6.2 Capacitors

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6.2 Capacitors

A capacitor is a passi ve element designed to store ener gy in its elec tric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communications, computers, and power systems. For example, they are used in the tuning circuits of radio recei vers and as dynamic memory elements in computer systems.

A capacitor is typically constructed as depicted in Fig. 6.1.

A capacitor consists of two conducting plates separated by an insulator (or dielectric).

In many practical applications, the plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica.

Historical

Michael Faraday (1791–1867), an Engl ish c hemist a nd phy sicist, was probably the greatest experimentalist who ever lived.

Born near London, Faraday realized his boyhood dream by work ing with the great chemist Sir Humphry Davy at the Royal Institu tion, where he worked for 54 years. He made several contributions in all areas of physical science and coined such words as electrolysis, anode, and cathode. His discovery of electromagnetic induction in 1831 was a major breakthrough in engineering because it provided a way of generating electricity. The electric motor and generator operate on this principle. The unit of capacitance, the farad, was named in his honor.

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When a voltage source v is connected to the capacitor, as in Fig. 6.2, the source deposits a positive charge q on one plate and a negative charge −q on the other . The capacitor is said to store the electric char ge. The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that

$$q = Cv \tag{6.1}$$

where C, the constant of proportionality , is kno wn as the capacitance of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we may derive the following definition.

Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F).

Note from Eq. (6.1) that 1 farad = 1 coulomb/volt.

Although the capacitance C of a capacitor is the ratio of the charge q per plate to the applied voltage v, it does not depend on q or v. It depends on the physical dimensions of the capacitor. For example, for the parallelplate capacitor shown in Fig. 6.1, the capacitance is given by

$$C = \frac{\epsilon A}{d} \tag{6.2}$$

where A is the surf ace area of each plate, d is the distance between the plates, and ϵ is the permittivity of the dielectric material between the plates. Although Eq. (6.2) applies to only parallel-plate capacitors, we may infer from it that, in general, three f actors determine the v alue of the capacitance:

• 1. The surface area of the plates—the lar ger the area, the greater the capacitance.
• 2. The spacing between the plates—the smaller the spacing, the greater the capacitance.

Figure 6.2 A capacitor with applied voltage v.

Alternatively, capacitance is the amount of charge stored per plate for a unit voltage difference in a capacitor.

Capacitor voltage rating and capacitance are typically inversely rated due to the relationships in Eqs. (6.1) and (6.2). Arcing occurs if d is small and V is high.

Figure 6.3

Circuit symbols for capacitors: (a) fixed capacitor, (b) variable capacitor.

3. The permitti vity of the material—the higher the permitti vity, the greater the capacitance.

Capacitors are commercially a vailable in different values and types. Typically, capacitors have values in the picofarad (pF) to microfarad (μF) range. They are described by the dielectric material the y are made of and by whether they are of fixed or variable type. Figure 6.3 shows the circuit symbols for fixed and variable capacitors. Note that according to the passive sign convention, if v > 0 and i > 0 or if v < 0 and i < 0, the capacitor is being charged, and if v · i < 0, the capacitor is discharging.

Figure 6.4

Figure 6.5 Variable capacitors: (a) trimmer capacitor, (b) filmtrim capacitor. Courtesy of Johanson.

According to Eq. (6.4), for a capacitor to carry current, its voltage must vary with time. Hence, for constant voltage, ⁱ = 0.

Figure 6.4 shows common types of fixed-value capacitors. Polyester capacitors are light in weight, stable, and their change with tempera ture is predictable. Instead of polyester, other dielectric materials such as mica and polystyrene may be used. Film capacitors are rolled and housed in metal or plastic films. Electrolytic capacitors produce very high capacitance. Figure 6.5 shows the most common types of variable capacitors. The capacitance of a trimmer (or padder) capacitor is often placed in parallel with another capacitor so that the equi valent capacitance can be varied slightly. The capacitance of the variable air capacitor (meshed plates) is varied by turning the shaft. Variable capacitors are used in radio receivers allowing one to tune to various stations. In addition, capacitors are used to block dc, pass ac, shift phase, store energy, start motors, and suppress noise.

