Why Are Capacitors Important?
What Is A Capacitor?
Voltage Current Relationships In Capacitors
Energy In Capacitors
What Is Impedance?
Using Impedance
Some Impedance Laws/Combinations

Why Are Capacitors Important?

       The capacitor is a widely used electrical component.  It has several features that make it useful and important:

What Is A Capacitor?

        Capacitors are two-terminal electrical elements.  Capacitors are essentially two conductors, usually conduction plates - but any two conductors - separated by an insulator - a dielectric - with conection wires connected to the two conducting plates.

        Capacitors occur naturally. On printed circuit boards two wires running parallel to each other on opposite sides of the board form a capacitor. That's a capacitor that comes about inadvertently, and we would normally prefer that it not be there. But, it's there.  It has electrical effects, and it will affect your circuit.  You need to understand what it does.

        At other times, you specifically want to use capacitors because of their frequency dependent behavior. There are lots of situations where we want to design for some specific frequency dependent behavior. Maybe you want to filter out some high frequency noise from a lower frequency signal. Maybe you want to filter out power supply frequencies in a signal running near a 60 Hz line. You're almost certainly going to use a circuit with a capacitor.

        Sometimes you can use a capacitor to store energy.  In a subway car, an insulator at a track switch may cut off power from the car for a few feet along the line. You might use a large capacitor to store energy to drive the subway car through the insulator in the power feed.

        Capacitors are used for all these purposes, and more. In this chapter you're going to start learning about this important electrical component. Remember capacitors do the following and more.


        You need to know what you should get from this lesson on capacitors.  Here's the story.

      Capacitors and inductors are both elements that can store energy in purely electrical forms. These two elements were both invented early in electrical history. The capacitor appeared first as the legendary Leyden jar, a device that consisted of a glass jar with metal foil on the inside and outside of the jar, kind of like the picture below. This schematic/picture shows a battery attached to leads on the Leyden jar capacitor.

        Although this device first appeared in Leyden, a city in the Netherlands sometime before 1750. It was discovered by E. G. von Kleist and Pieter van Musschenbroek. Although it has been around for about 250 years, it has all of the elements of a modern capacitor, including:

        The way the Leyden jar operated was that charge could be put onto both foil elements. If positive charge was put onto the inside foil, and negative charge on the outside foil, then the two charges would tend to hold each other in place. Modern capacitors are no different and usually consist of two metallic or conducting plates that are arranged in a way that permits charge to be bound to the two plates of the capacitor. A simple physical situation is the one shown at the right.

        If the top plate contains positive charge, and the bottom plate contains negative charge, then there is a tendency for the charge to be bound on the capacitor plates since the positive charge attracts the negative charge (and thereby keeps the negative charge from flowing out of the capacitor) and in turn, the negative charge tends to hold the positive charge in place. Once charge gets on the plates of a capacitor, it will tend to stay there, never moving unless there is a conductive path that it can take to flow from one plate to the other.
There is also a standard circuit symbol for a capacitor. The figure below shows a sketch of a physical capacitor, the corresponding circuit symbol, and the relationship between Q and V. Notice how the symbol for a capacitor captures the essence of the two plates and the insulating dielectric between the plates.

        Now, consider a capacitor that starts out with no charge on either plate. If the capacitor is connected to a circuit, then the same charge will flow into one plate as flows out from the other. The net result will be that the same amount of charge, but of opposite sign, will be on each plate of the capacitor. That is the usual situation, and we usually assume that if an amount of charge, Q, is on the positive plate then -Q is the amount of charge on the negative plate.

        The essence of a capacitor is that it stores charge.  Because they store charge they have the properties mentioned earlier - they store energy and they have frequency dependent behavior. When we examine charge storage in a capacitor we can understand other aspects of the behavior of capacitors.

        In a capacitor charge can accumulate on the two plates. Normally charge of opposite polarity accumulates on the two plates, positive on one plate and negative on the other. It is possible for that charge to stay there. The positive charge on one plate attracts and holds the negative charge on the other plate. In that situation the charge can stay there for a long time.

        That's it for this section. You now know pretty much what a capacitor is. What you need to learn yet is how the capacitor is used in a circuit - what it does when you use it. That's the topic of the next section. If you can learn that then you can begin to learn some of the things that you can do with a capacitor. Capacitors are a very interesting kind of component. Capacitors are one large reason why electrical engineers have to learn calculus, especially about derivatives.  In the next section you'll learn how capacitors influence voltage and current.

