Chapter 12: Capacitors Under DC Conditions

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?

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.

The value of capacitance of a capacitor depends on three factors:

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:

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.

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.

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 (E_s) is instantly felt across the entire circuit. Graph (A) of the figure shows an instantaneous rise at time tp from zero to source voltage (E_s = 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 (i_c). 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 e_r, 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.

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 (i_d) 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 (e_c) 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 (e_c) 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?

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 (C_T) of capacitors connected in series is similar to solving for the total resistance (R_T) of resistors connected in parallel.

Note the similarity between the formulas for R_T and C_T:

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 C_T, C1, C2, C3,…​ C_n 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).

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. C_T, 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:

C_T = C₁ + C₂ + C₃ + …​ C_n

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?