CAPACITORS / CAPACITANCE · CAPACITORS / CAPACITANCE Enzo Paterno -Capacitance -Capacitor types -Capacitors in series & parallel -RC Circuit Charging phase -RC ... negatively charged

ECET11

1

CAPACITORS / CAPACITANCE

Enzo Paterno

- Capacitance - Capacitor types - Capacitors in series & parallel - RC Circuit Charging phase - RC Circuit Discharging phase - RC Circuit Steady State model - Source Conversions - Superposition Theorem - Thevenin Equivalent Circuit Source: Introductory Circuit Analysis – Boylestad 10th Ed.

ECET11 CAPACITORS

2 Enzo Paterno

A capacitor is a circuit element that consists of two conducting plates separated by a non-conducting, (i.e. dielectric), material. Terminals are connected to the plates.

A capacitor is said to be a storage device in that it stores energy in the form of electric field. During its normal operation, a capacitor stores charges across its plates. The difference in potential energy across the two plates results in a voltage across the capacitor. Power is delivered to a capacitor and this energy is stored unlike the resistor whereby power is dissipated in the form of heat (power loss).

C Capacitance [Farads] Capacitance is a measure of a Capacitor’s ability to store Charge on its plates.

(t)

ECET11 ELECTRIC FIELD

3 Enzo Paterno

When a potential difference of V volts is applied across the two plates separated by a distance of d, the electric field strength, E, between the plates is determined by:

][ mV

dVE =

A capacitor stores energy in the form of electrical field (i.e. voltage) that is established by the opposite charges on the two plates.

v d

Ɛ The electric field strength, is represented graphically by electric flux lines (i.e. lines of force). These lines of force always extend from a positively charged body to a negatively charged body. They extend Perpendicular to the charged surfaces and never intersect.

Electric Flux Lines

ECET11

4 Enzo Paterno

ELECTRIC FIELD

The force exerted between two charged particles is given by Coulomb’s law:

221

dQkQF =

The electric field strength, ε, at a given point is the force, F, acting on a unit positive charge (Q1 = 1C) at that point:

ECET11 CAPACITOR OPERATION

5 Enzo Paterno

When the capacitor is in a neutral state, both plates have an equal number of free electrons.

•(Insulator)


6 Enzo Paterno

(Electric flux lines)

When a voltage is applied across the plates, electrons are attracted by the positive side of the battery and electrons are repelled by the negative side of the battery. For every electron that leaves one side of the plate, an electron travels to the other side of the plate.

e- e- The electrons can not pass through the dielectric so that plate A starts storing electrons and plate B effectively starts storing positive charges. This process lasts until the plates are fully charged to a potential difference of the source V. The result is a positive charge on plate B and a negative charge on plate A.

B A

V Charging phase process


7 Enzo Paterno

CURRENT APPEARS TO BE FLOWING FROM THE NEGATIVE SIDE OF THE BATTERY TOWARD THE POSITIVE SIDE THROUGH THE CAPACITOR.

IN FACT IT IS NOT SO, BECAUSE NO ELECTRONS CAN FLOW THROUGH THE INSULATOR MATERIAL OR GAP BETWEEN THE PLATES (DIELECTRIC)

THERE ARE IN FACT TWO SEPARATE CURRENTS FROM THE BATTERY TO THE CAPACITOR FROM THE CAPACITOR TO THE BATTERY


8 Enzo Paterno

If the capacitor is disconnected from the source, it retains the stored charge for a certain period of time (depends on capacitor type and amount of voltage).

Capacitors are not perfect. Ideal insulators have infinite resistance. Actual insulators have some very high resistance. The electrons on plate A will attract to plate B - Leakage current takes place. Once all the electrons have been neutralized the capacitor has lost all of its charge.

Capacitors cannot be used as a replacement of batteries as they slowly discharge over time. However, they are used to provide a short life supply of voltage (i.e. surge protectors, smoke alarms)

A B

ECET11 CAPACITANCE

9 Enzo Paterno

Capacitance is a measure of a capacitor’s ability to store charge on its plates. As such, the amount of charge a capacitor can store per unit of voltage across its plate is called capacitance (Q↑, C ↑ ).

Michael Faraday

VQC = C

QV =

VCQ =C = Capacitance FARADS [ F ] Q = Charge COULOMBS [ C ] V = Voltage VOLTS [ V ]

A capacitor is said to have a 1 Farad capacitance when 1 Coulomb of charge is deposited on the plates with A 1 volt potential difference across it.

