Capacitor Dielectric

Capacitors and Capacitance

Capacitor: any two conductors,

one with charge +

Q

, other

with charge –

Q

+Q

–Q

Uses: storing and releasing electric charge/energy.

Most electronic capacitors: micro-Farads (mF),

pico-Farads (pF) — 10–12 F

New technology:

compact 1 F capacitors

Potential DIFFERENCE between

conductors =

V

Q =

C

V

where C = capacitance

Units of capacitance:

Capacitance

•

_{Capacitance depends only}

on GEOMETRICAL

factors and on the

MATERIAL that

separates the two

conductors

•

_{e.g. Area of conductors,}

separation, whether the

space in between is filled

with air, plastic, etc.

+Q

–Q

(We first focus on capacitors where gap is filled by AIR!)

Electrolytic (1940-70) Electrolytic (new)

Paper (1940-70)

Tantalum (1980 on) Ceramic (1930 on) Mica_(1930-50) Variable

air, mica

Parallel Plate Capacitor

+Q

-Q

E field between the plates: (Gauss’ Law)

Relate E to potential difference V:

What is the capacitance C ?

Area of each plate = A

Separation = d charge/area = s = Q/A







d

x

d

E

V

0





A

Qd

dx

A

Q

d

0 0



0









d

A

V

Q

C







0

A

Q

E

0

0





s

_



Capacitance and Your iPod!

d

A

V

Q

Parallel Plate Capacitor — Example

•

_{A huge parallel plate capacitor}

consists of two square metal plates of

side 50 cm, separated by an air gap of

1 mm

•

_{What is the capacitance?}

Lesson: difficult to get large values of capacitance without special

tricks!

C = ₀

A/d

= (8.85 x 10

–12

F/m)(0.25 m

2

)/(0.001 m)

= 2.21 x 10

–9

F

Isolated

Parallel Plate Capacitor

• _{A parallel plate capacitor of}

capacitance C is charged using a battery.

• _{Charge = Q, potential difference = V.} • _{Battery is then disconnected.}

• _{If the plate separation is INCREASED,}

does Potential Difference V: (a) Increase?

(b) Remain the same? (c) Decrease?

+Q –Q

• _{Q is fixed!}

• C decreases (=₀A/d) • _{Q=CV; V increases.}

Parallel Plate Capacitor & Battery

• _{A parallel plate capacitor of}

capacitance C is charged using a battery.

• _{Charge = Q, potential difference = V.} • _{Plate separation is INCREASED while}

battery remains connected.

+Q –Q

• _{V is fixed by battery!} • C decreases (=₀A/d) • _{Q=CV; Q decreases} • E = Q/ ₀A decreases

Does the Electric Field Inside: (a) Increase?

(b) Remain the Same? (c) Decrease?

Spherical Capacitor

What is the electric field inside

the capacitor? (Gauss’ Law) Radius of outer _{plate = b} Radius of inner plate = a

Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell

Relate E to potential difference between the plates:

2 0

4

r

Q

E











b a

r

d

E

V





b a b a

r

kQ

dr

r

kQ















₂







 



b

a

Spherical Capacitor

What is the capacitance?

C = Q/V = Radius of outer _{plate = b}

Radius of inner plate = a

Concentric spherical shells: Charge +Q on inner shell, –Q on outer shell

Isolated sphere: let b >> a,







 



b

a

Q

Q

1

1

4



₀

)

(

4

₀

a

b

ab







a

C



4



₀

Cylindrical Capacitor

What is the electric field in

between the plates? Gauss’ Law!

Relate E to potential difference between the plates:

Radius of outer plate = b

Radius of inner plate = a

cylindrical Gaussian surface of radius r Length of capacitor = L

+Q on inner rod, –Q on outer shell

rL

Q

E

0

2











b a

r

d

E

V





b a b a

L

r

Q

dr

rL

Q



















0 0

2

ln

2







_











a

b

L

Q

ln

2



₀

Summary

• _{Any two charged conductors form a capacitor.} • _{Capacitance : C= Q/V}

• _{Simple Capacitors:}

Parallel plates: C = ₀ A/d

Spherical: C = 4 ₀ ab/(b-a)

Capacitors in Parallel

• _{A wire is an equipotential}

surface!

