Condensadores e Inductancias

transcript

6.1Capacitors

A capacitor is a circuit element that consists of two conducting surfaces separated by a non-conducting, or dielectric, material. A simplified capacitor and its electrical symbol are shownin Fig. 6.1.

There are many different kinds of capacitors, and they are categorized by the type ofdielectric material used between the conducting plates. Although any good insulator canserve as a dielectric, each type has characteristics that make it more suitable for particularapplications.

For general applications in electronic circuits (e.g., coupling between stages of amplifica-tion), the dielectric material may be paper impregnated with oil or wax, mylar, polystyrene,mica, glass, or ceramic.

Ceramic dielectric capacitors constructed of barium titanates have a large capacitance-to-volume ratio because of their high dielectric constant. Mica, glass, and ceram-ic dielectric capacitors will operate satisfactorily at high frequencies.

Aluminum electrolytic capacitors, which consist of a pair of aluminum plates separatedby a moistened borax paste electrolyte, can provide high values of capacitance in small vol-umes. They are typically used for filtering, bypassing, and coupling, and in power suppliesand motor-starting applications. Tantalum electrolytic capacitors have lower losses and morestable characteristics than those of aluminum electrolytic capacitors. Fig. 6.2 shows a varietyof typical discrete capacitors.

In addition to these capacitors, which we deliberately insert in a network for specificapplications, stray capacitance is present any time there is a difference in potential betweentwo conducting materials separated by a dielectric. Because this stray capacitance can cause

246 C H A P T E R 6 C A P A C I T A N C E A N D I N D U C T A N C E

Figure 6.2

Some typical capacitors.(Courtesy of Mark Nelms andJo Ann Loden)

d

Dielectric

(a)

Cq(t)v(t)

(b)

++

--

i=dqdt

A

Figure 6.1

A capacitor and itselectrical symbol.

Note the use of the passivesign convention.

[ h i n t ]

S E C T I O N 6 . 1 C A P A C I T O R S 247

unwanted coupling between circuits, extreme care must be exercised in the layout of elec-tronic systems on printed circuit boards.

Capacitance is measured in coulombs per volt or farads. The unit farad (F) is named afterMichael Faraday, a famous English physicist. Capacitors may be fixed or variable and typi-cally range from thousands of microfarads (F) to a few picofarads (pF).

Capacitor technology, initially driven by the modern interest in electric vehicles, is rapidlychanging, however. For example, the capacitor on the left in the photograph in Fig. 6.3 is adouble-layer capacitor, which is rated at 2.5 V and 100 F. An aluminum electrolyticcapacitor, rated at 25 V and is shown on the right in this photograph. The elec-trolytic capacitor can store 21.25 joules (J). The double-layercapacitor can store Lets connect ten of the 100-F capacitors inseries for an equivalent 25-V capacitor. The energy stored in this equivalent capacitor is3125 J. We would need to connect 147 electrolytic capacitors in parallel to store that muchenergy.

It is interesting to calculate the dimensions of a simple equivalent capacitor consisting of twoparallel plates each of area A, separated by a distance d as shown in Fig. 6.1. We learned inbasic physics that the capacitance of two parallel plates of area A, separated by distance d, is

where o , the permitivity of free space, is F/m. If we assume the plates are sep-arated by a distance in air of the thickness of one sheet of oil-impregnated paper, which isabout m, then

and since 1 square mile is equal to square meters, the area is

A443 square miles

which is the area of a medium-sized city! It would now seem that the double-layer capacitorin the photograph is much more impressive than it originally appeared. This capacitor is actu-ally constructed using a high surface area material such as powdered carbon which is adheredto a metal foil. There are literally millions of pieces of carbon employed to obtain therequired surface area.

Suppose now that a source is connected to the capacitor shown in Fig. 6.1; then positivecharges will be transferred to one plate and negative charges to the other. The charge on thecapacitor is proportional to the voltage across it such that

q=Cv 6.1

where C is the proportionality factor known as the capacitance of the element in farads.The charge differential between the plates creates an electric field that stores energy.

Because of the presence of the dielectric, the conduction current that flows in the wires thatconnect the capacitor to the remainder of the circuit cannot flow internally between theplates. However, via electromagnetic field theory it can be shown that this conduction cur-rent is equal to the displacement current that flows between the plates of the capacitor and ispresent any time that an electric field or voltage varies with time.

Our primary interest is in the currentvoltage terminal characteristics of the capacitor.Since the current is

i =dqdt

2.59 * 106

A = 1.148 * 109 m2

100 F =A8.85 * 10-12BA

1.016 * 10-4

1.016 * 10-4

8.85 * 10- 12

C =eo A

d

0.5 * 100 * 2.52 = 312.5 J.0.5 * 6.8 * 10-2 * 252 =

68,000 F, Figure 6.3A 100-F double-layercapacitor and a 68,000-Felectrolytic capacitor.(Courtesy of Mark Nelms andJo Ann Loden)


then for a capacitor

which for constant capacitance is

6.2

Eq. (6.2) can be rewritten as

Now integrating this expression from t=q to some time t and assuming v(q)=0yields

6.3

where v(t) indicates the time dependence of the voltage. Eq. (6.3) can be expressed as twointegrals, so that

6.4

where is the voltage due to the charge that accumulates on the capacitor from timet=q to time t=t0 .

