Tne Capacitor 281 61 The Capacitor A capacitor is a dynamic element involving the time variation of an electric field produced by a voltage. Figure 6-1(a) shows the parallel plate capacitor, which is the simplest physical form of a capacitive device. Figure 6-1 also shows two alternative circuit symbols. Photos of actual devices are shown in Figure 6-2. Note: Capacitor manufacturers typically do not use nanofarads and prefer to rate their capacitors as fractions of microfarads or multiples of picofarads. Some standard values for commercially available capacitors are found on the inside rear cover. Electrostatics shows that a uniform electric ficld €(1) exists between the metal plates in Figure 6-1(a) when a voltage exists across the capacitor.” The electric field produces charge separation with equal and opposite charges appearing on the capacitor plates. When the separation d is small compared with the dimension of the plates, the electric field between the plates is ey - er A where ¢ is the permittivity of the dielectric, A is the area of the plates, and q(t) is the magnitude of the electric charge on each plate. The relationship between the electric field and the voltage across the capacitor vc(t) is given by en - a Substituting Bq, (6-2) into Eq, (61) and solving for the charge q(t) yields at = SJocta oa ‘The proportionality constant inside the brackets in this equation is the capacitance C of the capacitor. That is, by definition, ca (4) cA a ‘The unit of capacitance is the farad (F), a term that honors the British physicist Michael Faraday. Values of capacitance range from a few pF (10 F) in semi- conductor devices to tens of mF (10~*F) in industrial capacitor banks. Using Eq (6-4), the defining relationship for the capacitor becomes a(t) = Cve(t) 5) Figure 6-3(a) graphically displays the element constraint in Eq, (6-5). The graph points out that the capacitor is a linear element since the defining relationship between voltage and charge is a straight line through the origin. I-V RELATIONSHIP To express the element constraint in terms of voltage and current, we differentiate Eq, (6-5) with respect to time 1 dats) _ alCrel) dr dt An electric feld i a vector quantity. In Figure 6-1(a) the Held is confined to the space between the two plates and is perpendicular to the plates, “Dielectric [Ra=[~S Meta plates @ © FIGURE 6-1 The capacitor: (a) Parallel plate device. (b) Circuit symbols. @ » (@) ) FIGURE 6-2 Photos of real ‘capacitors: (a) Ceramic: (b) Electrolytic. (c) Air-tunable (@) Film capacitors. (e) High- voltage. (f) Trim capacitors282 CHAPTER 6 CAPACITANCE AND INDUCTANGE FIGURE 6-2 (Continued) co) vel) i ) FIGURE 6-3 (a) Graph of the defining relationship of a linear capacitor. (6) Circuit symbol showing capacitor voltage and current Since C is constant and ic(t) is the time derivative of q(t), we obtain a capacitor Hv relationship in the form duel?) ie( = cD 5 (66) ‘The relationship assumes that the reference marks for the current and voltage follow the passive sign convention shown in Figure 6-3(b) The time derivative in Eq, (6-6) means the current is zero when the voltage across the capacitor is constant, and vice versa. In other words, the capacitor acts like an open circuit (jc = 0) when dc excitations are applied. The capacitor is a dynamic clement because the current is zero unless the voltage is changing. However, a discontinuous change in voltage would require an infinite current, which is physically impossible. Therefore, the capacitor voltage must be a continuous function of time. Equation (6) relates the capacitor current to the rate of change of the capacitor voltage. To express the voltage in terms of the current, we multiply both sides of Eq (6-6) by df, solve for the differential duc, and integrate: [ae - B [ieee Selecting the integration limits requires some discussion. We assume that at some time f the voltage across the capacitor uc(to) is known and we want to determine the voltage at some later time 1 > fp. Therefore, the integration limits are [eos [rete where x is a dummy integration variable. Integrating the left side of this equation yields