• _{Capacitors in parallel have}

SAME potential difference but NOT ALWAYS same charge!

• V_AB = V_CD = V • Q_total = Q₁ + Q₂ • C_eqV = C₁V + C₂V

• C_eq = C₁ + C₂

• _{Equivalent parallel}

capacitance = sum of capacitances

A B

C D

C₁ C₂ Q₁

Q₂

C_eq Q_total

Capacitors in Series

• Q₁ = Q₂ = Q (WHY??)

• V_AC = V_AB + V_BC A B C

C_{1 C2}

Q₁ Q₂

C_eq Q

2

1 C

Q C

Q C

Q

eq

 

2 1

1 1

1

C C

C_eq  

SERIES:

• _{Q is same for all capacitors}

Energy Storage in Capacitors

• _{Since capacitors store electric charge, they store electric potential}

energy.

• _{Consider a capacitor with capacitance}

C

_{, potential difference}

V

and charge

q

.

• _{The work}

dW

_{required to transfer an elemental charge}

dq

_{to the}

capacitor:

dq

C

q

Vdq

dW





The work required to charge capacitor from q=0 to q=Q:

2 2 2 0 2 0 0

2

1

2

)

(

2

2

1

CV

C

CV

C

Q

q

C

dq

C

q

Vdq

W

Q Q Q

















Energy Stored by a Capacitor

•

_{By the work-energy theorem, the potential}

energy stored by a capacitor is equal to the work

done in placing a charge on it.

2

2

2

Stored

Energy

2

2

_CV

_QV

C

Q

U









•

_{Energy may be stored in an electric field.}

•

_{Many electrical and electronic devices use}

Stored Energy Density

of a Charged Capacitor

2

2

1

CV

U



Stored Energy:

Parallel-Plate Capacitor:

Stored Energy:

E

d

E

Ad

d

A

U

0 2 2 ₀ 2

2

1

2

1









Stored Energy Density

in the Electric Field:

2 0

2

1

E

Ad

U

Volume

U

u

_E









d

V

E

d

A



What happens to a material when an electric field

is applied across it (i.e., in a capacitor)? How does

the electric field change, and how does the

charge/area change?



What are the 3 primary contributions to the

dielectric constant.



How fast would a dielectric respond?



Describe the 4 primary dielectric breakdown

mechanisms.



How can the breakdown strength be improved?



What are ferroelectrics?



What are other applications of ferroelectrics?

Capacitance

• _{Two electrodes separated by a gap define a}

capacitor.

• _{When a bias is applied across the capacitor}

plates, one charges positively, the other negatively.

• _{The amount of charge that the capacitor}

can store (Q) is proportional to the bias (V) times how good the capacitor is, the ‘capacitance’ (C).

• _{The capacitance is related to the area of}

the plates (A), their separation (d), and the

Dielectric Constant (εε_o) of the dielectric between the plates

• Dielectric constant of vacuum; ε_o =

8.85x10-12 _{F/m=55.2 Me/(V*m)}

d

A

C





o



 





e V

m m V

d A

Q o e V m* *

2 *

Why does charge built up?

There is generally not a built-in electric field between the

plates of an unbiased capacitor.

When an electric field is applied, any charged carriers or

species within the material will respond.

For a conductor or semiconductor, e

-

will flow to the + plate,

and possibly also holes will flow to the - plate.

Current is

carried=no charge buildup.

For an insulator, there aren’t a significant number of free

carriers. There are highly ionic species, however, but they

aren’t very mobile at low temperatures.

No appreciable

current is carried=charge buildup.

Polarization in Insulators

Positively charged species in insulators shift/rotate/align toward the negative electrode and negatively charged species shift/rotate/align towards the positive electrode; creating dipoles. The dipole moment density is termed the Polarization (P) and has the units of C/m2.

