5S Graphic World: February 2017

Friday 24 February 2017

Photoshop cs6 tutorial....full pack...

5S Graphic Design World SIS logo 3designs Professional Logo Design - Ado...

Professional Logo Design - Adobe Illustrator cs6

Here is another LOGO Designing Tutorial. In this video we have used 3D Extrude and Bevel, Gradient, Perspective to create this LOGO Design. Whole preparation process for designing a Professional Logo has been explained with necessary tips.

ABOUT VIDEO!

Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.

Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Professional Logo Design - Adobe Illustrator cs6

Here is another LOGO Designing Tutorial. In this video we have used 3D Extrude and Bevel, Gradient, Perspective to create this LOGO Design. Whole preparation process for designing a Professional Logo has been explained with necessary tips.

ABOUT VIDEO!

Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.

Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Professional Logo Design - Adobe Illustrator cs6

Here is another LOGO Designing Tutorial. In this video we have used 3D Extrude and Bevel, Gradient, Perspective to create this LOGO Design. Whole preparation process for designing a Professional Logo has been explained with necessary tips.

ABOUT VIDEO!

Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.

Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Professional Logo Design - Adobe Illustrator cs6

Here is another LOGO Designing Tutorial. In this video we have used 3D Extrude and Bevel, Gradient, Perspective to create this LOGO Design. Whole preparation process for designing a Professional Logo has been explained with necessary tips.

ABOUT VIDEO!

Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.

Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Wednesday 22 February 2017

5S Graphic Design World SIS logo 2 designs Professional Logo Design - Ad...

Professional Logo Design - Adobe Illustrator cs6

Here is another LOGO Designing Tutorial. In this video we have used 3D Extrude and Bevel, Gradient, Perspective to create this LOGO Design. Whole preparation process for designing a Professional Logo has been explained with necessary tips.

ABOUT VIDEO!

Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.

Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Professional Logo Design - Adobe Illustrator cs6

ABOUT VIDEO!

Professional Logo Design - Adobe Illustrator cs6

ABOUT VIDEO!

Professional Logo Design - Adobe Illustrator cs6

ABOUT VIDEO!

Tuesday 21 February 2017

5S Graphic Design World SIS logo designs Professional Logo Design - Adob...

Monday 20 February 2017

SIS IS BACK

5S Graphic Design World SIS logo designs Professional Logo Design - Adob...

5S Graphic Design World SIS logo designs Professional Logo Design - Adobe Illustrator cs6

Sunday 19 February 2017

5S Graphic Design World SIS logo designs Professional Logo Design - Adob...

ABOUT VIDEO!
Creating a 3D logo is a process that requires extensive knowledge in the use of Adobe Illustrator program, but with the help of some tricks and using simple strokes can get an amazing 3D effect. This tutorial is made and, if somewhere there is a part that relates to the art of drawing, believe me, I could not escape. I hope you will learn something you did not know, thus improving their technique. Working tutorials which are complicated as they Clever Mark’s, you are learn much more, because these are solutions that are final. No need to switch between progressive lessons that are not applicable in practice.
Logo design is a matter of ideas and the need to connect multiple business parameters of the company and its visual recognition should also be beautiful and ingenious, maybe just in this tutorial is an element that will give you an idea as inspiration.

Saturday 18 February 2017

SIS IS BACK

My weakness is that I always think everyone is a good person, therefore, I trust people so easily but am forcefully working at it to change that perception of mine.

I have lots of strengths but for the sake of this channel, I will just mention what is required amongst them for the purpose of this my channel. Loyalty, integrity, adaptable to changing work environment, hardworking, able to work with minimum supervision, able to work under stress and a good interpersonal skills with people from all levels being it customers, superiors and colleagues.

Friday 17 February 2017

How to make HT Transformers - High Tension Transformers and HT Transformer

Adobe Illustrator Tutorials | How To Make Logo Design 01 by SIS

Thursday 9 February 2017

LinkCollider - Free SEO Tools Plus Social Media Sharing

LinkCollider - Free SEO Tools Plus Social Media Sharing: Increase website traffic by using SEO tools by LinkCollider. LinkCollider is a free SEO tool that uses link building and social media sites to improve SEO.

Wednesday 8 February 2017

Capacitors

Capacitors

10.1 INTRODUCTION
Thus far, the only passive device appearing in the text has been the resistor. We will now consider two additional passive devices called the capacitor and the inductor (the inductor is discussed in detail in Chapter 12), which are quite different from the resistor in purpose, operation, and construction. Unlike the resistor, both elements display their total characteristics only when a change in voltage or current is made in the circuit in which they exist. In addition, if we consider the ideal situation, they do not dissipate energy as does the resistor but store it in a form that can be returned to the circuit whenever required by the circuit design. Proper treatment of each requires that we devote this entire chapter to the capacitor and, as mentioned above, Chapter 12 to the inductor. Since electromagnetic effects are a major consideration in the design of inductors, this topic will be covered in Chapter 11.
10.2 THE ELECTRIC FIELD
Recall from Chapter 2 that a force of attraction or repulsion exists between two charged bodies. We shall now examine this phenomenon in greater detail by considering the electric ﬁeld that exists in the region around any charged body. This electric ﬁeld is represented by electric ﬂux lines, which are drawn to indicate the strength of the electric ﬁeld at any point around the charged body; that is, the denser the lines of ﬂux, the stronger the electric ﬁeld. In Fig. 10.1, the electric ﬁeld strength is stronger at position a than at position b because the ﬂux lines are denser at a than at b. The symbol for electric ﬂux is the Greek letter w (psi). The ﬂux per unit area (ﬂux density) is represented by the capital letter D and is determined by
(ﬂux/unit area) (10.1)D w A
Capacitors
376  CAPACITORS
The larger the charge Q in coulombs, the greater the number of ﬂux lines extending or terminating per unit area, independent of the surrounding medium. Twice the charge will produce twice the ﬂux per unit area. The two can therefore be equated:
(coulombs, C) (10.2)
By deﬁnition, the electric ﬁeld strength at a point is the force acting on a unit positive charge at that point; that is,
(newtons/coulomb, N/C) (10.3)
The force exerted on a unit positive charge (Q2 1 C), by a charge Q1, r meters away, as determined by Coulomb’s law is
F kQ r 1 2 Q2 kQ r 1 2 (1) k r Q 2 1 (k 9 109 N.m2/C2)
Substituting this force F into Eq. (10.3) yields

(N/C) (10.4)
We can therefore conclude that the electric ﬁeld strength at any point distance r from a point charge of Q coulombs is directly proportional to the magnitude of the charge and inversely proportional to the distance squared from the charge. The squared term in the denominator will result in a rapid decrease in the strength of the electric ﬁeld with distance from the point charge. In Fig. 10.1, substituting distances r1 and r2 into Eq. (10.4) will verify our previous conclusion that the electric ﬁeld strength is greater at a than at b.
Electric ﬂux lines always extend from a positively charged body to a negatively charged body, always extend or terminate perpendicular to the charged surfaces, and never intersect.
For two charges of similar and opposite polarities, the ﬂux distribution would appear as shown in Fig. 10.2.
k r Q 2 1
kQ1/r2 1
F Q2
Q F
w Q
r1 +
r2 a b
Positive charge
Electric flux lines
Flux lines radiate outward for positive charges and inward for negative charges.
FIG. 10.1 Flux distribution from an isolated positive charge.
(b)
(a)
+ +
+ –
FIG. 10.2 Electric ﬂux distribution: (a) like charges; (b) opposite charges.
CAPACITANCE  377
The attraction and repulsion between charges can now be explained in terms of the electric ﬁeld and its ﬂux lines. In Fig. 10.2(a), the ﬂux lines are not interlocked but tend to act as a buffer, preventing attraction and causing repulsion. Since the electric ﬁeld strength is stronger (ﬂux lines denser) for each charge the closer we are to the charge, the more we try to bring the two charges together, the stronger will be the force of repulsion between them. In Fig. 10.2(b), the ﬂux lines extending from the positive charge are terminated at the negative charge. A basic law of physics states that electric ﬂux lines always tend to be as short as possible. The two charges will therefore be drawn to each other. Again, the closer the two charges, the stronger the attraction between the two charges due to the increased ﬁeld strengths.
10.3 CAPACITANCE
Up to this point we have considered only isolated positive and negative spherical charges, but the analysis can be extended to charged surfaces of any shape and size. In Fig. 10.3, for example, two parallel plates of a conducting material separated by an air gap have been connected through a switch and a resistor to a battery. If the parallel plates are initially uncharged and the switch is left open, no net positive or negative charge will exist on either plate. The instant the switch is closed, however, electrons are drawn from the upper plate through the resistor to the positive terminal of the battery. There will be a surge of current at ﬁrst, limited in magnitude by the resistance present. The level of ﬂow will then decline, as will be demonstrated in the sections to follow. This action creates a net positive charge on the top plate. Electrons are being repelled by the negative terminal through the lower conductor to the bottom plate at the same rate they are being drawn to the positive terminal. This transfer of electrons continues until the potential difference across the parallel plates is exactly equal to the battery voltage. The ﬁnal result is a net positive charge on the top plate and a negative charge on the bottom plate, very similar in many respects to the two isolated charges of Fig. 10.2(b). This element, constructed simply of two parallel conducting plates separated by an insulating material (in this case, air), is called a capacitor. Capacitance is a measure of a capacitor’s ability to store charge on its plates—in other words, its storage capacity.
A capacitor has a capacitance of 1 farad if 1 coulomb of charge is deposited on the plates by a potential difference of 1 volt across the plates.
The farad is named after Michael Faraday (Fig. 10.4), a nineteenthcentury English chemist and physicist. The farad, however, is generally too large a measure of capacitance for most practical applications, so the microfarad (10 6)orpicofarad (10 12)ismore commonly used. Expressed as an equation, the capacitance is determined by
C farads (F) Q coulombs (C) (10.5) V volts (V)
Different capacitors for the same voltage across their plates will acquire greater or lesser amounts of charge on their plates. Hence the capacitors have a greater or lesser capacitance, respectively. A cross-sectional view of the parallel plates is shown with the distribution of electric ﬂux lines in Fig. 10.5(a). The number of ﬂux lines per
C Q V
R
e
e
e e
E
e
e
e
+++ –––
V = E
+
–
FIG. 10.3 Fundamental charging network.

Fringing
+++++++++
–––––––––
(a)

+++++++++
(b)
–––––––––
FIG. 10.5 Electric ﬂux distribution between the plates of a capacitor: (a) including fringing; (b) ideal.
FIG. 10.4 Michael Faraday.
Courtesy of the Smithsonian Institution Photo No. 51,147
English (London) (1791–1867) Chemist and Electrical Experimenter Honory Doctorate, Oxford University, 1832
An experimenter with no formal education, he began his research career at the Royal Institute in London as a laboratory assistant. Intrigued by the interaction between electrical and magnetic effects, he discovered electromagnetic induction, demonstrating that electrical effects can be generated from a magnetic ﬁeld (the birth of the generator as we know it today). He also discovered self-induced currents and introduced the concept of lines and ﬁelds of magnetic force. Having received over one hundred academic and scientiﬁc honors, he became a Fellow of the Royal Society in 1824 at the young age of 32.
unit area (D) between the two plates is quite uniform. At the edges, the ﬂux lines extend outside the common surface area of the plates, producing an effect known as fringing. This effect, which reduces the capacitance somewhat, can be neglected for most practical applications. For the analysis to follow, we will assume that all the ﬂux lines leaving the positive plate will pass directly to the negative plate within the common surface area of the plates [Fig. 10.5(b)]. If a potential difference of V volts is applied across the two plates separated by a distance of d, the electric ﬁeld strength between the plates is determined by
(volts/meter, V/m) (10.6)
The uniformity of the ﬂux distribution in Fig. 10.5(b) also indicates that the electric ﬁeld strength is the same at any point between the two plates. Many values of capacitance can be obtained for the same set of parallel plates by the addition of certain insulating materials between the plates. In Fig. 10.6(a), an insulating material has been placed between a set of parallel plates having a potential difference of V volts across them. Since the material is an insulator, the electrons within the insulator are unable to leave the parent atom and travel to the positive plate. The positive components (protons) and negative components (electrons) of each atom do shift, however [as shown in Fig. 10.6(a)], to form dipoles. When the dipoles align themselves as shown in Fig. 10.6(a), the material is polarized. A close examination within this polarized material will indicate that the positive and negative components of adjoining dipoles are neutralizing the effects of each other [note the dashed area in Fig. 10.6(a)]. The layer of positive charge on one surface and the negative charge on the other are not neutralized, however, resulting in the establishment of an electric ﬁeld within the insulator [ dielectric; Fig. 10.6(b)]. The net electric ﬁeld between the plates ( resultant air dielectric) would therefore be reduced due to the insertion of the dielectric. The purpose of the dielectric, therefore, is to create an electric ﬁeld to oppose the electric ﬁeld set up by free charges on the parallel plates. For this reason, the insulating material is referred to as a dielectric, di for “opposing” and electric for “electric ﬁeld.” In either case—with or without the dielectric—if the potential across the plates is kept constant and the distance between the plates is ﬁxed, the net electric ﬁeld within the plates must remain the same, as determined by the equation V/d. We just ascertained, however, that the net electric ﬁeld between the plates would decrease with insertion of the dielectric for a ﬁxed amount of free charge on the plates. To compensate and keep the net electric ﬁeld equal to the value determined by V and d, more charge must be deposited on the plates. [Look ahead to Eq. (10.11).] This additional charge for the same potential across the plates increases the capacitance, as determined by the following equation:
CD
For different dielectric materials between the same two parallel plates, different amounts of charge will be deposited on the plates. But
QD V
V d
378  CAPACITORS
(a)

(b)
+++++++
– –
– – – – –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ – +++++++
– –
– – – – –
+
–
V (volts)
+
–
Dielectric
d
V (volts)
+
–
(Dielectric = air)
+++++++
– –
– – – – –

