EE 217

Lecture 2 Boundary Conditions

1s1415 Revision August 2015

Solution to Maxwell’s Equations

• Unknown: 𝑬𝑬, 𝑫𝑫 or 𝑯𝑯, 𝑩𝑩 at a certain region in space • Maxwell’s equations Differential equations • Solution to differential equations

– General solution – Particular solution

• Boundary (initial) conditions

2

Boundary Conditions

Electrostatic Boundary Conditions

3

Fields at a Boundary

• Fields can be split into normal and tangential components at the boundary

4 Pozar. Microwave Engineering. 4th ed. p. 12.

Gauss’s Law at Boundary

• 𝒏𝒏� Normal vector • Δ𝑆𝑆 Area of circle • 𝜌𝜌𝑆𝑆 Surface charge density (C/m2) • 𝐷𝐷𝑛𝑛𝑛, 𝐷𝐷𝑛𝑛𝑛 Electric flux normal to boundary

5 Pozar. Microwave Engineering. 4th ed. p. 13.

Gauss’s Law at Boundary

� 𝑫𝑫 ⋅ 𝑑𝑑𝒔𝒔𝑆𝑆

= 𝑄𝑄 = � 𝜌𝜌𝑑𝑑𝜌𝜌𝑉𝑉

• Let ℎ → 0, tangential component through cylinder wall becomes irrelevant

6 Pozar. Microwave Engineering. 4th ed. p. 13.

Gauss’s Law at Boundary

� 𝑫𝑫 ⋅ 𝑑𝑑𝒔𝒔𝑆𝑆

= 𝑄𝑄 = � 𝜌𝜌𝑑𝑑𝜌𝜌𝑉𝑉

𝐷𝐷𝑛𝑛𝑛Δ𝑆𝑆 − 𝐷𝐷𝑛𝑛𝑛Δ𝑆𝑆 = 𝜌𝜌𝑆𝑆Δ𝑆𝑆 𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 − 𝑫𝑫𝟏𝟏 = 𝜌𝜌𝑆𝑆

7

Pozar. Microwave Engineering. 4th ed. p. 13.

Gauss’s Law at Boundary

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• Normal electric flux at boundary

𝐷𝐷𝑛𝑛𝑛 − 𝐷𝐷𝑛𝑛𝑛 = 𝜌𝜌𝑆𝑆 𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 − 𝑫𝑫𝟏𝟏 = 𝜌𝜌𝑆𝑆

• Normal magnetic flux at boundary

– There is no free magnetic charge

𝐵𝐵𝑛𝑛𝑛 = 𝐵𝐵𝑛𝑛𝑛 𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 𝒏𝒏� ⋅ 𝑩𝑩𝟏𝟏

Faraday’s Law at Boundary

• 𝒏𝒏� Normal vector • Δ𝑙𝑙 Length of rectangular path • 𝑀𝑀𝑆𝑆𝑛𝑛 Magnetic surface current density (V/m) • 𝐸𝐸𝑡𝑡𝑛, 𝐸𝐸𝑡𝑡𝑛 Electric field tangential to boundary

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Faraday’s Law at Boundary

� 𝑬𝑬 ⋅ 𝑑𝑑𝒍𝒍𝐶𝐶

= −𝜕𝜕𝜕𝜕𝜕𝜕� 𝑩𝑩 ⋅ 𝑑𝑑𝒔𝒔𝑆𝑆

− � 𝑴𝑴 ⋅ 𝑑𝑑𝒔𝒔𝑆𝑆

• Let ℎ → 0, ∫𝑩𝑩 ⋅ 𝑑𝑑𝒔𝒔 → 0 𝐸𝐸𝑡𝑡𝑛Δ𝑙𝑙 − 𝐸𝐸𝑡𝑡𝑛Δ𝑙𝑙 = −𝑀𝑀𝑆𝑆Δ𝑙𝑙 𝑬𝑬𝟐𝟐 − 𝑬𝑬𝟏𝟏 × 𝒏𝒏� = 𝑴𝑴𝑺𝑺

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Faraday’s Law at Boundary

11

• Tangential electric field at boundary

𝐸𝐸𝑡𝑡𝑛 − 𝐸𝐸𝑡𝑡𝑛 = −𝑀𝑀𝑆𝑆 𝑬𝑬𝟐𝟐 − 𝑬𝑬𝟏𝟏 × 𝒏𝒏� = 𝑴𝑴𝑺𝑺 • Tangential magnetic field at boundary

