- 1. Capacitance
- 2. The Electrostatic Field
- 3. Voltage Rating of Capacitors
- 4. Capacitor Losses
- 5. Charging and Discharging a Capacitor
- 6. Charge and Discharge of an RC Series Circuit
- 7. RC Time Constant
- 8. Universal Time Constant Chart
- 9. Capacitors in Series and Parallel
In the previous chapter you learned that inductance is the property of a coil that causes electrical energy to be stored in a magnetic field about the coil. The energy is stored in such a way as to oppose any change in current. CAPACITANCE is similar to inductance because it also causes a storage of energy. A CAPACITOR is a device that stores electrical energy in an ELECTROSTATIC FIELD. The energy is stored in such a way as to oppose any change in voltage. Just how capacitance opposes a change in voltage is explained later in this chapter. However, it is first necessary to explain the principles of an electrostatic field as it is applied to capacitance.
Ql. Define the terms "capacitor" and "capacitance."
2. The Electrostatic Field
You previously learned that opposite electrical charges attract each other while like electrical charges repel each other. The reason for this is the existence of an electrostatic field. Any charged particle is surrounded by invisible lines of force, called electrostatic lines of force. These lines of force have some interesting characteristics:
They are polarized from positive to negative.
They radiate from a charged particle in straight lines and do not form closed loops.
They have the ability to pass through any known material.
They have the ability to distort the orbits of tightly bound electrons.
Examine Figure 1. This figure represents two unlike charges surrounded by their electrostatic field. Because an electrostatic field is polarized positive to negative, arrows are shown radiating away from the positive charge and toward the negative charge. Stated another way, the field from the positive charge is pushing, while the field from the negative charge is pulling. The effect of the field is to push and pull the unlike charges together.
Figure 2, two like charges are shown with their surrounding electrostatic field. The effect of the electrostatic field is to push the charges apart.
If two unlike charges are placed on opposite sides of an atom whose outermost electrons cannot escape their orbits, the orbits of the electrons are distorted as shown in Figure 3. Figure 3(A) shows the normal orbit. Part (B) of the figure shows the same orbit in the presence of charged particles. Since the electron is a negative charge, the positive charge attracts the electrons, pulling the electrons closer to the positive charge. The negative charge repels the electrons, pushing them further from the negative charge. It is this ability of an electrostatic field to attract and to repel charges that allows the capacitor to store energy.
Toward which charge will the electron move ?
2.1. The Simple Capacitor
A simple capacitor consists of two metal plates separated by an insulating material called a dielectric, as illustrated in Figure 4. Note that one plate is connected to the positive terminal of a battery; the other plate is connected through a closed switch (S1) to the negative terminal of the battery. Remember, an insulator is a material whose electrons cannot easily escape their orbits. Due to the battery voltage, plate A is charged positively and plate B is charged negatively. (How this happens is explained later in this chapter.) Thus an electrostatic field is set up between the positive and negative plates. The electrons on the negative plate (plate B) are attracted to the positive charges on the positive plate (plate A).
Notice that the orbits of the electrons in the dielectric material are distorted by the electrostatic field. The distortion occurs because the electrons in the dielectric are attracted to the top plate while being repelled from the bottom plate. When switch S1 is opened, the battery is removed from the circuit and the charge is retained by the capacitor. This occurs because the dielectric material is an insulator, and the electrons in the bottom plate (negative charge) have no path to reach the top plate (positive charge). The distorted orbits of the atoms of the dielectric plus the electrostatic force of attraction between the two plates hold the positive and negative charges in their original position. Thus, the energy which came from the battery is now stored in the electrostatic field of the capacitor. Two slightly different symbols for representing a capacitor are shown in Figure 5 and in Figure 6. Notice that each symbol is composed of two plates separated by a space that represents the dielectric. The curved plate in Figure 6 of the figure indicates the plate should be connected to a negative polarity.
O4. What are the basic parts of a capacitor?
