L4 Capacitance
Dr Graeme Burt
Jan 2011
01/11
ENGR 227
1
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
Q  CV
(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
01/11
ENGR 227
2
Capacitance between parallel plates
•
For infinite parallel plates carrying
charge σ C/m2 we know that the electric
field is uniform
E

0
•
d
(1)
•
If the separation between the plates is d then the potential difference
is
d
V  Ed 
0
(2)
•
Therefore the capacitance per square metre is
C
01/11

V

0
d
F .m2
(3)
ENGR 227
3
Capacitance between parallel plates
•
If the plates each have area A then
(ignoring the effect of fringing fields)
C
0 A
d
d
•
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
E
01/11
D
 0 r


 0 r
•
ENGR 227
(1)
4
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
•
E1  D 1   1
1
ε2
•
(8)
•
(9)
In region 2
E2  D  2    2
V2  E2 d 2   d 2  2
•
ε
d2
V1  E1d1   d1 1
•
(7)
d1
•
(10)
Total potential difference = V1 + V2
1
 d
 1
d 
1 
C

  1  2     
V1  V2   d1 d 2   A1 A 2   C1 C2 
   
 1  2 
01/11
ENGR 227
A
A
1
5
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
•
•
d
E1  D1 1  1 1
•
(7)
V1  E1d  1 d 1
•
(8)
In region 2
ε
ε2
1
E2  D2  2   2  2
•
V2  E2 d   2 d  2
•
(9)
(10)
Potential difference is the same V1 = V2 so
charges must be different
 1 A1   2 A2   1 A1  1  2 A2   2

C


V
01/11
 1d
 2d
ENGR 227
A11 A2 2


 C1  C2
d
d
6
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
01/11
ENGR 227
5mm
1mm
3mm
V = 100V
V=0
+ +
+
+
+ +
+++ + + + + + + + + ++
- - - - ---- - - - - - - - - ---- - - - -
7
Finite Parallel Plates
•
Assume there is a ground plane
half way between the square
plates of width, w.
Es   
•
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
01/11
ENGR 227
w
d=2h
t
h 
w2 
V
1 

2

 0 e   w  h  
8
Finite Parallel Plates
Remember we approximate this as a strip
embedded in an infinite dielectric of
lower permittivity.
e 
 r 1  r 1 
2

h
1



2  w
0.55
w
d=2h
t
The voltage is hence
h 
w2 
V
1 

 0 e   w  h 2 
And the capacitance per unit length is
C
 0 e w
 0 e
q w



V
V
  w 2   h
hw 
h 1  
  w  h    w  w  h 2 

 

 
01/11
ENGR 227
9
Capacitance between coaxial conductors
•
 q 
V 
 ln  b a 
 2 0 
•
(5)
The capacitance per unit length is
a
2 0
q
C 
V ln  b a 
•
b
From (L2:10) the potential difference between
concentric cylinders carrying charges ±q C/m is
(6)
Using the parameters (a = 0.5mm, b = 2.5mm, : C = 35
pF/m
For long transmission lines it is standard to give capacitance per unit
length, rather than capacitance.
01/11
ENGR 227
10
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
 q
V  
 2 0 r
•

 ln b a 

b
(5)
a
The capacitance per unit length is
C
q 2 0 r

V ln b a 
(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
01/11
ENGR 227
11
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
12
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.
Potential, V at y=0 and x=d-a




1
q
1
ln 

ln



2 0  p  d  a  2 0  p  d  a 
 d  pa
q
V
ln 

2 0  p  d  a 
V
q
Hence the capacitance per unit length is,
If d>>a this can be
simplified to
01/11
C
q

2V
C
q

2V
 0
d  pa
ln 

 pd a
 0
 2d 
ln 

 a 
ENGR 227
13
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.
As in the last lecture the ground plane
acts as a mirror so the geometry is
equivalent to two parallel strips.
w
h
01/11
ENGR 227
t
14
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)
C
•
h

2h
 r  0 wL
h
Hence the capacitance per unit length is
C
•
 r 0 A
w
 r 0 w
h
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.
01/11
ENGR 227
15
Microstrip Transmission Line
Remember we approximate this as a strip
embedded in an infinite dielectric of
lower permittivity.
e 
 r 1  r 1 
2

h
1



2  w
0.55
The voltage is hence
V
h 
w 
1



2 0 e  w  h 
w
And the capacitance per unit length is
C
01/11
h
t
2 0 e w
2 0 e
q w



w  h
h 
V
V

h 1 

 

w

h
w
w

h

 

ENGR 227
16
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.
w
h
t
The effective dielectric constant of the strip is
e 
 r 1  r 1 
2

h
1



2  w
0.55
10.9  1 10.9  1  1 
e 

1  
2
2  1
0.55
Hence the capacitance is
C
01/11
w
V

A better approximation is to
approximate the strip as a
cylindrical wire but that is
beyond this course.
2 0 e

h
h





 w wh 
 9.33
2  8.85 1012  9.33
1 1 
 

 1 11 
ENGR 227
 0.11 nF
17
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
Q
q
Q2
W   dq 
C
2C
0
•
Hence the stored energy is
•
UE=Q2/(2C) =QV/2 = CV2/2
01/11
ENGR 227
18
Parasitic capacitance
•
Electrostatic coupling between wires in a circuit
•
Earthed metallic screens are used to eliminate unwanted coupling
01/11
ENGR 227
19
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
20
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
21
Coupled Microstrip
• For two coupled microstrip
lines with a ground plane the
potentials equal
• V1=Q1/C1G + Q2/C12
C12
Line 1
C1G
Line 2
C2G
• V2=Q2/C2G + Q1/C21
• Hence the matrix is
V1  1/ C1G 1/ C12   Q1 
 Q 
V   1/ C
1/
C
21
2G   2 
 2 
• This is simplified as C12=C21
01/11
ENGR 227
22
Coupled Microstrip
d
w
h
t
• The capacitances C1G and C2G
can be calculated using
previous formulae.
Line 1
C1G
• 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.
01/11
C12
ENGR 227
Line 2
C2G
23
Coupled Microstrip
d
w
h
t
• Previously we say that the capacitance of two parallel wires
was
 r  0
q
C

2V
 4d 
ln 

 t 
• If half is in dielectric and half is in air
C
01/11
q

2V
 r  0
 4d 
2 ln 

 t 

 0
 4d 
2 ln 

 t 
ENGR 227
24