To obtain the current-voltage relationship of the capacitor , we take the derivative of both sides of Eq. (6.1). Since

$$i = \frac{dq}{dt} \tag{6.3}$$

differentiating both sides of Eq. (6.1) gives

$$i = C \frac{dv}{dt}$$

(6.4)

This is the current-v oltage relationship for a capacitor , assuming the passive sign convention. The relationship is illustrated in Fig. 6.6 for a capacitor whose capacitance is independent of v oltage. Capacitors that satisfy Eq. (6.4) are said to be linear. For a nonlinear capacitor, the plot of the current-voltage relationship is not a straight line. Although some capacitors are nonlinear, most are linear. We will assume linear capaci tors in this book.

The v oltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (6.4). We get

$$v(t) = \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau$$

(6.5)

or

$$v(t) = \frac{1}{C} \int_{t_0}^t i(\tau) \, d\tau + v(t_0)$$

\n(6.6)

where v(t0) = q(t0)∕C is the voltage across the capacitor at time t0. Equation (6.6) shows that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory—a property that is often exploited.

The instantaneous power delivered to the capacitor is

$$p = vi = Cv \frac{dv}{dt}$$

(6.7)

The energy stored in the capacitor is therefore

$$w = \int_{-\infty}^{t} p(\tau) d\tau = C \int_{-\infty}^{t} v \frac{dv}{d\tau} d\tau = C \int_{v(-\infty)}^{v(t)} v dv = \frac{1}{2} C v^{2} \Big|_{v(-\infty)}^{v(t)} \quad (6.8)$$

We note that v(−∞) = 0, because the capacitor was uncharged at t = −∞. Thus,

$$w = \frac{1}{2} C v^2 \tag{6.9}$$

Using Eq. (6.1), we may rewrite Eq. (6.9) as

$$w = \frac{q^2}{2C} \tag{6.10}$$

Equation (6.9) or (6.10) represents the energy stored in the electric field that exists between the plates of the capacitor. This energy can be retrieved, since an ideal capacitor cannot dissipate energy. In fact, the word capacitor is derived from this element’s capacity to store energy in an electric field.

We should note the following important properties of a capacitor:

1. Note from Eq. (6.4) that when the v oltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus,

A capacitor is an open circuit to dc.

However, if a battery (dc voltage) is connected across a capacitor, the capacitor charges.

An alternative way of looking at this is using Eq. (6.9), which indicates that energy is proportional to voltage squared. Since injecting or extracting energy can only be done over some finite time, voltage cannot change instantaneously across a capacitor.

Figure 6.6 Current-voltage relationship of a capacitor.

Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not possible.

Figure 6.8 Circuit model of a nonideal capacitor.

Example 6.1

2. The voltage on the capacitor must be continuous.

The voltage on a capacitor cannot change abruptly.

The capacitor resists an abrupt change in the v oltage across it. According to Eq. (6.4), a discontinuous change in v oltage requires an infinite current, which is physically impossible. For example, the voltage across a capacitor may tak e the form shown in Fig. 6.7(a), whereas it is not physically possible for the capacitor voltage to take the form shown in Fig. 6.7(b) because of the abrupt changes. Con versely, the current through a capacitor can change instantaneously.

• 3. The ideal capacitor does not dissipate ener gy. It tak es power from the circuit when storing ener gy in its field and returns previously stored energy when delivering power to the circuit.
• 4. A real, nonideal capacitor has a parallel-model leakage resistance, as shown in Fig. 6.8. The leakage resistance may be as high as 100 MΩ and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book.

(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor.

Solution:

(a) Since q = Cv,

$$q = 3 \times 10^{-12} \times 20 = 60 \text{ pC}$$

(b) The energy stored is

$$w = \frac{1}{2} Cv^2 = \frac{1}{2} \times 3 \times 10^{-12} \times 400 = 600 \text{ pJ}$$

What is the voltage across a 4.5- μF capacitor if the charge on one plate is 0.12 mC? How much energy is stored? Practice Problem 6.1

Answer: 26.67 V, 1.6 mJ.