2.2. The Farad
Capacitance is measured in units called FARADS. A one-farad capacitor stores one coulomb (a unit of charge (Q) equal to 6.28 x 10'° electrons) of charge when a potential of 1 volt is applied across the terminals of the capacitor. This can be expressed by the formula:
The farad is a very large unit of measurement of capacitance. For convenience, the microfarad (abbreviated μF) or the picofarad (abbreviated WF) is used. One (1.0) microfarad is equal to 0.000001 farad or 1 x 10° farad, and 1.0 picofarad is equal to 0.000000000001 farad or 1.0 x 10-12 farad. Capacitance is a physical property of the capacitor and does not depend on circuit characteristics of voltage, current, and resistance. A given capacitor always has the same value of capacitance (farads) in one circuit as in any other circuit in which it is connected.
2.3. Factors Affecting the Value of Capacitance
The value of capacitance of a capacitor depends on three factors:
The area of the plates.
The distance between the plates.
The dielectric constant of the material between the plates.
PLATE AREA affects the value of capacitance in the same manner that the size of a container affects the amount of water that can be held by the container. A capacitor with the large plate area can store more charges than a capacitor with a small plate area. Simply stated, "the larger the plate area, the larger the capacitance".
The second factor affecting capacitance is the DISTANCE BETWEEN THE PLATES. Electrostatic lines of force are strongest when the charged particles that create them are close together. When the charged particles are moved further apart, the lines of force weaken, and the ability to store a charge decreases.
The third factor affecting capacitance is the DIELECTRIC CONSTANT of the insulating material between the plates of a capacitor. The various insulating materials used as the dielectric in a capacitor differ in their ability to respond to (pass) electrostatic lines of force. A dielectric material, or insulator, is rated as to its ability to respond to electrostatic lines of force in terms of a figure called the DIELECTRIC CONSTANT. A dielectric material with a high dielectric constant is a better insulator than a dielectric material with a low dielectric constant. Dielectric constants for some common materials are given in the following list:
5 to 10
3 to 6
2.5 to 35
2.5 to 8
Notice the dielectric constant for a vacuum. Since a vacuum is the standard of reference, it is assigned a constant of one. The dielectric constants of all materials are compared to that of a vacuum. Since the dielectric constant of air has been determined to be approximately the same as that of a vacuum, the dielectric constant of AIR is also considered to be equal to one.
The formula used to compute the value of capacitance is:
Example: Find the capacitance of a parallel plate capacitor with paraffin paper as the dielectric.
By examining the above formula you can see that capacitance varies directly as the dielectric constant and the area of the capacitor plates, and inversely as the distance between the plates.
3. Voltage Rating of Capacitors
In selecting or substituting a capacitor for use, consideration must be given to (1) the value of capacitance desired and (2) the amount of voltage to be applied across the capacitor. If the voltage applied across the capacitor is too great, the dielectric will break down and arcing will occur between the capacitor plates. When this happens the capacitor becomes a short-circuit and the flow of direct current through it can cause damage to other electronic parts. Each capacitor has a voltage rating (a working voltage) that should not be exceeded.
The working voltage of the capacitor is the maximum voltage that can be steadily applied without danger of breaking down the dielectric. The working voltage depends on the type of material used as the dielectric and on the thickness of the dialectic. (A high-voltage capacitor that has a thick dielectric must have a relatively large plate area in order to have the same capacitance as a similar low-voltage capacitor having a thin dielectric.) The working voltage also depends on the applied frequency because the losses, and the resultant heating effect, increase as the frequency increases.
A capacitor with a voltage rating of 500 volts dc cannot be safely subjected to an alternating voltage or a pulsating direct voltage having an effective value of 500 volts. Since an alternating voltage of 500 volts (rms) has a peak value of 707 volts, a capacitor to which it is applied should have a working voltage of at least 750 volts. In practice, a capacitor should be selected so that its working voltage is at least 50 percent greater than the highest effective voltage to be applied to it.
4. Capacitor Losses
Power loss in a capacitor may be attributed to dielectric hysteresis and dielectric leakage. Dielectric hysteresis may be defined as an effect in a dielectric material similar to the hysteresis found in a magnetic material. It is the result of changes in orientation of electron orbits in the dielectric because of the rapid reversals of the polarity of the line voltage. The amount of power loss due to dielectric hysteresis depends upon the type of dielectric used. A vacuum dielectric has the smallest power loss.
