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 (8-B). 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.
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.
A voltage divider containing resistance and capacitance is connected in a circuit by means of a switch, as shown at the top of figure (9). 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 t0 (time zero, when the switch is first closed) to time t1 (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
(9) is closed at t0, the source voltage
instantly felt across the entire circuit. Graph (A) of the figure shows
an instantaneous rise at time t0 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
charging current is MAXIMUM. As time elapses toward time
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
the voltage drop (er) 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
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 0 volts to 6 volts across the circuit, while the voltage developed across the capacitor in graph (D) took the entire time interval from time to 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 t0 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 t0 to t1 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 table (1).
In figure (10) 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 t1, 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 (er) 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 (ec) across the capacitor is decreasing until at time t1 the voltage (ec) reaches zero.
By comparing graph (A) with graph (D) of figure (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 S1 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.
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 (11). Note that the charge curve is like the curve in figure 3-9, graph (D), and the discharge curve like the curve in figure (9), 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:
UNIVERSAL TIME CONSTANT CHART
Because the impressed voltage and the values of R and C or Figure (11). - RC time constant. and L in a circuit are usually known, a UNIVERSAL TIME CONSTANT CHART (fig. 3-12) 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 is a 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 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 (12) and proceed as follows:
Note that the vertical line crosses the
horizontal scale at about .22 RC as illustrated below:
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 µs 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.
Transpose the RC time constant formula as follows:
Substitute the R and RC values into the formula:
The graphs shown in figure (11) and (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.
(Use the universal time constant chart in figure (12).)
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,... 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 0.008µF is slightly smaller than the smallest capacitor (0.01µF).
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 (14) 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.
Figure (14). - Parallel capacitive circuit.
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.