VCF

111 =

The charge across a capacitor is proportional to the voltage across it:

)()( tCVtQ =

ECET11 CAPACITANCE

10 Enzo Paterno

Capacitance can be computed as follows:

Michael Faraday

dAC orεε=

C = CAPACITANCE [ F ] A = AREA OF THE PLATES [ m2 ] d = DISTANCE BETWEEN PLATES [ m ] ε = DIELECTRIC PERMITTIVITY

Permittivity for vacuum εo= 8.85 x 10-12 F/m

ECET11 CAPACITANCE

11 Enzo Paterno

Michael Faraday

AQ

AQ

AdQd

AQdV

dAC

CQV

dV

ε

εε

ε

ε

=∴

==

=⇒

==

=

Ԑ

Ԑ

Ԑ Eq 1:

Sub Eq 3 into Eq 2:

Eq 2: Eq 3:

Eq 4:

Sub Eq 4 into Eq 1:

ECET11 CAPACITANCE

12 Enzo Paterno

Michael Faraday

ECET11 CAPACITOR VOLTAGE BREAKDOWN

13 Enzo Paterno

Every capacitor has a voltage rating. A user must not exceed this voltage rating (i.e. voltage breakdown). Voltage breakdown is the voltage required to break the bonds within the dielectric that will create current flow in the dielectric through the plates

When breakdown occurs, the characteristic of the capacitor becomes that of a conductor (i.e. capacitor shorted)

Clouds

Earth

Atmosphere

+++++++ Lightning is an example of breakdown:

The potential between the clouds and the earth is so high that charge can pass from one to the other through the atmosphere (which acts as the dielectric)

a. C = 3(5 uF) = 15 uF

b. C = ½ (0.1 uF) = 0.05 uF

c. C = 2.5(20 uF) = 50 uF

d. C = (5)(4)/(1/8) (1000 pF) (160)(1000 pF) = 0.16 uF

dAxC rε121085.8 −=

14

15

16

b.

VQC =or

17

Assume the breakdown voltage for this capacitor is 200 V/mil.

ECET11 CAPACITORS

18 Enzo Paterno

There are many types of capacitors and they are characterized by the type of dielectric material used between the conducting plates.

Ceramic Disc

Polyester Film

Tantalum Polypropylene Film

Mylar

ECET11 CAPACITORS

19 Enzo Paterno


SMT Capacitor

Electrolytic

Radial Lead

Axial Lead

Electrolytic

Variable

Non-polarized

Variable Capacitors

ECET11 CAPACITORS

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Ceramic Disk Capacitors

ECET11 CAPACITORS

21 Enzo Paterno


Axial Lead Plastic Film Dielectric Tubular Capacitors

ECET11 CAPACITORS

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Tantalum Electrolytic Capacitors

ECET11 CAPACITOR CODES

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100,000 pF = 100 nF = .1uF

Ceramic Disc


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Electrolytic and large types of Capacitors usually have the value printed on them (i.e. 470uF 25V). Most of the smaller caps have two or three numbers printed on them, The Value is in pF. 105 means 10 x 105 = 1.000.000pF = 1000 nF = 1 uF. Letters added to the value represent the tolerance and in some cases represent the temperature coefficient. A 474J ceramic capacitor means 47 x 104 = 470.000pF and J= 5% tolerance. (470.000pF = 470nF = 0.47uF).

Other capacitors may just have 0.1 or 0.01 printed on them. If so, this means a value in uF. Thus 0.1 means just 0.1 uF. Values marked as 50 or 330 means just pF.


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ECET11 CAPACITOR CURRENT & VOLTAGE RELATIONSHIP

27 Enzo Paterno

∫=∴

=

=⇒=∴

==

dttiC

tv

dttiC

tdv

dttdvCti

dttCvdti

C v(t)q(t) dt

tdqti

)(1)(

)(1)( :getWe

)()())(()(

capacitor afor and )()(

If v(t) = k i(t) = 0, then the capacitor acts like an open to dc (i.e. blocks dc - Steady State). For dc circuit analysis, or Steady State circuit analysis, a capacitor is modeled as an open.

Capacitive coupling circuit

5v

Gnd

ECET11

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RC CIRCUIT – STEADY STATE ANALYSIS

When a capacitor has reached Steady State it is modeled as an open circuit.

Find the current I when the circuit has reached steady state.

I I

mAkvI 8

648

==

ECET11 CAPACITOR CURRENT & VOLTAGE

29 Enzo Paterno

If vC(t) = K: o dv(t)/dt = 0, p(t) = 0

If vC(t) is an instantaneous change:

o dv(t)/dt ∞ and p(t) ∞

Power can only be finite, as such the voltage across a capacitor cannot change instantaneously.

