Chapter 10 Capacitors Capacitance 1 10 1 Capacitance

Chapter 10 Capacitors & Capacitance 1

10. 1 Capacitance (p. 386) Capacitor • Consists of 2 conductors separated by an insulator. • Basic form: parallel-plate capacitor [Fig. 10 -1, page 386] which consists of two metal plates separated by dielectric (e. g. air). Charging capcitor • When DC source is connected to two metal plates [Fig. 10 -2, page 386], electrons are removed from +ve plate and an equal number of electrons will be deposited on -ve plate. • This leaves top plate positively charges and bottom plate negatively charged. • Capacitor can store charge, i. e. remains charged after DC source is removed. 2

Discharging Capacitor • By shorting a wire/resistor across the two leads. Definition of Capacitance • Charge stored depends on applied voltage Q = CV or C = Q/V unit: Farads, F • C is defined as the capacitance of the capacitor • Example 10 -1, [page 387] a. How much charge is stored on a 10 - F capacitor when it is connected to a 24 -volt source? b. The charge on a 20 -n. F capacitor is 1. 7 C. What is its voltage? Solution a. From Equation 10 -1, Q = CV. Thus, Q = (10 x 10 -6 F)(24 V) = 240 C. b. Rearranging Equation 10 -1, V = Q/C = (1. 7 x 10 -6 C)/(20 x 10 -9 F) = 85 V. 3

10. 2 Factors Affecting Capacitance (p. 387) Effect of area • The more charge put on a capacitor‘s plate for a given voltage, the greater will be its capacitance. • Capacitance is directly proportional to plate area. [Fig. 10 -4, page 387] Effect of Spacing • As space between plated decreases, the force of attraction increases and pulls more electrons from one plate to other. • Capacitance is inversely proportional to plate spacing. Effect of Dielectric • Capacitance varies for different materials [Table 10 -1, page 388]. • This factor is called relative dielectric onstant or relative perrmittivity of a material [Fig. 10 -6, page 388] 4

Capacitance of a Parallel Capacitor (p. 389) C= A/d (F) A = plate area d = spacing = absolute dielectric constant (F/m) For air or vacuum, = o = 8. 85 x 10 -12 F/m Other materials, expressed as the product of the relative dielectric constant, r times o , i. e. = o r 5

Example 10 -3, [page 389] A parallel-plate capacitor with air dielectric has a value of C = 12 p. F. What is the capacitance of a capacitor that has the following: a. The same separation and dielectrric but five times the plate area? b. The same dielectric but four times the area and one fifth the plate spacing? c. A dry paper dielectric, six times the plate area, and twice the plate spacing? Solution a. Since the plate area has increased by a factor of five and everything else remains the same, C increases by a factor of five. Thus, C = 5(12 p. F) = 60 p. F. b. With four times the plate area, C increases by a factor of four. With one fifth the plate spacing, C increases by a factor of five. Thus, C = (4)(5)(12 p. F) = 240 p. F. c. Dry paper increases C by a factor of 2. 2. The increases in plate area increases C by a factor of six. Doubling the plate spacing reduces C by one half. Thus, C = (2. 2)(6)(½)(12 p. F) = 79. 2 p. F. 6

Quiz For a given capacitor, it stores Q 1 charges when voltage V 1 applied across its two ends. It will store Q 2 charges when voltage V 2 applied across its two ends. If V 1 = 10 V and V 2 = 5 V, Q 1 is greater than Q 2 by 2 C. Find the a capacitance of the capacitor. 7

10. 3 Electric Fields (p. 390) Electric flux • Electric fields are force fields that exist in the region surrounding charged bodies. [Fig. 10 -8, page 391] • The direction of the field is defined as the direction of force on a positive charge. It is directed outward from the +ve charge and inward toward the -ve charge. • Field lines never cross and density of lines indicates the strength of the field. Fig. 10 -8: (a) Field about a pair of positive and negative charges (b) Field of parallel plate capacitor 8

Electric field of parallel-plate capacitor is uniform across the gap with some fringing near edges. Electric flux lines are represented by Electric Field Intensity, • Strength of an electric field, i. e. force that the field exerts on a small, +ve test charge, Qt = F / Qt unit: newtons / coulomb (N/C) F = Q Qt / (4 r 2) for a point charge, Q Hence = Q / (4 r 2) 9

Electric Flux Density, D (p. 391) • represent the density fo flux lnes in space • independent of the medium D = or D = Total flux / area = /A • The number of flux lines emanating from a charge is equal to the charge itself, i. e. = Q Figure 10 -9 - In the SI system, total flux equals charge Q. [page 392] 10

