Click here to purchase the entire book in PDF format. Capacitors Let's go back a couple of chapters to the concept of a water pump sending water out its output though a pipe which has a constriction in it back to the input of the pump. We equated this system with a battery pushing current through a wire and resistor. Now, we're replacing the restriction in the water pipe with a couple of waterbeds. Stay with me here - this will make sense, I promise. If the input of the water pump is connected to one of the waterbeds and the output of the pump is connected to the other waterbed, and the output waterbed is placed on top of the input waterbed, what will happen? Well, if we assume that the two waterbeds have the same amount of water in them before we turn on the pump (therefore the water pressure in the two are the same... sort of...) , then, after the pump is turned on, the water is drained from the bottom waterbed and placed in the top waterbed. This means that we have a change in the pressure difference between the two beds (The upper waterbed having the higher pressure). This difference will increase until we run out of water for the pump to move. The work the pump is doing is assisted by the fact that, as the top waterbed gets heavier, the water is pushed out of the bottom waterbed. Now, what does this have to do with electricity? We're going to take the original circuit with the resistor and the battery and we're going to add a device called a capacitor in series with the resistor. A capacitor is a device with two metal plates that are placed very close together, but without touching. There's a wire coming off of each of the two plates. Each of these plates, then can act as a resevoir for electrons - we can push extra ones into a plate, making it negative (by connecting the negative terminal of a battery to it), or we can take electrons out, making the plate positive (by connecting the positive terminal of the battery to it). Remember though that electrons, and the lack-of-electrons (holes) are mutually attracted to each other. As a result, the extra electrons in the negative plate are attracted to the holes in the positive plate. This means that the electrons and holes line up on the sides of the plates closest to the opposite plate - trying desperately to get across the gap. The narrower the gap, the more attraction, therefore the more electrons and holes we can pack in the plates. Also, the bigger the plates, the more electrons and holes we can get in there. This device has the capacity to store quantities of electrons and holes - that's why we call them capacitors. The value of the capacitor, measured in Farads (abbreviated F) is a measure of its capacity to hold electrons (we'll leave it at that for now). That capacitance is determined by two things essentially - both physical attributes of the device. The first is the size of the plates - the bigger the plates, the bigger the capacitance. For big caps, we take a couple of sheets of metal foil with a piece of non-conductive material sandwiched between them (called the dielectric) and roll the whole thing up like a sleeping bag being stored for a hike - put the whole thing in a little can and let the wires stick out of the ends (or one end). The second attribute controlling the capacitance is the gap between the plates (the smaller the gap, the bigger the capacitance). The reason we use these capacitors is because of a little property that they have which could almost be considered a problem - you can't dump all the electrons you want to through the wire into the plate instantaneously. It takes a little bit of time, especially if we restrict the current flow a bit with a resistor. Let's take a circuit as an example. We'll connect a switch, a resistor, and a capacitor all in series with a battery, as is shown in Figure 2.11. Figure 2.11: A battery, switch, resistor, and capacitor, all in series. Just before we close the switch, let's assume that the two plates of the capacitor have the same number of electrons and holes in them - therefore they are at the same potential - so the voltage across the capacitor is 0 V (In other words, they have the same electrical pressure.). When we close the switch, the electrons in the negative terminal want to flow to the top plate of the cap to meet the holes flowing into the bottom plate. Therefore, when we first close the switch, we get a surge of current through the circuit which gradually decreases as the voltage across the capacitor is increased. The more the capacitor fills with holes and electrons. the higher the voltage across it, and therefore the smaller the voltage across the resistor - this in turn means a smaller current. If we were to graph this change in the flow of current over time, it would look like the red line in Figure 2.12: Figure 2.12: The change in current flowing through the resistor and into the top plate of the capacitor after the switch is closed. As you can see, the longer in time after the switch has been closed, the smaller the current. The graph of the change in voltage over time would be exactly opposite to this, plotted as the blue line in Figure 2.12. In case you want to be a geek and calculate these values, you can use the following equations - I'm just putting these two in for reference purposes, not because you need to know or understand them: $$V_{c} = V_{in} \left ( 1 - e^{- \frac{t}{RC}} \right )$$ (3.78) $$I_{c} = \frac{V_{in}}{R} \left ( 1 - e^{- \frac{t}{RC}} \right )$$ (3.79) Where $V_{c}$ is the instantaneous voltage across the capacitor, $V_{in}$ is the instantaneous voltage applied to the whole circuit, $e$ is approximately 2.718, $R$ is the resistance of the resistor in $\Omega$, $C$ is the capacitance of the capacitor in Farads, $I_{c}$ is the instantaneous current flowing through the resistor and into the capacitor and $t$ is time in seconds. You may notice that in most books, the time axis of the graph is not marked in seconds but in something that looks like a $\tau$ - it's called tau (that's a Greek letter and not a Chinese word, in case you're thinking that I'm going to make a joke about Winnie the Pooh... It's also pronounced differently - say ``tao'' not ``dao''). Tau is the symbol for something called a time constant, which is determined by the value of the capacitor and the resistor, as in Equation 2.80: $$\tau = RC$$ (3.80) As you can see, if either the resistance or the capacitance is increased, the RC time constant goes up. ``But what's a time constant?'' I hear you cry... Well, a time constant is the time it takes for the voltage to reach 63.2% of the voltage applied to the capacitor. After 2 time constants, we've gone up 63.2% and then 63.2% of the remaining 36.8%, which means we're at 86.5%... Once we get to 5 time constants, we're at 99.3% of the voltage and we can consider ourselves to have reached our destination. (In fact, we never really get there - we just keep approaching the voltage forever.) So, this is all very well if or voltage source is providing us with a suddenly applied DC, but what would happen if we replaced our battery with a square wave and monitored the voltage across and the current flowing into the capacitor? Well, the output would look something like Figure 2.13 (assuming that the period of the square wave = 10 time constants). Figure 2.13: The change in voltage across a capacitor over time if the DC source in Figure 2.11 were changed to a square wave generator. The red plot is the output of the square wave generator, the blue plot is the voltage across the capacitor. What's going on? Well, the voltage is applied to the capacitor, and it starts charging, initially demanding lots of current through the resistor, but asking for less and less all the time. When the voltage drops to the lower half of the square wave, the capacitor starts charging (or discharging) to the new value, initally demanding lots of current in the opposite direction and slowly reaching the voltage. Since I said that the period of the square wave is 10 time constants, the voltage of the capacitor just reaches the voltage of the function generator (5 time constants...) when the square wave goes to the other value. Consider that, since the circuit is rounding off the square edges of the initially applied square wave, it must be doing something to the frequency response - but we'll worry about that later. Let's now apply an AC sine wave to the input of the same circuit and look at what's going on at the output. The voltage of the function generator is always changing, and therefore the capacitor is always being asked to change the voltage across it. However, it is not changing nearly as quickly as it was with the square wave. If the change in voltage over time is quite slow (therefore, a low frequency sine wave) the current required to bring the capacitor to its new (but always changing) voltage will be small. The higher the frequency of the sine wave at the input, the more quickly the capacitor must change to the new voltage, therefore the more current it demands. Therefore, the current flowing through the circuit is dependent on the frequency - the higher the frequency, the higher the current. If we think of this another way, we could pretend that the capacitor is a resistor which changes in value as the frequency changes - the lower the frequency, the bigger the resistor, because the smaller the current. This isn't really what's going on, but we'll work that out in a minute. The lower the frequency, the lower the current - the smaller the capacitor the lower the current (because it needs less current to change to the new voltage than a bigger capacitor). Therefore, we have a new equation which describes this relationship: $$X_{C} = \frac{1}{2 \pi f C} = \frac{1}{\omega C}$$ (3.81) Where f is the frequency in Hz, C is the capacitance in Farads, and $\pi$ is 3.14159264... What's $X_{C}$? It's something called the capacitive reactance of the capacitor, and it's expressed in $\Omega$. It's not the same as resistance for two reasons - firstly, resistance burns power (lost as heat) if it's resisting the flow of current; when current is impeded by capacitic reactance, there is no power lost. It's also different from a resistor becasue there is a different relationship between the voltage and the current flowing through (or into) the device. For resistors, Ohm's Law tells us that V=IR, therefore if the resistor stays the same and the voltage goes up, the current goes up at the same time. Therefore, we can say that, when an AC voltage is applied to a resistor, the flow of current through the resistor is in phase with the voltage. (when V is 0, I is 0, when V is maximum, I is maximum and so on). In a capacitive circuit (one where the reactance of the capacitor is much greater than the resistance of the resistor and the two are in series...) the current preceeds the voltage (remember the time constant curves - voltage changes slowly, current changes quickly...) by 90$^{\circ }$. This also means