(Dielectric)
(Resultant)
d
FIG. 10.6 Effect of a dielectric on the ﬁeld distribution between the plates of a capacitor: (a) alignment of dipoles in the dielectric; (b) electric ﬁeld components between the plates of a capacitor with a dielectric present.
CAPACITANCE  379
w Q, so the dielectric is also determining the number of ﬂux lines between the two plates and consequently the ﬂux density (D w/A) since A is ﬁxed. The ratio of the ﬂux density to the electric ﬁeld intensity in the dielectric is called the permittivity of the dielectric:
e (farads/meter, F/m) (10.7)
It is a measure of how easily the dielectric will “permit” the establishment of ﬂux lines within the dielectric. The greater its value, the greater the amount of charge deposited on the plates, and, consequently, the greater the ﬂux density for a ﬁxed area. For a vacuum, the value of e (denoted by eo) is 8.85 10 12 F/m. The ratio of the permittivity of any dielectric to that of a vacuum is called the relative permittivity, er. It simply compares the permittivity of the dielectric to that of air. In equation form,
er (10.8)
The value of e for any material, therefore, is
e ereo
Note that er is a dimensionless quantity. The relative permittivity, or dielectric constant, as it is often called, is provided in Table 10.1 for various dielectric materials. Substituting for D and in Eq. (10.7), we have
e
But C
and, therefore, e
Cd A
Q V
Qd VA
Q/A V/d
w/A V/d
D
e eo
D
TABLE 10.1 Relative permittivity (dielectric constant) of various dielectrics.
Dielectric er (Average Values)
Vacuum 1.0 Air 1.0006 Teﬂon 2.0 Paper, parafﬁned 2.5 Rubber 3.0 Transformer oil 4.0 Mica 5.0 Porcelain 6.0 Bakelite 7.0 Glass 7.5 Distilled water 80.0 Barium-strontium titanite (ceramic) 7500.0
and C e (F) (10.9)
or C eoer 8.85 10 12er (F) (10.10)
where A is the area in square meters of the plates, d is the distance in meters between the plates, and er is the relative permittivity. The capacitance, therefore, will be greater if the area of the plates is increased, or the distance between the plates is decreased, or the dielectric is changed so that er is increased. Solving for the distance d in Eq. (10.9), we have
D
and substituting into Eq. (10.6) yields
V d
But Q CV, and therefore
(V/m) (10.11)
which gives the electric ﬁeld intensity between the plates in terms of the permittivity e, the charge Q, and the surface area A of the plates. Thus, we have the ratio
er
or (10.12)
which, in words, states that for the same set of parallel plates, the capacitance using a dielectric (of relative permittivity er) is er times that obtained for a vacuum (or air, approximately) between the plates. This relationship between er and the capacitances provides an excellent experimental method for ﬁnding the value of er for various dielectrics.
EXAMPLE 10.1 Determine the capacitance of each capacitor on the right side of Fig. 10.7.
Solutions: a. C 3(5 mF) 15 mF b. C (0.1 mF) 0.05 mF
c. C 2.5(20 mF) 50 mF d. C (5) (1000 pF) (160)(1000 pF) 0.16 mF 4 (1/8)
1 2
C erCo
e eo
C eA/d Co eoA/d
Q eA
CV eA
V eA/C
eA C
A d
A d
A d
380  CAPACITORS
CAPACITANCE  381
EXAMPLE 10.2 For the capacitor of Fig. 10.8: a. Determine the capacitance. b. Determine the electric ﬁeld strength between the plates if 450 V are applied across the plates. c. Find the resulting charge on each plate.
Solutions:
a. Co 59.0 10 12 F
59 pF
b. 300 103 V/m
c. C or
Q CV (59.0 10 12 F)(450 V) 26.550 10 9 C 26.55 nC
EXAMPLE 10.3 A sheet of mica 1.5 mm thick having the same area as the plates is inserted between the plates of Example 10.2. a. Find the electric ﬁeld strength between the plates. b. Find the charge on each plate. c. Find the capacitance.
Q V
450 V 1.5 10 3 m
V d
(8.85 10 12 F/m)(0.01 m2) 1.5 10 3 m
eo A d
d
C = 5 µF
d
3A
C = ?
C = 0.1 µF
(a)
2dd
C = ?
(b)
C = 20 µF C = ?
r = 2.5 (paraffined paper)
o
(c)
C = 1000 pF C = ? A
d
(d)
o
1 8 d
4A
r = 5 (mica)
A
FIG. 10.7 Example 10.1.
FIG. 10.8 Example 10.2.
d = 1.5 mmQ (+)
A = 0.01 m2 Q(–) o
382  CAPACITORS
Solutions: a. is ﬁxed by

300 103 V/m
b. or
Q e A ereo A (5)(8.85 10 12 F/m)(300 103 V/m)(0.01 m2) 132.75 10 9 C 132.75 nC (ﬁve times the amount for air between the plates)
c. C erCo (5)(59 10 12 F) 295 pF
10.4 DIELECTRIC STRENGTH
For every dielectric there is a potential that, if applied across the dielectric, will break the bonds within the dielectric and cause current to ﬂow. The voltage required per unit length (electric ﬁeld intensity) to establish conduction in a dielectric is an indication of its dielectric strength and is called the breakdown voltage. When breakdown occurs, the capacitor has characteristics very similar to those of a conductor. A typical example of breakdown is lightning, which occurs when the potential between the clouds and the earth is so high that charge can pass from one to the other through the atmosphere, which acts as the dielectric. The average dielectric strengths for various dielectrics are tabulated in volts/mil in Table 10.2 (1 mil 0.001 in.). The relative permittivity appears in parentheses to emphasize the importance of considering both factors in the design of capacitors. Take particular note of bariumstrontium titanite and mica.
Q eA
450 V 1.5 103 m
V d
TABLE 10.2 Dielectric strength of some dielectric materials.
Dielectric Strength (Average Value), in Dielectric Volts/Mil (er)
Air 75 (1.0006) Barium-strontium titanite (ceramic) 75 (7500) Porcelain 200 (6.0) Transformer oil 400 (4.0) Bakelite 400 (7.0) Rubber 700 (3.0) Paper, parafﬁned 1300 (2.5) Teﬂon 1500 (2.0) Glass 3000 (7.5) Mica 5000 (5.0)
TYPES OF CAPACITORS  383
EXAMPLE 10.4 Find the maximum voltage that can be applied across a 0.2-mF capacitor having a plate area of 0.3 m2. The dielectric is porcelain. Assume a linear relationship between the dielectric strength and the thickness of the dielectric.
Solution:
C 8.85 10 12er
or d 7.965 10 5 m
79.65 mm
Converting micrometers to mils, we have 79.65 mm 3.136 mils Dielectric strength 200 V/mil Therefore, (3.136 mils) 627.20 V
10.5 LEAKAGE CURRENT
Up to this point, we have assumed that the ﬂow of electrons will occur in a dielectric only when the breakdown voltage is reached. This is the ideal case. In actuality, there are free electrons in every dielectric due in part to impurities in the dielectric and forces within the material itself. When a voltage is applied across the plates of a capacitor, a leakage current due to the free electrons ﬂows from one plate to the other. The current is usually so small, however, that it can be neglected for most practical applications. This effect is represented by a resistor in parallel with the capacitor, as shown in Fig. 10.9(a), whose value is typically more than 100 megohms (M ). Some capacitors, however, such as the electrolytic type, have high leakage currents. When charged and then disconnected from the charging circuit, these capacitors lose their charge in a matter of seconds because of the ﬂow of charge (leakage current) from one plate to the other [Fig. 10.9(b)].
10.6 TYPES OF CAPACITORS
Like resistors, all capacitors can be included under either of two general headings: ﬁxed or variable. The symbol for a ﬁxed capacitor is , and
200 V mil
1000 mils 1 in.
39.371 in. m
10 6 m mm
(8.85)(6)(0.3 m2) (1012)(0.2 10 6 F)
8.85erA 1012C
A d
FIG. 10.9 Demonstrating the effect of the leakage current.
FIG. 10.10 Basic structure of a stacked mica capacitor.
(a)
(b)
Rleakage
C
+ –
Foil
Foil
Foil
Foil
Mica
Mica
Mica
for a variable capacitor, . The curved line represents the plate that is usually connected to the point of lower potential.
Fixed Capacitors
Many types of ﬁxed capacitors are available today. Some of the most common are the mica, ceramic, electrolytic, tantalum, and polyesterﬁlm capacitors. The typical ﬂat mica capacitor consists basically of mica sheets separated by sheets of metal foil. The plates are connected to two electrodes, as shown in Fig. 10.10. The total area is the area of
384  CAPACITORS
one sheet times the number of dielectric sheets. The entire system is encased in a plastic insulating material as shown for the two central units of Fig. 10.11. The mica capacitor exhibits excellent characteristics under stress of temperature variations and high voltage applications (its dielectric strength is 5000 V/mil). Its leakage current is also very small (Rleakage about 1000 M ). Mica capacitors are typically between a few picofarads and 0.2 mF, with voltages of 100 V or more. The ability to “roll” the mica to form the cylindrical shapes of Fig. 10.11 is due to a process whereby the soluble contaminants in natural mica are removed, leaving a paperlike structure resulting from the cohesive forces in natural mica. It is commonly referred to as reconstituted mica, although the terminology does not mean “recycled” or “secondhand” mica. For some of the units in the photograph, different levels of capacitance are available between different sets of terminals. The ceramic capacitor is made in many shapes and sizes, two of which are shown in Fig. 10.12. The basic construction, however, is about the same for each, as shown in Fig. 10.13. A ceramic base is coated on two sides with a metal, such as copper or silver, to act as the two plates. The leads are then attached through electrodes to the plates. An insulating coating of ceramic or plastic is then applied over the plates and dielectric. Ceramic capacitors also have a very low leakage current (Rleakage about 1000 M ) and can be used in both dc and ac networks. They can be found in values ranging from a few picofarads to perhaps 2 mF, with very high working voltages such as 5000 V or more. In recent years there has been increasing interest in monolithic (single-structure) chip capacitors such as those appearing in Fig. 10.14(a) due to their application on hybrid circuitry [networks using both discrete and integrated circuit (IC) components]. There has also been increasing use of microstrip (strip-line) circuitry such as the one in Fig. 10.14(b). Note the small chips in this cutaway section. The L and H of Fig. 10.14(a) indicate the level of capacitance. For example, if in black ink, the letter H represents 16 units of capacitance (in picofarads), or 16 pF. If blue ink is used, a multiplier of 100 is applied, resulting in 1600 pF. Although the size is similar, the type of ceramic material controls the capacitance level. The electrolytic capacitor is used most commonly in situations where capacitances of the order of one to several thousand microfarads
FIG. 10.11 Mica capacitors. (Courtesy of Custom Electronics Inc.)
FIG. 10.12 Ceramic disc capacitors: (a) photograph; (b) construction.
Lead wire soldered to silver electrode
Solder
Ceramic dielectric
Dipped phenolic coating
Silver electrodes deposited on top and bottom of ceramic disc
(b)(a)
TYPES OF CAPACITORS  385
are required. They are designed primarily for use in networks where only dc voltages will be applied across the capacitor because they have good insulating characteristics (high leakage current) between the plates in one direction but take on the characteristics of a conductor in the other direction. Electrolytic capacitors are available that can be used in ac circuits (for starting motors) and in cases where the polarity of the dc voltage will reverse across the capacitor for short periods of time. The basic construction of the electrolytic capacitor consists of a roll of aluminum foil coated on one side with an aluminum oxide, the aluminum being the positive plate and the oxide the dielectric. A layer of paper or gauze saturated with an electrolyte is placed over the aluminum oxide on the positive plate. Another layer of aluminum without the oxide coating is then placed over this layer to assume the role of the negative plate. In most cases the negative plate is connected directly to the aluminum container, which then serves as the negative terminal for external connections. Because of the size of the roll of aluminum foil, the overall area of this capacitor is large; and due to the use of an oxide as the dielectric, the distance between the plates is extremely small. The negative terminal of the electrolytic capacitor is usually the one with no visible identiﬁcation on the casing. The positive is usually indicated by such designs as , , , and so on. Due to the polarity requirement, the symbol for an electrolytic capacitor will normally appear as . Associated with each electrolytic capacitor are the dc working voltage and the surge voltage. The working voltage is the voltage that can be applied across the capacitor for long periods of time without breakdown. The surge voltage is the maximum dc voltage that can be applied for a short period of time. Electrolytic capacitors are characterized as having low breakdown voltages and high leakage currents (Rleakage about 1 M ). Various types of electrolytic capacitors are shown in Fig. 10.15. They can be found in values extending from a few microfarads to several thousand microfarads and working voltages as high as 500 V. However, increased levels of voltage are normally associated with lower values of available capacitance.
FIG. 10.13 Multilayer, radial-lead ceramic capacitor.
FIG. 10.14 Monolithic chip capacitors. (Courtesy of Vitramon, Inc.)
Dipped phenolic coating
Lead wire soldered to electrode pickup
Solder
Electrode pickupMetal electrodes
Ceramic dielectric
(Alternately deposited layers of ceramic dielectric material and metal electrodes fired into a single homogeneous block)
386  CAPACITORS
There are fundamentally two types of tantalum capacitors: the solid and the wet-slug. In each case, tantalum powder of high purity is pressed into a rectangular or cylindrical shape, as shown in Fig. 10.16. Next the anode ( ) connection is simply pressed into the resulting structures, as shown in the ﬁgure. The resulting unit is then sintered (baked) in a vacuum at very high temperatures to establish a very porous material. The result is a structure with a very large surface area in a limited volume. Through immersion in an acid solution, a very thin manganese dioxide (MnO2) coating is established on the large, porous surface area. An electrolyte is then added to establish contact between the surface area and the cathode, producing a solid tantalum capacitor. If an appropriate “wet” acid is introduced, it is called a wet-slug tantalum capacitor.
FIG. 10.15 Electrolytic capacitors: (a) Radial lead with extended endurance rating of 2000 h at 85°C. Capacitance range: 0.1–15,000 mF with a voltage range of 6.3 to 250 WV dc (Courtesy of Illinois Capacitor, Inc.). (b) Solid aluminum electrolytic capacitors available in axial, resin-dipped, and surface-mount conﬁgurations to withstand harsh environmental conditions (Courtesy of Philips Components, Inc.).
FIG. 10.16 Tantalum capacitor. (Courtesy of Union Carbide Corp.)
FIG. 10.17 Polyester-ﬁlm capacitor.
The last type of ﬁxed capacitor to be introduced is the polyester-ﬁlm capacitor, the basic construction of which is shown in Fig. 10.17. It consists simply of two metal foils separated by a strip of polyester material such as Mylar®. The outside layer of polyester is applied to act as an insulating jacket. Each metal foil is connected to a lead that extends either axially or radially from the capacitor. The rolled construction results in a large surface area, and the use of the plastic dielectric results in a very thin layer between the conducting surfaces.

