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Chapter 10

Capacitors & Capacitance

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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.

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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-nF 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.

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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]

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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

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Example 10-3, [page 389]

A parallel-plate capacitor with air dielectric has a value of C = 12 pF. 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 pF) = 60 pF.

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)(12pF) = 240 pF.

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 pF) = 79.2 pF.

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Quiz

For a given capacitor, it stores Q1 charges when voltage V1 applied across its two ends. It will store Q2 charges when voltage V2 applied across its two ends.

If V1 = 10V and V2 = 5V, Q1 is greater than Q2 by 2C. Find the a capacitance of the capacitor.

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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

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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 te

st charge, Qt

= F / Qt unit: newtons / coulomb (N/C)

F = Q Qt / (4 r2) for a point charge, Q

Hence = Q / (4 r2)

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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]

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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 = F d

Define voltage, V = W / Qt for a test charge Qt

= F d / Qt

= V / d as = F / Qt

i.e. electric field strength between capacitor plates

= voltage across the plates divided by the distance between them.

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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 = 80V/50x103V/m

= 1.6x10-3 m

b. Since = V/d, will double to 100 000 V/m.

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10.4 Dielectrics (p. 393)

Voltage breakdown

• If voltage applied across capacitor is increased beyond a critical value, force on electronics is so great that 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 = 3kV / mm

• Table 10-2, [page 393]

MATERIAL kV/mm

Air 3

Ceramic (high r) 3

Mica 40

Mylar 16

Oil 15

Polystyrene 24

Rubber 18

Teflon 60

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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 2.5cm and a ceramic dielectric with r = 7500 experiences breakdown at 2400 V. What is C?

Solution From Table 10-2 dielectric strength = 3 kV/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) (0.025m)2/(8x10-4m)

= 51.9 nF

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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]

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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 = RS / XC where XC = 1/2fC

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)

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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.

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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 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

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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

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10.7 Capacitors in parallel and series

• capacitors in parallel [fig. 10-20, page 400]

. Voltage is the same across each.

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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

a. Cr = C1 + C2 + c3 = 10F + 15 F + 100 F = 125 F

b. QT = CTV = (125 F)(50 V) = 6.25 mC

c. Q1 = C1V = (10 . Q1 = C1V = (10 F)(50 V) = 0.5 mC)(50 V) = 0.5 mC

Q2 = C2V = (15 F)(50 V) = 0.75 mC

Q3 = C3V = (100 F)(50 V) = 5.0 mC

Note Q1 + Q2 + Q3 = (0.5 + 0.75 + 5.0) mC = 6.25 mC, which checks with (b).

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Capacitors in series [Fig. 10-21, page 401]

• Charge is the same on each.

NCCCrC1...

2

1

1

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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/C1 + 1/C2 + 1/C3 = 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 x10-6F)(50 V) = 0.5 mC

c. V1 = Q/C1= (0.5 x 10-3 C) / (30 x 10-6F) = 16.7V

V2 = Q/C2 = (0.5x10-3C)/(60x10-6F) = 8.3V

V3 = Q/C3 = (0.5x10-3C)/(20x10-6F) = 25.0V

Check: V1 + V2 + V3 = 16.7 + 8.3 + 25 = 50 V.

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EXAMPLE 10-8 [page 403]

For the circuit of Figure 10-23(a), determine CT

Refer to Figure 10-23: Systematic reduction.

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Quiz: Consider capacitors of 1F, 1.5F, and 10F. If equivalent capacitance is equal to 10.6F. 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 1.5F 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

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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 10V in 1ms, falls linearly to -10V at t=3ms, remains constant until t=4ms, rises to 10V at t=5ms, and remains constant thereafter.

a. Determine the slope of vc.

b. Determine the current and sketch its graph.

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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 10-6F) (10 000 V/s) = 50 mA.

1 ms to 3 ms: i = -(5 x 10-6F) (10 000 V/s) = -50 mA.

3 ms to 4 ms: i = (5 x 10-6F) (0 V/s) = 0 mA.

4 ms to 5 ms: i = (5 x 10-6F) (20 000 V/s) = 100 mA.

Refer to Figure 10-27, page 406

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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 V2

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.)

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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.

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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