Figure 10 -10 - Work moving test charge Qt is force times distance. [page 392] Work required to move the charge against the force (F) through distance (d) W = Fd Define voltage, V = = = F d / Qt V/d W / Qt for a test charge Qt as = F / Qt i. e. electric field strength between capacitor plates = voltage across the plates divided by the distance between them. 11

Capacitance, C = = = Q/V /V A D / d (D / ) (A / d) A/d • EXAMPLE 10 -4, [page 392] Suppose that the electric field intensity between the plates of a capacitor is 50 000 V/m when 80 V is applied a. What is the plate spacing if the dielectric is air? If the dielectric is ceramic? b. What is if the plate spacing is halved? Solution a. = V/ d, independent of dielectric. Thus, d = V/ξ = 80 V/50 x 103 V/m = 1. 6 x 10 -3 m b. Since = V/d, will double to 100 000 V/m. 12

10. 4 Dielectrics (p. 393) Voltage breakdown • If voltage applied across capacitor is increased beyond a critical value, force on electronics is so great they are torn from orbit. • This is called dielectric breakdown. • Electric field intensity at breakdown is called dielectric strength of a material. E. g. air = 3 k. V / mm • Table 10 -2, [page 393] MATERIAL k. V/mm Air 3 Ceramic (high r) 3 Mica 40 Mylar 16 Oil 15 Polystyrene 24 Rubber 18 Teflon 60 13

Voltage rating • because of dielectric breakdown, capacitors are rated for max operating voltage (called working voltage). EXAMPLE 10 -5, [page 394] A capacitor with plate dimensions of 2. 5 cm by 2. 5 cm and a ceramic dielectric with r = 7500 experiences breakdown at 2400 V. What is C? Solution From Table 10 -2 dielectric strength = 3 k. V/mm. Thus, d = 2400 V/3000 V/mm = 0. 8 mm = 8 x 10 -4 m. So C= r o A/d = (7500) (8. 85 x 10 -12) (0. 025 m)2/(8 x 10 -4 m) = 51. 9 n. F 14

10. 5 Nonideal Effect (p. 394) Leakage current • charged capacitor will discharge after disconnected from source • small leakage current will pass through dielectric when capacitor connected to a source. • Charge leaks through the dielectric [Fig. 10 -12, page 394] R = hundreds of Mohm • Figure 10 -12 - Leakage current. [page 394] 15

Equivalent series resistance • resistance (RS) may develop in capacitor’s leads as its internal connections begins to fail • cause problems in high-frequency circuits • dissipation factor D = R S / XC where XC = 1/2 f. C Dielectric absorption • after shorting two leads of a capacitor, a residual voltage remains • Disadv. : upset circuit voltage levels Temperature coefficient • +ve / zero / -ve temperature coefficient imply capacitance increases / no change / decreases with increasing temp. • in parts per million (ppm) 16

10. 6 Types of Capacitors (p. 395) Fixed capacitor [Fig. 10 -13, 10 -14, page 396] Variable capacitor [Fig. 10 -19, page 399] Figure 10 -13: Stacked capacitor construction. The stack is compressed, leads attached, and the unit coated with epoxy resin or other insulting material. 17

Fixed capacitor Ceramic: relative permittivity = 30 to 7500 • extremely high permittivity permit small packaging but characteristics vary wifely with temp. and operating voltage. Use in limited temp applications where small size and cost are important • many surface mount capacitors use ceramic dielectrics Plastic film: • film/foil [Fig. 10 -14, page 396] use metal foil separated by plastic film • metallized-film have foil material vaccum deposited directly onto plastic film Mica: • low in cost with low leakage and good stability 18

Electrolytic • provide large apacitance up to hundred thousand microfarads • relatively low cost • leakage is relatively high and breakdown voltage is relatively low Surface mount: extremely small and provide high packaging density Variable capacitor • used in radio tuning circuits • stationary plates and set of movable plates which are ganged together and mounted on a shaft • as shaft is rotated, the effective area change 19

10. 7 Capacitors in parallel and series • capacitors in parallel [fig. 10 -20, page 400]. Voltage is the same across each. 20

Example 10 -6, [page 401] Example 10 -6 A 10 - F, a 15 - F, and a 100 - F capacitor are connected in parallel across a 50 -V source. Determine the following: a. Total capacitance. b. Total charge stored. c. Charge on each capacitor. Solution m. C a. Cr = C 1 + C 2 + c 3 = 10 F + 15 F + 100 F = 125 F b. QT = CTV = (125 F)(50 V) = 6. 25 m. C c. Q 1 = C 1 V = (10 F)(50 V) = 0. 5 m. C)(50 V) = 0. 5 Q 2 = C 2 V = (15 F)(50 V) = 0. 75 m. C Q 3 = C 3 V = (100 F)(50 V) = 5. 0 m. C Note Q 1 + Q 2 + Q 3 = (0. 5 + 0. 75 + 5. 0) m. C = 6. 25 m. C, which checks with (b). 21

