01/11 ENGR 227 1

L4 Capacitance

Dr Graeme Burt

Jan 2011

01/11 ENGR 227 2

Capacitance

• Consider two isolated, uncharged, electrodes A and B immersed in a linear dielectric material

• Let charge +Q be transferred from electrode B to electrode A.

• The charge on B is now –Q and the potential difference between the electrodes is V

• Since the system is linear the potential difference is proportional to the charge transferred so we can write

(14)

where C (the Capacitance) is a constant which depends only on the geometry of the electrodes and the dielectric. C is measured in Farads (F).

• A circuit component designed to provide a specific capacitance is a capacitor

• Capacitance occurs wherever two conductors are at different potentials from one another

CVQ

01/11 ENGR 227 3

Capacitance between parallel plates

• For infinite parallel plates carrying charge σ C/m2 we know that the electric field is uniform

• (1)

0

E

d

• If the separation between the plates is d then the potential difference is

(2)

• Therefore the capacitance per square metre is

(3)

20 .C F mV d

0

dV Ed

01/11 ENGR 227 4

Capacitance between parallel plates

• If the plates each have area A then (ignoring the effect of fringing fields)

• Michael Faraday first discovered that the capacitance increases when a dielectric is placed between the plates.

• For infinite parallel plates with dielectric of relative permittivity εr between them carrying charge σ C/m2 we know that the electric field is uniform

• (1) rr

DE

00

d 0 AC

d

01/11 ENGR 227 5

Capacitor with layers of dielectric in series

• Parallel plates of area A (ignoring fringing fields) with charge σA

• D is the same everywhere

• In region 1

• (7)

• (8)

• In region 2

• (9)

• (10)

• Total potential difference = V1 + V2

111 DEε2

222 DE

11111 ddEV

22222 ddEV

1

21

1

2

2

1

1

2

2

1

121

11

CCA

d

A

d

dd

A

VV

AC

d1

d2

ε

1

01/11 ENGR 227 6

Capacitor with layers of dielectric in parallel

• Parallel plates of area A1 covered with dielectric 1 and area A2 covered by dielectric 2 (ignoring fringing fields) with charges σ1A1 and σ2A2

• In region 1

• (7)

• (8)

• In region 2

• (9)

• (10)

• Potential difference is the same V1 = V2 so charges must be different

1 1 1 1 1E D

ε2

2 2 2 2 2E D

1 1 1 1V E d d

2 2 2 2V E d d

1 1 2 2 1 1 1 2 2 2 1 1 2 21 2

1 2

A A A A A AC C C

V d d d d

d

ε

1

01/11 ENGR 227 7

Fringing field

• When a pair of electrodes is finite the field spreads out into the adjoining region

• Example: Infinite strip conductor over an infinite conducting plane

• When the fringing field is ignored E is uniform under the strip and zero elsewhere

• If the charges are kept constant they redistribute themselves moving outwards and round the back of the strip

• The potential difference between the electrodes has now decreased and the capacitance has increased

5mm 1mm

3mm

V = 100V

V = 0

+ + + + + + + + + + + + +

+ + + + + +

- - - - - - - - - - - - - - - - - - - - - - - -

01/11 ENGR 227 8

Finite Parallel Plates

• Assume there is a ground plane half way between the square plates of width, w.

• At the strip

• If w>>h we can assume the field cover a width of w+h at the ground plane.

• Eg (w+h)2 = w 2/

• Eg = w2/[ (w+h)2]

• Along the centre-line the electric field is roughly constant hence the voltage in is roughly, V=Eav d

d=2h t

w

sE

2

2

0

1e

h wV

w h

01/11 ENGR 227 9

Finite Parallel Plates

0.551 1

12 2

r re

h

w

d=2h t

w Remember we approximate this as a strip embedded in an infinite dielectric of lower permittivity.

The voltage is hence

And the capacitance per unit length is

0 0

2

21

e ewq wC

V V w h hwh

w h w w h

2

2

0

1e

h wV

w h

01/11 ENGR 227 10

Capacitance between coaxial conductors

• From (L2:10) the potential difference between concentric cylinders carrying charges ±q C/m is

(5)

• The capacitance per unit length is

(6)

• Using the parameters (a = 0.5mm, b = 2.5mm, : C = 35 pF/m

0

ln2

qV b a

02

ln

qC

V b a

b

a

For long transmission lines it is standard to give capacitance per unit length, rather than capacitance.

01/11 ENGR 227 11

Capacitance between concentric cylinders

• If the coaxial line is filled with a dielectric

• From (L2:10) the potential difference between concentric cylinders carrying charges ±q C/m is

(5)

• The capacitance per unit length is

(6)

• Using the same parameters as before (a = 0.5mm, b = 2.5mm, εr = 2.25): C = 78 pF/m

• For the same geometry with air spacing (εr = 1.0): C = 35 pF/m

abq

Vr

ln2 0

abV

qC r

ln

2 0

b

a

01/11 ENGR 227 12

Capacitance between concentric cylinders

• This is the same as for the case of parallel dielectric regions in the parallel plates, hence

• C = C1 + C2

• If only half the circumference is filled with dielectric the equipotentials must still be concentric cylinders. V is unchanged and the charge is the sum of the charges on the two parts

C = (78 + 35)/2 = 57 pF/m

01/11 ENGR 227 13

Parallel Wire Transmission Line

• Imagine we have two wires of radius a separated by a distance 2d

• This can be approximated by line charges of charge, Q=qL placed at p2=4d2-a2.