Solder
Cathode ( – )
MnO2 coat
Carbon
Solder
Lead wire
Anode ( + )
Tantalum
Tantalum wire
Metal foils
Polyester (plastic) film
Data such as capacitance and working voltage are printed on the outer wrapping if the polyester capacitor is large enough. Color coding is used on smaller devices (see Appendix D). A band (usually black) is sometimes printed near the lead that is connected to the outer metal foil. The lead nearest this band should always be connected to the point of
TYPES OF CAPACITORS  387
lower potential. This capacitor can be used for both dc and ac networks. Its leakage resistance is of the order of 100 M . An axial lead and radial lead polyester-ﬁlm capacitor appear in Fig. 10.18. The axial lead variety is available with capacitance levels of 0.1 mF to 18 mF, with working voltages extending to 630 V. The radial lead variety has a capacitance range of 0.01 mF to 10 mF, with working voltages extending to 1000 V.
FIG. 10.18 Polyester-ﬁlm capacitors: (a) axial lead; (b) radial lead. (Courtesy of Illinois Capacitor, Inc.)
FIG. 10.19 Variable air capacitors. [Part (a) courtesy of James Millen Manufacturing Co.; part (b) courtesy of Johnson Manufacturing Co.]
(a) (b)
Variable Capacitors
The most common of the variable-type capacitors is shown in Fig. 10.19. The dielectric for each capacitor is air. The capacitance in Fig. 10.19(a) is changed by turning the shaft at one end to vary the common area of the movable and ﬁxed plates. The greater the common area, the larger the capacitance, as determined by Eq. (10.10). The capacitance of the trimmer capacitor in Fig. 10.19(b) is changed by turning the screw, which will vary the distance between the plates (the common area is ﬁxed) and thereby the capacitance.
(a) (b)
388  CAPACITORS
Measurement and Testing
A digital reading capacitance meter appears in Fig. 10.20. Simply place the capacitor between the provided clips with the proper polarity, and the meter will display the level of capacitance. The best check of a capacitor is to use a meter designed to perform the necessary tests. However, an ohmmeter can identify those in which the dielectric has deteriorated (especially in paper and electrolytic capacitors). As the dielectric breaks down, the insulating qualities decrease to a point where the resistance between the plates drops to a relatively low level. After ensuring that the capacitor is fully discharged, place an ohmmeter across the capacitor, as shown in Fig. 10.21. In a polarized capacitor, the polarities of the meter should match those of the capacitor. A low-resistance reading (zero ohms to a few hundred ohms) normally indicates a defective capacitor. The above test of leakage is not all-inclusive, since some capacitors will break down only when higher voltages are applied. The test, however, does identify those capacitors that have lost the insulating quality of the dielectric between the plates.
Standard Values and Recognition Factor
The standard values for capacitors employ the same numerical multipliers encountered for resistors. The most common have the same numerical multipliers as the most common resistors, that is, those available with the full range of tolerances (5%, 10%, and 20%) as shown in Table 3.8. They include 0.1 mF, 0.15 mF, 0.22 mF, 0.33 mF, 0.47 mF, and 0.68 mF, and then 1 mF, 1.5 mF, 2.2 mF, 3.3 mF, 4.7 mF, and so on. Figure 10.22 was developed to establish a recognition factor when it comes to the various types of capacitors. In other words, it will help you to develop the skills to identify types of capacitors, their typical range of values, and some of the most common applications. The ﬁgure is certainly not all-inclusive, but it does offer a ﬁrst step in establishing a sense for what to expect for various applications.
Marking Schemes
Due to the small size of some capacitors, various marking schemes have been adopted to provide the capacitance level, the tolerance, and, if possible, the maximum working voltage. In general, however, the size of the capacitor is the ﬁrst indicator of its value. The smaller units are typically in picofarads (pF) and the larger units in microfarads (mF). Keeping this simple fact in mind will usually provide an immediate indication of the expected capacitance level. On larger mF units, the value can usually be printed on the jacket with the tolerance and maximum working voltage. However, smaller units need to use some form of abbreviation as shown in Fig. 10.23. For very small units such as appearing in Fig. 10.23(a), the value is recognized immediately as in pF, with the K an indicator of a 10% tolerance level. Too often the K is read as a multiplier of 10 3, and the capacitance read as 20,000 pF or 20 nF. For the unit of Fig. 10.23(b), there was room for a lowercase “n” to represent a multiplier of 10 9. The presence of the lowercase “n” in combination with the small size is clear indication that this is a 200-nF capacitor. To avoid unnecessary confusion, the letters used for tolerance do not include N or U or P, so any form of these letters will usually sug
FIG. 10.20 Digital reading capacitance meter. (Courtesy of BK PRECISION, Maxtec International Corp.)
FIG. 10.21 Checking the dielectric of an electrolytic capacitor.

+ –
+++
– – –
TYPES OF CAPACITORS  389
Type: Miniature Axial Electrolytic Typical Values: 0.1 F to 15,000 F Typical Voltage Range: 5 V to 450 V Capacitor tolerance: ±20% Applications: Polarized, used in DC power supplies, bypass filters, DC blocking. µµ
Type: Miniature Radial Electrolyte Typical Values: 0.1 F to 15,000 F Typical Voltage Range: 5 V to 450 V Capacitor tolerance: ±20% Applications: Polarized, used in DC power supplies, bypass filters, DC blocking. µµ
Type: Ceramic Disc Typical Values: 10 pF to 0.047 F Typical Voltage Range: 100 V to 6 kV Capacitor tolerance: ±5%, ±10% Applications: Non-polarized, NPO type, stable for a wide range of temperatures. Used in oscillators, noise filters, circuit coupling, tank circuits. µ
Type: Surface Mount Type (SMT) Typical Values: 10 pF to 10 F Typical Voltage Range: 6.3 V to 16 V Capacitor tolerance: ±10% Applications: Polarized and nonpolarized, used in all types of circuits, requires a minimum amount of PC board real estate. µ
Type: Silver Mica Typical Value: 10 pF to 0.001 F Typical Voltage Range: 50 V to 500 V Capacitor tolerance: ±5% Applications: Non-polarized, used in oscillators, in circuits that require a stable component over a range of temperatures and voltages. µ
Type: Mylar Paper Typical Value: 0.001 F to 0.68 F Typical Voltage Range: 50 V to 600 V Capacitor tolerance: ±22% Applications: Non-polarized, used in all types of circuits, moisture resistant. µµ
Type: Tuning variable Typical Value: 10 pF to 600 pF Typical Voltage Range: 5 V to 100 V Capacitor tolerance: ±10% Applications: Non-polarized, used in oscillators, radio tuning circuit.
Type: Trimmer Variable Typical Value: 1.5 pF to 600 pF Typical Voltage Range: 5 V to 100 V Capacitor tolerance: ±10% Applications: Non-polarized, used in oscillators, tuning circuits, AC filters.
Type: AC/DC Motor Run Typical Value: 0.25 F to 1200 F Typical Voltage Range: 240 V to 660 V Capacitor tolerance: ±10% Applications: Non-polarized, used in motor run-start, high-intensity lighting supplies, AC noise filtering. µµ
+ + + 22 F 50 V µ
R M C .0 5 Z 1 0 0 V Z 5 V
470K Y5P
200 F 10 V µ +
+
+
84-44 640
DIELEKTROL No PCB'S CAPACITOR MADE IN USA
61L1019 10 f 90C 400VAC 60HZ PROTECTED P922 B 10000AFC 16-10000 µ
.1 ± 10% 600WV EM
10 0V E R IE 81 07
.33
35+
+
Type: Dipped Tantalum (solid and wet) Typical Values: 0.047 F to 470 F Typical Voltage Range: 6.3 V to 50 V Capacitor tolerance: ±10%, ±20% Applications: Polarized, low leakage current, used in power supplies, high frequency noise filters, bypass filter. µµ
FIG. 10.22 Summary of capacitive elements.
(a)
20 K
(b)
200n J
(d)
339M
(c)
223F
FIG. 10.23 Various marking schemes for small capacitors.
390  CAPACITORS
gest the multiplier level. The J represents a 5% tolerance level. For capacitors such as appearing in Fig. 10.23(c), the ﬁrst two numbers are actual digits of the value, while the third number is the power of a multiplier (or number of zeros to be added). The F represents a 1% tolerance level. Multipliers of 0.01 use an 8, while 9 is used for 0.1 as shown for the capacitor of Fig. 10.23(d) where the M represents a 20% tolerance level.
10.7 TRANSIENTS IN CAPACITIVE NETWORKS: CHARGING PHASE
Section 10.3 described how a capacitor acquires its charge. Let us now extend this discussion to include the potentials and current developed within the network of Fig. 10.24 following the closing of the switch (to position 1). You will recall that the instant the switch is closed, electrons are drawn from the top plate and deposited on the bottom plate by the battery, resulting in a net positive charge on the top plate and a negative charge on the bottom plate. The transfer of electrons is very rapid at ﬁrst, slowing down as the potential across the capacitor approaches the applied voltage of the battery. When the voltage across the capacitor equals the battery voltage, the transfer of electrons will cease and the plates will have a net charge determined by Q CVC CE. Plots of the changing current and voltage appear in Figs. 10.25 and 10.26, respectively. When the switch is closed at t 0 s, the current jumps to a value limited only by the resistance of the network and then decays to zero as the plates are charged. Note the rapid decay in current level, revealing that the amount of charge deposited on the plates per unit time is rapidly decaying also. Since the voltage across the plates is directly related to the charge on the plates by vC q/C, the rapid rate with which charge is initially deposited on the plates will result in a rapid increase in vC. Obviously, as the rate of ﬂow of charge (I) decreases, the rate of change in voltage will follow suit. Eventually, the ﬂow of charge will stop, the current I will be zero, and the voltage will cease to change in magnitude—the charging phase has passed. At this point the capacitor takes on the characteristics of an open circuit: a voltage drop across the plates without a ﬂow of charge “between” the plates. As demonstrated in Fig. 10.27, the voltage across the capacitor is the source voltage since i iC iR 0 A and vR iRR (0)R 0 V. For all future analysis:
A capacitor can be replaced by an open-circuit equivalent once the charging phase in a dc network has passed.
Looking back at the instant the switch is closed, we can also surmise that a capacitor behaves as a short circuit the moment the switch is closed in a dc charging network, as shown in Fig. 10.28. The current i iC iR E/R, and the voltage vC E vR E iRR E (E/R)R E E 0 V att 0 s. Through the use of calculus, the following mathematical equation for the charging current iC can be obtained:
(10.13)i C E R e t/RC
R
+
E
–
+ – vR = 0 V
VC = E
+
–
iC = 0 A
FIG. 10.27 Open-circuit equivalent for a capacitor following the charging phase.
R
+
E
–
+ – vR = E
vC = 0 V
+
–
iC = iR = E R
FIG. 10.28 Short-circuit equivalent for a capacitor (switch closed, t 0).
FIG. 10.24 Basic charging network.
+ vR –
2
R
+
–
vC
1
iC
C
e
E
+
–
Rapid decay
Small change in iC
R E
iC
0 t
FIG. 10.25 iC during the charging phase.
Small increase in vC
E
vC
0 t
Rapid increase
FIG. 10.26 vC during the charging phase.
TRANSIENTS IN CAPACITIVE NETWORKS: CHARGING PHASE  391
The factor e t/RC is an exponential function of the form e x,where x t/RC and e 2.71828 . . . . A plot of e x for x ≥ 0 appears in Fig. 10.29. Exponentials are mathematical functions that all students of electrical, electronic, or computer systems must become very familiar with. They will appear throughout the analysis to follow in this course, and in succeeding courses. Our current interest in the function e x is limited to values of x greater than zero, as noted by the curve of Fig. 10.25. All modern-day scientiﬁc calculators have the function ex. To obtain e x, the sign of x must be changed using the sign key before the exponential function is keyed in. The magnitude of e x has been listed in Table 10.3 for a range of values of x. Note the rapidly decreasing magnitude of e x with increasing value of x.
TABLE 10.3 Selected values of e x.
x 0 e x e 0 1
x 1 e 1 0.3679
x 2 e 2 0.1353
x 5 e 5 0.00674
x 10 e 10 0.0000454
x 100 e 100 3.72 10 441 e100
1 e10
1 e5
1 e2
1 2.71828 . . .
1 e
1 1
1 e0
0.3679
0.1353
0.0497
0.0067
1
0 1 2 345 x
e–x
0.0183
FIG. 10.29 The e x function (x ≥ 0).
TABLE 10.4 iC versus t (charging phase).
t Magnitude
0 100% 1t 36.8% 2t 13.5% 3t 5.0% 4t 1.8% 5t 0.67%
Less than
6t 0.24%
1% of maximum
    
←
TABLE 10.5 Change in iC between time constants.
(0 → 1)t 63.2% (1 → 2)t 23.3% (2 → 3)t 8.6% (3 → 4)t 3.0% (4 5)t 1.2% (5 → 6)t 0.4% ← Less than 1%
The factor RC in Eq. (10.13) is called the time constant of the system and has the units of time as follows: RC t Its symbol is the Greek letter t (tau), and its unit of measure is the second. Thus,
(seconds, s) (10.14)
If we substitute t RC into the exponential function e t/RC, we obtain e t/t.Inone time constant, e t/t e t/t e 1 0.3679, or the function equals 36.79% of its maximum value of 1. At t 2t, e t/t e 2t/t e 2 0.1353, and the function has decayed to only 13.53% of its maximum value. The magnitude of e t/t and the percentage change between time constants have been tabulated in Tables 10.4 and 10.5, respectively. Note that the current has dropped 63.2% (100% 36.8%) in the ﬁrst time constant but only 0.4% between the ﬁfth and sixth time constants. The rate of change of iC is therefore quite sensitive to the time constant determined by the network parameters R and C. For this reason, the universal time constant chart of Fig. 10.30 is provided to permit a more accurate estimate of the value of the function e x for speciﬁc time intervals related to the time constant. The term universal is used because the axes are not scaled to speciﬁc values.
t RC
Q V
V Q/t
Q V
V I
392  CAPACITORS
Returning to Eq. (10.13), we ﬁnd that the multiplying factor E/R is the maximum value that the current iC can attain, as shown in Fig. 10.25. Substituting t 0 s into Eq. (10.13) yields iC e t/RC E R e 0 E R
verifying our earlier conclusion. For increasing values of t, the magnitude of e t/t, and therefore the value of iC, will decrease, as shown in Fig. 10.31. Since the magnitude of iC is less than 1% of its maximum after ﬁve time constants, we will assume the following for future analysis:
The current iC of a capacitive network is essentially zero after ﬁve time constants of the charging phase have passed in a dc network.
Since C is usually found in microfarads or picofarads, the time constant t RC will never be greater than a few seconds unless R is very large. Let us now turn our attention to the charging voltage across the capacitor. Through further mathematical analysis, the following equation for the voltage across the capacitor can be determined:
(10.15)
Note the presence of the same factor e t/RC and the function (1 e t/RC) appearing in Fig. 10.30. Since e t/t is a decaying function, the factor (1 e t/t) will grow toward a maximum value of 1 with time, as shown in Fig. 10.30. In addition, since E is the multiplying factor, we can conclude that, for all practical purposes, the voltage vC is E volts
vC E(1 e t/RC)
E R
36.8%
13.5%
5%
1.8% 0 1 2 345 t
iC =
0.67%
E R
e–t/
E R
iC

FIG. 10.31 iC versus t during the charging phase.
0 1 2 3 4 5 6
1.0
y
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
1
y = e–t/
y = 1 – e –t/
0.368 (close to ) 1 3
0.632 (close to ) 2 3
t

FIG. 10.30 Universal time constant chart.
TRANSIENTS IN CAPACITIVE NETWORKS: CHARGING PHASE  393
after ﬁve time constants of the charging phase. A plot of vC versus t is provided in Fig. 10.32. If we keep R constant and reduce C, the product RC will decrease, and the rise time of ﬁve time constants will decrease. The change in transient behavior of the voltage vC is plotted in Fig. 10.33 for various values of C. The product RC will always have some numerical value, even though it may be very small in some cases. For this reason:
The voltage across a capacitor cannot change instantaneously.
In fact, the capacitance of a network is also a measure of how much it will oppose a change in voltage across the network. The larger the capacitance, the larger the time constant, and the longer it takes to charge up to its ﬁnal value (curve of C3 in Fig. 10.33). A lesser capacitance would permit the voltage to build up more quickly since the time constant is less (curve of C1 in Fig. 10.33). The rate at which charge is deposited on the plates during the charging phase can be found by substituting the following for vC in Eq. (10.15):
vC
and charging (10.16)
indicating that the charging rate is very high during the ﬁrst few time constants and less than 1% after ﬁve time constants. The voltage across the resistor is determined by Ohm’s law:
vR iRR RiC R E R e t/t
or (10.17)
A plot of vR appears in Fig. 10.34. Applying Kirchhoff’s voltage law to the circuit of Fig. 10.24 will result in
vC E vR
Substituting Eq. (10.17):
vC E Ee t/t Factoring gives vC E(1 e t/t), as obtained earlier.
EXAMPLE 10.5 a. Find the mathematical expressions for the transient behavior of vC, iC, and vR for the circuit of Fig. 10.35 when the switch is moved to position 1. Plot the curves of vC, iC, and vR. b. How much time must pass before it can be assumed, for all practical purposes, that iC 0 A and vC E volts?
Solutions: a. t RC (8 103 )(4 10 6 F) 32 10 3 s 32 ms By Eq. (10.15), vC E(1 e t/t) 40(1 e t/(32 10 3))
vR Ee t/t
q CvC CE(1 e t/t)
q C
vC = E(1 – e–t/
E
vC
0 t 1 2 3 4 5
63.2%
86.5%
95% 98.2% 99.3%

