6. dielectrics and capacitance -...
TRANSCRIPT
6. Dielectrics and
Capacitance
May 2015
Dept. of Electronics Engineering
Sogang University
Microwave & Millimeter-wave Lab. 1
2
Teaching points
Capacitance/Capacitors
1. Capacitance
- Definition
- Examples
- Working principles
2. Poisson’s and Laplace’s equations
- examples
3
Capacitance: definition
Electric flux from directed from M2 to M1: potential difference between
two conductors exits
- let the potential difference between M2 to M1 as V0
define the capacitance between two conductors as
𝐶 =𝑄
𝑉0F(Farad=C/V)
or
𝐶 = 𝑆 𝜖0𝐸∙𝑑 𝑆
− −+𝐸∙𝑑𝐿
The capacitance is independent of 𝑄 𝑎𝑛𝑑 𝑉0
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Example 1 – parallel plate
Uniform sheets of surface charge
- Electric field between plates as
𝐸 =𝜌𝑠
𝜖 𝑎𝑧
the flux density
𝐷 = 𝜌𝑠 𝑎𝑧
Potential difference between plates
𝑉0 = − −+𝐸 ∙ 𝑑𝐿 = − 𝑑
0 𝜌𝑠
𝜖𝑑𝑧 =
𝜌𝑠
𝜖𝑑
The capacitance is independent of 𝑄 𝑎𝑛𝑑 𝑉0
Q = 𝜌𝑠𝑆, 𝑉0 =𝜌𝑠
𝜖𝑑 C =
𝑄
𝑉0=
𝜖𝑆
𝑑
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Example 1 – parallel plate(2)
Total energy stored in the capacitor:
𝑊𝐸 =1
2 𝑣𝑜𝑙 𝜖𝐸
2 =1
2 0𝑆 0𝑑 𝜖𝜌𝑠
2
𝜖2𝑑𝑧 𝑑𝑆 =
1
2
𝜖𝑆
𝑑
𝜌𝑠2𝑑2
𝜖2
or
𝑊𝐸 =1
2𝐶𝑉0
2 =1
2𝑄𝑉0 =
1
2
𝑄2
𝐶
With a fixed V0 the stored energy increases as the dielectric
constant of the medium increases.
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Example 2 – Coaxial cables and Spheres
Coaxial capacitor:
𝐸 =𝜌𝐿
2𝜋𝜖𝜌 𝑉0 = − 𝑏
𝑎𝐸𝑑𝐿 =
𝜌𝐿
2𝜋𝜖𝑙𝑛 𝑏/𝑎 =
𝑄
2𝜋𝜖𝐿𝑙𝑛 𝑏/𝑎 𝐶 =
2𝜋𝜖𝐿
𝑙𝑛 𝑏/𝑎
Concentric spheres (b>a)
𝐸 =𝑄
4𝜋𝜖𝑟2 𝑎𝑟 𝑉𝑎𝑏 =
𝑄
4𝜋𝜖
1
𝑎−
1
𝑏 𝐶 =
𝑄
𝑉𝑎𝑏=
4𝜋𝜖1
𝑎−1
𝑏
In case the outer sphere infinitely large, the capacitance becomes
𝐶 = 4𝜋𝜖𝑎
For a diameter of 1 cm,
𝐶 = 0.556 𝑝𝐹 in free space
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Example 3 – dielectric interfaces: series
The normal component of D is equal to the surface charge
density: 𝐷𝑧 = 𝜌𝑠
Capacitance:
x
xy
z
Szz EE 2211
1
1
2
2
1122
0dddEdELdEV SS
zzd
SSSdDQ ssS )ˆ)(ˆ( zz
1
1
2
21
1
2
2
dd
S
dd
S
V
QC
SS
s
The tangential component of E is continuous:
Capacitance:
equipotential on the conductor surfaces.