The energy stored in the capacitor can be derived from the power that is delivered to theelement. This power is given by the expression

6.5

and hence the energy stored in the electric field is

6.6

since v(t=q)=0. The expression for the energy can also be written using Eq. (6.1) as

6.7

Eqs. (6.6) and (6.7) represent the energy stored by the capacitor, which, in turn, is equal tothe work done by the source to charge the capacitor.

Now lets consider the case of a dc voltage applied across a capacitor. From Eq. (6.2), wesee that the current flowing through the capacitor is directly proportional to the time rate ofchange of the voltage across the capacitor. A dc voltage does not vary with time, so the cur-rent flowing through the capacitor is zero. We can say that a capacitor is an open circuit to

wC(t) =1

2

q2(t)

C

=12

Cv2(t) J

= C3v(t)

v(-q)v(x)dv(x) =

1

2Cv2(x) 2 v(t)

v(-q)

wC(t) = 3t

-qCv(x)

dv(x)dx

dx = C3t

-qv(x)

dv(x)dx

dx

p(t) = v(t)i(t) = Cv(t)dv(t)dt

vAt0B

= v(t0) +1

C 3t

t0

i(x)dx

v(t) =1

C 3t0

-qi(x)dx +

1

C 3t

t0i(x)dx

v(t) =1

C 3t

-qi(x)dx

dv =1

Ci dt

i = Cdvdt

i =d

dt(Cv)


If the charge accumulated on two parallel conductors charged to 12 V is 600 pC, what is thecapacitance of the parallel conductors?

Using Eq. (6.1), we find that

C =Q

V=(600)A10-12B

12= 50 pF

SOLUTION

EXAMPLE

6.1

dc or blocks dc. Capacitors are often utilized to remove or filter out an unwanted dc volt-age. In analyzing a circuit containing dc voltage sources and capacitors, we can replace thecapacitors with an open circuit and calculate voltages and currents in the circuit using ourmany analysis tools.

Note that the power absorbed by a capacitor, given by Eq. (6.5), is directly proportional tothe time rate of change of the voltage across the capacitor. What if we had an instantaneouschange in the capacitor voltage? This would correspond to and infinite power. In Chapter 1, we ruled out the possibility of any sources of infinite power. Since we only havefinite power sources, the voltage across a capacitor cannot change instantaneously. This willbe a particularly helpful idea in the next chapter when we encounter circuits containingswitches. This idea of continuity of voltage for a capacitor tells us that the voltage acrossthe capacitor just after a switch moves is the same as the voltage across the capacitor justbefore that switch moves.

The polarity of the voltage across a capacitor being charged is shown in Fig. 6.1b. In theideal case, the capacitor will hold the charge for an indefinite period of time, if the source isremoved. If at some later time an energy-absorbing device (e.g., a flash bulb) is connectedacross the capacitor, a discharge current will flow from the capacitor and, therefore, thecapacitor will supply its stored energy to the device.

dvdt = q

The voltage across a 5-F capacitor has the waveform shown in Fig. 6.4a. Determine thecurrent waveform.

Note that

= 0 8 ms t

=-24

2 * 10-3t + 96 6 t 6 8 ms

v(t) =24

6 * 10-3t 0 t 6 ms

Figure 6.4

Voltage and current wave-forms for a

5-F capacitor.

SOLUTION

EXAMPLE

6.2

20

i(t) (mA)

60

0 6 t (ms)8

(b)

v(t) (V)

24 V

0 6 8 t (ms)

(a)


Determine the energy stored in the electric field of the capacitor in Example 6.2 at t=6 ms.

Using Eq. (6.6), we have

At t=6 ms,

= 1440 J

w(6 ms) =12A5 * 10-6B(24)2

w(t) =1

2Cv2(t)

EXAMPLE

6.3

The current in an initially uncharged 4-F capacitor is shown in Fig. 6.5a. Let us derive thewaveforms for the voltage, power, and energy and compute the energy stored in the electricfield of the capacitor at t=2 ms.

The equations for the current waveform in the specific time intervals are

Since v(0)=0, the equation for v(t) in the time interval 0 t 2 ms is

and hence,v(2 ms) = 103A2 * 10-3B2 = 4 mV

v(t) =1

(4)A10-6B 3t

08A10-3Bx dx = 103t2

= 0 4 ms 6 t = -8 * 10-6 2 ms t 4 ms

i(t) =16 * 10-6t2 * 10-3

0 t 2 ms

EXAMPLE

6.4

Using Eq. (6.2), we find that

and

Therefore, the current waveform is as shown in Fig. 6.4b and i(t)=0 for t>8 ms.

i(t) = 0 8 ms t

= -60 mA 6 t 6 8 ms i(t) = 5 * 10-6A-12 * 103B 6 t 8 ms

= 20 mA 0 t 6 ms = 5 * 10-6A4 * 103B 0 t 6 ms

i(t) = Cdv (t)dt

SOLUTION

SOLUTION

E6.1 A 10-F capacitor has an accumulated charge of 500 nC. Determine the voltage across thecapacitor.

Learning AssessmentANSWER: 0.05 V.