)
FIG. 10.32 vC versus t during the charging phase.
FIG. 10.33 Effect of C on the charging phase.
E
vC
0 t
C1 C2
C3
C3 > C2 > C1, R fixed
vR = Ee–t/t
E
vR
0 t 1t 2t 3t 4t 5t
36.8%
13.5%
5%
1.8% 0.67%
FIG. 10.34 vR versus t during the charging phase.
R
8 k
C 4 mF vC + –
+ vR – (t = 0)
1
2
E
+
–
40 V
iC
FIG. 10.35 Example 10.5.
394  CAPACITORS
By Eq. (10.13),
iC E R e t/t e t/(32 10 3) (5 10 3)e t/(32 10 3)
By Eq. (10.17),
vR Ee t/t 40e t/(32 10 3)
The curves appear in Fig. 10.36. b. 5t 5(32 ms) 160 ms
Once the voltage across the capacitor has reached the input voltage E, the capacitor is fully charged and will remain in this state if no further changes are made in the circuit. If the switch of Fig. 10.24 is opened, as shown in Fig. 10.37(a), the capacitor will retain its charge for a period of time determined by its leakage current. For capacitors such as the mica and ceramic, the leakage current (ileakage vC/Rleakage) is very small, enabling the capacitor to retain its charge, and hence the potential difference across its plates, for a long time. For electrolytic capacitors, which have very high leakage currents, the capacitor will discharge more rapidly, as shown in Fig. 10.37(b). In any event, to ensure that they are completely discharged, capacitors should be shorted by a lead or a screwdriver before they are handled.
40 V 8 k
R
+
–
vR = 0 V1
2
E
i = 0
vC E
(a)
2
vR = 0 V
i = 0
+
– vC Rleakage
ileakage
(b)
2
– vR +
+
– vC
R
= E
iC = iR = idischarge
FIG. 10.37 Effect of the leakage current on the steady-state behavior of a capacitor.
FIG. 10.38 Demonstrating the discharge behavior of a capacitive network.
40
vC (V)
0 t 1t 2t 3t 4t 5t
t = 32 ms
5.0
iC (mA)
0 t 1t 2t 3t 4t 5t
t = 32 ms
40
vR (V)
0 t 1t 2t 3t 4t 5t
t = 32 ms
FIG. 10.36 Waveforms for the network of Fig. 10.35.
10.8 DISCHARGE PHASE
The network of Fig. 10.24 is designed to both charge and discharge the capacitor. When the switch is placed in position 1, the capacitor will charge toward the supply voltage, as described in the last section. At any point in the charging process, if the switch is moved to position 2, the capacitor will begin to discharge at a rate sensitive to the same time constant t RC. The established voltage across the capacitor will create a ﬂow of charge in the closed path that will eventually discharge the capacitor completely. In essence, the capacitor functions like a battery with a decreasing terminal voltage. Note in particular that the current iC has reversed direction, changing the polarity of the voltage across R. If the capacitor had charged to the full battery voltage as indicated in Fig. 10.38, the equation for the decaying voltage across the capacitor would be the following:
discharging (10.18)v C Ee t/RC
DISCHARGE PHASE  395
which employs the function e x and the same time constant used above. The resulting curve will have the same shape as the curve for iC and vR in the last section. During the discharge phase, the current iC will also decrease with time, as deﬁned by the following equation:
discharging
(10.19)
The voltage vR vC, and
discharging
(10.20)
The complete discharge will occur, for all practical purposes, in ﬁve time constants. If the switch is moved between terminals 1 and 2 every ﬁve time constants, the wave shapes of Fig. 10.39 will result for vC, iC, and vR. For each curve, the current direction and voltage polarities were deﬁned by Fig. 10.24. Since the polarity of vC is the same for both the charging and the discharging phases, the entire curve lies above the axis. The current iC reverses direction during the charging and discharging phases, producing a negative pulse for both the current and the voltage vR. Note that the voltage vC never changes magnitude instantaneously but that the current iC has the ability to change instantaneously, as demonstrated by its vertical rises and drops to maximum values.
vR Ee t/RC
iC E R e t/RC
E
vC
0 t Pos. 1 15t Pos. 2 Pos. 1 Pos. 2
E iC
0 t Pos. 1
Pos. 2
Pos. 1
Pos. 2
R
E
vR
0 t Pos. 1
Pos. 2
Pos. 1
Pos. 2
E R
10t5 t
Charging Discharging
5t 15t
5t 15t
– E
–
10t
10t
FIG. 10.39 The charging and discharging cycles for the network of Fig. 10.24.
8 k
2
– vR +
+
– vC
R
= E = 40 V
iC
C 4 F
FIG. 10.40 Example 10.6.
EXAMPLE 10.6 After vC in Example 10.5 has reached its ﬁnal value of 40 V, the switch is thrown into position 2, as shown in Fig. 10.40. Find the mathematical expressions for the transient behavior of vC, iC,
396  CAPACITORS
and vR after the closing of the switch. Plot the curves for vC, iC, and vR using the deﬁned directions and polarities of Fig. 10.35. Assume that t 0 when the switch is moved to position 2.
Solution:
t 32 ms
By Eq. (10.18),
vC Ee t/t 40e t/(32 10 3)
By Eq. (10.19),
iC E R e t/t (5 10 3)e t/(32 10 3)
By Eq. (10.20),
vR Ee t/t 40e t/(32 10 3)
The curves appear in Fig. 10.41.
The preceding discussion and examples apply to situations in which the capacitor charges to the battery voltage. If the charging phase is disrupted before reaching the supply voltage, the capacitive voltage will be less, and the equation for the discharging voltage vC will take on the form
(10.21)
where Vi is the starting or initial voltage for the discharge phase. The equation for the decaying current is also modiﬁed by simply substituting Vi for E; that is,
(10.22)
Use of the above equations will be demonstrated in Examples 10.7 and 10.8.
EXAMPLE 10.7 a. Find the mathematical expression for the transient behavior of the voltage across the capacitor of Fig. 10.42 if the switch is thrown into position 1 at t 0 s. b. Repeat part (a) for iC. c. Find the mathematical expressions for the response of vC and iC if the switch is thrown into position 2 at 30 ms (assuming that the leakage resistance of the capacitor is inﬁnite ohms). d. Find the mathematical expressions for the voltage vC and current iC if the switch is thrown into position 3 at t 48 ms. e. Plot the waveforms obtained in parts (a) through (d) on the same time axis for the voltage vC and the current iC using the deﬁned polarity and current direction of Fig. 10.42.
iC V R i e t/t Iie t/t
vC Vie t/RC
R2 200 k
R1
100 k
E 10 V
C 0.05 mF
21 3
iC
vC + –
FIG. 10.42 Example 10.7.
40
vC (V)
0 t 1t 2t 3t 4t 5t
t = 32 ms
40
vR (V)
t1 t 4t 5t
0
iC (mA)
5.0
t = 32 ms
t1 t 4t 5t
0
t = 32 ms
2t 3t
2t 3t
FIG. 10.41 The waveforms for the network of Fig. 10.40.
DISCHARGE PHASE  397
Solutions: a. Charging phase: vC E(1 e t/t) t R1C (100 103 )(0.05 10 6 F) 5 10 3 s 5 ms vC 10(1 e t/(5 10 3))
b. iC e t/t
e t/(5 10 3)
iC (0.1 10 3)e t/(5 10 3)
c. Storage phase:
vC E 10 V iC 0 A
d. Discharge phase (starting at 48 ms with t 0 s for the following equations): vC Ee t/t′ t′ R2C (200 103 )(0.05 10 6 F) 10 10 3 s 10 ms vC 10e t/(10 10 3) iC e t/t′
e t/(10 10 3)
iC (0.05 10 3)e t/(10 10 3)
e. See Fig. 10.43.
10 V 200 103
E R2
10 V 100 103
E R1
FIG. 10.43 The waveforms for the network of Fig. 10.42.
10 V
vC (V)
0 t (ms) 10 20 25 30 40 50 60 70 80 90 100 5t 5t' (t' = 2t)
48 98
0.1
iC (mA)
0 t (ms) 10 20 25 30 40
50 60 70 80 90 100
5t
(t' = 2t)
48 98
–0.05
5t'
398  CAPACITORS
EXAMPLE 10.8 a. Find the mathematical expression for the transient behavior of the voltage across the capacitor of Fig. 10.44 if the switch is thrown into position 1 at t 0 s.
R2 1 k 20 VE C 10 mF
2 1 iC
vC + –
R1
5 k
R3
3 k
FIG. 10.45 The charging phase for the network of Fig. 10.44.
10 mF
3 k
+
–
iC
Cv C + –
R3
1 k R 2
2
12.64 V

b. Repeat part (a) for iC. c. Find the mathematical expression for the response of vC and iC if the switch is thrown into position 2 at t 1t of the charging phase. d. Plot the waveforms obtained in parts (a) through (c) on the same time axis for the voltage vC and the current iC using the deﬁned polarity and current direction of Fig. 10.44.
Solutions: a. Charging phase: Converting the current source to a voltage source will result in the network of Fig. 10.45.
FIG. 10.46 Network of Fig. 10.45 when the switch is moved to position 2 at t 1t1.
vC E(1 e t/t1) t1 (R1 R3)C (5 k 3 k )(10 10 6 F) 80 ms vC 20(1 e t/(80 10 3))
b. iC e t/t1
e t/(80 10 3)
iC (2.5 10 3)e t/(80 10 3) c. At t 1t1, vC 0.632E 0.632(20 V) 12.64 V, resulting in the network of Fig. 10.46. Then vC Vie t/t2 with
20 V 8 k
E R1 R3
R2 1 k
4 mA
I C 10 mF
2 1 iC
vC + –
R1 5 k
R3
3 k
FIG. 10.44 Example 10.8.
INITIAL VALUES  399
t2 (R2 R3)C (1 k 3 k )(10 10 6 F) 40 ms and vC 12.64e t/(40 10 3) At t 1t1, iC drops to (0.368)(2.5 mA) 0.92 mA. Then it switches to
iC Iie t/t2
e t/t2 e t/(40 10 3)
iC 3.16 10 3e t/(40 10 3)
d. See Fig. 10.47.
12.64 V 1 k 3 k
Vi R2 R3
vC (V)
0 t (ms) 80 160 240 320 400
1 5
12.64 V
20 V
iC (mA)
0 t (ms) 320 400
5 2
0.92
2.5
2