el) = 00) +5 [elec 7) In practice, the time fo is established by a physical event such as closing a switch or the start of a particular clock pulse. Nothing is lost in the integration in Eq, (6-7) if we arbitrarily define fo to be zero, Using fo = 0 in Eq, (6-7) yields Lp vele) = ve(0) +2 [ eto)as ws Equation (6-8 isthe integral form of the capacitor i-» constraint. Both the integral form and the derivative form in Eq. (6-6) assume that the reference marks for current and voltage follow the passive sign convention in Figure 6-3(b). PoWER AND ENERGY ‘With the passive sign convention the capacitor power is elt) = ie()ve() wa ‘Using Eq, (6-6) to eliminate io(t) from Eq. (6-9) yields the capacitor power in the form devel) aT (6-10) pe(t) = Cvel?)Tue Caracitor 283 This equation shows that the power can be either positive or negative because the capacitor voltage and its time rate of change can have opposite signs. With the passive sign convention, a positive sign means the element absorbs power, while a negative sign means the clement delivers power. The ability to deliver power implies that the capacitor can store energy. To determine the stored energy, we note that the expression for power in Eq. (6-10) is a perfect derivative. Since power is the time rate of change of energy, the quantity inside the brackets must be the energy stored in the capacitor. Mathemati- cally, we can infer from Eq, (6-10) that the energy at time t is, welt) = zouk) + constant The constant in this equation is the value of stored energy at some instant £ when sic(*) =0. At such an instant the electri field is zero; hence the stored energy is also zero. As a result, the constant is zero and we write the capacitor energy as welt) = Fondly) wm The stored energy is never negative, since it is proportional to the square of the voltage. The capacitor absorbs power from the circuit when storing energy and returns previously stored energy when delivering power to the circuit The relationship in Eq. (6-11) also implies that voltage is a continuous function of time, since an abrupt change in the voltage implies a discontinuous change in energy. Since power is the time derivative of energy, a discontinuous change in energy implies infinite power, which is physically impossible. The capacitor voltage is called a state variable because it determines the energy state of the element. To summarize, the capacitor is a dynamic circuit element with the following properties: 1. The current through the capacitor is zero unless the voltage is changing. The capacitor acts like an open circuit to de excitations. 2. The voltage across the capacitor is a continuous function of time. A discontinuous change in capacitor voltage would require infimite current and power, which is physically impossible. 3. The capacitor absorbs power from the circuit when storing energy and returns previously stored energy when delivering power. The net energy transfer is nonnegative, indicating that the capacitor is a passive element, The following examples illustrate these properties. EXAMPLE 6-1 ‘The voltage in Figure 6—4(a) appears across a 14-yF capacitor. Find the current through the capacitor. SOLUTION: The capacitor current is proportional to the time rate of change of the voltage. For 0 3 ms, the voltage is constant, so its slope is zero; hence the current is zero. The resulting current waveform is shown in Figure 6~4(b). Note that the voltage across the capacitor (the state variable) is continuous, but the capacitor current can be, and in this case is, discontinuous. . Exercise 6-1 ie( (A) ALF capacitor has no voltage across it atr given asic(®) =2u(t) ~ 3u(t—2)-+ u(t 4) WA. A current flowing through the capacitor is the voltage across the capacitor atr— 4s. Answer: vel) =2V. EXAMPLE 6-2 o 1 2 3 4 5 Te The io(t) in Figure 6-5(a) is given by vel) (¥) ict) = ole VT ua Find the voltage across the capacitor if vc(0) = OV. SOLUTION: Using the capacitor i-v relationship in integral form, welt) = ve(0)+& f ieee o) 1 ft) witegye — Te, 048 [ite dx He wiTey) FIGURE 6-5 ety ‘The graph in Figure 6-5(b) shows that the voltage is continuous while