In the time interval 2 ms t 4 ms,

The waveform for the voltage is shown in Fig. 6.5b.Since the power is p(t)=v(t)i(t), the expression for the power in the time interval

0 t 2 ms is p(t)=8t3. In the time interval 2 ms t 4 ms, the equation for thepower is

The power waveform is shown in Fig. 6.5c. Note that during the time interval 0 t 2 ms,the capacitor is absorbing energy and during the interval 2 ms t 4 ms, it is deliveringenergy.

The energy is given by the expression

w(t) = 3t

t0

p(x)dx + wAt0B

= 16A10-6Bt - 64A10-9B p(t) = -(8)A10-6B A-2t + 8 * 10-3B

= -2t + 8 * 10-3

v(t) =1

(4)A10-6B 3t

2A10-3B- A8B A10-6Bdx + A4B A10-3B

0Time(ms)

Current (A)

10

5

5

10

15

0.5 1 1.5 2.5 3 3.5

(a)

2 4

Time(ms)

Voltage (mV)

00.5

1

1.5

3

3.5

0.5 1 1.5 2 2.5 3 3.5 4

2

2.5

4

(b)

Time(ms)

Energy (pJ)

00.5 1 1.5 2 2.5 3 3.5 4

5

10

15

20

25

30

35

(d)

Time(ms)

Power (nW)

0

50

2.5 3 4

3.5

60

40302010

102030

(c)

0.5 1 1.5

2

Figure 6.5

Waveforms used inExample 6.4.


E6.2 The voltage across a 2-F capacitor is shown in Fig. E6.2. Determine the waveform for thecapacitor current.

Figure E6.2

Learning Assessments

ANSWER:v(t) (V)

12

0 1 2 3 4 t (ms)5 6

12i(t) (mA)

60 1 3 4 t (ms)52

6

E6.3 Compute the energy stored in the electric field of the capacitor in Learning AssessmentE6.2 at t=2 ms.

ANSWER: w=144 J.

In the time interval 0 t 2 ms,

Hence,w(2 ms)=32 pJ

In the time interval 2 t 4 ms,

From this expression we find that w(2 ms)=32 pJ and w(4 ms)=0. The energy wave-form is shown in Fig. 6.5d.

= A8 * 10-6Bt2 - A64 * 10-9Bt + 128 * 10-12 = C A8 * 10-6Bx2 - A64 * 10-9Bx D t2* 10-3 + 32 * 10-12

w(t) = 3t

2* 10-3C A16 * 10-6Bx - A64 * 10-9B D dx + 32 * 10-12

w(t) = 3t

08x3 dx = 2t4

E6.4 The voltage across a 5-F capacitor is shown in Fig. E6.4. Find the waveform for the current in the capacitor. How much energy is stored in the capacitor at t=4 ms.

Figure E6.4

10

5

5

v(t) (V)

t (ms)1 2 3 4 5 6

7

8

9

10


ANSWER: 312.5 nJ.

ANSWER: 250 J.

Figure E6.5

t (ms)

25

25

1 2 3

4

5 6

7

8 9 10

i(t) (mA)

t (ms)

10

10

1

2

3

4

5 6 7 8 9 10

i(t) (A)

t (ms)

15

5

35

15

1 2 3 4 5 6 7

8

9 10

v(t) (V)

E6.5 The waveform for the current in a 1-nF capacitor is Fig. E6.5. If the capacitor has an initialvoltage of 5V, determine the waveform for the capacitor voltage. How much energy is stored inthe capacitor at t=6 ms?


Figure 6.6

Two inductors and their electrical symbol

Figure 6.7Some typical inductors.

(Courtesy of Mark Nelmsand Jo Ann Loden)

6.2Inductors

An inductor is a circuit element that consists of a conducting wire usually in the form of a coil.Two typical inductors and their electrical symbol are shown in Fig. 6.6. Inductors are typicallycategorized by the type of core on which they are wound. For example, the core material maybe air or any nonmagnetic material, iron, or ferrite. Inductors made with air or nonmagneticmaterials are widely used in radio, television, and filter circuits. Iron-core inductors are used inelectrical power supplies and filters. Ferrite-core inductors are widely used in high-frequencyapplications. Note that in contrast to the magnetic core that confines the flux, as shown inFig. 6.6b, the flux lines for nonmagnetic inductors extend beyond the inductor itself, as illus-trated in Fig. 6.6a. Like stray capacitance, stray inductance can result from any element carry-ing current surrounded by flux linkages. Fig. 6.7 shows a variety of typical inductors.

From a historical standpoint, developments that led to the mathematical model we employto represent the inductor are as follows. It was first shown that a current-carrying conductorwould produce a magnetic field. It was later found that the magnetic field and the current thatproduced it were linearly related. Finally, it was shown that a changing magnetic field pro-duced a voltage that was proportional to the time rate of change of the current that producedthe magnetic field; that is,

6.8

The constant of proportionality L is called the inductance and is measured in the unit henry,named after the American inventor Joseph Henry, who discovered the relationship. As seenin Eq. (6.8), 1 henry (H) is dimensionally equal to 1 volt-second per ampere.