Cq(t) v(t)since J )(

21 )(

21)(

0)( that Assuming )(21)(

)()()()(

dt(t)dw p(t) :know We

:be tofound iscapacitor ain storedenergy The

)()()()()(

:is deliveredpower The

22

)(

)(

2

c

===

=−∞⇒=

==

=

==

−∞

∞−∞− ∫∫

CtqtCvtw

vtCvtw

dtdt

tdvtvCdttptw

dttdvtCitvtitp

C

v

vC

C

τ

ττ

ECET11 CAPACITORS

30 Enzo Paterno

The voltage across a 5-uF capacitor has the waveform shown below. Determine the current waveform.

mAxti

tms

mAxti

mst

mAxti

mst

020105)(

8@

60224105)(

86@

20624105)(

60@

6

6

6

==

≤

−=−

=

≤≤

==

≤≤

−

−

−

dttdvCti )()( =

ECET11 CAPACITORS - Example

31 Enzo Paterno

The voltage across a 5-uF capacitor has the waveform shown below. Determine the Energy at t = 6ms.

)(21)( 2 tCvtwC =

26 )24)(105(21)6( −= xmswC

uJmswC 1440)6( =

ECET11 CAPACITORS IN SERIES

32 Enzo Paterno

Using KVL: )()()()()( 321 tvtvtvtvtv N++++=

∫= dttiC

tvi

i )(1)( ∫∑

=

=

dttiC

tvn

I ii )(1)(

1∫= dtti

Ctv

S

)(1)(

N

n

I iS CCCCCC111111

3211++++=

= ∑

=

Capacitors in series act like resistors in parallel

Current is the same through each element in a series circuit

C1 C2

21

21

CCCCCT +

=

ECET11 CAPACITORS IN PARALLEL

Using KCL: )()()()()( 321 tititititi N++++=

dttdvC

dttdvC

dttdvC

dttdvCti N

)()()()()( 321 ++++=

Capacitors in parallel act like resistors in series

Enzo Paterno 33

N

N

iiP

N

ii

CCCCCC

dttdvCti

++++=

=

=

∑

∑

=

=

3211

1

)()(

Voltage is the same across each branch in a parallel circuit

ECET11 TESLA COIL

34 Enzo Paterno

• A high voltage power supply charges up a capacitor C1. • When the capacitor reaches a high enough voltage, the spark gap (switch) fires. • When the spark gap fires, the energy stored up in the capacitor dumps into a 1:100 step-up transformer. • The primary (L1) is about 10 turns of heavy wire. The secondary (L2) is about 1000 turns of thin wire ( 1:100 ratio). Feed in 10,000 volts, get out 1,000,000 volts. It all happens at a rate of over 120 times per second, often generating multiple discharges in many directions

ECET11 CAPACITORS IN SERIES & PARALLEL

35 Enzo Paterno

Tesla coil designs require a capacitor with large voltage breakdown. Placing capacitors is series increases the voltage breakdown (add each voltage breakdown). However, placing capacitors in series reduces the total capacitance. This banks of capacitors in series are placed in parallel.

Use Power factor correction (PFC) capacitors only. These are used to correct the power factor of the AC connected to the Neon Sign Transformer (NST).

ECET11 CAPACITORS IN SERIES

36 Enzo Paterno http://www.teslacoildesign.com/

CAPACITORS IN SERIES & PARALLEL

MMC: Multi Mini Capacitor

http://www.teslacoildesign.com/

http://www.teslacoildesign.com/

ECET11 TRANSIENT ANALYSIS - CAPACITOR CHARGING PHASE

37 Enzo Paterno

t = 0

vC (0) = 0

@ t = 0+ switch is closed vC (0+) = 0 RC

t

R

RCt

C

RCt

CR

EetV

eEtV

eREtCI

tVtVEt

−

−

−

=

−=

=

+=⇒>

)(

1)(

)(

)()(0@

When

vC (0) = 0

Capacitor is modeled as a short Capacitor is modeled as an open

The RC Circuit provides a 1st order Differential equation. Solving this equation results in:


Capacitor Charging

volta

ge

time

curr

ent

time

E E/R


39 Enzo Paterno

t = 0

vC (0) = 0

E

E/R

E

Charging Phase

RCt

R EetV−

=)(

RCt

eREtCI

−

=)(

−=

−RC

t

C eEtV 1)(

][sRC=τ


40 Enzo Paterno

t = 0

vC (0) = Vi

@ t = 0+ switch is closed vC (0+) = Vi

( )

