Key Concepts to Understand Capacitors

RPS Regular Practice Sheets

Time: … Hrs RPS. No. – 12

CAPACITORS

Why Are Capacitors Important?

The capacitor is a widely used electrical component. It has several

features that make it useful and important:

A capacitor can store energy, so capacitors are often found

in power supplies.

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.

Circuits with capacitors exhibit frequency-dependent behavior so

that circuits that amplify certain frequencies selectively can be

built.

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 connection 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.

Store energy

Change their behavior with frequency

Come about naturally in circuits and can change a circuit's behavior

Goals

You need to know what you should get from this lesson on

capacitors. Here's the story.

Given a capacitor,

o Be able to write and use the voltage-current relationship for

the capacitor,

o Be able to compute the current through a capacitor when you

know the voltage across a capacitor.

Given a capacitor that is charged,

o Be able to compute the amount of energy that is stored in the

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.

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:

Two conducting plates. That's the metallic foil in the Leyden jar.

An insulator that separates the plates so that they make no

electrical contact. That's the glass jar - the Leyden jar.

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:

V is the voltage across the plates.

You will need to define a polarity for that voltage. We've defined the

voltage above. You could reverse the "+" and "-".

Q is the charge on the plate with the "+" on the voltage polarity

definition.

C is a constant - the capacitance of the capacitor.

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 coulombs, 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.

Question

Q1 If the voltage across a capacitor is decreasing (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 4.7f

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.

Questions

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

seconds?

Q3. Is the current constant in the time interval from t = 10 milli seconds

to the last time shown?

If current is proportional to the time derivative of voltage, there is

only one time segment, from t = 0 to 1 millisecond, where the

voltage derivative is non-zero, so that's the only time there is any

current that is non-zero.

After one millisecond has elapsed, the voltage derivative goes to

zero, so there isn't any current then. If there isn't any current,

then the voltage stays constant because no charge is flowing in or

out. Remember, current is charge flow!

The voltage derivative is constant from t = 0 to 1 millisecond. If

that's true, then the current is constant in that period.

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.

You're reading this on a computer, and the computer keeps track of

the date and time. It does that by keeping a small part of the

computer running when you think that the computer is turned off.

There's a small battery that stores the energy to keep the clock

running when everything else is turned off.

If you own a stereo or television that you have to plug into the wall

plug, then you should realize that the wall plug voltage becomes zero

120 times a second. When that happens, the system keeps running

because there are capacitors inside the system that store energy to

carry you through those periods when the line voltage isn't large

enough to keep things going!

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.

When relatively small amounts of energy are needed.

Where batteries are not desired because they might deteriorate.

For larger power/short duration applications - as in power supply

filters, or to keep power up long enough for a computer to shut

down gracefully when the line power fails.

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)

Given the expression for the power:

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

And given the expression for the current:

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

We can use the expression for current in the power expression:

P(t) = (C dv(t)/dt) v(t)

We can recognize that power is simply rate of energy input.

P(t) = dE/dt = (C dv(t)/dt) v(t)

Now, the derivative of energy can be integrated to find the total

energy input.

P(t) = dE/dt = (C dv(t)/dt) v(t)

gives

Now, assuming that the initial voltage is zero (there is no energy

stored in the capacitor initially, we find that the energy stored in a

capacitor is proportional to the capacitance and to the square of

the voltage across the capacitor.

Ec = (1/2)CV2

The expression for the energy stored in a capacitor resembles

other energy storage formulae.

For kinetic energy, with a mass, M, and a velocity, v.

EM = (1/2)MV2

For potential energy, with a spring constant, K, and an elongation, x.

ESpring = (1/2)Kx2

Since the square of the voltage appears in the energy formula, the

energy stored is always positive. You can't have a negative amount

of energy in the capacitor. That means you can put energy into the

capacitor, and you can take it out, but you can't take out more than

you put in.

Power in to the capacitor can be negative. Voltage can be positive

while current is negative. Imagine a capacitor that is charged. You

could charge a capacitor by putting a battery across the capacitor,

for example. Then, if you placed a resistor across the capacitor,

charge would leave the capacitor - current would flow out of the

capacitor - and the energy in the capacitor would leave the

capacitor only to become heat energy in the resistor. When energy

leaves the capacitor, power is negative.