Voltage-Current Relationships In Capacitors

        There is a relationship between the charge on a capacitor and the voltage across the capacitor.  The relationship is simple. For most dielectric/insulating materials, charge and voltage are linearly related.

Q = C V
where: You will need to define a polarity for that voltage. We've defined the voltage above. You could reverse the "+" and "-".         The relationship between the charge on a capacitor and the voltage across the capacitor is linear with a constant, C, called the capacitance.

Q = C V

        When V is measured in volts, and Q is measured in couloumbs, then C has the units of farads. Farads are really coulombs/volt.

        The relationship, Q = C V, is the most important thing you can know about capacitance. There are other details you may need to know at times, like how the capacitance is constructed, but the way a capacitor behaves electrically is determined from this one basic relationship.

       Shown to the right is a circuit that has a voltage source, Vs, a resistor, R, and a capacitor, C. If you want to know how this circuit works, you'll need to apply KCL and KVL to the circuit, and you'll need to know how voltage and current are related in the capacitor. We have a relationship between voltage and charge, and we need to work with it to get a voltage current relationship. We'll look at that in some detail in the next section.

        The basic relationship in a capacitor is that the voltage is proportional to the charge on the "+" plate. However, we need to know how current and voltage are related. To derive that relationship you need to realize that the current flowing into the capacitor is the rate of charge flow into the capacitor. Here's the situation. We'll start with a capacitor with a time-varying voltage, v(t), defined across the capacitor, and a time-varying current, i(t), flowing into the capacitor. The current, i(t), flows into the "+" terminal taking the "+" terminal using the voltage polarity definition. Using this definition we have:

ic(t) = C dvc(t)/dt

        This relationship is the fundamental relationship between current and voltage in a capacitor. It is not a simple proportional relationship like we found for a resistor. The derivative of voltage that appears in the expression for current means that we have to deal with calculus and differential equations here - whether we want to or not.


Q1  If the voltage across a capacitor is descreasing (and voltage and current are defined as above) is the current positive of negative?

        This derivative kind of relationship also has some implications for what happens in a capacitor, and we are going to spend some time exploring that relationship. Clearly, we need to understand what this relationship implies, and then we need to learn how it affects things when we write circuit equations using KVL and KCL.

        We'll start by considering a time varying voltage across a capacitor. To have something specific, let's say that we have a 4.7mf capacitor, and that the voltage across the capacitor is the voltage time function shown below. That voltage rises from zero to ten volts in one millisecond, then stays constant at ten volts.  Before you go on try to determine what the current through the capacitor looks like, then answer these questions.


Q2.  Is the current constant in the time interval from t = 0 to t = 10 msec?

Q3.  Is the current constant in the time interval from t = 10 msec to the last time shown?

        Now, you should be able to compute the current.

Energy In Capacitors

        Storing energy is very important. You count on the energy stored in your gas tank if you drove a car to school or work today. That's an obvious case of energy storage. There are lots of other places where energy is stored. Many of them are not as obvious as the gas tank in a car. Here are a few.

        Capacitors can't really be used to store a lot of energy, but there are many situations in which a capacitor's ability to store energy becomes important. In this lesson we will discuss how much energy a capacitor can store.

Capacitors are often used to store energy.

        To calculate how much energy is stored in a capacitor, we start by looking at the basic relationship between voltage and current in a capacitor.

i(t) = C dv(t)/dt

        Once we have this relationship, we can calculate the power - the rate of flow of energy into the capacitor - by multiplying the current flowing through the capacitor by the voltage across the capacitor.

P(t) = i(t)v(t)

P(t) = i(t)v(t)
i(t) = C dv(t)/dt
P(t) = (C dv(t)/dt) v(t)
P(t) = dE/dt = (C dv(t)/dt) v(t)
P(t) = dE/dt = (C dv(t)/dt) v(t)
Ec = (1/2)CV2
EM = (1/2)MV2
ESpring = (1/2)Kx2

Frequency Dependent Behavior For A Capacitor

       We start with a capacitor with a sinusoidal voltage across it.


        We will assume that the voltage across the capacitor is sinusoidal:
vC(t) = Vmax sin(wt)
        Knowing the voltage across the capacitor allows us to calculate the current:
iC(t) = C dvC(t)/dt = wC Vmax cos(wt) = Imax cos(wt)

where  Imax = wC Vmax

Comparing the expressions for the voltage and current we note the following.