Example 6.2 The voltage across a 5-μF capacitor is

v(t) = 10 cos 6000t V

Calculate the current through it.

Solution:

By definition, the current is

$$i(t) = C \frac{dv}{dt} = 5 \times 10^{-6} \frac{d}{dt} (10 \cos 6000t)$$

= -5 × 10^-6 × 6000 × 10 sin 6000t = -0.3 sin 6000t A

If a 10-μF capacitor is connected to a voltage source with

v(t) = 20 cos(200t) V

determine the current through the capacitor.

Answer: −40 sin (200t) mA.

Determine the voltage across a 2-μF capacitor if the current through it is Example 6.3

$$i(t) = 6e^{-3000t} \text{ mA}$$

Assume that the initial capacitor voltage is zero.

Solution:

Since

$$v = \frac{1}{C} \int_0^t i \, d\tau + v(0)$$

and $v(0) = 0$ ,
\n

$$v = \frac{1}{2 \times 10^{-6}} \int_0^t 6e^{-3000\tau} \, d\tau \cdot 10^{-3}$$

\n

$$= \frac{3 \times 10^3}{-3000} e^{-3000\tau} \Big|_0^t = (1 - e^{-3000t}) \text{ V}$$

The current through a 100-μF capacitor is i(t) = 50 sin 120πt mA. Calculate the voltage across it at t = 1 ms and t = 5 ms. Take v(0) = 0.

Answer: 93.14 mV, 1.736 V.

Determine the current through a 200- *^μ*F capacitor whose voltage is Example 6.4 shown in Fig. 6.9.

Solution:

The voltage waveform can be described mathematically as

$$v(t) = \begin{cases} 50t \text{ V} & 0 < t < 1 \\ 100 - 50t \text{ V} & 1 < t < 3 \\ -200 + 50t \text{ V} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}$$

Since i = C dv∕dt and C = 200 μF, we take the derivative of v to obtain

$$i(t) = 200 \times 10^{-6} \times \begin{cases} 50 & 0 < t < 1 \\ -50 & 1 < t < 3 \\ 50 & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}$$ $$= \begin{cases} 10 \text{ mA} & 0 < t < 1 \\ -10 \text{ mA} & 1 < t < 3 \\ 10 \text{ mA} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases}$$

Thus, the current waveform is as shown in Fig. 6.10.

Practice Problem 6.3

Figure 6.9 For Example 6.4.

Practice Problem 6.2

For Practice Prob. 6.4.

An initially uncharged 1-mF capacitor has the current shown in Fig. 6.11 across it. Calculate the voltage across it at t = 2 ms and t = 5 ms.

Answer: 100 mV, 400 mV.

Example 6.5 Obtain the ener gy stored in each capacitor in Fig. 6.12(a) under dc conditions.

Solution:

Under dc conditions, we replace each capacitor with an open circuit, as shown in Fig. 6.12(b). The current through the series combination of the 2-kΩ and 4-kΩ resistors is obtained by current division as

$$i = \frac{3}{3 + 2 + 4}(6 \text{ mA}) = 2 \text{ mA}$$

Hence, the voltages v1 and v2 across the capacitors are

$$v_1 = 2000i = 4 \text{ V} \qquad v_2 = 4000i = 8 \text{ V}$$

and the energies stored in them are

$$w_1 = \frac{1}{2}C_1v_1^2 = \frac{1}{2}(2 \times 10^{-3})(4)^2 = 16 \text{ mJ}$$ $$w_2 = \frac{1}{2}C_2v_2^2 = \frac{1}{2}(4 \times 10^{-3})(8)^2 = 128 \text{ mJ}$$

Figure 6.12

For Example 6.5.

Practice Problem 6.5

Figure 6.13 For Practice Prob. 6.5.

Under dc conditions, find the energy stored in the capacitors in Fig. 6.13.

Answer: 20.25 mJ, 3.375 mJ.