Dielectric leakage occurs in a capacitor as the result of LEAKAGE CURRENT through the dielectric. Normally it is assumed that the dielectric will effectively prevent the flow of current through the capacitor. Although the resistance of the dielectric is extremely high, a minute amount of current does flow. Ordinarily this current is so small that for all practical purposes it is ignored. However, if the leakage through the dielectric is abnormally high, there will be a rapid loss of charge and an overheating of the capacitor.
The power loss of a capacitor is determined by loss in the dielectric. If the loss is negligible and the capacitor returns the total charge to the circuit, it is considered to be a perfect capacitor with a power loss of zero.
QO9. Name two types of power losses associated with a capacitor.
Define the term "working voltage" of a capacitor.
What should be the working voltage of a capacitor in a circuit that is operating at 600 volts ?
5. Charging and Discharging a Capacitor
In order to better understand the action of a capacitor in conjunction with other components, the charge and discharge actions of a purely capacitive circuit are analyzed first. For ease of explanation the capacitor and voltage source shown in figure 3-6 are assumed to be perfect (no internal resistance), although this is impossible in practice.
In Figure 7, an uncharged capacitor is shown connected to a four-position switch. With the switch in position | the circuit is open and no voltage is applied to the capacitor. Initially each plate of the capacitor is a neutral body and until a difference of potential is impressed across the capacitor, no electrostatic field can exist between the plates.
To CHARGE the capacitor, the switch must be thrown to position 2, which places the capacitor across the terminals of the battery. Under the assumed perfect conditions, the capacitor would reach full charge instantaneously. However, the charging action is spread out over a period of time in the following discussion so that a step-by-step analysis can be made.
At the instant the switch is thrown to position 2 (Figure 8), a displacement of electrons occurs simultaneously in all parts of the circuit. This electron displacement is directed away from the negative terminal and toward the positive terminal of the source (the battery). A brief surge of current will flow as the capacitor charges.
If it were possible to analyze the motion of the individual electrons in this surge of charging current, the following action would be observed. See Figure 9.
At the instant the switch is closed, the positive terminal of the battery extracts an electron from the bottom conductor. The negative terminal of the battery forces an electron into the top conductor. At this same instant an electron is forced into the top plate of the capacitor and another is pulled from the bottom plate. Thus, in every part of the circuit a clockwise DISPLACEMENT of electrons occurs simultaneously.
As electrons accumulate on the top plate of the capacitor and others depart from the bottom plate, a difference of potential develops across the capacitor. Each electron forced onto the top plate makes that plate more negative, while each electron removed from the bottom causes the bottom plate to become more positive. Notice that the polarity of the voltage which builds up across the capacitor is such as to oppose the source voltage. The source voltage (emf) forces current around the circuit of Figure 9 in a clockwise direction. The emf developed across the capacitor, however, has a tendency to force the current in a counterclockwise direction, opposing the source emf. As the capacitor continues to charge, the voltage across the capacitor rises until it is equal to the source voltage. Once the capacitor voltage equals the source voltage, the two voltages balance one another and current ceases to flow in the circuit.
In studying the charging process of a capacitor, you must be aware that NO current flows THROUGH the capacitor. The material between the plates of the capacitor must be an insulator. However, to an observer stationed at the source or along one of the circuit conductors, the action has all the appearances of a true flow of current, even though the insulating material between the plates of the capacitor prevents the current from having a complete path. The current which appears to flow through a capacitor is called DISPLACEMENT CURRENT.
When a capacitor is fully charged and the source voltage is equaled by the counter electromotive force (cemf) across the capacitor, the electrostatic field between the plates of the capacitor is maximum. (Look again at Figure 4.) Since the electrostatic field is maximum the energy stored in the dielectric is also maximum.