021 EEE
dEdEdELdEVd 021
0
2201102121 21 SESESdDSdDQQQ
SS
d
S
d
S
V
QC 2211
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Example 4 – dielectric interfaces: parallel
21 EE
2211 // zz DD
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Poisson’s equation and Lapalce’s equation
- Differential forms :
𝛻 ∙ 𝐷 = 𝜌𝑣 Gauss’s law
𝐸 = −𝛻𝑉 𝐷 = 𝜖𝐸
𝛻 ∙ 𝐷 = 𝛻 ∙ 𝜖𝐸 = −𝜖𝛻 ∙ 𝛻𝑉 = −𝜖𝛻2𝑉 = 𝜌𝑣
𝛻2𝑉 = −𝜌𝑣
𝜖: Poisson’s equation
𝛻 ∙ 𝐸 =𝜕𝐸𝑥𝜕𝑥
+𝜕𝐸𝑦𝜕𝑦
+𝜕𝐸𝑧𝜕𝑧
𝛻𝑉 =𝜕𝑉
𝜕𝑥 𝑎𝑥 +
𝜕𝑉
𝜕𝑦 𝑎𝑦 +
𝜕𝑉
𝜕𝑧 𝑎𝑧
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Poisson’s equation and Lapalce’s equation(2)
𝛻 ∙ 𝛻𝑉 =𝜕
𝜕𝑥
𝜕𝑉
𝜕𝑥+
𝜕
𝜕𝑦
𝜕𝑉
𝜕𝑦+
𝜕
𝜕𝑧
𝜕𝑉
𝜕𝑧=𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2
which leads to
𝛻 ∙ 𝛻𝑉 = 𝛻2𝑉 =𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2= −
𝜌𝑣𝜖
Poisson’s equation
In case 𝜌𝑣 = 0,
𝛻2𝑉 =𝜕2𝑉
𝜕𝑥2+𝜕2𝑉
𝜕𝑦2+𝜕2𝑉
𝜕𝑧2= 0
Laplace’s equation
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Poisson’s equation and Lapalce’s equation(3)
Laplace equation in other coordinate systems:
- cylindrical
𝛻2𝑉 =1
𝜌
𝜕
𝜕𝜌𝜌𝜕𝑉
𝜕𝜌+
1
𝜌2𝜕2𝑉
𝜕𝜑2+𝜕2𝑉
𝜕𝑧2= 0
- spherical
𝛻2𝑉 =1
𝑟2𝜕
𝜕𝑟𝑟2
𝜕𝑉
𝜕𝑟+
1
𝑟2𝑠𝑖𝑛2𝜃
𝜕
𝜕𝑟𝑠𝑖𝑛𝜃
𝜕𝑉
𝜕𝜃+
1
𝑟2𝑠𝑖𝑛2𝜃
𝜕2𝑉
𝜕𝜑2= 0
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Poisson’s equation and Lapalce’s equation(4)
Examples: parallel plate capacitor
𝑉 𝑧 = 𝑑 = 𝑉0𝑉 𝑧 = 0 = 0
the Laplace equation:
𝜕2𝑉
𝜕𝑧2= 0
𝑉 𝑧 = 𝐴𝑧 + 𝐵
𝐵 = 0, 𝐴 = 𝑉0/𝑑 𝑉 𝑧 = 𝑉0𝑧/𝑑 𝐸 = −𝛻𝑉 = −𝑉0/𝑑 𝑎𝑧
𝐷 = 𝜖𝐸 = −𝜖𝑉0
𝑑 𝑎𝑧 = −𝜌𝑠 𝑎𝑧 𝐶 =
𝑄
𝑉0=
𝜖𝑆
𝑑
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Poisson’s equation and Lapalce’s equation(5)
Examples: coaxial capacitor
𝑉 𝜌 = 𝑎 = 𝑉0, 𝑉 𝜌 = 𝑏 = 0
Laplace’equation becomes
1
𝜌
𝜕
𝜕𝜌𝜌𝜕𝑉
𝜕𝜌= 0, 𝜌
𝜕𝑉
𝜕𝜌= 𝐴 𝑉 𝜌 = 𝐴𝑙𝑛𝜌 + 𝐵
Apply the conditions: 𝑉 𝜌 = 𝑉0ln(𝑏
𝜌)/ln(
𝑏
𝑎)
Electric field: 𝐸 = −𝛻𝑉 =𝑉0
𝜌ln( 𝑏 𝑎) 𝑎𝜌
Charge density: 𝐷𝑛 =𝜖𝑉0
𝑎ln( 𝑏 𝑎)= 𝜌𝑠 𝑄 = 2𝜋𝑎𝜌𝑠 =
2𝜋𝑎𝜖𝑉0
𝑎ln( 𝑏 𝑎)
Capacitance/m: 𝐶 = 2𝜋𝜖ln( 𝑏 𝑎)
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Capacitance
•The magnitude of an electric field is proportional to charges, and voltages are proportional to electric field.
Hence, charges are proportional to voltages. This proportionality constant is called capacitance.
CVQVQQEV
VV
d
d
d
V
QC SS
aE
sE
aE
SS
S
d
d
SC
d
S
dz
da
dz
da
d
d
V
QC
d
S
S
S
d
S
S
S
S
S
00
1ˆˆ
)ˆ(ˆ
ˆ
zz
zz
sE
aE
zEz
x
Example: Capacitance of a parallel plate capacitor
2
00 2
1f
VQ
total CVdCdqWff
Electrostatic energy
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Homework
Problems,
6.3, 6.6, 6.9, 6.12, 6.22, 6.26, 6.41