Following the development of the mathematical equations for the capacitor, we find thatthe expression for the current in an inductor is

6.9i(t) =1

L 3t

-qv(x)dx

v(t) = Ldi(t)dt

v(t)

(a)

+

-

i(t)

(b) (c)

i(t)

v(t)

+

-

i(t)

Flux lines

Flux lines

L

S E C T I O N 6 . 2 I N D U C T O R S 255

which can also be written as

6.10

The power delivered to the inductor can be used to derive the energy stored in the element.This power is equal to

6.11

Therefore, the energy stored in the magnetic field is

Following the development of Eq. (6.6), we obtain

6.12

Now lets consider the case of a dc current flowing through an inductor. From Eq. (6.8),we see that the voltage across the inductor is directly proportional to the time rate of changeof the current flowing through the inductor. A dc current does not vary with time, so the volt-age across the inductor is zero. We can say that an inductor is a short circuit to dc. In ana-lyzing a circuit containing dc sources and inductors, we can replace any inductors with shortcircuits and calculate voltages and currents in the circuit using our many analysis tools.

Note from Eq. (6.11) that an instantaneous change in inductor current would require infi-nite power. Since we dont have any infinite power sources, the current flowing through aninductor cannot change instantaneously. This will be a particularly helpful idea in the nextchapter when we encounter circuits containing switches. This idea of continuity of currentfor an inductor tells us that the current flowing through an inductor just after a switch movesis the same as the current flowing through an inductor just before that switch moves.

wL(t) =12

Li2(t) J

wL(t) = 3t

-qcL di(x)

dxd i(x)dx

= cL di(t)dtd i(t)

p(t) = v(t)i(t)

i(t) = iAt0B + 1L 3t

t0

v(x)dx

EXAMPLE

6.5

Find the total energy stored in the circuit of Fig. 6.8a.

6 3

9 V 6 3 A

C1=20 F C2=50 F

L2=4 mHL1=2 mH

(a)

6 3

9 V 6 3 A

A

(b)

IL1 IL2

VC2+

-

VC1+

-

Figure 6.8

Circuits used inExample 6.5.


The inductor, like the resistor and capacitor, is a passive element. The polarity of the volt-age across the inductor is shown in Fig. 6.6.

Practical inductors typically range from a few microhenrys to tens of henrys. From a cir-cuit design standpoint it is important to note that inductors cannot be easily fabricated on anintegrated circuit chip, and therefore chip designs typically employ only active electronicdevices, resistors, and capacitors that can be easily fabricated in microcircuit form.

This circuit has only dc sources. Based on our earlier discussions about capacitors and induc-tors and constant sources, we can replace the capacitors with open circuits and the inductorswith short circuits. The resulting circuit is shown in Fig. 6.8b.

This resistive circuit can now be solved using any of the techniques we have learned inearlier chapters. If we apply KCL at node A, we get

Applying KVL around the outside of the circuit yields

Solving these equations yields and The voltages and canbe calculated from the currents:

The total energy stored in the circuit is the sum of the energy stored in the two inductorsand two capacitors:

The total stored energy is 13.46 mJ.

wC2 =1

2A50 * 10-6B(10.8)2 = 2.92 mJ

wC1 =1

2A20 * 10-6B(16.2)2 = 2.62 mJ

wL2 =1

2A4 * 10-3B(1.8)2 = 6.48 mJ

wL1 =1

2A2 * 10-3B(-1.2)2 = 1.44 mJ

VC2 = 6IL2 = 6(1.8) = 10.8 VVC1 = -6IL1 + 9 = 16.2 V

VC2VC1IL2 = 1.8 A.IL1 = -1.2 A

6IL1 + 3IL2 + 6IL2 = 9

IL2 = IL1 + 3

EXAMPLE

6.6SOLUTION

The current in a 10-mH inductor has the waveform shown in Fig. 6.9a. Determine thevoltage waveform.

Using Eq. (6.8) and noting that

andi(t)=0 4 ms


we find that

and

and v(t)=0 for t>4 ms. Therefore, the voltage waveform is shown in Fig. 6.9b.

= -100 mV

v(t) = A10 * 10-3B -20 * 10-32 * 10-3

2 t 4 ms

= 100 mV

v(t) = A10 * 10-3B 20 * 10-32 * 10-3

0 t 2 ms

The current in a 2-mH inductor is

i(t)=2 sin 377t A

Determine the voltage across the inductor and the energy stored in the inductor.

From Eq. (6.8), we have

and from Eq. (6.12),

= 0.004 sin2 377t J

=1

2A2 * 10-3B(2 sin 377t)2

wL(t) =1

2Li2(t)

= 1.508 cos 377t V

= A2 * 10-3B ddt(2 sin 377t)

v(t) = Ldi(t)dt

The voltage across a 200-mH inductor is given by the expression

Let us derive the waveforms for the current, energy, and power.

The waveform for the voltage is shown in Fig. 6.10a. The current is derived from Eq. (6.10) as

= 0 t 6 0 = 5te-3t mA t 0

= 5 e e-3x-3

2 t0- 3 c- e-3x

9(3x + 1) d t

0f

= 5 e 3t

0e-3x dx - 33

t

0xe-3x dx f

i(t) =103

200 3t

0(1 - 3x)e-3x dx

= 0 t 6 0

v(t) = (1 - 3t)e-3tmV t 0

SOLUTION

EXAMPLE

6.7

SOLUTION

EXAMPLE

6.8


A plot of the current waveform is shown in Fig. 6.10b.The power is given by the expression

The equation for the power is plotted in Fig. 6.10c.The expression for the energy is

This equation is plotted in Fig. 6.10d.