1
240
–3.16
10.9 INITIAL VALUES
In all the examples examined in the previous sections, the capacitor was uncharged before the switch was thrown. We will now examine the effect of a charge, and therefore a voltage (V Q/C), on the plates at the instant the switching action takes place. The voltage across the capacitor at this instant is called the initial value, as shown for the general waveform of Fig. 10.48. Once the switch is thrown, the transient
FIG. 10.47 The waveforms for the network of Fig. 10.44.
400  CAPACITORS
Initial conditions Vi Transient response
Steady-state region
Vf
0 t
vC
FIG. 10.48 Deﬁning the regions associated with a transient response.
phase will commence until a leveling off occurs after ﬁve time constants. This region of relatively ﬁxed value that follows the transient response is called the steady-state region, and the resulting value is called the steady-state or ﬁnal value. The steady-state value is found by simply substituting the open-circuit equivalent for the capacitor and ﬁnding the voltage across the plates. Using the transient equation developed in the previous section, an equation for the voltage vC can be written for the entire time interval of Fig. 10.48; that is, vC Vi (Vf Vi)(1 e t/t)
However, by multiplying through and rearranging terms:
vC Vi Vf V f e t/t Vi V ie t/t Vf V f e t/t V ie t/t
we ﬁnd
(10.23)
If you are required to draw the waveform for the voltage vC from the initial value to the ﬁnal value, start by drawing a line at the initial and steady-state levels, and then add the transient response (sensitive to the time constant) between the two levels. The example to follow will clarify the procedure.
EXAMPLE 10.9 The capacitor of Fig. 10.49 has an initial voltage of 4 V.
vC Vf (Vi Vf)e t/t
iC
R2
1.2 k
R1
2.2 k
E C 24 V 3.3 F
+
– 4 Vv C + –
FIG. 10.49 Example 10.9.
INITIAL VALUES  401
a. Find the mathematical expression for the voltage across the capacitor once the switch is closed. b. Find the mathematical expression for the current during the transient period. c. Sketch the waveform for each from initial value to ﬁnal value.
Solutions: a. Substituting the open-circuit equivalent for the capacitor will result in a ﬁnal or steady-state voltage vC of 24 V. The time constant is determined by
t (R1 R2)C (2.2 k 1.2 k )(3.3 mF) 11.22 ms with 5t 56.1 ms
Applying Eq. (10.23): vC Vf (Vi Vf)e t/t 24 V (4 V 24 V)e t/11.22ms and vC 24 V 20 Ve t/11.22ms
b. Since the voltage across the capacitor is constant at 4 V prior to the closing of the switch, the current (whose level is sensitive only to changes in voltage across the capacitor) must have an initial value of 0 mA. At the instant the switch is closed, the voltage across the capacitor cannot change instantaneously, so the voltage across the resistive elements at this instant is the applied voltage less the initial voltage across the capacitor. The resulting peak current is
Im 5.88 mA
The current will then decay (with the same time constant as the voltage vC) to zero because the capacitor is approaching its opencircuit equivalence. The equation for iC is therefore: iC 5.88 mAe t/11.22ms
c. See Fig. 10.50.
20 V 3.4 k
24 V 4 V 2.2 k 1.2 k
E VC R1 R2
4 V
24 V
vC
0
iC 5.88 mA
56.1 ms
56.1 ms
t
t0
5
FIG. 10.50 vC and iC for the network of Fig. 10.49.
402  CAPACITORS
The initial and ﬁnal values of the voltage were drawn ﬁrst, and then the transient response was included between these levels. For the current, the waveform begins and ends at zero, with the peak value having a sign sensitive to the deﬁned direction of iC in Fig. 10.49. Let us now test the validity of the equation for vC by substituting t 0 s to reﬂect the instant the switch is closed. e t/t e 0 1 and vC 24 V 20 Ve t/t 24 V 20 V 4 V
When t 5t,
e t/t 0 and vC 24 V 20 Ve t/t 24 V 0 V 24 V
10.10 INSTANTANEOUS VALUES
On occasion it will be necessary to determine the voltage or current at a particular instant of time that is not an integral multiple of t, as in the previous sections. For example, if vC 20(1 e t/(2 10 3)) the voltage vC may be required at t 5 ms, which does not correspond to a particular value of t. Figure 10.30 reveals that (1 e t/t) is approximately 0.93 at t 5 ms 2.5t, resulting in vC 20(0.93) 18.6 V. Additional accuracy can be obtained simply by substituting t 5 ms into the equation and solving for vC using a calculator or table to determine e 2.5. Thus, vC 20(1 e 5ms/2ms) 20(1 e 2.5) 20(1 0.082) 20(0.918) 18.36 V
The results are close, but accuracy beyond the tenths place is suspect using Fig. 10.30. The above procedure can also be applied to any other equation introduced in this chapter for currents or other voltages. There are also occasions when the time to reach a particular voltage or current is required. The procedure is complicated somewhat by the use of natural logs (loge, or ln), but today’s calculators are equipped to handle the operation with ease. There are two forms that require some development. First, consider the following sequence: vC E(1 e t/t)
v E C 1 e t/t
1 v E C e t/t loge 1 v E C logee t/t loge 1 v E C t t
INSTANTANEOUS VALUES  403
and t t loge 1 v E C but loge x y loge y x Therefore, t t loge E E vC (10.24)
The second form is as follows:
vC Ee t/t
v E C e t/t
loge v E C logee t/t
loge v E C and t t loge v E C
or t t loge v E C (10.25)
For iC (E/R)e t/t:
t t loge iC E R (10.26)
For example, suppose that vC 20(1 e t/(2 10 3))
and the time to reach 10 V is required. Substituting into Eq. (10.24), we have t (2 ms)loge (2 ms)loge2 (2 ms)(0.693) 1.386 ms Using Fig. 10.30, we ﬁnd at (1 e t/t) vC /E 0.5 that t 0.7t 0.7(2 ms) 1.4 ms, which is relatively close to the above.
Mathcad
It is time to see how Mathcad can be applied to the transient analysis described in this chapter. For the ﬁrst equation described in Section 10.10, vC 20(1 e t/(2 10 3))
the value of t must be deﬁned before the expression is written, or the value can simply be inserted in the equation. The former approach is
20 V 20 V 10 V
t t
key on calculatorINF
404  CAPACITORS
often better because changing the deﬁned value of t will result in an immediate change in the result. In other words, the value can be used for further calculations. In Fig. 10.51 the value of t was deﬁned as 5 ms. The equation was then entered using the e function from the Calculator palette obtained from View-Toolbars-Calculator. Be sure to insert a multiplication operator between the initial 20 and the main left bracket. Also, be careful that the control bracket is in the correct place before placing the right bracket to enclose the equation. It will take some practice to ensure that the insertion bracket is in the correct place before entering a parameter, but in time you will ﬁnd that it is a fairly direct procedure. The 3 is placed using the shift operator over the number 6 on the standard keyboard. The result is displayed by simply entering v again, followed by an equal sign. The result for t 1 ms can now be obtained by simply changing the deﬁned value for t. The result of 7.869 V will appear immediately.
FIG. 10.51 Applying Mathcad to the transient R-C equations.
For the second equation of Section 10.10, vC 20(1 e 5ms/2ms)
the equation for t can be entered directly as shown in the bottom of Fig. 10.51. The ln from the Calculator is for a base e calculation, while log is for a base 10 calculation. The result will appear the instant the equal sign is placed after the t on the bottom line. The text you see on the screen to deﬁne each operation is obtained by clicking on Insert-Text Region and then simply typing in the text material. The boldface was obtained by simply clicking on the text material and swiping the text to establish a black background. Then select B from the toolbar, and the boldface will appear.
THÉVENIN EQUIVALENT: t RThC  405
10.11 THÉVENIN EQUIVALENT: t RThC
Occasions will arise in which the network does not have the simple series form of Fig. 10.24. It will then be necessary ﬁrst to ﬁnd the Thévenin equivalent circuit for the network external to the capacitive element. ETh will then be the source voltage E of Eqs. (10.15) through (10.20), and RTh will be the resistance R. The time constant is then t RThC.
EXAMPLE 10.10 For the network of Fig. 10.52:
R230 k R1
60 k
E 21 V + –
R3
10 k
vC + – iC 12 C = 0.2 mF R4 10 k
FIG. 10.52 Example 10.10.
FIG. 10.53 Applying Thévenin’s theorem to the network of Fig. 10.52.
vC
RTh = 30 k
ETh = 7 V C = 0.2 mF
iC
+
–
FIG. 10.54 Substituting the Thévenin equivalent for the network of Fig. 10.52.
a. Find the mathematical expression for the transient behavior of the voltage vC and the current iC following the closing of the switch (position 1 at t 0 s). b. Find the mathematical expression for the voltage vC and current iC as a function of time if the switch is thrown into position 2 at t 9 ms. c. Draw the resultant waveforms of parts (a) and (b) on the same time axis.
Solutions: a. Applying Thévenin’s theorem to the 0.2-mF capacitor, we obtain Fig. 10.53: RTh R1 R2 R3 10 k
20 k 10 k RTh 30 k
ETh (21 V) 7 V
The resultant Thévenin equivalent circuit with the capacitor replaced is shown in Fig. 10.54. Using Eq. (10.23) with Vf ETh and Vi 0 V, we ﬁnd that vC Vf (Vi Vf)e t/t becomes vC ETh (0 V ETh)e t/t or vC ETh(1 e t/t) with t RC (30 k )(0.2 mF) 6 ms so that vC 7(1 e t/6ms)
For the current: iC E R Th e t/RC
e t/6ms
iC (0.233 10 3)e t/6ms
7 V 30 k
1 3
(30 k )(21 V) 30 k 60 k
R2E R2 R1
(60 k )(30 k ) 90 k
ETh:
R1
60 k
R3
10 k
R2 30 k EThE – + 21 V
RTh
RTh:
R1
60 k
R3
10 k
R2 30 k
EXAMPLE 10.11 The capacitor of Fig. 10.56 is initially charged to 40 V. Find the mathematical expression for vC after the closing of the switch.
406  CAPACITORS
b. At t 9 ms, vC ETh(1 e t/t) 7(1 e (9 10 3)/(6 10 3)) 7(1 e 1.5) 7(1 0.223) vC 7(0.777) 5.44 V and iC E R Th e t/t (0.233 10 3)e 1.5 (0.233 10 3)(0.223) iC 0.052 10 3 0.052 mA Using Eq. (10.23) with Vf 0 V and Vi 5.44 V, we ﬁnd that vC Vf (Vi Vf)e t/t′ becomes vC 0 V (5.44 V 0 V)e t/t′ 5.44e t/t′ with t′ R4C (10 k )(0.2 mF) 2 ms and vC 5.44e t/2ms
By Eq. (10.22),
Ii 0.054 mA
and iC Iie t/t (0.54 10 3)e t/2ms
c. See Fig. 10.55.
5.44 V 10 k
vC (V)
0 t (ms) 15 25 30 35 5
ETh = 7
20105
Vi = 5.44 V
iC (mA)
0 t (ms) 25 30 3520
10
5
0.052
– 0.54
5 '
15
0.233

5
5 '
FIG. 10.55 The resulting waveforms for the network of Fig. 10.52.
R1 7 k
R4 2 k
C
40 F + 40 V –
+ vC –
R2
5 k R3
18 k
E 120 V

FIG. 10.56 Example 10.11.
THÉVENIN EQUIVALENT: t RThC  407
Solution: The network is redrawn in Fig. 10.57.
40 mF40 V + – Thévenin C
R2
5 k
R1
7 k
R3 18 k
R4 2 k
E 120 V
FIG. 10.57 Network of Fig. 10.56 redrawn.
ETh:
ETh
80 V
RTh:
RTh 5 k 18 k (7 k 2 k ) 5 k 6 k 11 k
Therefore, Vi 40 V and Vf 80 V and t RThC (11 k )(40 mF) 0.44 s Eq. (10.23): vC Vf (Vi Vf)e t/t 80 V (40 V 80 V)e t/0.44s and vC 80 V 40 Ve t/0.44s
EXAMPLE 10.12 Forthe network of Fig. 10.58, ﬁnd the mathematical expression for the voltage vC after the closing of the switch (at t 0).
18 k (120 V) 18 k 7 k 2 k
R3E R3 R1 R4
I R1 = 6
R2
10
20 mA
C 500 mF vC + –
FIG. 10.58 Example 10.12.
Solution:
RTh R1 R2 6 10 16 ETh V1 V2 IR1 0 (20 10 3 A)(6 ) 120 10 3 V 0.12 V and t RThC (16 )(500 10 6 F) 8 ms so that vC 0.12(1 e t/8ms)
408  CAPACITORS
10.12 THE CURRENT iC
The current iC associated with a capacitance C is related to the voltage across the capacitor by
(10.27)
where dvC /dt is a measure of the change in vC in a vanishingly small period of time. The function dvC /dt is called the derivative of the voltage vC with respect to time t. If the voltage fails to change at a particular instant, then dvC 0
and iC C d d v t C 0
In other words, if the voltage across a capacitor fails to change with time, the current iC associated with the capacitor is zero. To take this a step further, the equation also states that the more rapid the change in voltage across the capacitor, the greater the resulting current. In an effort to develop a clearer understanding of Eq. (10.27), let us calculate the average current associated with a capacitor for various voltages impressed across the capacitor. The average current is deﬁned by the equation
(10.28)
where indicates a ﬁnite (measurable) change in charge, voltage, or time. The instantaneous current can be derived from Eq. (10.28) by letting t become vanishingly small; that is,
iCinst lim ∆t→0
C v t C C d d v t C In the following example, the change in voltage vC will be considered for each slope of the voltage waveform. If the voltage increases with time, the average current is the change in voltage divided by the change in time, with a positive sign. If the voltage decreases with time, the average current is again the change in voltage divided by the change in time, but with a negative sign.
EXAMPLE 10.13 Find the waveform for the average current if the voltage across a 2-mF capacitor is as shown in Fig. 10.59.
Solutions: a. From 0 ms to 2 ms, the voltage increases linearly from 0 V to 4 V, the change in voltage v 4 V 0 4 V (with a positive sign since the voltage increases with time). The change in time t 2 ms 0 2 ms, and
iCav C v t C (2 10 6 F) 4 10 3 A 4 mA 4 V 2 10 3 s
iCav C v t C
iC C d d v t C
THE CURRENT iC  409
vC (V)
0
t (ms) 9 10 11 12
t1
4
5678
1
234t 3t 2 t
v3 v2 v
FIG. 10.59 Example 10.13.
b. From 2 ms to 5 ms, the voltage remains constant at 4 V; the change in voltage v 0. The change in time t 3 ms, and
iCav C v t C C 0 t 0
c. From 5 ms to 11 ms, the voltage decreases from 4V to 0V.The change in voltage v is, therefore, 4 V 0 4V(with a negative sign since the voltage is decreasing with time). The change in time t 11 ms 5 ms 6 ms, and
iCav C v t C (2 10 6 F) 1.33 10 3 A 1.33 mA
d. From 11 ms on, the voltage remains constant at 0 and v 0, so iCav 0. The waveform for the average current for the impressed voltage is as shown in Fig. 10.60.
4 V 6 10 3 s
Note in Example 10.13 that, in general, the steeper the slope, the greater the current, and when the voltage fails to change, the current is zero. In addition, the average value is the same as the instantaneous value at any point along the slope over which the average value was found. For example, if the interval t is reduced from 0 → t1 to t2 t3, as noted in Fig. 10.59, v/ t is still the same. In fact, no matter how small the interval t, the slope will be the same, and therefore the current iCav will be the same. If we consider the limit as t → 0, the slope will still remain the same, and therefore iCav iCinst at any instant of
iC (mA)
0
t (ms)9 10 11 12
4
5 6 7 81 234
–1.33
FIG. 10.60 The resulting current iC for the applied voltage of Fig. 10.59.
410  CAPACITORS
time between 0 and t1. The same can be said about any portion of the voltage waveform that has a constant slope. An important point to be gained from this discussion is that it is not the magnitude of the voltage across a capacitor that determines the current but rather how quickly the voltage changes across the capacitor. An applied steady dc voltage of 10,000 V would (ideally) not create any ﬂow of charge (current), but a change in voltage of 1 V in a very brief period of time could create a signiﬁcant current. The method described above is only for waveforms with straight-line (linear) segments. For nonlinear (curved) waveforms, a method of calculus (differentiation) must be employed.
10.13 CAPACITORS IN SERIES AND PARALLEL
Capacitors, like resistors, can be placed in series and in parallel. Increasing levels of capacitance can be obtained by placing capacitors in parallel, while decreasing levels can be obtained by placing capacitors in series. For capacitors in series, the charge is the same on each capacitor (Fig. 10.61):
(10.29)
Applying Kirchhoff’s voltage law around the closed loop gives
E V1 V2 V3
However, V Q C
so that
Using Eq. (10.29) and dividing both sides by Q yields
(10.30)
which is similar to the manner in which we found the total resistance of a parallel resistive circuit. The total capacitance of two capacitors in series is
(10.31)
The voltage across each capacitor of Fig. 10.61 can be found by ﬁrst recognizing that
QT Q1 or CT E C1V1
Solving for V1:
V1
CTE C1
CT C1 C 1C2 C2
C 1 T C 1 1 C 1 2 C 1 3
Q3 C3
Q2 C2
Q1 C1
QT CT
QT Q1 Q2 Q3
V2
E
+
–
QT
V3V 1 + – + – + – Q2 Q3Q 1
FIG. 10.61 Series capacitors.
CAPACITORS IN SERIES AND PARALLEL  411
and substituting for CT:
V1 E (10.32)
A similar equation will result for each capacitor of the network. For capacitors in parallel, as shown in Fig. 10.62, the voltage is the same across each capacitor, and the total charge is the sum of that on each capacitor:
(10.33)
However, Q CV
Therefore, CT E C1V1 C2V2 C3V3 but E V1 V2 V3
Thus, (10.34)
which is similar to the manner in which the total resistance of a series circuit is found.
EXAMPLE 10.14 For the circuit of Fig. 10.63: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the voltage across each capacitor.
Solutions:
a.