( ) RCt

iR

RCt

iC

RCt

i

CR

eVEtV

eVEEtV

eR

VEtCI

tVtVEt

−

−

−

−=

−−=

−=

+=⇒>

)(

)(

)(

)()(0@

When

vC (0) = Vi

Capacitor is modeled as a battery Capacitor is modeled as an open

The RC Circuit provides a 1st order Differential equation. Solving this equation results in:

RVE

CIi−

=

Vi

iVE −


41 Enzo Paterno

t = 0

RC=τ

vC (0) = Vi R

VE i−

E

VR iVE −

( ) RCt

iR eVEtV−

−=)(

( ) RCt

iC eVEEtV−

−−=)(

( ) RCt

iC eEVEtV−

−+=)(or

RCt

i eR

VEtCI−−

=)(


42 Enzo Paterno

VR , IC

VR follows this curve f(∞ ) = f(5τ )

Steady State Definition


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44

ECET11 TRANSIENT ANALYSIS - CAPACITOR DISCHARGE PHASE

45 Enzo Paterno

@ P1 Charging phase @ P2 Discharging phase

t = 0 vC (0) = 0

t = to

t = to

RCt

C EetV−

=)(

RCt

R EetV−

−=)(

RCt

c eREtI

−−=)(

E

RE

−

E−

Note the current reversal

Note the voltage reversal

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TRANSIENT ANALYSIS - CAPACITOR DISCHARGE PHASE

•CV -

ECET11

47 Enzo Paterno

If the charging phase is disrupted before reaching the supply voltage, the capacitive voltage will be less then E. We call it Vf .

TRANSIENT ANALYSIS - CAPACITOR DISCHARGE PHASE

•CV

•CV

ECET11 SOURCE CONVERSION

48 Enzo Paterno

A voltage source can be converted to a current source and vice-versa producing equal behaviors across its load.

ECET11

49 Enzo Paterno

SOURCE CONVERSION

ECET11 TRANSIENT ANALYSIS

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ECET11 SUPERPOSITION

54 Enzo Paterno

Given a linear circuit, (i.e. described by a set of linear algebraic equations), the superposition analysis technique provides a mean to determine a voltage drop or current by calculating the contribution of each source acting independently and algebraically adding each contribution. Procedure: 1. Remove all sources except one of them by

replacing current sources with an open replacing voltage sources with a short retaining all internal resistance

2. Calculate the desired voltage drop or branch current from that source paying

close attention to polarity or direction

3. Repeat steps 1 and 2 for each additional source acting independently

4. Algebraically add each sources contribution

ECET11 SUPERPOSITION – Example 1

55 Enzo Paterno

For the circuit below, find VA using superposition:

VA’

VA

VS1 Contribution:

vV

kkkkkk

V

A

A

74.4

08.12.18.17.2

)8.1)(7.2(10

'

'

+=

++=

+

ECET11

56 Enzo Paterno

For the circuit below, find VA using superposition: VA

VA’’

vV

kkkkkk

V

A

A

37.11

83.08.12.17.2

)2.1)(7.2(36

''

''

−=

++=

-

VS2 Contribution:

SUPERPOSITION – Example 1

ECET11

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vvvVVV AAA 63.637.1174.4''' −=−=+=


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58 Enzo Paterno

For the circuit below, find VA and VB using superposition:

VA VB


ECET11

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VS1 Contribution: R2 // R4 = 4705 Ω RL + 4705 Ω = 6705 Ω R3 // 6705 Ω = 5021 Ω

vV

kkkV

A

A

26.22

021.54)021.5(40

'

'

=

+=

VA’ VB

’

vV

kkkV

B

B

62.15

705.42)705.4(26.22

'

'

=

+=


ECET11

60 Enzo Paterno

VS2 Contribution: R1// R3 = 3.33k Ω RL + 3.33k Ω = 5.33k Ω R4 // 5.33k Ω = 4997 Ω

vV

kkkV

B

B

49.12

997.45)997.4(25

''

''

=

+=

vV

kkkV

A

A

80.7

33.32)33.3(49.12

''

''

=

+=

VA’’ VB

’’


ECET11

61 Enzo Paterno

vVA 26.22' = vVB 62.15' =

vVB 49.12'' =vVA 80.7'' =

VA VB

vVVV AAA 06.30''' =+= vVVV BBB 11.28''' =+=


ECET11

62 Enzo Paterno


VA

For the circuit below, find VA using superposition:

ECET11

63 Enzo Paterno


IS1 Contribution: R2 // (R3+R4) = 172 Ω

mAI

I

A

A

18.20

172680)172(100

'