the current is discontinuous . Exercise 6-2 (a) The voltage across a 10-uF capacitor is 25jsin 2000f|u(1)V, Derive an expression for the current through the capacitor. (b) At 1=0 the voltage across a 100-pF capacitor is —SV. The current through the capacitor is 10|u(0) ~ u(t ~ 10] waA. What isthe voltage across the capacitor for > 0? Answers (a) c(t) = 0.5{c0s2000%|u(t) A (b) scl#) = 5+ 10V for 00.1 ms Exercise 6-3 For 120 the voltage across a 200-pF capacitor is Se. (a) What is the charge on the capacitor at r= 0 and ¢= | 20? (b) Derive an expression for the current through the capacitor for t>0, (c) For 10 is the device absorbing or delivering power? Answers (a) InCandoc. (b) ic(t) = ~4e WA (6) DeliveringTue Caracitor 285 EXAMPLE 6-3 Figure 6-6(a) shows the voltage across a 0.5-4F capacitor. Find the capacitor’s energy and power. vel (V) fc() (2A) FIGURE 6-6 w}- 25 5 of 1 o1 [2 fsa s 1 (ms) oft. i i 25 fesse o12s 45 (03) @ tO) wel (4) i. Stored energy / Z i 1 1 1 a o12a 3s 4 8 @ rms) SOLUTION: ‘The current through the capacitor was found in Example 6-1. The power waveform is the point-by-point product of the voltage and current waveforms. The energy is found either by integrating the power waveform or by calculating "C|vc(1)|* point by point. The current, power, and energy are shown in Figures 6-6(b), 6-6(c), and 6-6(d). Note that the capacitor energy increases wheniit is absorbing power [pe(t) > 0] and decreases when delivering power [pe(t) < 0]. . Exercise 6-4 Find the power and energy for the capacitors in Exercise 6-2. Answers: (a) pelt) = 6.25|sind000rlu(e) W vwe(#) = 3.125sin# 2000 mI () pelt) = 0.05 + 10% mW ford << 0.1 ms pelt) =0 for # > 0.1 ms Well) = 1.25 —5 x 104 45 x 10% al for 0.< 1 < 0.1 ms L2Snfforr > 0.1me286 CHAPTER 6 CAPACITANCE AND INDUCTANGE (0 (A) ° ia 16) © co etd OW) Pel cr TK. wel 0) were 3 ° 16) 0 @ FIGURE 6-7 EXAMPLE 6-4 The current through a capacitor is given by ie(t) = hole! |u() A Find the capacitor’s energy and power. SOLUTION: ‘The current and voltage were found in Example 6-2 and are shown in Figures 6-7(a) and 6-1(b). The power waveform is found as the product of current and voltage: Pelt) = ie(ve() pte [ll ur = loo 7) [EF —@ V7) Bley, te _ ¢ mite “cee ‘The waveform of the power is shown in Figure 6-7(c). The energy is Loy —(eTel gute elt) =Aevacy = LOT! ev teye welt) = 5 Crele) =F y ‘The time history of the energy is shown in Figure 6-7(d). In this example, both power and energy are always positive. . Exercise 6-5 Find the power and energy for the capacitor in Exercise 63. Answers P(t) = —20e-° pW welt) = 2.52% ny Pen EXAMPLE 6-5 A sample-and-hold circuit is usually found at the input to an analog-to-digital converter (ADC). The purpose of the circuit is to sample a time-varying input waveform at a specified instant and then hold that value constant until conversion to digital form is complete. This example discusses the role of a capacitor in such a circuit. The basic sample-and-hold circuit in Figure 6-8(a) includes an input buffer, a digitally controlled electronic switch, a holding capacitor, and an output buffer. The input buffer is a voltage follower whose output replicates the analog input vs(t) and supplies charging current to the capacitor. The output buffer is also a voltage follower whose output replicates the capacitor voltage, To see how the circuit operates, we describe one cycle of the sample-and-hold process. At time ty shown in Figure 6-8(b), the digital control vg(t) goes high, which causes the switch to close. Thereafter, the input buffer supplies a charging current ic(t) to drive the capacitor voltage to the level of the analog input. At time f2 shown in Figure 6-8(b) the digital control goes low, the switch opens, and thereafter the capacitor current ic(t) = 0. Zero current means that capacitor voltage is constant since duc dt is zero. In sum, closing the switch causes the capacitor voltage to track the input and opening the switch causes the capacitor voltage to hold a sample of the input, Figure 6-8(b) shows several more cycles of the sample-and-hold process, Samples of the input waveform are acquired during the