– Apply Ampere’s law

𝐻𝐻𝑡𝑡𝑛 − 𝐻𝐻𝑡𝑡𝑛 = 𝐽𝐽𝑆𝑆 𝒏𝒏� × 𝑯𝑯𝟐𝟐 − 𝑯𝑯𝟏𝟏 = 𝑱𝑱𝑺𝑺

Boundary Conditions

12

𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 − 𝑫𝑫𝟏𝟏 = 𝜌𝜌𝑆𝑆

𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 𝒏𝒏� ⋅ 𝑩𝑩𝟏𝟏

𝑬𝑬𝟐𝟐 − 𝑬𝑬𝟏𝟏 × 𝒏𝒏� = 𝑴𝑴𝑺𝑺

𝒏𝒏� × 𝑯𝑯𝟐𝟐 − 𝑯𝑯𝟏𝟏 = 𝑱𝑱𝑺𝑺

2 Lossless Dielectric Materials

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𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 − 𝑫𝑫𝟏𝟏 = 𝜌𝜌𝑆𝑆 = 0

𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 𝒏𝒏� ⋅ 𝑩𝑩𝟏𝟏

𝑬𝑬𝟐𝟐 − 𝑬𝑬𝟏𝟏 × 𝒏𝒏� = 𝑴𝑴𝑺𝑺 = 0

𝒏𝒏� × 𝑯𝑯𝟐𝟐 − 𝑯𝑯𝟏𝟏 = 𝑱𝑱𝑺𝑺 = 0

2 Lossless Dielectric Materials

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𝒏𝒏� ⋅ 𝑫𝑫𝟏𝟏 = 𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐

𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 𝒏𝒏� ⋅ 𝑩𝑩𝟏𝟏

𝒏𝒏� × 𝑬𝑬𝟏𝟏 = 𝒏𝒏� × 𝑬𝑬𝟐𝟐

𝒏𝒏� × 𝑯𝑯𝟏𝟏 = 𝒏𝒏� × 𝑯𝑯𝟐𝟐 • Normal components of 𝑫𝑫 and 𝑩𝑩 are continuous • Tangential components of 𝑬𝑬 and 𝑯𝑯 are continuous

PC (1) to Lossless Dielectric (2)

• Electrons (charges) inside a perfect conductor? – Electrons will repel each other, will try to get as far away as

possible from each other – But they can’t get out of the conductor – No charges remain inside, all charges are on the surface – No electric fields inside conductor (try Gauss’s Law)

15 Conductor cross section Conductor cross section

PC (1) to Lossless Dielectric (2)

• Electrons (charges) inside a perfect conductor? – What about current? – Electrons will repel each other, will try to get as far away as

possible from each other – No current inside, only surface current – No magnetic field inside conductor (try Ampere’s law)

16 Conductor cross section Conductor cross section

PC (1) to Lossless Dielectric (2)

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𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 − 𝑫𝑫𝟏𝟏 = 𝜌𝜌𝑆𝑆, 𝑫𝑫𝟏𝟏 = 0

𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 𝒏𝒏� ⋅ 𝑩𝑩𝟏𝟏, 𝑩𝑩𝟏𝟏 = 0

𝑬𝑬𝟐𝟐 − 𝑬𝑬𝟏𝟏 × 𝒏𝒏� = 𝑴𝑴𝑺𝑺 = 0, 𝑬𝑬𝟏𝟏 = 0

𝒏𝒏� × 𝑯𝑯𝟐𝟐 − 𝑯𝑯𝟏𝟏 = 𝑱𝑱𝑺𝑺, 𝑯𝑯𝟏𝟏 = 0

PC (1) to Lossless Dielectric (2)

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𝒏𝒏� ⋅ 𝑫𝑫𝟐𝟐 = 𝜌𝜌𝑆𝑆

𝒏𝒏� ⋅ 𝑩𝑩𝟐𝟐 = 0,

𝒏𝒏� × 𝑬𝑬𝟐𝟐 = 0

𝒏𝒏� × 𝑯𝑯𝟐𝟐 = 𝑱𝑱𝑺𝑺 • Perfect conductor A.K.A. electric wall • Tangential components of 𝑬𝑬 are “shorted out”

Boundary Conditions

Boundary Conditions for AC Fields

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Continuity for AC Fields

• Time-varying case: 𝑬𝑬 and 𝑯𝑯 are interdependent

𝛻𝛻 × 𝑬𝑬 = −𝜕𝜕𝑩𝑩𝜕𝜕𝜕𝜕

−𝑴𝑴𝒔𝒔

𝛻𝛻 × 𝑯𝑯 =𝜕𝜕𝑫𝑫𝜕𝜕𝜕𝜕

+ 𝑱𝑱𝒔𝒔

• Normal and tangential components are interdependent

• At a general boundary, – Can solve tangential component continuity – Solution can be checked using normal components – 𝑫𝑫 can be discontinuous! How?