= 0 t 6 0 = 2.5t2e-6t J t 0

w(t) =1

2Li2(t)

= 0 t 6 0 = 5t(1 - 3t)e-6t W t 0

p(t) = v(t)i(t)

1.0

0.8

0.6

0.4

0.2

0

0.21.5 2 2.5 3 3.5 Time (s)

Time (s)

Voltage (mV)

0.5 1

00.5 1 1.5 2 2.5 3 3.5

Time (s)

Current (mA)

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.1

2.5 Time (s)

Power (W)

0.2

0.15

0.1

0.05

0

0.050.5

1

1.5 2

2.5

Energy (nJ)

35

25

20

15

5

10

30

0.5 1 1.5 2

40

0

(a) (b)

(c) (d)

Figure 6.10

Waveforms used inExample 6.8.


E6.8 The current in a 2-H inductor is shown in Fig. E6.8. Find the waveform for the inductorvoltage. How much energy is stored in the inductor at t=3 ms?

Figure E6.8

ANSWER: 25 J.

t (ms)

5

10

5

1

2

3 54

6

7 8 109 11 12

i(t) (mA)

t (ms)

53.33

10

1

2

3 5

4 6

7 8 10

9

11

12

v(t) (V)

E6.6 The current in a 5-mH inductor has the waveform shown in Fig. E6.6. Compute the waveform for the inductor voltage.

Learning Assessments

ANSWER:

E6.7 Compute the energy stored in the magnetic field of the inductor in Learning AssesmentE6.6 at t=1.5 ms.

i(t) (mA)

10

20

0 1 2 3 4t (ms)

t (ms)

100

v(t) (mV)

50

0 1 2 3 4

Figure E6.6

ANSWER: W=562.5 nJ.


E6.9 The voltage across a 0.1-H inductor is shown in Fig. E6.9. Compute the waveform for thecurrent in the inductor if i(0)=0.1A. How much energy is stored in the inductor at t=7 ms?

Figure E6.9

E6.10 Find the energy stored in the capacitor and inductor in Fig. E6.10.

Figure E6.10

ANSWER: 1.125 mJ.

t (ms)

10

5

5

10

1

2

3

4

5

6

7 8

9

10

v(t) (V)

t (ms)

0.1

0.25

0.2

0.1

1 2 3 4 5 6

7

8 9 10

i(t) (A)

1 H10 nF

12 V

3 k

6 k

8 mA

2 mA

2 k

+-

ANSWER: 0.72 J, 0.5 J.


CAPACITOR AND INDUCTOR SPECIFICATIONS There are a couple of important parame-ters that are used to specify capacitors and inductors. In the case of capacitors, the capac-itance value, working voltage, and tolerance are issues that must be considered in theirapplication. Standard capacitor values range from a few pF to about 50 mF. Capacitorslarger than 1 F are available but will not be discussed here. Table 6.1 is a list of standardcapacitor values, which are typically given in picofarads or microfarads. Although bothsmaller and larger ratings are available, the standard working voltage, or dc voltage rat-ing, is typically between 6.3 V and 500 V. Manufacturers specify this working voltagesince it is critical to keep the applied voltage below the breakdown point of the dielec-tric. Tolerance is an adjunct to the capacitance value and is usually listed as a percent-age of the nominal value. Standard tolerance values are ; 5%, ; 10%, and ; 20%.Occasionally, tolerances for single-digit pF capacitors are listed in pF. For example,5 pF ; 0.25 pF.

The two principal inductor specifications are inductance and resistance. Standard com-mercial inductances range from about 1 nH to around 100 mH. Larger inductances can, ofcourse, be custom built for a price. Table 6.2 lists the standard inductor values. The currentrating for inductors typically extends from a few dozen mAs to about 1 A. Tolerances aretypically 5% or 10% of the specified value.