Capacitors in series [Fig. 10 -21, page 401] • Charge is the same on each. 22

EXAMPLE 10 -7 (p. 402) Refer to Figure 10 -22(a): a. Determine Cr. b. If 50 V is applied across the capacitors, determine Q. d. Determine the voltage on each capacitor. Solution a. 1/Cr = 1/C 1 + 1/C 2 + 1/C 3 = 1/30μF + 1/60μF + 1/20μF = 0. 0333 x 106 + 0. 0167 x 106 + 106 = 0. 1 x 106 Therefore, Cr = 1/(0. 1 x 10 -6) = 10 F b. Q = CTV = (10 x 10 -6 F)(50 V) = 0. 5 m. C c. V 1 = Q/C 1= (0. 5 x 10 -3 C) / (30 x 10 -6 F) = 16. 7 V V 2 = Q/C 2 = (0. 5 x 10 -3 C)/(60 x 10 -6 F) = 8. 3 V V 3 = Q/C 3 = (0. 5 x 10 -3 C)/(20 x 10 -6 F) = 25. 0 V Check: V 1 + V 2 + V 3 = 16. 7 + 8. 3 + 25 = 50 V. 23

EXAMPLE 10 -8 [page 403] For the circuit of Figure 10 -23(a), determine CT Refer to Figure 10 -23: Systematic reduction. 24

Quiz: Consider capacitors of 1 F, 1. 5 F, and 10 F. If equivalent capacitance is equal to 10. 6 F. How are the capacitors connected? Solution: To yield 10. 6 F, the 10 F capacitor must be connected in parallel with some combination of the 1 F and 1. 5 F capacitors. Only the connection shown below works. 10. 8 Capacitor current and voltage [page 404] (Refer to Fig. 10 -25, page 405) During charging • movement of electrons constitutes current • current lasts for capacitor to be charged • no current pass through dielectric • capacitor voltage builds as charge deposited on plates • as capacitor voltage increases, charging current decreases 25

Capacitor V-1 relationship, [page 405] q = C VC ic = dq / dt = d (C VC) / dt ic = C d Vc / dt Current through a capacitor is equal to C times the rate of change of voltage across it. Example 10 -9 [page 406] A signal generator applies voltage to a 5 -µF capacitor with a wavefrom as in Figure 10 -27(a). The voltage rises linearly from 0 to 10 V in 1 ms, falls linearly to -10 V at t=3 ms, remains constant until t=4 ms, rises to 10 V at t=5 ms, and remains constant thereafter. a. Determine the slope of vc. b. Determine the current and sketch its graph. 26

Solution a. We need the slope of vc during each time interval where slope = rise/run = v/ t. 0 ms to 1 ms: v = 10 V; t = 1 ms; Therefore, slope = 10 V/1 ms 10 000 V/s. 1 ms to 3 ms: Slope = -20 V/2 ms = -10 000 V/s. 3 ms to 4 ms: Slope = 0 V/s. 4 ms to 5 ms: Slope = 20 V/1 ms = 20 000 V/s. b. ic = Cdvc/dt = C times slope. Thus, 0 ms to 1 ms: i = (5 x 10 -6 F) (10 000 V/s) = 50 m. A. 1 ms to 3 ms: i = -(5 x 10 -6 F) (10 000 V/s) = -50 m. A. 3 ms to 4 ms: i = (5 x 10 -6 F) (0 V/s) = 0 m. A. 4 ms to 5 ms: i = (5 x 10 -6 F) (20 000 V/s) = 100 m. A. Refer to Figure 10 -27, page 406 27

10. 9 Energy stored by a capacitor [page 407] • an ideal capacitor does not dissipate power • when power is transferred to a capacitor, all of it is stored as energy in the capacitor’s electric field Stored energy, W = 1/2 C V 2 10. 10 Capacitor failures and troubleshooting [page 408] Failures: • short internally • leads open • dielectric leaky (Noted: if electrolytic capacitor is connected with its polarity reversed, it may explode. ) 28

Capacitors fail because of • misapplication • excessive voltage, current, or temperature • aging Basic testing with an ohmmeter • out-of-circuit tests with analog ohmmeter - detect opens and shorts - leaky dielectrics (Noted: discharge capacitor first before measurement) • for normal capacitor, the ohmmeter reading should be low initially and gradually increase to infinity. 29

Capacitor testers [page 408] • some digital multimeter can measure capacitance • LCR (inductance, capacitance, resistance) analyzer can determine capacitance as well as detect opens and short 30