Vmax/Imax = 1/ wC

        Now, with these observations in hand, it is possible to see that there may be an algebraic way to express all of these facts and relationships.  The method reduces to the following.

        We need to do two things here.  First, we can illustrate what we mean with an example.  Secondly, we need to justify the claim above.  We will look at an example first, and we will do two examples.  The first example is jsut the capacitor - all by itself.  The second example will be one that you have considered earlier, a simple RC low-pass filter.

Example 1 - The Capacitor

        In a capacitor with sinusoidal voltage and currents, we have:


        We represent the voltage with a complex variable, V.  Considering this as a complex variable, it has a magnitude of Vmaxand and angle of 0o.  We would write:
V = Vmax/0o
        Similarly, we can get a representation for the current.  However, first note:
iC(t) = wC Vmax cos(wt) = Imax cos(wt) = Imax sin(wt + 90o)
(Here you must excuse the mixing of radians and degrees in the argument of the sine.  The only excuse is that everyone does it!)  Anyhow, we have:
I = Imax/90o = j Imax = jwC Vmax
Where j is the square root of -1.

        Then we would write:

V/I = Vmax/jwC Vmax = 1/jwC
and the quantity 1/jwC is called the impedance of the capacitor.  In the next section we will apply that concept to a small circuit - one you should have seen before.

        Before moving to the next section, a little reflection is in order.  Here are some points to think about.

Using Impedance

        In the last section we began to talk about the concept of impedance.  Let us do that a little more formally.  We begin by defining terms.

        A sinusoidally varying signal (vC(t) = Vmax sin(wt) for example) will be represented by a phasor, V, that incorporates the magnitude and phase angle of the signal as a magnitude and angle in a complex number.  Examples include these taken from the last section.  (Note that these phasors have nothing to do with any TV program about outer space.)

vC(t) = Vmax sin(wt)

is represented by a phasor V = Vmax/0o

iC(t) = Imax sin(wt + 90o)

is represented by a phasor I = Imax/90o

va(t) = VA sin(wt + f)

is represented by a phasor Va = VA/f

        Next, we can use the relationships for voltage and current phasors to analyze a circuit.  Here is the circuit.

        Now, this circuit is really a frequency dependent voltage divider, and it is analyzed differently in another lesson.  However, here we will use phasors.  At the end of this analysis, you should compare how difficult it is using phasors to the method in the other lesson.

        We start by noting that the current in the circuit - and there is only one current - has a phasor representation:

I = Imax/0o

We will use the current phase as a reference, and measure all other phases from the current's phase.  That's an arbitrary decision, but that's the way we will start.

        Next we note that we can compute the voltage across the capacitor.

VC = I/jwC

This expression relates the current phasor to the phasor that represents the voltage across the capacitor.  The quantity 1/jwC is the impedance of the capacitor.  In the last section  we justified this relationship.

        We can also compute the phasor for the voltage across the resistor.

This looks amazingly like Ohm's law, and it is, in fact, Ohm's law, but it is in phasor form.  For that matter, the relationship between voltage and current phasors in a capacitor - just above - may be considered a generalized form of Ohm's law!

        Now, we can also apply Kirchhoff's Voltage Law (KVL) to compute the phasor for the input voltage.

VIN = VR + VC = IR + I/jwC = I(R + 1/jwC)

        You should note the similarities in what happens here and what happens when you have two resistors in series.


        Consider a series circuit of a resistor and capacitor.  The series impedance is:

Z = R + 1/jwC

That's the same as we showed just above.  The impedance can be used to predict relationships between voltage and current.  Assume that the voltage across the series connection is given by:

vSeries(t) = Vmax cos(wt)

That corresponds to having a voltage phasor of:

V = Vmax/0o

We also know that the impedance establishes a relationship between the voltage and current phasors in the series circuit.  In particular, the voltage phasor is the product of the current phasor and the impedance.

V = I Z

For our particular impedance, we have:

V = I*(R + 1/jwC)

So, we can solve for the current phasor:

I = V / (R + 1/jwC)

Now, we know the voltage phasor and we know the impedance so we can compute the current phasor.  Let us look at some particular values.


Then: And, the total impedance is: This impedance value can also be expressed in polar notation: Now, compute the current phasor: Substituting values, we find: And, we need to examine exactly what this means for the current as a function of time.  But that isn't very difficult.  We can write out the expression for the current from what we have above.