chapter 9 capacitors - ecourse2.ccu.edu.tw

Chapter 9 Capacitors

ElectronicCircuits&Electronics Shao-YuLien 1

9.1 The Basic Capacitor

• A capacitor is a passive electrical component that storeselectrical charge and has the property of capacitance

• A capacitor is constructed of two parallel conductive platesseparated by an insulating material called dielectric

• Connecting leads are attached to the parallel plates



• In the neutral state, both plates of a capacitor have an equal number of free electrons

• When the capacitor is connected to a dc voltage source through a resistor, electrons are removed from plate A, and an equal number of deposited on plate B

• Electrons flow only through the connecting leads and the source. No electrons flow through the dielectric of the capacitor

• The movement of electrons ceases when the voltage across the capacitor equals the source voltage

• If the capacitor is disconnected from the source, it retains the stored charge, and still have voltage across it



Definition. The amount of charge that capacitor can store per unit of voltage across its plates is its capacitance, designated as C. That is, capacitance is a measurement of a capacitor’s ability to store charge

𝐶 =𝑄𝑉

– Q: Charge– V: Voltage

Q=CV

𝑉 =𝑄𝐶



• The farad (F) is the basic unit of capacitance.Definition. One farad is the amount of capacitance when one coulomb of charge is stored with one volt across the plates

Example. (a) A certain capacitor stores 50 𝜇C when 10 V are applied across its plates. What is the capacitance?(b) A 2.2 𝜇F capacitor has 100 V across its plates. How much charge does it store?(c) Determine the voltage across a 100 pF capacitor that is storing2 𝜇C of charge.



• (a) C=Q/V=(50 𝜇C)/(10 V)=5 𝜇F• (b) Q=CV=(2.2 𝜇F)(10 V)=220 𝜇C• (c) V=Q/C=(2 𝜇C)/(100 pF)=20 kV



• How a capacitor stores energy?• A capacitor stores energy in an electric field established by the

opposite charges stored on the two plates • The energy stored in the electric field is directly related to the

size of the capacitor and to the square of the voltage

W='(CV2



• Every capacitor has a limit on the amount of voltage that it can withstand across its plates

Definition. The voltage rating is the maximum dc voltage that can be applied without risk of damage to the device

• If this maximum voltage (breakdown voltage or working voltage) is exceeded, permanent damage to the capacitor can result

• The breakdown voltage of a capacitor is determined by the dielectric strength of the dielectric material used



• The temperature coefficient indicates the amount and directionof a change in capacitance value with temperature– Positive temperature coefficient: capacitance increases with an increase

in temperature, or decrease with a decrease in temperature– Negative temperature coefficient: capacitance decrease with an increase

in temperature, or increase with a decrease in temperature

• Temperature coefficients typically are specified in ppm/℃



• No insulating material is perfect• The dielectric of any capacitor will conduct some very small

amount of current• Thus, the charge on a capacitor will eventually leak off

• The parallel resistor Rleak represents the extremely high resistance



• The following parameters are important in establishing the capacitance and the voltage rating of a capacitor

• Plate area: capacitance is directly proportional to the physical size of the plates as determined by the plate area

• Plat separation: capacitance is inversely proportional to the distance between the plates



• Dielectric: capacitance is directly proportional to the dielectric constant– The measure of a material’s ability to establish an electric field is called

the dielectric constant (relative permittivity)– The dielectric constant is a ratio of the absolute permittivity of a

material 𝜀 to the absolute permittivity of a vacuum 𝜀+𝜀, =

𝜀𝜀+

– 𝜀+=8.85×10-12 F/m

• An exact formula for calculating the capacitance is

𝐶 =𝐴𝜀,𝜀+𝑑

– A is in m2, d is in m, and C is in F


9.2 Types of Capacitors

• Fixed capacitors– Mica capacitors – Ceramic capacitors– Plastic-film capacitors– Electrolytic capacitors

• Variable capacitors



• There are two types of mica capacitors: stacked-foil and silver-mica

• Mica capacitors are generally with capacitance values ranging from 1 pF to 0.1 𝜇F, and the voltage ratings from 100 V dc to 2500 V dc.

• The dielectric constant is typically 5



• The ceramic dielectrics provide very high dielectric constant(typically 1200)



• Most plastic-film capacitors have capacitance less than 1 𝜇F



• Electrolytic capacitors are polarized so that one plate is positiveand the other is negative

• These capacitors are generally used for high capacitance valuesfrom 1 𝜇F to 200,000 𝜇F

• They have very low breakdown voltage (< 350 V)




• Adjustable capacitors that normally have slotted screw-type adjustments and are used for very fine adjustments in a circuit are called trimmers

• The varactor is a semiconductive device that exhibits a capacitance characteristic that is varied by changing the voltage across its terminals

9.3 Series Capacitors

Lemma. When two capacitors, with capacitances C1 and C2, are connected in series with a dc voltage source, the total capacitance is