0.005 106 0.02 106 0.1 106 0.125 106
and CT 8 mF
b. QT Q1 Q2 Q3 QT CTE (8 10 6 F)(60 V) 480 mC
c. V1 2.4 V
V2 9.6 V
V3 48.0 V
and E V1 V2 V3 2.4 V 9.6 V 48 V 60 V (checks)
480 10 6 C 10 10 6 F
Q3 C3
480 10 6 C 50 10 6 F
Q2 C2
480 10 6 C 200 10 6 F
Q1 C1
1 0.125 106
1 10 10 6 F
1 50 10 6 F
1 200 10 6 F
1 C3
1 C2
1 C1
1 CT
CT C1 C2 C3
QT Q1 Q2 Q3
1/C1 1/C1 1/C2 1/C3
FIG. 10.62 Parallel capacitors.
E
+
–
QT
V1 Q1+ –
V2 Q2+ –
V3 Q3+ –
E
+
–
CT
C2 C3C 1
200 mF 50 mF 10 mF
60 V
FIG. 10.63 Example 10.14.
412  CAPACITORS
EXAMPLE 10.15 For the network of Fig. 10.64: a. Find the total capacitance. b. Determine the charge on each plate. c. Find the total charge.
Solutions: a. CT C1 C2 C3 800 mF 60 mF 1200 mF 2060 mF b. Q1 C1E (800 10 6 F)(48 V) 38.4 mC Q2 C2E (60 10 6 F)(48 V) 2.88 mC Q3 C3E (1200 10 6 F)(48 V) 57.6 mC c. QT Q1 Q2 Q3 38.4 mC 2.88 mC 57.6 mC 98.88 mC
EXAMPLE 10.16 Find the voltage across and charge on each capacitor for the network of Fig. 10.65.
Solution:
C′T C2 C3 4 mF 2 mF 6 mF
CT 2 mF
QT CT E (2 10 6 F)(120 V) 240 mC
An equivalent circuit (Fig. 10.66) has
QT Q1 Q′T and, therefore, Q1 240 mC
and V1 80 V
Q′T 240 mC
and, therefore,
V′T 40 V
and Q2 C2V′T (4 10 6 F)(40 V) 160 mC Q3 C3V′T (2 10 6 F)(40 V) 80 mC
EXAMPLE 10.17 Find the voltage across and charge on capacitor C1 of Fig. 10.67 after it has charged up to its ﬁnal value.
240 10 6 C 6 10 6 F
Q′T C′T
240 10 6 C 3 10 6 F
Q1 C1
(3 mF)(6 mF) 3 mF 6 mF
C1C′T C1 C′T
E
CT
C1 C2 C3
800 mF 60 mF 1200 mF
QT
48 V
FIG. 10.64 Example 10.15.
E = 120 V
C1
C2 C3
3 mF
4 mF 2 mF
+
–
FIG. 10.65 Example 10.16.
E = 120 V
C1
C'T
3 mF
6 mF
+
–
Q1 +– V1
V'T + –Q' T
Q3
V'T + –C 3
Q2
C2
FIG. 10.66 Reduced equivalent for the network of Fig. 10.65.
R1
4 E = 24 V C1 = 20 mF + – Q1 R2 8 VC + –
FIG. 10.67 Example 10.17.
ENERGY STORED BY A CAPACITOR  413
Solution: As previously discussed, the capacitor is effectively an open circuit for dc after charging up to its ﬁnal value (Fig. 10.68). Therefore,
VC 16 V Q1 C1VC (20 10 6 F)(16 V) 320 mC
EXAMPLE 10.18 Find the voltage across and charge on each capacitor of the network of Fig. 10.69 after each has charged up to its ﬁnal value.
(8 )(24 V) 4 8
Solution:
VC2 56 V
VC1 16 V Q1 C1VC1 (2 10 6 F)(16 V) 32 mC Q2 C2VC2 (3 10 6 F)(56 V) 168 mC
10.14 ENERGY STORED BY A CAPACITOR
The ideal capacitor does not dissipate any of the energy supplied to it. It stores the energy in the form of an electric ﬁeld between the conducting surfaces. A plot of the voltage, current, and power to a capacitor during the charging phase is shown in Fig. 10.70. The power curve can be obtained by ﬁnding the product of the voltage and current at selected instants of time and connecting the points obtained. The energy stored is represented by the shaded area under the power curve. Using calculus, we can determine the area under the curve: WC CE2
In general, (J) (10.35) WC 1 2 CV2
1 2
(2 )(72 V) 2 7
(7 )(72 V) 7 2
FIG. 10.68 Determining the ﬁnal (steady-state) value for vC.
4 E = 24 V 8 VC + –
FIG. 10.69 Example 10.18.
E = 72 V R2 7
R1
2
R3 8
C2 = 3 mF
C1 = 2 mF
+
–
R2 7
R1
2
R3 8
+
–
+ VC1 –
E = 72 V
+ VC2 –
I = 0
0 t
E
R E
v, i, p
vC p = vC iC iC
FIG. 10.70 Plotting the power to a capacitive element during the transient phase.
414  CAPACITORS
where V is the steady-state voltage across the capacitor. In terms of Q and C, WC C 2
or (J) (10.36)
EXAMPLE 10.19 For the network of Fig. 10.69, determine the energy stored by each capacitor.
Solution:
For C1,
WC 1 2 CV2
1 2 (2 10 6 F)(16 V)2 (1 10 6)(256) 256 mJ
For C2,
WC 1 2 CV2
1 2 (3 10 6 F)(56 V)2 (1.5 10 6)(3136) 4704 mJ Due to the squared term, note the difference in energy stored because of a higher voltage.
10.15 STRAY CAPACITANCES
In addition to the capacitors discussed so far in this chapter, there are stray capacitances that exist not through design but simply because two conducting surfaces are relatively close to each other. Two conducting wires in the same network will have a capacitive effect between them, as shown in Fig. 10.71(a). In electronic circuits, capacitance levels exist between conducting surfaces of the transistor, as shown in Fig. 10.71(b). As mentioned earlier, in Chapter 12 we will discuss another element called the inductor, which will have capacitive effects between the windings [Fig. 10.71(c)]. Stray capacitances can often lead to serious errors in system design if they are not considered carefully.
10.16 APPLICATIONS
This Applications section for capacitors includes both a description of the operation of one of the less expensive, throwaway cameras that have become so popular today and a discussion of the use of capacitors in the line conditioners (surge protectors) that have found their way into most homes and throughout the business world. Additional examples of the use of capacitors will appear throughout the chapter to follow.
WC 2 Q C 2
Q C
1 2
(a)
P P EC
B Cce
N
(b)
Cbe Cbc
(c)
Conductors
FIG. 10.71 Examples of stray capacitance.
APPLICATIONS  415
Flash Lamp
The basic circuitry for the ﬂash lamp of the popular, inexpensive, throwaway camera of Fig. 10.72(a) is provided in Fig. 10.72(b), with the physical circuitry appearing in Fig. 10.72(c). The labels added to Fig. 10.72(c) identify broad areas of the design and some individual components. The major components of the electronic circuitry include a large 160-mF, 330-V, polarized electrolytic capacitor to store the necessary charge for the ﬂash lamp, a ﬂash lamp to generate the required light, a dc battery of 1.5 V, a chopper network to generate a dc voltage in excess of 300 V, and a trigger network to establish a few thousand volts for a very short period of time to ﬁre the ﬂash lamp. There are both a 22-nF capacitor in the trigger network as shown in Fig. 10.72(b) and (c) and a third capacitor of 470 pF in the high-frequency oscillator of the chopper network. In particular, note that the size of each capacitor is directly related to its capacitance level. It should certainly be of some interest that a single source of energy of only 1.5 V dc can be converted to one of a few thousand volts (albeit for a very short period of time) to ﬁre the ﬂash lamp. In fact, that single, small battery has sufﬁcient power for the entire run of ﬁlm through the camera. Always keep in mind that energy is related to power and time by W Pt (VI)t.

FIG. 10.72(a) Flash camera: general appearance.
FIG. 10.72(b) Flash camera: basic circuitry.
(b)
V
t0
V
t0
V
t0
V
t0
V
t0
t0
1.5 V
R1
On/Off
Flash switch (top of camera)
+
1.5 V
Highfrequency oscillator
A
G
22 nF
SCR
K
Flash unit
Trigger network
300 V dc
Sense
Rs
Rr 300 V + –
Highfrequency transformer Auto transformer
Vt
4000-V spikes
160 µF
Flash lamp
1.5 V
300 V
Lamp flashes
Flash button (face of camera)
R2
60-V neon light
Rn
416  CAPACITORS

!"
#
$ %!

&' ( "

))

#

*

!
"

+

"!

FIG. 10.72(c) Flash camera: internal construction.
That is, a high level of voltage can be generated for a deﬁned energy level so long as the factors I and t are sufﬁciently small. When you ﬁrst use the camera, you are directed to press the ﬂash button on the face of the camera and wait for the ﬂash-ready light to come on. As soon as the ﬂash button is depressed, the full 1.5 V of the dc battery are applied to an electronic network (a variety of networks can perform the same function) that will generate an oscillating waveform of very high frequency (with a high repetitive rate) as shown in Fig. 10.72(b). The high-frequency transformer will then signiﬁcantly increase the magnitude of the generated voltage and will pass it on to a half-wave rectiﬁcation system (introduced in earlier chapters), resulting in a dc voltage of about 300 V across the 160-mF capacitor to charge the capacitor (as determined by Q CV). Once the 300-V level is reached, the lead marked “sense” in Fig. 10.72(b) will feed the information back to the oscillator and will turn it off until the output dc voltage drops to a low threshold level. When the capacitor is fully charged, the neon
light in parallel with the capacitor will turn on (labeled “ﬂash-ready lamp” on the camera) to let you know that the camera is ready to use. The entire network from the 1.5-V dc level to the ﬁnal 300-V level is called a dc-dc converter. The terminology chopper network comes from the fact that the applied dc voltage of 1.5 V was chopped up into one that changes level at a very high frequency so that the transformer can perform its function. Even though the camera may use a 60-V neon light, the neon light and series resistor Rn must have a full 300 V across the branch before the neon light will turn on. Neon lights are simply bulbs with a neon gas that will support conduction when the voltage across the terminals reaches a sufﬁciently high level. There is no ﬁlament, or hot wire as in a light bulb, but simply conduction through the gaseous medium. For new cameras the ﬁrst charging sequence may take 12 s to 15 s. Succeeding charging cycles may only take some 7 s or 8 s because the capacitor will still have some residual charge on its plates. If the ﬂash unit is not used, the neon light will begin to drain the 300-V dc supply with a drain current in microamperes. As the terminal voltage drops, there will come a point where the neon light will turn off. For the unit of Fig. 10.72, it takes about 15 min before the light turns off. Once off, the neon light will no longer drain the capacitor, and the terminal voltage of the capacitor will remain fairly constant. Eventually, however, the capacitor will discharge due to its own leakage current, and the terminal voltage will drop to very low levels. The discharge process is very rapid when the ﬂash unit is used, causing the terminal voltage to drop very quickly (V Q/C) and, through the feedback-sense connection signal, causing the oscillator to start up again and recharge the capacitor. You may have noticed when using a camera of this type that once the camera has its initial charge, there is no need to press the charge button between pictures—it is done automatically. However, if the camera sits for a long period of time, the charge button will have to be depressed again; but you will ﬁnd that the charge time is only 3 s or 4 s due to the residual charge on the plates of the capacitor. The 300 V across the capacitor are insufﬁcient to ﬁre the ﬂash lamp. Additional circuitry, called the trigger network, must be incorporated to generate the few thousand volts necessary to ﬁre the ﬂash lamp. The resulting high voltage is one reason that there is a CAUTION note on each camera regarding the high internal voltages generated and the possibility of electrical shock if the camera is opened. The thousands of volts required to ﬁre the ﬂash lamp require a discussion that introduces elements and concepts beyond the current level of the text. However, this description is sensitive to this fact and should be looked upon as simply a ﬁrst exposure to some of the interesting possibilities available from the right mix of elements. When the ﬂash switch at the bottom left of Fig. 10.72(a) is closed, it will establish a connection between the resistors R1 and R2. Through a voltage divider action, a dc voltage will appear at the gate (G) terminal of the SCR (silicon-controlled rectiﬁer—a device whose state is controlled by the voltage at the gate terminal). This dc voltage will turn the SCR “on” and will establish a very low resistance path (like a short circuit) between its anode (A) and cathode (K) terminals. At this point the trigger capacitor, which is connected directly to the 300 V sitting across the capacitor, will rapidly charge to 300 V because it now has a direct, lowresistance path to ground through the SCR. Once it reaches 300 V, the charging current in this part of the network will drop to 0 A, and the
APPLICATIONS  417
418  CAPACITORS
SCR will open up again since it is a device that needs a steady current in the anode circuit to stay on. The capacitor then sits across the parallel coil (with no connection to ground through the SCR) with its full 300 V and begins to quickly discharge through the coil because the only resistance in the circuit affecting the time constant is the resistance of the parallel coil. As a result, a rapidly changing current through the coil will generate a high voltage across the coil for reasons to be introduced in Chapter 12. When the capacitor decays to zero volts, the current through the coil will be zero amperes, but a strong magnetic ﬁeld has been established around the coil. This strong magnetic ﬁeld will then quickly collapse, establishing a current in the parallel network that will recharge the capacitor again. This continual exchange between the two storage elements will continue for a period of time, depending on the resistance in the ciruit. The more the resistance, the shorter the “ringing” of the voltage at the output. This action of the energy “ﬂying back” to the other element is the basis for the “ﬂyback” effect that is frequently used to generate high dc voltages such as needed in TVs. In Fig. 10.72(b), you will ﬁnd that the trigger coil is connected directly to a second coil to form an autotransformer (a transformer with one end connected). Through transformer action the high voltage generated across the trigger coil will be increased further, resulting in the 4000 V necessary to ﬁre the ﬂash lamp. Note in Fig. 10.72(c) that the 4000 V are applied to a grid that actually lies on the surface of the glass tube of the ﬂash lamp (not internally connected or in contact with the gases). When the trigger voltage is applied, it will excite the gases in the lamp, causing a very high current to develop in the bulb for a very short period of time and producing the desired bright light. The current in the lamp is supported by the charge on the 160-mF capacitor which will be dissipated very quickly. The capacitor voltage will drop very quickly, the photo lamp will shut down, and the charging process will begin again. If the entire process didn’t occur as quickly as it does, the lamp would burn out after a single use.
Line Conditioner (Surge Protector)
In recent years we have all become familiar with the line conditioner as a safety measure for our computers, TVs, CD players, and other sensitive instrumentation. In addition to protecting equipment from unexpected surges in voltage and current, most quality units will also ﬁlter out (remove) electromagnetic interference (EMI) and radio-frequency interference (RFI). EMI encompasses any unwanted disturbances down the power line established by any combination of electromagnetic effects such as those generated by motors on the line, power equipment in the area emitting signals picked up by the power line acting as an antenna, and so on. RFI includes all signals in the air in the audio range and beyond which may also be picked up by power lines inside or outside the house. The unit of Fig. 10.73 has all the design features expected in a good line conditioner. Figure 10.73(a) reveals that it can handle the power drawn by six outlets and that it is set up for FAX/MODEM protection. Also note that it has both LED (light-emitting diode) displays which reveal whether there is fault on the line or whether the line is OK and an external circuit breaker to reset the system. In addition, when the surge protector is on, a red light will be visible at the power switch.

FIG. 10.73(a) Line conditioner: general appearance.
APPLICATIONS  419
(b)
Black (feed)
Ground
White (return)
Ground (bare or green)
Black
White
Black (feed)
Circuit breaker
Line conditioner casing On/Off switch
On/Off lamp
Switch assembly
White (return)
228 µH
2 nF 2 nF
2 nF 2 nF
228 µH
2 nF
180 V MOV
Black (feed)
Green LED
Red LED
Green LED
Protection present
Line fault
Line on Voltage monitor network
White (return)
Bare or green (ground)
Outlets
etc.