time intervals labeled tc. During these intervals the control signal is high, the switch is closed, and the capacitor charges or discharges in order to track the analog input voltage. Analog- to-digital conversion of the circuit output voltage takes place during the timeTue Inouctor 287 intervals tapc. During these intervals the control signal is low, the switch is open, and the capacitor holds the output voltage constant, Sample-and-hold circuits are available as monolithic integrated circuits (see Figure 6-8(c)) that include the two buffers, the electronic switch, but not the holding capacitor. The capacitor is supplied externally, and its selection involves a trade-off. In an ideal sample-and-hold circuit, the capacitor voltage tracks the input when the switch is closed (sample mode) and holds the value indefinitely when the switch is open (hold mode). In real circuits the input buffer has a maximum output current, which means that some time is needed to charge the capacitor in the sample mode, Minimizing this sample acquisition time argues for a small capacitor. On the other hand, in the hold mode the output buffer draws a small current which gradually discharges the capacitor, causing the output voltage to slowly decrease. Minimizing this output droop calls for a large capacitor. Thus, selecting the capacitance of the holding capacitor involves a compromise between the sample acquisition time and the output voltage droop in the hold mode. v9 vo vel8)= vol) TT 1 oe i vs) eo vl) @) © FIGURE 6-8 6-2 THe Inpuctor The inductor is a dynamic circuit element involving the time variation of the magnetic field produced by a current. Magnetostatics shows that a magnetic flux surrounds a wire carrying an electric current. When the wire is wound into a coil, the lines of flux concentrate along the axis of the coil, as shown in Figure 6-9(a), In a linear magnetic medium, the flux is proportional to both the current and the number of turns in the coil. Therefore, the total flux is b(t) = kiN. (0) (19 where ky is a constant of proportionality. The magnetic flux intercepts or links the turns of the coil. The flux linkage in a coil is represented by the symbol \, with units of webers (Wb), named after the German scientist Wilhelm Weber (1804~1891). The flux linkage is proportional to the number of turns in the coil and to the total magnetic flux, so A(t) is Mo) = NO) (13) Substituting Eq, (6-12) into Eq. (6-13) gives X(t) = [ead hin(e) (14) ‘The proportionality constant inside the brackets in this equation is the inductance 1. of the coil. That is, by definition L= kN (6-15) ‘The unit of inductance is the henry (H) (plural: henrys), a name that honors American scientist Joseph Henry. Figure 6-9(b) shows the circuit symbol for an inductor. Photos of some examples of actual devices are shown in Figure 6-10. Some standard values for commercially available inductors are found in the inside rear cover. © A nu GL (b) FIGURE 6-9 (a) Magnetic ‘flux surrounding a current- carrying coil. (b) Cirewit symbol showing inductor current and voltage.288 CHAPTER 6 CAPACITANCE AND INDUCTANGE FIGURE 6-11. Graph of i) W \ the dfing relationship of 4,6) \ linear inductor. 1 cL A) @ —_~en— Using Eq, (6-15), the defining relationship for the inductor becomes © MA) = Liv) (616) Figure 6-11 graphically displays the inductor’ element constraint in Eq, (6-16). The graph points out that the inductor is a linear element since the defining relationship is a straight line through the origin. T-V RELATIONSHIP Equation (6-16) is the inductor element constraint in terms of current and flux linkage, To obtain the element characteristic in terms of voltage and current, we differentiate Eq. (6-16) with respect to time: ‘blerining com © ano) _ ditin(e) ee vo - © ‘The inductance L is a constant. According to Faraday’s law, the voltage across the G inductor is equal to the time rate of change of flux linkage, Therefore, we obtain an inductor i-v relationship in the form ti.