If the switch is now opened as shown in Figure 10, the electrons on the upper plate are isolated. The electrons on the top plate are attracted to the charged bottom plate. Because the dielectric is an insulator, the electrons can not cross the dielectric to the bottom plate. The charges on both plates will be effectively trapped by the electrostatic field and the capacitor will remain charged indefinitely. You should note at this point that the insulating dielectric material in a practical capacitor is not perfect and small leakage current will flow through the dielectric. This current will eventually dissipate the charge. However, a high quality capacitor may hold its charge for a month or more.
To review briefly, when a capacitor is connected across a voltage source, a surge of charging current flows. This charging current develops a cemf across the capacitor which opposes the applied voltage. When the capacitor is fully charged, the cemf is equal to the applied voltage and charging current ceases. At full charge, the electrostatic field between the plates is at maximum intensity and the energy stored in the dielectric is maximum. If the charged capacitor is disconnected from the source, the charge will be retained for some period of time. The length of time the charge is retained depends on the amount of leakage current present. Since electrical energy is stored in the capacitor, a charged capacitor can act as a source emf.
To DISCHARGE a capacitor, the charges on the two plates must be neutralized. This is accomplished by providing a conducting path between the two plates as shown in Figure 10. With the switch in position (4) the excess electrons on the negative plate can flow to the positive plate and neutralize its charge. When the capacitor is discharged, the distorted orbits of the electrons in the dielectric return to their normal positions and the stored energy is returned to the circuit. It is important for you to note that a capacitor does not consume power. The energy the capacitor draws from the source is recovered when the capacitor is discharged.
(b) the capacitor is discharging.
6. Charge and Discharge of an RC Series Circuit
Ohm’s law states that the voltage across a resistance is equal to the current through the resistance times the value of the resistance. This means that a voltage is developed across a resistance ONLY WHEN CURRENT FLOWS through the resistance.
A capacitor is capable of storing or holding a charge of electrons. When uncharged, both plates of the capacitor contain essentially the same number of free electrons. When charged, one plate contains more free electrons than the other plate. The difference in the number of electrons is a measure of the charge on the capacitor. The accumulation of this charge builds up a voltage across the terminals of the capacitor, and the charge continues to increase until this voltage equals the applied voltage. The charge in a capacitor is related to the capacitance and voltage as follows:
in which Q is the charge in coulombs, C the capacitance in farads, and E the emf across the capacitor in volts.
6.1. Charge Cycle
A voltage divider containing resistance and capacitance is connected in a circuit by means of a switch, as shown at the top of Figure 12. Such a series arrangement is called an RC series circuit.
In explaining the charge and discharge cycles of an RC series circuit, the time interval from time to (time zero, when the switch is first closed) to time t; (time one, when the capacitor reaches full charge or discharge potential) will be used. (Note that switches S1 and S2 move at the same time and can never both be closed at the same time.)
When switch S1 of the circuit in Figure 12 is closed at t0, the source voltage (Es) is instantly felt across the entire circuit. Graph (A) of the figure shows an instantaneous rise at time tp from zero to source voltage (Es = 6 volts). The total voltage can be measured across the circuit between points 1 and 2. Now look at graph (B) which represents the charging current in the capacitor (ic). At time t0, charging current is MAXIMUM. As time elapses toward time t1, there is a continuous decrease in current flowing into the capacitor. The decreasing flow is caused by the voltage buildup across the capacitor. At time t1, current flowing in the capacitor stops. At this time, the capacitor has reached full charge and has stored maximum energy in its electrostatic field. Graph (C) represents the voltage drop (e) across the resistor (R). The value of er, is determined by the amount of current flowing through the resistor on its way to the capacitor. At time t0 the current flowing to the capacitor is maximum. Thus, the voltage drop across the resistor is maximum (E = IR). As time progresses toward time t1, the current flowing to the capacitor steadily decreases and causes the voltage developed across the resistor (R) to steadily decrease. When time t1, is reached, current flowing to the capacitor is stopped and the voltage developed across the resistor has decreased to zero.