1𝐶1

=1𝐶'+1𝐶(

CT= 343534635

Proof.• Suppose that the two capacitors are initially uncharged• When the switch is closed, the current is the same at all points in a

series circuitI=Q/t



• In a given period of time, the same amount of charges isdeposited on the plates of both capacitors

• The total charges move through the circuit is QT=Q1=Q2ElectronicCircuits&Electronics Shao-YuLien 21


• After two capacitors have been completed charged, the current ceased

• The voltage across each capacitor areV1=Q1/C1

V2=Q2/C2

• By Kirrchhoff’s voltage law,VS=V1+V2

• Since QT=Q1=Q2=Q,𝑄𝐶1

=𝑄𝐶'+𝑄𝐶(



• Therefore, we obtain1𝐶1

=1𝐶'+1𝐶(

• When capacitors are connected in series, the total capacitances is less than the smallest capacitance value, since the effective plate separation increases



Example. Find the total capacitance in the following circuit

CT= 343534635

=('++9:)(<<+9:)'++9:6<<+9:

=76.7 pF



• The formula for total capacitance for any number of capacitors in series is developed as follows

1𝐶1

=1𝐶'+1𝐶(+⋯+

1𝐶>

𝐶1 =1

1𝐶'+ 1𝐶(+⋯+ 1

𝐶>



Example. Determine the total capacitance in the following circuit

𝐶1 ='

4?46 4?56 4?@

= '4

4ABC64

D.FBC64

G.5BC=2.30 𝜇F



Lemma. The voltage across any individual capacitor can be determined by the following formula

𝑉H =𝐶1𝐶H

𝑉I

– Cx: any capacitor in series– Vx: voltage across Cx

• The largest-value capacitor in a series connection will have the smallest voltage across it. The smallest-value capacitor will have the largest voltage across it.



Example. Find the voltage across each capacitors in the following circuit

• 𝐶1 ='

4?46 4?56 4?@

= '4

A.4BC64

A.DFBC64

A.55BC=0.06 𝜇F



• The voltages are as follows

𝑉' =3J34

𝑉I =+.+KL:+.'L:

25𝑉 =15.0 V

𝑉( =3J35

𝑉I =+.+KL:+.OPL:

25𝑉 =3.19 V

𝑉< =3J3@

𝑉I =+.+KL:+.((L:

25𝑉 =6.82 V


9.4 Parallel Capacitors

Lemma. When two capacitors, with capacitances C1 and C2, are connected in parallel with a dc voltage source, the total capacitance is

CT=C1+C2

Proof.• When the switch is closed, a total amount of charges QT move

through the circuit• Part of the total charges are stored by C1 and part by C2



IT=I1+I2

QT/t=Q1/t+Q2/tQT=Q1+Q2



• Since Q=CV,CTVS=C1VS+C2VS

• We therefore obtain CT=C1+C2



Example. What is the total capacitance in the following circuit? What is the voltage across each capacitor?

CT=C1+C2=330 pF+220 pF=550 pF



• The voltage across each capacitor in parallel is equal to thesource voltage

VS=V1=V2=5 V



• The formula for total capacitance for any number of capacitors in parallel is developed as follows

CT=C1+C2+…+Cn



Example. Determine CT in the following circuit

CT=C1+C2+C3+C4+C5+C6

= 0.01 𝜇F+0.022 𝜇F+0.01 𝜇F+0.047 𝜇F+0.022 𝜇F= 0.133 𝜇F


9.5 Capacitors in DC Circuits

• A capacitor will charge when it is connected to a dc voltagesource



• Charging curve with percentages of the final voltage



• When a conductor is connected across a charged capacitor, thecapacitor will discharge



• Discharging curve with percentages of the initial voltage



• Charging of a capacitor

– Recall that V=Q/C and I=Q/t, Q=It– According to the Kirchhoff’s voltage law,

iR+'3 ∫ 𝑖 𝑑𝑡=VS ⇒ UV

UW𝑅𝐶+i=0



• Initial condition:At t=0, vc(0)=0• Solution of the differential equation: i(t)=C1e-t/RC

• Two rules about capacitors in dc circuits– A capacitor appears as a short to an instantaneous change in voltage– A capacitor appears as an open to constant voltage



• The voltage across the resistorvR(t)=i(t)R=Vse-t/RC

• The voltage across the capacitorvc(t)=VS-vR(t)=VS-Vse-t/RC



• Discharging of a capacitor

– According to the Kirchhoff’s voltage law,

iR+'3 ∫ 𝑖 𝑑𝑡=VS ⇒ UV

UW𝑅𝐶+i=0



• Initial condition:At t=0, vc(0)=V0

• Solution of the differential equation: i(t)=C1e-t/RC



• The voltage across the resistorvR(t)=i(t)R=-V0e-t/RC

• The voltage across the capacitorvc(t)=-vR(t)=V0e-t/RC



• The time constant of a series RC circuit determines the rate atwhich the capacitor charges or discharges

