "Linear Integrator Circuit"), such as in controlling the horizontal sweep of an oscilloscope. Look at "

Special Effects" caused by variable Constant Current Sources

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RCTimeConstants.html - SfE-DCS, ddf - 11/20/2001