You should remember that capacitance opposes a change in voltage. This is shown by comparing graph (A) to graph (D). In graph (A) the voltage changed instantly from O volts to 6 volts across the circuit, while the voltage developed across the capacitor in graph (D) took the entire time interval from time t0, to time t1, to reach 6 volts. The reason for this is that in the first instant at time t0, maximum current flows through R and the entire circuit voltage is dropped across the resistor. The voltage impressed across the capacitor at 0 is zero volts. As time progresses toward t1, the decreasing current causes progressively less voltage to be dropped across the resistor (R), and more voltage builds up across the capacitor (C). At time t1, the voltage felt across the capacitor is equal to the source voltage (6 volts), and the voltage dropped across the resistor (R) is equal to zero. This is the complete charge cycle of the capacitor.
As you may have noticed, the processes which take place in the time interval tp to t; in a series RC circuit are exactly opposite to those in a series LR circuit.
For your comparison, the important points of the charge cycle of RC and LR circuits are summarized in Figure 13.
6.2. Discharge Cycle
In Figure 14 at time t0, the capacitor is fully charged. When S1 is open and S2 closes, the capacitor discharge cycle starts. At the first instant, circuit voltage attempts to go from source potential (6 volts) to zero volts, as shown in graph (A). Remember, though, the capacitor during the charge cycle has stored energy in an electrostatic field.
Because S2 is closed at the same time S1 is open, the stored energy of the capacitor now has a path for current to flow. At t0, discharge current (id) from the bottom plate of the capacitor through the resistor (R) to the top plate of the capacitor (C) is maximum. As time progresses toward t;, the discharge current steadily decreases until at time t1 it reaches zero, as shown in graph (B).
The discharge causes a corresponding voltage drop across the resistor as shown in graph (C). At time t0, the current through the resistor is maximum and the voltage drop (e,) across the resistor is maximum. As the current through the resistor decreases, the voltage drop across the resistor decreases until at t1 it has reached a value of zero. Graph (D) shows the voltage across the capacitor (ec) during the discharge cycle. At time t0 the voltage is maximum and as time progresses toward time t1, the energy stored in the capacitor is depleted. At the same time the voltage across the resistor is decreasing, the voltage (e) across the capacitor is decreasing until at time t1 the voltage (ec) reaches zero.
By comparing graph (A) with graph (D) of figure 3-10, you can see the effect that capacitance has on a change in voltage. If the circuit had not contained a capacitor, the voltage would have ceased at the instant Sl was opened at time t0. Because the capacitor is in the circuit, voltage is applied to the circuit until the capacitor has discharged completely at t1. The effect of capacitance has been to oppose this change in voltage.
capacitor is charging ?
capacitor is equal to the battery voltage?
7. RC Time Constant
The time required to charge a capacitor to 63 percent (actually 63.2 percent) of full charge or to discharge it to 37 percent (actually 36.8 percent) of its initial voltage is known as the TIME CONSTANT (TC) of the circuit. The charge and discharge curves of a capacitor are shown in Figure 15. Note that the charge curve is like the curve in Figure 12, graph (D), and the discharge curve like the curve in Figure 12, graph (B).
The value of the time constant in seconds is equal to the product of the circuit resistance in ohms and the circuit capacitance in farads. The value of one time constant is expressed mathematically as t = RC. Some forms of this formula used in calculating RC time constants are:
8. Universal Time Constant Chart
Because the impressed voltage and the values of R and C or R and L in a circuit are usually known, a UNIVERSAL TIME CONSTANT CHART (Figure 16) can be used to find the time constant of the circuit. Curve A is a plot of both capacitor voltage during charge and inductor current during growth. Curve B isa plot of both capacitor voltage during discharge and inductor current during decay.
The time scale (horizontal scale) is graduated in terms of the RC or L/R time constants so that the curves may be used for any value of R and C or L and R. The voltage and current scales (vertical scales) are graduated in terms of percentage of the maximum voltage or current so that the curves may be used for any value of voltage or current. If the time constant and the initial or final voltage for the circuit in question are known, the voltages across the various parts of the circuit can be obtained from the curves for any time after the switch is closed, either on charge or discharge. The same reasoning is true of the current in the circuit.
The following problem illustrates how the universal time constant chart may be used.
