Why Are Capacitors Important?
What Is A Capacitor?Voltage Current Relationships In CapacitorsEnergy In CapacitorsWhat Is Impedance?Using ImpedanceSome Impedance Laws/Combinations
The capacitor is a widely used electrical component. It has several features that make it useful and important:
l A capacitor can store energy, so capacitors
are often found in power supplies.
l A capacitor has a voltage that is proportional
to the charge (the integral of the current) that
is stored in the capacitor, so a capacitor can
be used to perform interesting computations
in op-amp circuits, for example.
l Circuits with capacitors exhibit frequency-
dependent behavior so that circuits that
amplify certain frequencies selectively can
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.
lGiven a capacitor,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.
uBe able to write and use the voltage-current
relationship for the capacitor,
uBe able to compute the current through a
capacitor when you know the voltage across
lGiven a capacitor that is charged,
uBe able to compute the amount of energy that
is stored in the 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:
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.
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 Vwhere:
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.
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.
Q3. Is the current constant in the time interval from t = 10 msec to the last time shown?
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 are often used to store energy.
i(t) = C dv(t)/dt
P(t) = i(t)v(t)
start with a capacitor with a sinusoidal voltage across it.
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.
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.
Example 1 - The Capacitor
In a capacitor with sinusoidal voltage and currents, we have:
V = Vmax/0oSimilarly, 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 VmaxWhere j is the square root of -1.
Then we would write:
V/I = Vmax/jwC Vmax = 1/jwCand 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.
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 I = Imax/90o
is represented by a phasor V = Vmax/0o
iC(t) = Imax sin(wt + 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/jwCThis 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.
VR = IRThis 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.
lIf you have a resistor, R, and a capacitor, C, in series, the current phasor can be computed by dividing the input voltage phasor by the sum of R and 1/jwC.
lIf you have two resistors in series (call them R1and R2), the current can be computed by dividing the input voltage by the sum of R1and R2.
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
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.
R = 1.0 kWThen:
C = .1mf = 10-7 f
f = 1 kHz, so w = 2p 103
Vmax = 20 v
ZR = 1.0 kWAnd, the total impedance is:
ZC = 1/(jwC) = 1/(j2p 103 10-7 ) = j 1.59 kW
Z = ZR + ZC = (1.0 + j 1.59) kWThis impedance value can also be expressed in polar notation:
Z = 1.878 /62oNow, compute the current phasor:
I = V / (R + 1/jwC)Substituting values, we find:
I = V / Z = Vmax/0o / 1.878 /62o =20/0o / 1.878 /62oAnd, 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.
I = V / Z = (20 / 1.878) /-62o = 10.65 /-62oamps
iC(t) = 10.65 cos(wt - 62o) amps