TABLE 6.1 Standard capacitor values

pF pF pF pF F F F F F F F

1 10 100 1000 0.010 0.10 1.0 10 100 1000 10,000

12 120 1200 0.012 0.12 1.2 12 120 1200 12,000

1.5 15 150 1500 0.015 0.15 1.5 15 150 1500 15,000

18 180 1800 0.018 0.18 1.8 18 180 1800 18,000

2 20 200 2000 0.020 0.20 2.0 20 200 2000 20,000

22 220 2200 0.022 0.22 2.2 22 220 2200 22,000

27 270 2700 0.027 0.27 2.7 27 270 2700 27,000

3 33 330 3300 0.033 0.33 3.3 33 330 3300 33,000

4 39 390 3900 0.039 0.39 3.9 39 390 3900 39,000

5 47 470 4700 0.047 0.47 4.7 47 470 4700 47,000

6 51 510 5100 0.051 0.51 5.1 51 510 5100 51,000

7 56 560 5600 0.056 0.56 5.6 56 560 5600 56,000

8 68 680 6800 0.068 0.68 6.8 68 680 6800 68,000

9 82 820 8200 0.082 0.82 8.2 82 820 8200 82,000

TABLE 6.2 Standard inductor values

nH nH nH H H H mH mH mH

1 10 100 1.0 10 100 1.0 10 100

1.2 12 120 1.2 12 120 1.2 12

1.5 15 150 1.5 15 150 1.5 15

1.8 18 180 1.8 18 180 1.8 18

2 20 200 2.0 20 200 2.0 20

2.2 22 220 2.2 22 220 2.2 22

2.7 27 270 2.7 27 270 2.7 27

3 33 330 3.3 33 330 3.3 33

4 39 390 3.9 39 390 3.9 39

5 47 470 4.7 47 470 4.7 47

6 51 510 5.1 51 510 5.1 51

7 56 560 5.6 56 560 5.6 56

8 68 680 6.8 68 680 6.8 68

9 82 820 8.2 82 820 8.2 82


The capacitor in Fig. 6.11a is a 100-nF capacitor with a tolerance of 20%. If the voltagewaveform is as shown in Fig. 6.11b, let us graph the current waveform for the minimum andmaximum capacitor values.

The maximum capacitor value is 1.2C=120 nF, and the minimum capacitor value is0.8C=80 nF. The maximum and minimum capacitor currents, obtained from the equation

are shown in Fig. 6.11c.

i(t) = Cdv(t)dt

SOLUTION

EXAMPLE

6.10

We wish to find the possible range of capacitance values for a 51-mF capacitor that has atolerance of 20%.

The minimum capacitor value is 0.8C=40.8 mF, and the maximum capacitor value is1.2C=61.2 mF.

SOLUTION

EXAMPLE

6.9

As indicated in Chapter 2, wire-wound resistors are simply coils of wire, and therefore itis only logical that inductors will have some resistance. The major difference between wire-wound resistors and inductors is the wire material. High-resistance materials such asNichrome are used in resistors, and low-resistance copper is used in inductors. The resistanceof the copper wire is dependent on the length and diameter of the wire. Table 6.3 lists theAmerican Wire Gauge (AWG) standard wire diameters and the resulting resistance per footfor copper wire.

TABLE 6.3 Resistance per foot of solid copper wire

AWG No. Diameter (in.) m/ft

12 0.0808 1.59

14 0.0641 2.54

16 0.0508 4.06

18 0.0400 6.50

20 0.0320 10.4

22 0.0253 16.5

24 0.0201 26.2

26 0.0159 41.6

28 0.0126 66.2

30 0.0100 105

32 0.0080 167

34 0.0063 267

36 0.0049 428

38 0.0039 684

40 0.0031 1094


(a) (b)

v(t)

(V

)

0 1 2 3 4 5 6 7Time (s)

4

3

2

1

0

1

2

3

4

(c)

Volta

ge (V

)

Cur

rent

(mA

)0 1 2 3 4 5 6 7

Time (s)

4

3

2

1

0

1

2

3

4

800

600

400

200

0

200

400

600

800v(t)i(t) at Cmini(t) at Cmax

v(t)

i(t)

C

Figure 6.11

Circuit and graphs usedin Example 6.10.

The inductor in Fig. 6.12a is a 100-H inductor with a tolerance of 10%. If the currentwaveform is as shown in Fig. 6.12b, let us graph the voltage waveform for the minimumand maximum inductor values.

The maximum inductor value is 1.1L=110 H, and the minimum inductor value is0.9L = 90 H. The maximum and minimum inductor voltages, obtained from the equation

are shown in Fig. 6.12c.

v(t) = Ldi(t)dt

SOLUTION

EXAMPLE

6.11


SERIES CAPACITORS If a number of capacitors are connected in series, their equiva-lent capacitance can be calculated using KVL. Consider the circuit shown in Fig. 6.13a. Forthis circuit

v(t)=v1(t)+v2(t)+v3(t)+p+vN(t) 6.13but

6.14vi(t) =1

Ci 3t

t0

i(t)dt + viAt0B

(c)

i(t)

(m

A)

v(t)

(V

)

0 10 20 30 40 50 60

Time (s)

150

100

50

0

50

100

150

4

3

2

1

0

1

2

i(t)

v(t) at Lminv(t) at Lmax

(b)

i(t)

(m

A)

0 10 20 30 40 50 60

Time (s)

150

100

50

0

50

100

150

(a)

v(t)i(t) L

+

-

Figure 6.12Circuit and graphsused in Example 6.11.

6.3Capacitorand InductorCombinations

(a)

v1(t)

vN(t)

CN

CSC1 C2 C3v(t)

i(t) + -

- +

+

-

v2(t)+ -

v3(t)+ -

(b)

v(t)

i(t)

+

-

Figure 6.13

Equivalent circuit forN series-connected

capacitors.

S E C T I O N 6 . 3 C A P A C I T O R A N D I N D U C T O R C O M B I N A T I O N S 265

Determine the equivalent capacitance and the initial voltage for the circuit shown in Fig. 6.14.

Note that these capacitors must have been charged before they were connected in series orelse the charge of each would be equal and the voltages would be in the same direction.

The equivalent capacitance is

where all capacitance values are in microfarads.Therefore, CS=1 F and, as seen from the figure, Note that the total

energy stored in the circuit is

However, the energy recoverable at the terminals is

= 4.5 J

=1

2C1 * 10-6(-3)2 D

wCAt0B = 12 CS v2(t)

= 31 J

wAt0B = 12 C2 * 10-6(2)2 + 3 * 10-6(-4)2 + 6 * 10-6(-1)2 D

vAt0B = -3 V.