+ -Electron Cloud Electron Cloud + E Electronic polarization, occurs in all insulators

-+ + + -+ + E Ionic polarization occurs in all ionic solids: NaCl, MgO…

-- -+ -+ + -+ + E Molecular polarization, occurs in all insulating molecules; oils, polymers,

H₂O…

A

q

V

p

P





Electric Dipole Moment

Polarization

x

q

p





Relative Dielectric Constants

Generally, the less conducting and more polar a material is,

the greater will be its dielectric constant.

A Materials/Design Problem

How can we increase the charge stored in a parallel-plate capacitor? This is an extremely important problem in solid state computer memories (RAMs, DRAMs, SDRAMs) that are based on capacitors.

d

A

C





o

1. Use a material with a higher dielectric constant (ε), limited by material properties (see table next page).

2. Increase capacitor area (A), limited by how much space you have on the IC/device. But, one can always increase the projected lateral

area!!! This is a design problem.

3. Decrease plate spacing d. Limited by dielectric breakdown as the electric field across the plate increases with d.

4. Fast Read/Write speeds (typically GHz) limits the material that can be used (ionic/electronic polarization, SiO₂, Si₃N₄, TiO₂, HfO₂…).

25

Fundamental Laws of Electrostatics















V

ev S

C

dv

q

s

d

D

l

d

E

0

E

D





Conservative field Gauss’s law

Constitutive relation

ev

q

D

E













0

• The integral forms of the fundamental laws are more general because they apply over regions of space. The differential forms are only valid at a point.

• From the integral forms of the fundamental laws both the differential equations governing the field within a medium and the boundary conditions at the interface between two media can be derived.

26

Boundary Conditions

•

_{Within a homogeneous medium, there are no abrupt}

changes in

E or

D. However, at the interface between

two different media (having two different values of e), it

is obvious that one or both of these must change

abruptly.

•

_{To derive the boundary conditions on the normal and}

tangential field conditions, we shall apply the integral

form of the two fundamental laws to an infinitesimally

small region that lies partially in one medium and

partially in the other.

27

Boundary Conditions (Cont’d)

•

_{Consider two semi-infinite media separated by a}

boundary. A surface charge may exist at the interface.

Medium 1

Medium 2 x x

x x

r

_s

•

_{Locally, the boundary will look planar}

1



2



n

a

ˆ

2 2

,

D

E

1 1

,

D

E

28

Boundary Condition on Normal Component of D

• _{Consider an infinitesimal}

cylinder (pillbox) with cross-sectional area Ds and height Dh lying half in medium 1 and half in medium 2:

1



2



2 2

,

D

E

1 1

,

D

E

Ds

Dh/2

Dh/2

x x x x x x

r

s

n

a

ˆ

• Applying Gauss’s law to the pillbox, we have

s q RHS s D s D s d D s d D s d D LHS dv q s d D es n n side bottom top V ev S D  D  D         











2 1

0

• The boundary condition is

• If there is no surface charge

s n

n

D

D

₁



₂



r

n n

D

D

₁



₂

For non-conducting materials,

r_s = 0 unless an impressed

29

Boundary Condition on Tangential Comp. of E

• _{Consider an infinitesimal path}

abcd with width Dw and height Dh lying half in medium 1 and half in medium 2: 1



2



n

a

ˆ

2 2

,

D

E

1 1

,

D

E

Dh/2 Dh/2 Dw a b c d path along boundary the to al r tangenti unit vecto ˆ ˆ ˆ contour by the defined direction in the path lar to perpendicu r unit vecto ˆ     n s t s a a a abcd a n

a

ˆ

a b c d s

a

ˆ

t

a

ˆ

30

Boundary Condition on Tangential Comp. of E …..

•

_{Applying conservative law to the path, we have}



E E



w

w E h E h E w E h E h E l d E l d E l d E l d E LHS l d E t t t n n t n n a d d c c b b a C D   D  D  D  D  D  D            











2 1 1 2 1 2 2 1 2 2 2 2 0

•

The boundary condition is

E

1t



E

2t

• At any point on the boundary,

– the components of E1 and E2 tangential to the boundary are equal

– the components of D1 and D2 normal to the boundary are

discontinuous by an amount equal to any surface charge existing at that point