'

=

+=

20.18 mA IA’

VA

ECET11

64 Enzo Paterno


VS2 Contribution: R1 // (R3+R4) = 367 Ω

mAI

I

VI

A

A

AA

38.18

680220367

)367(20680

''

''

'''

=

+=

=18.38 mA

IA’’

VA

ECET11

65 Enzo Paterno


mAI A 18.20' =

mAI A 38.18'' =

VA A

I

mAI

mAmAIII AA

80.1

38.1818.20'''

=

−=+=

vV

mAV

A

A

224.1

)680(80.1

=

Ω=

1.224 v

ECET11

66 Enzo Paterno

THEVENIN’S THEOREM

This image cannot currently be displayed.

A two-terminal linear network connected to a load can be replaced with an equivalent circuit consisting of an independent voltage source called VTH in series with a resistor RTH such that the current-voltage relationship at the load is unchanged. Procedure: 1. Identify and remove the load 2. Label the load terminals 3. Look in the load terminals and calculate VTH 4. Remove all sources by replacing:

voltage sources with a short current sources with an open If the source has an internal resistance, keep the resistance in the circuit

5. Look in the load terminals and calculate RTH 6. Create a series circuit consisting of VTH, RTH, and the load 7. Calculate the load current or voltage as desired

ECET11

67 Enzo Paterno

THEVENIN’S THEOREM – Example 1


Identify Load, remove load, label terminals, & find VTH

a

b vV

kkkV

TH

TH

77.6

2.89.3)2.8(10

=

+=

ECET11

68 Enzo Paterno


Replace voltage sources with a short , current sources with an open & find RTH

a

b Ω=

++

=

kR

kkk

kkR

TH

TH

34.7

7.42.89.3

)2.8(9.3


ECET11

69 Enzo Paterno


Connect the load to the equivalent circuit across terminals a & b

a

b

6.77v

7.34 kΩ


ECET11

70 Enzo Paterno

This image cannot currently be displayed. a

6.77v

7.34 kΩ

vV

kkkV

RL

RL

10.2

3.334.7)3.3(77.6

=

+=

b


ECET11

71 Enzo Paterno

This image cannot currently be displayed. a

6.77v

7.34 kΩ

uAI

kkI

RL

RL

27.636

3.334.777.6

=

+=

b


ECET11

72 Enzo Paterno



vkmAV

mAIkkkmAI

TH

R

R

18)3(6

627

)2(27

1

3

3

==

=+

=

ECET11

73 Enzo Paterno



vVVVvV

kkkkV

THTHTH

TH

TH

268180.8

432)3(24

21

2

2

=+=+==

++=

ECET11

74 Enzo Paterno



Ω=

++

=

kR

kkk

kkR

TH

TH

3

136

)3(6

ECET11

75 Enzo Paterno



26v

3kΩ

ECET11

76 Enzo Paterno



26v

3kΩ

vV

kkkV

RL

RL

61.13

3.33)3.3(26

=

+=

ECET11

77 Enzo Paterno



mAI

kkI

RL

RL

12.4

3.3326

=

+=

26v

3kΩ

ECET11 RC CIRCUIT – THEVENIN EQUIVALENT

78 Enzo Paterno

ECET11

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RC CIRCUIT – THEVENIN EQUIVALENT

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FORMULAS

Charging Phase: @ vC (0) = Vi

)()( tVtVE CR +=

the voltage across a capacitor cannot change instantaneously

Discharging Phase:

dVE =Ɛ

VQC =

dAC orεε=

dtvCi

∆∆

=∆

)(21)( 2 tCvtwC =

)()()( tvtitp =

= ∑

=

n

I iS CC 1

11

= ∑

=

N

iiP CC

1

Charging Phase: @ vC (0) = 0

RCt

R EetV−

=)(

RCt

eREtCI

−

=)(

−=

−RC

t

C eEtV 1)(

RC=τ

( ) RCt

iR eVEtV−

−=)(

( ) RCt

iC eVEEtV−

−−=)(

( ) RCt

iC eEVEtV−

−+=)(

RCt

i eR

VEtCI−−

=)(

@ Steady State C is an open circuit

RCt

C EetV−

=)(

RCt

R EetV−

=)(

RCt

c eREtI

−=)(

Steady State: t ∞ = 5τ

Documents

CAPACITORS / CAPACITANCE · CAPACITORS / CAPACITANCE Enzo Paterno -Capacitance -Capacitor types -Capacitors in series & parallel -RC Circuit Charging phase -RC ... negatively charged