(0) ad w(t) <1 1) ‘The time derivative in Eq, (6-18) means that the voltage across the inductor is zero unless the current is time varying, Under de excitation the current is constant and i a1, = 0, so the inductor acts like a short circuit. The inductor is a dynamic element because only a changing current produces a nonzero voltage. However, a dis- ‘wor butlerinding.com continuous change in current would produce an infinite voltage, which is physically ©) impossible. Therefore, the current i;,(t) must be a continuous function of time t. ‘7 Equation (6-18) relates the inductor voltage to the rate of change of the inductor current. To express the inductor current in terms of the voltage, we multiply both sides of Eq. (6-18) by dt, solve for the differential di, and integrate: [ao=Ef aoe 619) ‘To set the limits of integration, we assume that the inductor current i (t) is known at woreuilerwindingcom _sotte time fp. Under this assumption the integration limits are o a0 te FIGURE 6-10. Beamplesof [0-7 [ore m9 real inductors: (a) Toroidal. (b) “Axial. (c) Choke. (d) Surface Where xis a dummy integration variable. The left side of Eq, (6-20) integrates to produce mount. (@) Air Col) Bobbin ry Wound ” i) = lt) +F [moras wa‘Tue Inpucror 289 The reference time f is established by some physical event, such as closing or opening a switch. Without losing any generality, we can assume fo = 0 and write Eq, (6-21) in the form ip iu) = (0) + if waar em Equation (6-22) is the integral form of the inductor i-v characteristic. Both the integral form and the derivative form in Eq, (6-18) assume that the reference marks for the inductor voltage and current follow the passive sign convention shown in Figure 6-9(b) PoWER AND ENERGY With the passive sign convention the inductor power is Puls) = iets) 2) Using Eq, (6-18) to eliminate »,() from this equation puts the inductor power in the form, diy) _ a 1a : = 5 ijt | om put) = [iv()] pelt) = lin) | LF a This expression shows that power can be positive or negative because the inductor current and its time derivative can have opposite signs. Therefore, like a capacitor, an inductor can both absorb and deliver power. The ability to deliver power indicates that the inductor can store energy To find the stored energy, we note that the power relation in Eq. (6-24) is a perfect derivative. Since power is the time rate of change of energy, the quantity inside the brackets must represent the energy stored in the magnetic field of the inductor, From Eq, (6-24), we infer that the energy at time f is wilt) = hit + constant Asis the case with capacitor energy, the constant in this expression is zero since it is, the energy stored at an instant ¢ at which i,.(t) = 0. As a result, the energy stored in the inductor is wl) = $n) 2) The energy stored in an inductor is never negative because it is proportional to the square of the current. The inductor stores energy when absorbing power and returns previously stored energy when delivering power, so that the net energy transfer is never negative. Equation (6-25) implies that inductor current is a continuous function of time because an abrupt change in current causes a discontinuity in the energy Since power is the time derivative of energy, an energy discontinuity implies infinite power, which is physically impossible. Current is called the state variable of the inductor because it determines the energy state of the element. Insummary, the inductor is a dynamic circuit element with the following properties 1 rhe voltage across the inductor is zero unless the current through the inductoris changing. The inductor acts like a short circuit for de excitations. 