!
" #
$

$

FIG. 10.73 (cont.) Line conditioner: (b) electrical schematic; (c) internal construction.
The schematic of Fig. 10.73(b) does not include all the details of the design, but it does include the major components that appear in most good line conditioners. First note in the photograph of Fig. 10.73(c) that the outlets are all connected in parallel, with a ground bar used to establish a ground connection for each outlet. The circuit board had to be ﬂipped over to show the components, so it will take some adjustment to relate the position of the elements on the board to the casing. The feed line or hot lead wire (black in the actual unit) is connected directly from the line to the circuit breaker. The other end of the circuit breaker is connected to the other side of the circuit board. All the large discs that you see are 2-nF/73 capacitors [not all have been included in Fig 10.73(b) for clarity]. There are quite a few capacitors to handle all the possibilities. For instance, there are capacitors from line to return (black wire to white wire), from line to ground (black to green), and from return to ground (white to ground). Each has two functions. The ﬁrst and most obvious function is to prevent any spikes in voltage that may come down the line because of external effects such as lightning from reaching the equipment plugged into the unit. Recall from this chapter that the voltage across capacitors cannot change instantaneously and in fact will act to squelch any rapid change in voltage across its terminals. The capacitor, therefore, will prevent the line to neutral voltage from changing too quickly, and any spike that tries to come down the line will have to ﬁnd another point in the feed circuit to fall across. In this way the appliances to the surge protector are well protected. The second function requires some knowledge of the reaction of capacitors to different frequencies and will be discussed in more detail in later chapters. For the moment, let it sufﬁce to say that the capacitor will have a different impedance to different frequencies, thereby preventing undesired frequencies, such as those associated with EMI and RFI disturbances, from affecting the operation of units connected to the line conditioner. The rectangular-shaped capacitor of 1 mF near the center of the board is connected directly across the line to take the brunt of a strong voltage spike down the line. Its larger size is clear evidence that it is designed to absorb a fairly high energy level that may be established by a large voltage—signiﬁcant current over a period of time that might exceed a few milliseconds. The large, toroidal-shaped structure in the center of the circuit board of Fig. 10.73(c) has two coils (Chapter 12) of 228 mH that appear in the line and neutral of Fig. 10.73(b). Their purpose, like that of the capacitors, is twofold: to block spikes in current from coming down the line and to block unwanted EMI and RFI frequencies from getting to the connected systems. In the next chapter you will ﬁnd that coils act as “chokes” to quick changes in current; that is, the current through a coil cannot change instantaneously. For increasing frequencies, such as those associated with EMI and RFI disturbances, the reactance of a coil increases and will absorb the undesired signal rather than let it pass down the line. Using a choke in both the line and the neutral makes the conditioner network balanced to ground. In total, capacitors in a line conditioner have the effect of bypassing the disturbances, whereas inductors block the disturbance. The smaller disc (blue) between two capacitors and near the circuit breaker is an MOV (metal-oxide varistor) which is the heart of most line conditioners. It is an electronic device whose terminal characteristics will change with the voltage applied across its terminals. For the normal range of voltages down the line, its terminal resistance will be
420  CAPACITORS
COMPUTER ANALYSIS  421
sufﬁciently large to be considered an open circuit, and its presence can be ignored. However, if the voltage is too large, its terminal characteristics will change from a very large resistance to a very small resistance that can essentially be considered a short circuit. This variation in resistance with applied voltage is the reason for the name varistor. For MOVs in North America where the line voltage is 120 V, the MOVs are 180 V or more. The reason for the 60-V difference is that the 120-V rating is an effective value related to dc voltage levels, whereas the waveform for the voltage at any 120-V outlet has a peak value of about 170 V. A great deal more will be said about this topic in Chapter 13. Taking a look at the symbol for an MOV in Fig. 10.73(b), you will note that it has an arrow in each direction, revealing that the MOV is bidirectional and will block voltages with either polarity. In general, therefore, for normal operating conditions, the presence of the MOV can be ignored; but, if a large spike should appear down the line, exceeding the MOV rating, it will act as a short across the line to protect the connected circuitry. It is a signiﬁcant improvement to simply putting a fuse in the line because it is voltage sensitive, can react much quicker than a fuse, and will display its low-resistance characteristics for only a short period of time. When the spike has passed, it will return to its normal open-circuit characteristic. If you’re wondering where the spike will go if the load is protected by a short circuit, remember that all sources of disturbance, such as lightning, generators, inductive motors (such as in air conditioners, dishwashers, power saws, and so on), have their own “source resistance,” and there is always some resistance down the line to absorb the disturbance. Most line conditioners, as part of their advertising, like to mention their energy absorption level. The rating of the unit of Fig. 10.73 is 1200 J which is actually higher than most. Remembering that W Pt EIt from the earlier discussion of cameras, we now realize that if a 5000-V spike came down the line, we would be left with the product It W/E 1200 J/5000 V 240 mAs. Assuming a linear relationship between all quantities, the rated energy level is revealing that a current of 100 A could be sustained for t 240 mAs/100 A 2.4 ms, a current of 1000 A for 240 ms, and a current of 10,000 A for 24 ms. Obviously, the higher the power product of E and I, the less the time element. The technical speciﬁcations of the unit of Fig. 10.73 include an instantaneous response time of 0 ns (questionable), with a phone line protection of 5 ns. The unit is rated to dissipate surges up to 6000 V and current spikes up to 96,000 A. It has a very high noise suppression ratio (80 dB; see Chapter 23) at frequencies from 50 kHz to 1000 MHz, and (a credit to the company) it has a lifetime warranty.
10.17 COMPUTER ANALYSIS
PSpice
Transient RC Response PSpice will now investigate the transient response for the voltage across the capacitor of Fig. 10.74. In all the examples in the text involving a transient response, a switch appeared in series with the source as shown in Fig. 10.75(a). When applying PSpice, we establish this instantaneous change in voltage level by
R
5 k
20 VE C
iC
8 µF
+
– vC
FIG. 10.74 Circuit to be analyzed using PSpice.
422  CAPACITORS
applying a pulse waveform as shown in Fig. 10.75(b) with a pulse width (PW) longer than the period (5t) of interest for the network. A pulse source is obtained through the sequence Place part keyLibraries-SOURCE-VPULSE-OK. Once in place, the label and all the parameters can be set by simply double-clicking on each to obtain the Display Properties dialog box. As you scroll down the list of attributes, you will see the following parameters deﬁned by Fig. 10.76:
V1 is the initial value. V2 is the pulse level. TD is the delay time. TR is the rise time. TF is the fall time. PW is the pulse width at the V2 level. PER is the period of the waveform.
All the parameters have been set as shown on the schematic of Fig. 10.77 for the network of Fig. 10.74. Since a rise and fall time of 0 s is unrealistic from a practical standpoint, 0.1 ms was chosen for each in this example. Further, since t RC (5 k ) (8 mF) 20 ms and 5t 200 ms, a pulse width of 500 ms was selected. The period was simply chosen as twice the pulse width. Now for the simulation process. First the New Simulation Proﬁle key is selected to obtain the New Simulation dialog box in which TransientRC is inserted for the Name and Create is chosen to leave the dialog box. The Simulation Settings-Transient RC dialog box will result, and under Analysis, the Time Domain (Transient) option is chosen under Analysis type. The Run to time is set at 200 ms so that only the ﬁrst ﬁve time constants will be plotted. The Start saving data after option will be 0 s to ensure that the data are collected immediately. The Maximum step size is 1 ms to provide sufﬁcient data points for a good plot. Click OK, and we are ready to select the Run PSpice key. The result will be a graph without a plot (since it has not been deﬁned yet) and an x-axis that extends from 0 s to 200 ms as deﬁned above. To obtain a plot of the voltage across the capacitor versus time, the following sequence is applied: Add Trace key-Add Traces dialog box-V1(C)-OK, and the plot of Fig. 10.78 will result. The color and thickness of the plot and the axis can be changed by placing the cursor on the plot line and performing a right click. A list will appear in which Properties should be selected; then a Trace Properties dialog box will appear in which the color and thickness of the line can be changed. Since the plot is against a black background, a better printout occurred when yellow was selected and the line was made thicker as shown in Fig. 10.78. Next, the cursor can be put on the axis, and another right click will allow you to make the axis yellow and thicker for a better printout. For comparison it seemed appropriate to plot the applied pulse signal also. This is accomplished by going back to Trace and selecting Add Trace followed by V(Vpulse: ) and OK. Now both waveforms appear on the same screen as shown in Fig. 10.78. In this case, the plot was left a greenish tint so it could be distinguished from the axis and the other plot. Note that it follows the left axis to the top and travels across the screen at 20 V. If you want the magnitude of either plot at any instant, simply select the Toggle cursor key. Then click on V1(C) at the bottom left of the screen. A box will appear around V1(C) that will reveal the spacing between the dots of the cursor on the screen. This is important when
(a)
(b) b
a +
–
e
0 t
20 V
b
a +
–
20 VE
t0
20 V
e
FIG. 10.75 Establishing a switching dc voltage level: (a) series dc voltage-switch combination; (b) PSpice pulse option.
V1
TR
V2
PW
TF
TD t0
PER
FIG. 10.76 The deﬁning parameters of PSpice VPULSE.
COMPUTER ANALYSIS  423
FIG. 10.77 Using PSpice to investigate the transient response of the series R-C circuit of Fig. 10.74.
FIG. 10.78 Transient response for the voltage across the capacitor of Fig. 10.74 when VPulse is applied.
424  CAPACITORS
more than one cursor is used. By moving the cursor to 200 ms, we ﬁnd that the magnitude (A1) is 19.865 V (in the Probe Cursor dialog box), clearly showing how close it is to the ﬁnal value of 20 V. A second cursor can be placed on the screen with a right click and then a click on the same V1(C) on the bottom of the screen. The box around V1(C) cannot show two boxes, but the spacing and the width of the lines of the box have deﬁnitely changed. There is no box around the Pulse symbol since it was not selected—although it could have been selected by either cursor. If we now move the second cursor to one time constant of 40 ms, we ﬁnd that the voltage is 12.633 V as shown in the Probe Cursor dialog box. This conﬁrms the fact that the voltage should be 63.2% of its ﬁnal value of 20 V in one time constant (0.632 20 V 12.4 V). Two separate plots could have been obtained by going to Plot-Add Plot to Window and then using the trace sequence again.
Average Capacitive Current As an exercise in using the pulse source and to verify our analysis of the average current for a purely capacitive network, the description to follow will verify the results of Example 10.13. For the pulse waveform of Fig. 10.59, the parameters of the pulse supply appear in Fig. 10.79. Note that the rise time is now 2 ms, starting at 0 s, and the fall time is 6 ms. The period was set at 15 ms to permit monitoring the current after the pulse had passed. Simulation is initiated by ﬁrst selecting the New Simulation Proﬁle key to obtain the New Simulation dialog box in which AverageIC is entered as the Name. Create is then chosen to obtain the Simulation
FIG. 10.79 Using PSpice to verify the results of Example 10.13.
COMPUTER ANALYSIS  425
Settings-AverageIC dialog box. Analysis is selected, and Time Domain(Transient) is chosen under the Analysis type options. The Run to time is set to 15 ms to encompass the period of interest, and the Start saving data after is set at 0 s to ensure data points starting at t 0 s. The Maximum step size is selected from 15 ms/1000 15 ms to ensure 1000 data points for the plot. Click OK, and the Run PSpice key is selected. A window will appear with a horizontal scale that extends from 0 to 15 ms as deﬁned above. Then the Add Trace key is selected, and I(C) is chosen to appear in the Trace Expression below. Click OK, and the plot of I(C) appears in the bottom of Fig. 10.80. This time it would be nice to see the pulse waveform in the same window but as a separate plot. Therefore, continue with Plot-Add Plot to WindowTrace-Add Trade-V(Vpulse: )-OK, and both plots appear as shown in Fig. 10.80.
FIG. 10.80 The applied pulse and resulting current for the 2-mF capacitor of Fig. 10.79.
The cursors can now be used to measure the resulting average current levels. First, select the I(C) plot to move the SEL>> notation to the lower plot. The SEL>> deﬁnes which plot for multiplot screens is active. Then select the Toggle cursor key, and left-click on the I(C) plot to establish the crosshairs of the cursor. Set the value at 1 ms, and the magnitude A1 is displayed as 4 mA. Right-click on the same plot, and a second cursor will result that can be placed at 6 ms to get a response of 1.33 mA (A2) as expected from Example 10.13. Both plots were again placed in the yellow color with a wider line by rightclicking on the curve and choosing Properties.
426  CAPACITORS
PROBLEMS
SECTION 10.2 The Electric Field
1. Find the electric ﬁeld strength at a point 2 m from a charge of 4 mC. 2. The electric ﬁeld strength is 36 newtons/coulomb (N/C) at a point r meters from a charge of 0.064 mC. Find the distance r.
SECTION 10.3 Capacitance
3. Find the capacitance of a parallel plate capacitor if 1400 mC of charge are deposited on its plates when 20 V are applied across the plates. 4. How much charge is deposited on the plates of a 0.05-mF capacitor if 45 V are applied across the capacitor?
5. Find the electric ﬁeld strength between the plates of a parallel plate capacitor if 100 mV are applied across the plates and the plates are 2 mm apart.
6. Repeat Problem 5 if the plates are separated by 4 mils. 7. A 4-mF parallel plate capacitor has 160 mC of charge on its plates. If the plates are 5 mm apart, ﬁnd the electric ﬁeld strength between the plates.
8. Find the capacitance of a parallel plate capacitor if the area of each plate is 0.075 m2 and the distance between the plates is 1.77 mm. The dielectric is air.
9. Repeat Problem 8 if the dielectric is parafﬁn-coated paper. 10. Find the distance in mils between the plates of a 2-mF capacitor if the area of each plate is 0.09 m2 and the dielectric is transformer oil.
11. The capacitance of a capacitor with a dielectric of air is 1200 pF. When a dielectric is inserted between the plates, the capacitance increases to 0.006 mF. Of what material is the dielectric made?
12. The plates of a parallel plate air capacitor are 0.2 mm apart and have an area of 0.08 m2, and 200 V are applied across the plates. a. Determine the capacitance. b. Find the electric ﬁeld intensity between the plates. c. Find the charge on each plate if the dielectric is air.
13. A sheet of Bakelite 0.2 mm thick having an area of 0.08 m2 is inserted between the plates of Problem 12. a. Find the electric ﬁeld strength between the plates. b. Determine the charge on each plate. c. Determine the capacitance.
SECTION 10.4 Dielectric Strength
14. Find the maximum voltage ratings of the capacitors of Problems 12 and 13 assuming a linear relationship between the breakdown voltage and the thickness of the dielectric.
15. Find the maximum voltage that can be applied across a parallel plate capacitor of 0.006 mF. The area of one plate is 0.02 m2 and the dielectric is mica. Assume a linear relationship between the dielectric strength and the thickness of the dielectric.
PROBLEMS  427
16. Find the distance in millimeters between the plates of a parallel plate capacitor if the maximum voltage that can be applied across the capacitor is 1250 V. The dielectric is mica. Assume a linear relationship between the breakdown strength and the thickness of the dielectric.
FIG. 10.83 Problem 20.
R
100 k
E 20 V C
iC
vC
+ –
5 mF vR + –
FIG. 10.81 Problems 17 and 18.
R2
3.3 k
E 100 V C
iC
vC
+– vR
+
–
R1
2.2 k
1 mF
FIG. 10.82 Problem 19.
+15 V
(t = 0 s)
C 0.1 µF µ
R 56 k
–10 V
vC + –
iC
R2 2 k 50 V
C
iC
2 mF
R1
3 k 1
2 3
+ vC –
FIG. 10.84 Problems 21 and 22.
SECTION 10.7 Transients in Capacitive Networks: Charging Phase
17. For the circuit of Fig. 10.81: a. Determine the time constant of the circuit. b. Write the mathematical equation for the voltage vC following the closing of the switch. c. Determine the voltage vC after one, three, and ﬁve time constants. d. WritetheequationsforthecurrentiC andthevoltagevR. e. Sketch the waveforms for vC and iC. 18. Repeat Problem 17 for R 1 M , and compare the results.
19. For the circuit of Fig. 10.82: a. Determine the time constant of the circuit. b. Write the mathematical equation for the voltage vC following the closing of the switch. c. Determine vC after one, three, and ﬁve time constants. d. WritetheequationsforthecurrentiC andthevoltagevR. e. Sketch the waveforms for vC and iC. 20. For the circuit of Fig. 10.83: a. Determine the time constant of the circuit. b. Write the mathematical equation for the voltage vC following the closing of the switch. c. Write the mathematical expression for the current iC following the closing of the switch. d. Sketch the waveforms of vC and iC.
SECTION 10.8 Discharge Phase
21. For the circuit of Fig. 10.84: a. Determine the time constant of the circuit when the switch is thrown into position 1. b. Find the mathematical expression for the voltage across the capacitor after the switch is thrown into position 1. c. Determine the mathematical expression for the current following the closing of the switch (position 1). d. Determine the voltage vC and the current iC if the switch is thrown into position 2 at t 100 ms. e. Determine the mathematical expressions for the voltage vC and the current iC if the switch is thrown into position 3 at t 200 ms. f. Plot the waveforms of vC and iC for a period of time extending from t 0 to t 300 ms.
428  CAPACITORS
R2 390 k 80 V C iC
1 2
vCE 10 pF + – R1 100 k
FIG. 10.85 Problem 23.
R
2.2 k
– vR +
C = 2000 mF
+ 40 V –
iC
+ vC –
FIG. 10.86 Problem 24.
FIG. 10.87 Problems 25 and 29.
iC
R1
4.7 k
vC + – E 10 V 10 F
–
+ C 3 V
FIG. 10.88 Problem 26.
–12 V
R1
10 k
40 V
R2
8.2 k
C
6.8 F 12 V+ –
vC+ –
iC