that the voltage across the resistor is 90$^{\circ }$ ahead of the voltage across the capacitor (because the voltage across the resistor is in phase with the current through it and into the capacitor). If this is a little tough to follow, try thinking of it a different way... Stand in a swimming pool up to your neck in water and take a dinner plate and hold it in your hands like you would hold the steering wheel of a car. Now start pushing and pulling the plate forwards and backwards - you'll notice that this is hard to do because the water in the pool resists the movement of the plate. Now think about the relationship between where the plate is (its displacement), its speed and direction of travel (its velocity), and whether you're pushing or pulling (your force). These three things are illustrated (but not to scale) in Figure 2.14. Notice that while the plate is moving away from you, you're pushing. This is true whether the plate is near to you or far from you - your force is dependent on the velocity of the plate. One other thing to ask is where all your hard work is going - it's being used to push water out of the way. In other words, you're doing a lot of work for nothing. Figure 2.14: The relationship between the displacement of a dinner plate, its velocity and the force you have to apply to keep it moving while you're in a swimming pool. Ignore the vertical scale of these three plots - just think about their shape and polarities. Positive displacement is further away from you. Positive velocity is moving away from you. Positive force is you pushing. Now get out of the pool and stand in front of a concrete wall with a spring sticking straight out of it. Glue the bottom of your dinner plate to the spring so that if you push the plate, it moves towards the wall and squeezes the spring. If you pull the plate, it moves towards you away from the wall, and expands the spring. Now think about the relationship between the plate's displacement and velocity and your force once more. This is illustrated in Figure 2.15. You'll notice that it's a little different than when you were in the swimming pool. Now, your force is dependent on the displacement of the plate (since you're doing all the work to overcome the force of the spring). If the plate has moved away from you, you're pushing to overcome the compression of the spring - this is true regardless of whether the plate is moving away from or towards you. In addition, the force that you're putting on the plate is not lost. You're storing energy in the spring instead of just losing it as you were in the swimming pool Figure 2.15: The relationship between the displacement of a dinner plate, its velocity and the force you have to apply to keep it moving if the plate is glued to a spring sticking out of a wall. Ignore the vertical scale of these three plots - just think about their shape and polarities. Positive displacement is further away from you. Positive velocity is moving away from you. Positive force is you pushing. Let's get back to the circuit we were talking about before all of this dinner plate stuff... As far as the function generator is concerned, it doesn't know whether the current it's being asked to supply is determined by resistance or reactance - all it sees is some THING out there, impeding the current flow differently at different frequencies (the lower the frequency, the higher the impedance). This impedance is not simply the addition of the resistance and the reactance, because the two are not in phase with each other - in fact they're 90$^{\circ }$ out of phase. The way we calculate the total impedance of the circuit is by finding the square root of the sum of the squares of the resistance and the reactance or : $$Z = \sqrt{R^{2} + X_{C}^{2}}$$ (3.82) Where $Z$ is the impedance of the RC combination, $R$ is the resistance of the resistor, and $X_{C}$ is the capacitive reactance, all expressed in $\Omega$. Remember back to Pythagoreas - that same equation above is the one we use to find the length of the hypotenuse of a right triangle (a triangle whose legs are 90$^{\circ }$ apart) when we know the lengths of the legs. Get it? Voltages are 90$^{\circ }$ apart, legs are 90$^{\circ }$ apart. If you don't get it, not to worry, it's explained in Section 2.5. Also, remember that, as frequency goes up, the $X_{C}$ goes down, and therefore the $Z$ goes down. If the frequency is 0 Hz (or DC) then the $X_{C}$ is $\infty$ $\Omega$, and the circuit is no longer closed - no current will flow. This will come in handy in the next chapter. As for the combination of capacitors in seies and parallel, it's exactly the same equations as for resistors except that they're opposite. If you put two capacitors in parallel - the total capacitance is bigger... in fact it's the addition of the two capacitances (because you're effectively making the plates bigger). Therefore, in order to calculate the total capacitance for a number of capacitors connected in parallel, you use Equation 2.83. $$C_{total} = C_{1} + C_{2} + ... + C_{n}$$ (3.83) If the capacitors are in series, then you use the equation $$\frac{1}{C_{total}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + ... + \frac{1}{C_{n}}$$ (3.84) Note that both of these equations are very similar to the ones for resistors, except that we use them ``backwards.'' That is to say that the equations for series resistors is the same as for parallel capacitors, and the one for parallel resistors is the same as for series capacitors.