0 0

0

1 1ln ln

2 2

ln2

q qV

p d a p d a

q d p aV

p d a

Potential, V at y=0 and x=d-a

Hence the capacitance per unit length is,

If d>>a this can be simplified to

0

0

2ln

22ln

qC

V d p a

p d a

qC

dV

a

01/11 ENGR 227 14

Microstrip

• The transmission line you are most likely to use frequently when working with electronics above 500 MHz is the microstrip line.

• It is critical you can work out the impedance for microstrip on your PCB’s

• For this we need the capacitance and the inductance.

• Here we will calculate the capacitance.

h t

w

As in the last lecture the ground plane acts as a mirror so the geometry is equivalent to two parallel strips.

01/11 ENGR 227 15

Capacitance of Microstrip Lines

• If w>>h we can approximate as parallel plates (except voltage is half so capacitance is doubled)

• If the strip has an area A=wL then (ignoring the effect of fringing fields)

• Hence the capacitance per unit length is

• For the microstrip the potential difference between the ground and the strip is half the potential difference of the parallel plates but the distance is also half hence the capacitance is the same.

2h

0 0r rA wLC

h h

w

0r wC

h

01/11 ENGR 227 16

Remember we approximate this as a strip embedded in an infinite dielectric of lower permittivity.

The voltage is hence

And the capacitance per unit length is

Microstrip Transmission Line

h t

w

0.551 1

12 2

r re

h

w

0

12 e

h wV

w h

0 02 2

1

e ewq wC

w h hV Vh

w h w w h

01/11 ENGR 227 17

The effective dielectric constant of the strip is

Hence the capacitance is

Example

• Calculate the capacitance of a microstrip line on a 1 mm thick substrate of GaAs (r=10.9) if the strip width is 1 mm.

0.55

0.55

1 11

2 2

10.9 1 10.9 1 11 9.33

2 2 1

r re

e

h

w

h t

w

12

02 8.85 10 9.332

0.11 nF1 1

1 1 1

ewC

h hV

w w h

A better approximation is to approximate the strip as a cylindrical wire but that is beyond this course.

01/11 ENGR 227 18

Energy Stored

• The energy stored in a capacitor is equal to the work done (for example by a battery to charge it)

• If at any time the charge on the plates is q and the voltage is V=q/C

• The work done in taking a charge dq from the negative plate to the positive plate is dW=Vdq=(q/C) dq

• Hence the stored energy is

• UE=Q2/(2C) =QV/2 = CV2/2

2

02

Qq Q

W dqC C

01/11 ENGR 227 19

Parasitic capacitance

• Electrostatic coupling between wires in a circuit

• Earthed metallic screens are used to eliminate unwanted coupling

01/11 ENGR 227 20

Capacitive Coupling

• In coaxial lines the outer conductor shields the fields from the outside enviroment.

• In parallel plates, parallel wires and microstrip the fields spread out into the area around it.

• If another unshielded transmission line is nearby the electric fields will overlap and will cause a capacitive coupling (inductive couplings are also possible and will be discussed in later lectures)

01/11 ENGR 227 21

Capacitance Matrix

• If we have two conductors which are capacitively coupled the voltage is given by

• V1= Q1/C11 + Q2/C12

• However if net charge (Q1+Q2) is zero, this reduces to the well known formula

• V1= Q/C12

• If we have more than two conductors or the net charge is non-zero things get more complicated as there is a capacitance between each pair of conductors.

• Take three charged conductors

• V1= Q1/C11 + Q2/C12 + Q3/C13

• As each charged conductor will have an impact on the other two conductors. This is written as a matrix

• [V]=[P][Q]

• Where P is a a matrix with each element equal to 1/Cmn

01/11 ENGR 227 22

Coupled Microstrip

• For two coupled microstrip lines with a ground plane the potentials equal

• V1=Q1/C1G + Q2/C12

• V2=Q2/C2G + Q1/C21

• Hence the matrix is

• This is simplified as C12=C21

C1G C2G

C12

Line 1 Line 2

1 121 1

21 22 2

1/ 1/

1/ 1/

G

G

C CV Q

C CV Q

01/11 ENGR 227 23

Coupled Microstrip

• The capacitances C1G and C2G

can be calculated using previous formulae.

• C12 is normally calculated using numerical codes but when d<h we can roughly approximate as two parallel wires, of diameter t, half in dielectric.

C1G C2G

C12

Line 1 Line 2

h

w

t

d

01/11 ENGR 227 24

Coupled Microstrip

• Previously we say that the capacitance of two parallel wires was

• If half is in dielectric and half is in air

0

42ln

rqC

dV

t

h

w

t

d

0 0

4 422ln 2ln

rqC

d dV

t t