FIG. 10.89 Problem 27.
6 V + –
C = 1000 µF µ
22. Repeat Problem 21 for a capacitance of 20 mF. *23. For the network of Fig. 10.85: a. Find the mathematical expression for the voltage across the capacitor after the switch is thrown into position 1. b. Repeat part (a) for the current iC. c. Find the mathematical expressions for the voltage vC and current iC if the switch is thrown into position 2 at a time equal to ﬁve time constants of the charging circuit. d. Plot the waveforms of vC and iC for a period of time extending from t 0 to t 30 ms.
24. The capacitor of Fig. 10.86 is initially charged to 40 V before the switch is closed. Write the expressions for the voltages vC and vR and the current iC for the decay phase.
25. The 1000-mF capacitor of Fig. 10.87 is charged to 6 V. To discharge the capacitor before further use, a wire with a resistance of 0.002 is placed across the capacitor. a. How long will it take to discharge the capacitor? b. What is the peak value of the current? c. Based on the answer to part (b), is a spark expected when contact is made with both ends of the capacitor?
SECTION 10.9 Initial Values
26. The capacitor in Fig. 10.88 is initially charged to 3 V with the polarity shown. a. Find the mathematical expressions for the voltage vC and the current iC when the switch is closed. b. Sketch the waveforms for vC and iC. *27. The capacitor of Fig. 10.89 is initially charged to 12 V with the polarity shown. a. Find the mathematical expressions for the voltage vC and the current iC when the switch is closed. b. Sketch the waveforms for vC and iC.
SECTION 10.10 Instantaneous Values 28. Given the expression vC 8(1 e t/(20 10 6)): a. Determine vC after ﬁve time constants. b. Determine vC after 10 time constants. c. Determine vC at t 5 ms.
PROBLEMS  429
R 33 k
12 VE
(t = 0 s)
C 20 mF System R = ∞ VL + –
FIG. 10.90 Problem 30.
R
20 VE
(t = 0 s)
C 200 mF System R = ∞
VL = 12 V to turn on
FIG. 10.91 Problem 31.
R2
12 k
R1
8 k
iC
E 80 V C 6 mF vC + –
FIG. 10.92 Problem 32.
R2 4 M 60 V C
iC
0.2 mF
R1
1 M 1
2 vCE
+ vR1 –
+
–
FIG. 10.93 Problem 33.
29. For the situation of Problem 25, determine when the discharge current is one-half its maximum value if contact is made at t 0 s. 30. For the network of Fig. 10.90, VL must be 8 V before the system is activated. If the switch is closed at t 0 s, how long will it take for the system to be activated?
*31. Design the network of Fig. 10.91 such that the system will turn on 10 s after the switch is closed.
32. For the circuit of Fig. 10.92: a. Find the time required for vC to reach 60 V following the closing of the switch. b. Calculate the current iC at the instant vC 60 V. c. Determine the power delivered by the source at the instant t 2t.
*33. For the network of Fig. 10.93: a. Calculate vC, iC, and vR1 at 0.5 s and 1 s after the switch makes contact with position 1. b. The network sits in position 1 10 min before the switch is moved to position 2. How long after making contact with position 2 will it take for the current iC to drop to 8 mA? How much longer will it take for vC to drop to 10 V?
+ –
C
0.2 mF
E 60 V
iC
+ vC –
DMM
FIG. 10.94 Problem 34.
34. For the system of Fig. 10.94, using a DMM with a 10-M internal resistance in the voltmeter mode: a. Determine the voltmeter reading one time constant after the switch is closed. b. Find the current iC two time constants after the switch is closed. c. Calculate the time that must pass after the closing of the switch for the voltage vC to be 50 V.
430  CAPACITORS
SECTION 10.11 Thévenin Equivalent: t RThC 35. Forthe system of Fig. 10.95, using a DMM with a 10-M internal resistance in the voltmeter mode: a. Determine the voltmeter reading four time constants after the switch is closed. b. Find the time that must pass before iC drops to 3 mA. c. Find the time that must pass after the closing of the switch for the voltage across the meter to reach 10 V.
36. For the circuit of Fig. 10.96: a. Find the mathematical expressions for the transient behavior of the voltage vC and the current iC following the closing of the switch. b. Sketch the waveforms of vC and iC.
*37. Repeat Problem 36 for the circuit of Fig. 10.97.
38. The capacitor of Fig. 10.98 is initially charged to 4 V with the polarity shown. a. Write the mathematical expressions for the voltage vC and the current iC when the switch is closed. b. Sketch the waveforms of vC and iC.
R
2 M
+ –
C 1 mFE 24 V
iC
FIG. 10.95 Problem 35.
R2 24 k
R1
8 k R3 4 k 20 VE
C
15 mF
+ vC – iC
FIG. 10.96 Problem 36.
R1
0.56 k
R2
3.9 k
R3
6.8 k
iC C 20 mF
vC + –
+ 4 V
5 mA
FIG. 10.97 Problems 37 and 55.
iC
R2
3.9 k
20 F
–
+
Cv C + – 4 VR 1 1.8 k E 36 V
FIG. 10.98 Problem 38.
iC
R2
1.5 k
2.2 F
+
–
Cv C + – 2 VR 1 6.8 k I = 4 mA
FIG. 10.99 Problem 39.
39. The capacitor of Fig. 10.99 is initially charged to 2 V with the polarity shown. a. Write the mathematical expressions for the voltage vC and the current iC when the switch is closed. b. Sketch the waveforms of vC and iC.
PROBLEMS  431
–20 V2 k +10 V 6.8 k C
39 F 3 V+ –
vC+ –
iC

FIG. 10.100 Problem 40.
100
v (V)
80
60
40
20
0 – 20 – 40 – 60
1 2 3 4 5 6 7 8 9 10 11 13 14 t (ms) 12 15
FIG. 10.101 Problem 41.
*40. The capacitor of Fig. 10.100 is initially charged to 3 V with the polarity shown. a. Write the mathematical expressions for the voltage vC and the current iC when the switch is closed. b. Sketch the waveforms of vC and iC.
SECTION 10.12 The Current iC 41. Find the waveform for the average current if the voltage across a 0.06-mF capacitor is as shown in Fig. 10.101.
42. Repeat Problem 41 for the waveform of Fig. 10.102.
*43. Given the waveform of Fig. 10.103 for the current of a 20-mF capacitor, sketch the waveform of the voltage vC across the capacitor if vC 0 V at t 0 s.
v (V)
3
2
1
0 –1 –2 –3
1 2 4 5 6 7 8 9 1011 12 13 14 15 t (ms) 163
FIG. 10.102 Problem 42.
iC
0 4 6 16 18 20 25 t (ms)
–120 mA
–80 mA
+40 mA
FIG. 10.103 Problem 43.
432  CAPACITORS
SECTION 10.13 Capacitors in Series and Parallel 44. Find the total capacitance CT between points a and b of the circuits of Fig. 10.104.
45. Find the voltage across and charge on each capacitor for the circuits of Fig. 10.105.
*46. For each conﬁguration of Fig. 10.106, determine the voltage across each capacitor and the charge on each capacitor.
3 mF 6 mF
7 mF
0.2 mF
CT
a
b
(a)
60 pF
CT
a
b
(b)
20 pF 10 pF 30 pF
FIG. 10.104 Problem 44.
10 V
(a)
C3 12 mF C1 6 mF C2 6 mF
40 V
(b)
C4 600 pF
C1
1200 pF
C3
400 pF
C2 200 pF
FIG. 10.105 Problem 45.
24 V
(a)
C4 72 mF
C2 10 mF
C3 9 mF
C1
9 mF
E 16 V
(b)
C5 35 mF
C3 2 mF
C4 14 mF
C1
7 mF
E
C2
42 mF
FIG. 10.106 Problem 46.
PROBLEMS  433
100 V
R2 40 k
R1
20 k
C2 12 mF
C1 8 mF
E
1
2
d
a
b
c
C3 12 mF
FIG. 10.107 Problem 47.
*47. For the network of Fig. 10.107, determine the following 100 ms after the switch is closed: a. Vab b. Vac c. Vcb d. Vda e. If the switch is moved to position 2 one hour later, ﬁnd the time required for vR2 to drop to 20 V. 48. For the circuits of Fig. 10.108, ﬁnd the voltage across and charge on each capacitor after each capacitor has charged to its ﬁnal value.
48 V
2 k
(a)
C2
0.08 mF
C1
0.04 mF
4 k
FIG. 10.108 Problem 48.
4 k
6 k
(b)
60 mF
C1
80 V
5 k
40 mF
C2
SECTION 10.14 Energy Stored by a Capacitor
49. Find the energy stored by a 120-pF capacitor with 12 V across its plates. 50. If the energy stored by a 6-mF capacitor is 1200 J, ﬁnd the charge Q on each plate of the capacitor. *51. An electronic ﬂashgun has a 1000-mF capacitor that is charged to 100 V. a. How much energy is stored by the capacitor? b. What is the charge on the capacitor? c. When the photographer takes a picture, the ﬂash ﬁres for 1/2000 s. What is the average current through the ﬂashtube? d. Find the power delivered to the ﬂashtube. e. After a picture is taken, the capacitor has to be recharged by a power supply that delivers a maximum current of 10 mA. How long will it take to charge the capacitor?
52. For the network of Fig. 10.109: a. Determine the energy stored by each capacitor under steady-state conditions. b. Repeat part (a) if the capacitors are in series.
3 k
6 k
24 V
4 k
6 mF 12 mF
FIG. 10.109 Problem 52.
434  CAPACITORS
SECTION 10.17 Computer Analysis
PSpice or Electronics Workbench
53. Using schematics: a. Obtain the waveforms for vC and iC versus time for the network of Fig. 10.35. b. Obtain the power curve (representing the energy stored by the capacitor over the same time interval), and compare it to the plot of Fig. 10.70. *54. Using schematics, obtain the waveforms of vC and iC versus time for the network of Fig. 10.49 using the IC option.
55. Verify your solution to Problem 37 (Fig. 10.97) using schematics.
Programming Language (C , QBASIC, Pascal, etc.) 56. Write a QBASIC program to tabulate the voltage vC and current iC for the network of Fig. 10.44 for ﬁve time constants after the switch is moved to position 1 at t 0 s. Use an increment of (1/5)t.
*57. Write a program to write the mathematical expression for the voltage vC for the network of Fig. 10.52 for any element values when the switch is moved to position 1.
*58. Given three capacitors in any series-parallel arrangement, write a program to determine the total capacitance. That is, determine the total number of possibilities, and ask the user to identify the conﬁguration and provide the capacitor values. Then calculate the total capacitance.
GLOSSARY
Breakdown voltage Another term for dielectric strength, listed below. Capacitance A measure of a capacitor’s ability to store charge; measured in farads (F). Capacitive time constant The product of resistance and capacitance that establishes the required time for the charging and discharging phases of a capacitive transient. Capacitive transient The waveforms for the voltage and current of a capacitor that result during the charging and discharging phases. Capacitor A fundamental electrical element having two conducting surfaces separated by an insulating material and having the capacity to store charge on its plates. Coulomb’s law An equation relating the force between two like or unlike charges. Dielectric The insulating material between the plates of a capacitor that can have a pronounced effect on the charge stored on the plates of a capacitor. Dielectric constant Another term for relative permittivity, listed below. Dielectric strength An indication of the voltage required for unit length to establish conduction in a dielectric.

Electric ﬁeld strength The force acting on a unit positive charge in the region of interest. Electric ﬂux lines Lines drawn to indicate the strength and direction of an electric ﬁeld in a particular region. Fringing An effect established by ﬂux lines that do not pass directly from one conducting surface to another. Leakage current The current that will result in the total discharge of a capacitor if the capacitor is disconnected from the charging network for a sufﬁcient length of time. Permittivity A measure of how well a dielectric will permit the establishment of ﬂux lines within the dielectric. Relative permittivity The permittivity of a material compared to that of air. Stray capacitance Capacitances that exist not through design but simply because two conducting surfaces are relatively close to each other. Surge voltage The maximum voltage that can be applied across a capacitor for very short periods of time. Working voltage The voltage that can be applied across a capacitor for long periods of time without concern for dielectric breakdown.

Popular Posts

Friday 24 February 2017

Wednesday 22 February 2017

Tuesday 21 February 2017

Monday 20 February 2017

Sunday 19 February 2017

Saturday 18 February 2017

Friday 17 February 2017

Thursday 9 February 2017

Wednesday 8 February 2017