2. The current through the inductor is a continuous function of time. A discontinuous change in inductor current would require infinite voltage ‘and power, which is physically impossible.290 CHAPTER 6 CAPACITANCE AND INDUCTANGE HA) 5 s ow 2 — FIGURE 6-1 (b) Voltage. p (es) aS ic) U o 12 (a) Current 3) (A) w sk 7. o o os ois we (VD i oa wos ” os to as FIGURE 6-13 3. The inductor absorbs power from the circuit when storing energy and delivers power to the circuit when returning previously stored energy. The net energy is, nonnegative, indicating that the inductor is a passive element. EXAMPLE 6-6 ‘The current through a 2-mH inductor is i,(f) = 4sin 1000¢ + 1 sin 3000" A. Find the resulting inductor voltage. SOLUTION: ‘The voltage is found from the derivative form of the i-» relationship: dit) dt u(t) = = 0.002[4 x 1000 cos 10001 +1 x 3000 cos 30007) = 8cos 10001 + 6 cos 3000 V ‘The current and voltage waveforms are shown in Figure 6-12. Note that the current and voltage each contain two sinusoids at different frequencies. However, the relative amplitudes of the two sinusoids are different. In i,() the ratio of the amplitude of the ac component at w = 3 krad/s to the ac component at w = 1 kradis is I-to-4, whereas in v(t) this ratio is 310-4, The fact that the ac responses of energy storage elements depend on frequency allows us to create frequency-selective signal processors called filters. Ml Exercise 6—6 — For 1 > 0, the voltage across a 4-mH inductor is u(t) = 20e"2" V. The initial current is i,(0) 0 (a) What is the current through the inductor for ¢ > 07 (b) What is the power for t > 0? (©) What is the energy for t > 0? Answers (a) i) =25(1 0 (b) Find the time t > 0 at which the inductor voltage passes through zero (©) Derive an expression for the inductor power for f > 0. (4) Find the time interval over which the inductor absorbs power and the interval over which it delivers power. Answers (a) v(t) = der 20 4 Berton (b) 1=0347ms (©) pu{t) = ~B0e8 + 240¢" 00" — 1606-18 mW (@) Absorbing for 0 < 1 < 0.347 ms, delivering for 1 > 0.347 ms EXAMPLE 6-8 The current through a 2.5-mH inductor is a damped sine (t) = 10¢ ° sin 2000r Plot the waveforms of the element current, voltage, power, and energy. SOLUTION: Following are the MATLAB code and the resulting plots for the current, voltage, power, and energy. The code uses symbolic variables, differentiation, and multipli- cation to calculate the expressions for the four signals. An appropriate time vector is then substituted into the symbolic expressions to create numerical vectors that can 30 be plotted. The subp1ot function allows multiple plots to be placed in a single figure window. In the plots shown in Figure 6-14, note that the current, voltage, and power alternate signs, whereas the energy signal is always positive. The MATLAB code uses a numerical time vector defined by the expression (tt 0:0.00001:0.004;1. The sinusoid has a period of Ty = 2/2000 = 0.00314, so plotting from t= 0 ms to f = 4 ms is sufficient to capture a full period of the sinusoid. 9, The time step size of 10 ps is a convenient choice and generated 401 points to plot, which produced a smooth curve for cach signal. 7 Volage MATLAB Code % Create symbolic variables syms t iL vL pL wL real % Create the inductor current signal IL = 10*exp(-500*t) #sin¢2000*t) ; ol} % Define the inductance Tine.) aps L=0,0025; © Eoerey % Calculate the voltage, power, and energy 0s vis LeaifeGl, "9; ss pL =iL*vi; 5 WL = L*ALA2/25 Sos % Create a time vector # oo tt = 0:0.00001:0.004; % Substitute the time vector into the symbolic expressions or iLtt = subs(iL, t, tt); Be aos « vite pLet subs(vL, t, tt subs(pL, t, tt); FIGURE 6-14292 CHAPTER 6 Ao.) Votage 0.09) THs 7H ® Tine 0) © x10 Boeuey TAA roy Tine) @ FIGURE 6-15 CAPACITANCE AND INDUCTANGE wLtt = subs(wh, t, tt); %Plot the results using the subplot function subplot (2,2,1) plot(tt,iltt, "b’, 'Linewidth',2); gridon ylabel(’i_L¢(t), (A)") title(' Current’) subplot (2,2,2) plot(tt,vLtt,'b’,’LineWidth’ ,2); gridon ylabel(’v_L(t), (V)") itle(' Voltage’) subplot (2,2,3) plot(tt,pLtt gridon xlabel(’Time, (s)')i ylabel('p_L¢t), (W)") title(' Power’) subplot (2,2,4) plot(tt,wLtt,'b’,’LineWidth’ ,2); gridon xlabel(’Time, (s)') ylabel('wLct), (J)") vitle(' Energy’) * "LineWiath’ 2); Exercise 6-8 A 50-mHT inductor has an initial current of i,(0)=0 A. The following voltage is applied across the inductor starting at *=0: 5 a) = ssn) ~ St) — 200) Y For 1 > 0, use MATLAB to determine the inductor current, power, and energy. Plot those three signals and the inductor voltage for 0 Node Resistance <> —- Conductance Voltage source + Current source Thevenin Norton Shorteireuit + Opencireuit Series > Parallel Capacitance Inductance Flux linkage Charge