Definition. The RC constant is a fixed time interval that equals theproduct of the resistance and the capacitance in a series RCcircuit

𝜏=RC



• The general expressions for either increasing or decreasingexponential curves of the voltage are given in the followingequations for instantaneous values

v=VF+(Vi-VF)e-t/𝜏

i=IF+(Ii-IF)e-t/𝜏

• Charging from zero (Vi=0)v=VF+(Vi-VF)e-t/𝜏= VF+(0-VF)e-t/𝜏= VF(1- e-t/𝜏)

• Discharging to zero (VF=0)v=VF+(Vi-VF)e-t/𝜏= 0+(Vi-0)e-t/𝜏= Vi e-t/𝜏



Example. In the following circuit, determine the capacitor voltage50 𝜇 s after the switch is closed if the capacitor is initiallyuncharged.



• The time constant is𝜏=RC=(8.2 kΩ)(0.01 𝜇F)=82 𝜇s

• The voltage to which the capacitor is fully charged is VF=50 V,and the initial voltage is 0 V

vc=VF(1- e-t/𝜏)=(50 V)(1-e-50 𝜇s/82 𝜇s)=22.8 V



Example. In the following circuit, determine the capacitor voltageat the time of 6 ms after the switch is closed



• The time constant is𝜏=RC=(10 kΩ)(2.2 𝜇F)=22 ms

• The initial capacitor voltage is Vi=10 V,vc=Vi e-t/𝜏=(10 V)(1-e-6 ms/22 ms)=7.61 V


9.6 Capacitors in AC Circuits

• When a capacitor is connected to a ac voltage source whose thepeak voltage is held at a constant– When the frequency increases, the current also increases– When the frequency decreases, the current also decreases



• An increase in the amount of current with a fixed amount ofvoltage indicates that opposition to the current has decreased

• The capacitor offers opposition to current and that opposition isinversely proportional to frequency

Definition. The opposition to sinusoidal current in a capacitor iscalled capacitive reactance• The symbol for capacitive is XC, and the unit is Ω



• When the capacitance is increased, the opposition to current(and thus the capacitive reactance) decreases

• The capacitive reactance is inversely proportional tocapacitance



• XC is proportional to 1/𝜔c

XC= '(\]3

Example. A sinusoidal voltage is applied to a capacitor. Thefrequency of the sine wave is 1.0 kHz. Determine the capacitivereactance.



XC= '(\]3

= '(\('.+×'+@ ^_)(+.++OP×'+`ab)

=33.9 kΩ



• With series capacitors, the total capacitive reactance XC(tot) isXC(tot)=XC1+XC2+…+XCn

• With parallel capacitors, the total reactance can be found by

𝑋3(WdW) =1

1𝑋3'

+ 1𝑋3(

+ ⋯+ 1𝑋3>



Example. What is the total capacitive reactance in each of thefollowing circuits?



• The reactances of the individual capacitors are the same forboth circuits

XC1= '(\]3'

= '(\(e.+f^_)(+.+'gb)

=3.18 kΩ

XC2= '(\]3(

= '(\(e.+f^_)(+.+Khgb)

=468 Ω

• In (a),XC(tot)=XC1+XC2= 3.18 kΩ+ 468 Ω=3.65 kΩ

• In (b),

𝑋3(WdW) ='

4i?4

6 4i?5

=408 Ω



Lemma (Ohm’s Law). The relationship between I, V, and XC is

𝐼 =𝑉𝑋3

Proof.

• Since the instantaneous current is 𝑖 𝑡 = Uk(W)UW

= 𝐶 Ul(W)UW

,current leads capacitor voltage by 90°



• In phasor notation, the above equation becomesI=j𝜔CV

𝑉𝐼=

1𝑗𝜔𝐶

= −𝑗𝑋3

• If we only emphasize on the magnitude, then

𝐼 =𝑉𝑋3



• For a number of capacitors in series, the voltage across a seriescapacitor is given by

𝑉H =𝐶WdW𝐶H

𝑉I

𝑉H =𝑋3p

𝑋3(WdW)𝑉I



Example. What is the voltage across C2 in the following circuit?



𝑉( =3qrq35

𝑉I=+.++hPgb+.+Khgb

10V=1.28V

𝑉( =v?5

v?(qrq)𝑉I=

OKhw<.Kefw

10V=1.28V



• Power in a capacitor



• True power (Ptrue): Ideally, all of the energy stored by acapacitor during the positive portion of the power cycle isreturned to the source during the negative. The true power iszero.

• Reactive power (Pr): The rate at which a capacitor stores orreturns energy is called its reactive power

Pr=VrmsIrms

Pr=(Vrms)2/ XC

Pr=(Irms)2 XC


chapter 9 capacitors - ecourse2.ccu.edu.tw

Documents