An RC circuit is to be designed in which a capacitor (C) must charge to 20 percent (0.20) of the maximum charging voltage in 100 microseconds (0.0001 second). Because of other considerations, the resistor (R) must have a value of 20,000 ohms. What value of capacitance is needed?
Find: The capacitance of capacitor C.
Solution: Because the only values given are in units of time and resistance, a variation of the formula to find RC time is used:
Find the value of RC by referring to the universal time constant chart in figure 3-12 and proceed as follows:
Locate the 20 point on the vertical scale at the left side of the chart (percentage).
Follow the horizontal line from this point to intersect curve A.
Follow an imaginary vertical line from the point of intersection on curve A downward to cross the RC scale at the bottom of the chart.
Note that the vertical line crosses the horizontal scale at about .22 RC as illustrated in Figure 17:
The value selected from the graph means that a capacitor (including the one you are solving for) will reach twenty percent of full charge in twenty-two one hundredths (.22) of one RC time constant. Remember that it takes 100 us for the capacitor to reach 20% of full charge. Since 100 μs is equal to .22 RC (twenty- two one-hundredths), then the time required to reach one RC time constant must be equal to:
Now use the following formula to find C:
To summarize the above procedures, the problem and solution are shown below without the step by step explanation.
The graphs shown in figure 3-11 and 3-12 are not entirely complete. That is, the charge or discharge (or the growth or decay) is not quite complete in 5 RC or 5 L/R time constants. However, when the values reach 0.99 of the maximum (corresponding to 5 RC or 5 L/R), the graphs may be considered accurate enough for all practical purposes.
charging voltage in 200 microseconds. The resistor to be used has a resistance of 40,000 ohms. What size capacitor must be used? (Use the universal time constant chart in figure 3-12.)
9. Capacitors in Series and Parallel
Capacitors may be connected in series or in parallel to obtain a resultant value which may be either the sum of the individual values (in parallel) or a value less than that of the smallest capacitance (in series).
9.1. Capacitors in Series
The overall effect of connecting capacitors in series is to move the plates of the capacitors further apart. This is shown in Figure 18. Notice that the junction between C1 and C2 has both a negative and a positive charge. This causes the junction to be essentially neutral. The total capacitance of the circuit is developed between the left plate of C1 and the right plate of C2. Because these plates are farther apart, the total value of the capacitance in the circuit is decreased. Solving for the total capacitance (CT) of capacitors connected in series is similar to solving for the total resistance (RT) of resistors connected in parallel.
Note the similarity between the formulas for RT and CT:
If the circuit contains more than two capacitors, use the above formula. If the circuit contains only two capacitors, use the below formula:
Note: All values for CT, C1, C2, C3,… Cn should be in farads. It should be evident from the above formulas that the total capacitance of capacitors in series is less than the capacitance of any of the individual capacitors.
Example: Determine the total capacitance of a series circuit containing three capacitors whose values are 0.01 μF, 0.25 μF, and 50,000 pF, respectively.
The total capacitance of 0.008 μF is slightly smaller than the smallest capacitor (0.01 μF).
9.2. Capacitors in Parallel
When capacitors are connected in parallel, one plate of each capacitor is connected directly to one terminal of the source, while the other plate of each capacitor is connected to the other terminal of the source. Figure 19 shows all the negative plates of the capacitors connected together, and all the positive plates connected together. CT, therefore, appears as a capacitor with a plate area equal to the sum of all the individual plate areas. As previously mentioned, capacitance is a direct function of plate area. Connecting capacitors in parallel effectively increases plate area and thereby increases total capacitance.
For capacitors connected in parallel the total capacitance is the sum of all the individual capacitances. The total capacitance of the circuit may by calculated using the formula:
CT = C1 + C2 + C3 + … Cn
where all capacitances are in the same units.
Example: Determine the total capacitance in a parallel capacitive circuit containing three capacitors whose values are 0.03 μF, 2.0 μF, and 0.25 μF, respectively.
μF and 0.1 μF) wired together in series?
21 μF, 0.1 μF and 2 μF) are connected in parallel?