When you use capacitors in a circuit and you analyze the

circuit you need to be careful about sign conventions.

Here are the conventions we used, and these conventions

were assumed in any results we got in this lesson.

Frequency Dependent Behavior For A Capacitor

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

where:

vC(t) = Voltage across the capacitor

iC(t) = Current through the capacitor

C = Capacitance (in farads)

We will assume that the voltage across the capacitor is sinusoidal:

vC(t) = Vmax sin(t)

Knowing the voltage across the capacitor allows us to calculate the

current:

iC(t) = C dvC(t)/dt = C Vmax cos(t) = Imax cos(t)

where Imax = C Vmax

Comparing the expressions for the voltage and current we note the

following.

The voltage and the current are both sinusoidal signals (a sine

function or a cosine function) at the same frequency.

The current leads the voltage. In other words, the peak of the

current occurs earlier in time than the peak of the voltage signal.

The current leads the voltage by exactly 90o. It will always be

exactly 90o in a capacitor.

The magnitude of the current and the magnitude of the voltage are

related:

Vmax/Imax = 1/ C

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.

If we have a circuit with sinusoidally varying voltages and currents

(as we would have in a circuit with resistors, capacitors and

inductors and sinusoidal voltage and current sources) we associate a

complex variable with every voltage and current in the circuit.

The complex variable for a voltage or current encodes the amplitude

and phase for that voltage or current.

The voltage and current variables can be used (using complex

algebra) to predict circuit behavior just as though the circuit were

a resistive circuit.

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 just 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:

where:

vC(t) = Voltage across the inductor

vC(t) = Vmax sin(t)

iC(t) = Current through the inductor

iC(t) = C Vmax cos(t) = Imax cos(t)

C = Capacitance (in farads)

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) = C Vmax cos(t) = Imax cos(t) = Imax sin(t + 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 = jC Vmax

Where j is the square root of -1.

Then we would write:

V/I = Vmax/jC Vmax = 1/jC

and the quantity 1/jC 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.

A phasor summarizes information about a sinusoidal signal.

Magnitude and phase information are encoded into the phasor.

Frequency information is not encoded, and there is a tacit

assumption that all signals are of the same frequency, which would

be the case in a linear circuit with sinusoidal voltage and current

sources.

We looked at a case where we encoded a signal Vmax sin(t) into a

phasor of Vmax/0o. That was completely arbitrary, and many others

would have encoded Vmax cos(t) into a phasor ofVmax/0o.

Phasors are intended only to show relative phase information, and it

doesn't matter which way you go.

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(t) 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(t)

is represented by a phasor V = Vmax/0o

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

is represented by a phasor I = Imax/90o

va(t) = VA sin(t + )

is represented by a phasor Va = VA/

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/jC

This expression relates the current phasor to the phasor that represents

the voltage across the capacitor. The quantity 1/jC 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 = IR

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/jC = I(R + 1/jC)

You should note the similarities in what happens here and what

happens when you have two resistors in series.

If 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/jC.

If 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.

Example

Consider a series circuit of a resistor and capacitor. The series

impedance is:

Z = R + 1/jC

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(t)

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/jC)

So, we can solve for the current phasor:

I = V / (R + 1/jC)

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.

Assume:

R = 1.0 k

C = .1f = 10-7 f

f = 1 kHz, so = 2 103

Vmax = 20 v

Then:

ZR = 1.0 k

ZC = 1/(jC) = 1/(j2 103 10-7 ) = j 1.59 k

And, the total impedance is:

Z = ZR + ZC = (1.0 + j 1.59) k

This impedance value can also be expressed in polar notation:

Z = 1.878 /62o

Now, compute the current phasor:

I = V / (R + 1/jC)

Substituting values, we find:

I = V / Z = Vmax/0o / 1.878 /62o =20/0o / 1.878 /62o

I = V / Z = (20 / 1.878) /-62o = 10.65 /-62oamps

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.

iC(t) = 10.65 cos(t - 62o) amps

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