1

CS=

1

2+

1

3+

1

6

SOLUTION

EXAMPLE

6.12

Therefore, Eq. (6.13) can be written as follows using Eq. (6.14):

6.15

6.16

where

and

6.17

Thus, the circuit in Fig. 6.13b is equivalent to that in Fig. 6.13a under the conditions statedpreviously.

It is also important to note that since the same current flows in each of the series capaci-tors, each capacitor gains the same charge in the same time period. The voltage across eachcapacitor will depend on this charge and the capacitance of the element.

1

CS= a

N

i= 1

1

Ci=

1

C1+

1

C2+ p +

1

CN

vAt0B = aN

i= 1viAt0B

=1

CS 3t

t0

i(t)dt + vAt0B

v(t) = a aN

i= 1

1

Cib 3

t

t0

i(t)dt + aN

i= 1viAt0B

4 V3 F

2 F

6 F

v(t)

+

+

-

-

2 V+ -

1 V+ -

Figure 6.14

Circuit containing multiplecapacitors with initialvoltages.

Capacitors in series combinelike resistors in parallel.

[ h i n t ]


Two previously uncharged capacitors are connected in series and then charged with a 12-Vsource. One capacitor is 30 F and the other is unknown. If the voltage across the 30-Fcapacitor is 8 V, find the capacitance of the unknown capacitor.

The charge on the 30-F capacitor is

Q=CV=(30 F)(8 V)=240 C

Since the same current flows in each of the series capacitors, each capacitor gains the samecharge in the same time period:

C =Q

V=

240 C4V

= 60 F

SOLUTION

EXAMPLE

6.13

PARALLEL CAPACITORS To determine the equivalent capacitance of N capacitorsconnected in parallel, we employ KCL. As can be seen from Fig. 6.15a,

6.18

6.19

where

Cp=C1+C2+C3+p+CN 6.20

= Cpdv(t)dt

= a aN

i= 1Ci b dv(t)dt

= C1dv(t)dt

+ C2dv(t)dt

+ C3dv(t)dt

+ p + CNdv(t)dt

i(t) = i1(t) + i2(t) + i3(t) + p + iN(t)

v(t)

+

-

i1(t)

i(t)

v(t)

+

-

i(t)

C1 C2 C3 CN

i2(t) i3(t) iN(t)

(a) (b)

Cp

Figure 6.15Equivalent circuit for

N capacitors connectedin parallel.

Determine the equivalent capacitance at terminals A-B of the circuit shown in Fig. 6.16.

Cp = 15 FSOLUTION

EXAMPLE

6.14

v(t)

+

-

A

B

4 F 6 F 2 F 3 F

Figure 6.16

Circuit containingmultiple capacitors

in parallel.

Capacitors in parallel combinelike resistors in series.

[ h i n t ]


E6.11 Two initially uncharged capacitors are connected as shown in Fig. E6.11. After a periodof time, the voltage reaches the value shown. Determine the value of C1.

Learning AssessmentsANSWER: C1=4 F.

E6.12 Compute the equivalent capacitance of the network in Fig. E6.12. ANSWER: Ceq=1.5 F.

24 V

6 V

+

-

+

-

12 F

C1

Figure E6.11

Ceq

3 F

2 F 4 F

2 F

3 F

12 FFigure E6.12

E6.13 Determine CT in Fig. E6.13. ANSWER: 1.667 F.

CT5 F 2 F8 F

6 F

3 F 6 F10 F

6 FA

B

6 F4 F

Figure E6.13

SERIES INDUCTORS If N inductors are connected in series, the equivalent inductanceof the combination can be determined as follows. Referring to Fig. 6.17a and using KVL, wesee that

v(t)=v1(t)+v2(t)+v3(t)+p+vN(t) 6.21and therefore,

6.22

6.23 = LSdi(t)dt

= a aN

i= 1Li b di(t)dt

v(t) = L1di(t)dt+ L2

di(t)dt+ L3

di(t)dt+ p + LN

di(t)dt


where

6.24

Therefore, under this condition the network in Fig. 6.17b is equivalent to that in Fig. 6.17a.

LS = aN

i= 1Li = L1 + L2 + p + LN

v(t)

L1

LN

L2 L3+

+

-

-i(t) v1(t)

- +vN(t)

+ -v2(t) + -v3(t)

v(t) LS

+

-

i(t)

(a) (b)

Figure 6.17

Equivalent circuitfor N series-connected

inductors.

Find the equivalent inductance of the circuit shown in Fig. 6.18.

The equivalent inductance of the circuit shown in Fig. 6.18 is

LS=1H+2H+4H=7H

SOLUTION

EXAMPLE

6.15

v(t) 4 H

1 H 2 H

+

-

Figure 6.18

Circuit containingmultiple inductors.

PARALLEL INDUCTORS Consider the circuit shown in Fig. 6.19a, which contains Nparallel inductors. Using KCL, we can write

i(t)=i1(t)+i2(t)+i3(t)+p+iN(t) 6.25However,

6.26

Substituting this expression into Eq. (6.25) yields

6.27

6.28 =1

Lp 3t

t0

v(x)dx + iAt0B

i(t) = a aN

j= 1

1

Ljb 3

t

t0

v(x)dx + aN

j= 1ijAt0B

ij(t) =1

Lj 3t

t0

v(x)dx + ijAt0B

Inductors in series combinelike resistors in series.

[ h i n t ]


Determine the equivalent inductance and the initial current for the circuit shown inFig. 6.20.

The equivalent inductance is

where all inductance values are in millihenrys:

Lp=2 mH

and the initial current is iAt0B = -1 A.

1

Lp=1

12+1

6+1

4

SOLUTION

EXAMPLE

6.16

where

6.29

and is equal to the current in Lp at t=t0. Thus, the circuit in Fig. 6.19b is equivalent tothat in Fig. 6.19a under the conditions stated previously.iAt0B

1

Lp=1

L1+1

L2+1

L3+ p +

1

LN

v(t)

+

-

i1(t)

i(t)

L1

i2(t)

L2

i3(t) iN(t)

L3 LN v(t)

+

-

i(t)

Lp

(a) (b)

Figure 6.19

Equivalent circuits forN inductors connectedin parallel.

v(t)

+

-

3 A

i(t)

12 mH

6 A

6 mH

2 A

4 mH

Figure 6.20

Circuit containingmultiple inductors withinitial currents.

The previous material indicates that capacitors combine like conductances, whereasinductances combine like resistances.

E6.14 Determine the equivalent inductance of the network in Fig. E6.14 if all inductors are 6 mH.

Learning AssessmentANSWER: 9.429 mH.

Leq

Figure E6.14

Inductors in parallel combinelike resistors in parallel.

[ h i n t ]


E6.15 Find LT in Fig. E6.15. ANSWER: 5 mH.

LT2 mH 6 mH 4 mH12 mHA

B

2 mH 4 mH 5 mH

2 mH 3 mH 2 mHFigure E6.15

CHIP CAPACITORS In Chapter 2, we briefly discussed the resistors that are used in mod-ern electronic manufacturing. An example of these surface mount devices was shown inFig. 2.41, together with some typical chip capacitors. As we will indicate in the material thatfollows, modern electronics employs primarily resistors and capacitors and avoids the use ofinductors when possible.

Surface-mounted chip capacitors account for the majority of capacitors used in electron-ics assembly today. These capacitors have a large range of sizes, from as small as 10 mils ona side up to 250 mils on a side. All ceramic chip capacitors consist of a ceramic dielectriclayer between metal plates. The properties of the ceramic and metal layers determine the typeof capacitor, its capacitance, and reliability. A cut-away view of a standard chip capacitor isshown in Fig. 6.21. The inner metal electrodes are alternately connected to the opposing sidesof the chip where metal terminators are added. These terminators not only make connectionto the inner electrodes, but also provide a solder base for attaching these chips to printed

Ceramic dielectric

Inner electrodes (Ni/Cu)

Tin

Nickel

Copper

Figure 6.21

Cross section of amultilayer ceramic chip

capacitor.

circuit boards. The number of alternating layers, the spacing between them, along with thedielectric constant of the ceramic material, will determine the capacitance value.

We indicated earlier that resistors are normally manufactured in standard sizes with specific power ratings. Chip capacitors are also manufactured in this manner, and Table 6.4provides a partial listing of these devices.

The standard sizes of chip capacitors are shown in Table 6.4.


TABLE 6.4 Ceramic chip capacitor standard sizes

Size Code Size (Mils) Power Rating (Watts)

0201 20 10 1/20

0402 40 20 1/16

0603 60 30 1/10

0805 80 50 1/8

1206 120 60 1/4

2010 200 100 1/2

2512 250 120 1

CHIP INDUCTORS A chip inductor consists of a miniature ceramic substrate with eithera wire wrapped around it or a thin film deposited and patterned to form a coil. They can beencapsulated or molded with a material to protect the wire from the elements or left unpro-tected. Chip inductors are supplied in a variety of types and values, with three typical con-figurations that conform to the standard chip package widely utilized in the printed circuitboard (PCB) industry.

The first type is the precision chip inductor where copper is deposited onto the ceramicand patterned to form a coil, as shown in Fig. 6.22.

Copper (Cu)Termination Base

EtchedCopperCoil

AluminaSubstrate

3 mTin (Sn)Outerplating

2 mNickel (Ni)Barrier

Figure 6.22

Precision chip inductor cross section.

TerminalElectrode

InternalMedium

Ferrite

E

WL

T

Figure 6.23

Ferrite chip inductor cross section

The second type is a ferrite chip inductor, which uses a series of coil patterns stackedbetween ferrite layers to form a multiplayer coil as shown in Fig. 6.23.

The third type is a wire-wound open frame in which a wire is wound around a ceramicsubstrate to form the inductor coil. The completed structure is shown in Fig. 6.24.

Each of these configurations displays different characteristics, with the wire-wound typeproviding the highest inductance values (10 nH4.7 uH)and reasonable tolerances

The ferrite chip inductor gives a wide range of values (47 nH33 uH) but has tol-erances in the 5% range. The precision chip inductor has low inductance values(1100 nH) but very good tolerances (/0.1 nH).

(12%).

B

K

C

EA

FF

G

D

Figure 6.24

Wire-wound chip inductorcross section

Condensadores e Inductancias

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