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Boundary Conditions

Good Conductors

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Good Conductor

• Good but not perfect – 𝜎𝜎 is large but finite

• Conduction current follows Ohm’s law: 𝑱𝑱 = 𝜎𝜎𝑬𝑬

• Ampere’s law (phasor simplification): 𝛻𝛻 × 𝑯𝑯 = 𝑗𝑗𝑗𝑗𝑫𝑫 + 𝑱𝑱 𝛻𝛻 × 𝑯𝑯 = 𝑗𝑗𝑗𝑗𝑗𝑗𝑬𝑬 + 𝜎𝜎𝑬𝑬

𝛻𝛻 × 𝑯𝑯 = 𝑗𝑗𝑗𝑗𝑗𝑗 + 𝜎𝜎 𝑬𝑬 ≈ 𝜎𝜎𝑬𝑬 – For a good conductor, 𝑗𝑗𝑗𝑗 ≪ 𝜎𝜎

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Good Conductor

• Using the 𝛻𝛻 ⋅ 𝛻𝛻 × 𝑨𝑨 = 0, 𝛻𝛻 ⋅ 𝛻𝛻 × 𝑯𝑯 = 𝑗𝑗𝑗𝑗𝑗𝑗 + 𝜎𝜎 𝛻𝛻 ⋅ 𝑬𝑬 = 0

𝛻𝛻 ⋅ 𝑬𝑬 = 0

• Thus, inside a good conductor, 𝛻𝛻 ⋅ 𝑗𝑗𝑬𝑬 = 𝛻𝛻 ⋅ 𝑫𝑫 = 𝜌𝜌 = 0

– Zero charge density

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Field Penetration

• Faraday’s law (phasor simplification), 𝛻𝛻 × 𝑬𝑬 = −𝑗𝑗𝑗𝑗𝑩𝑩 −𝑴𝑴 𝛻𝛻 × 𝑬𝑬 = −𝑗𝑗𝑗𝑗𝑗𝑗𝑯𝑯

• Using the 𝛻𝛻 × 𝛻𝛻 × 𝑨𝑨 = 𝛻𝛻 𝛻𝛻 ⋅ 𝑨𝑨 − 𝛻𝛻𝑛𝑨𝑨, 𝛻𝛻 × 𝛻𝛻 × 𝑬𝑬 = 𝛻𝛻 𝛻𝛻 ⋅ 𝑬𝑬 − 𝛻𝛻𝑛𝑬𝑬 = −𝑗𝑗𝑗𝑗𝑗𝑗 𝛻𝛻 × 𝑯𝑯

• Since 𝛻𝛻 ⋅ 𝑬𝑬 = 0, 𝛻𝛻𝑛𝑬𝑬 = 𝑗𝑗𝑗𝑗𝑗𝑗 𝛻𝛻 × 𝑯𝑯

• Substituting 𝛻𝛻 × 𝑯𝑯 = 𝜎𝜎𝑬𝑬, 𝛻𝛻𝑛𝑬𝑬 = 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝑬𝑬

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Field Penetration

• Leading to 𝜎𝜎 𝛻𝛻𝑛𝑬𝑬 = 𝜎𝜎 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝑬𝑬

𝛻𝛻𝑛𝑱𝑱 = 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝑱𝑱 • Similarly,

𝛻𝛻 × 𝑯𝑯 = 𝜎𝜎𝑬𝑬 𝛻𝛻 × 𝛻𝛻 × 𝑯𝑯 = 𝛻𝛻 𝛻𝛻 ⋅ 𝑯𝑯 − 𝛻𝛻𝑛𝑯𝑯 = 𝜎𝜎 𝛻𝛻 × 𝑬𝑬

𝛻𝛻𝑛𝑯𝑯 = 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝑯𝑯 – Very similar differential equation as 𝑱𝑱 and 𝑬𝑬

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Field Penetration

• Consider a plane conductor of infinite depth (𝑥𝑥) – Semi-infinite solid for 𝑥𝑥 > 0 – No field variations along 𝑦𝑦 and 𝑧𝑧 – We expect variation along 𝑥𝑥, as the fields penetrates

• Let 𝑬𝑬 be polarized in the 𝑧𝑧 direction 𝛻𝛻𝑛𝑬𝑬 = 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝑬𝑬

𝑑𝑑𝑛𝐸𝐸𝑧𝑧𝑑𝑑𝑥𝑥𝑛

= 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎𝐸𝐸𝑧𝑧 = 𝜏𝜏𝑛𝐸𝐸𝑧𝑧

• Looking at 𝜏𝜏,

𝜏𝜏𝑛 = 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎 → 𝜏𝜏 = 1 + 𝑗𝑗 𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎 =1 + 𝑗𝑗𝛿𝛿

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Field Penetration

• Solving for 𝐸𝐸𝑧𝑧 using 𝜏𝜏, 𝐸𝐸𝑧𝑧 = 𝐶𝐶𝑛𝑒𝑒−𝜏𝜏𝜏𝜏 + 𝐶𝐶𝑛𝑒𝑒𝜏𝜏𝜏𝜏

– Where 𝐶𝐶𝑛 and 𝐶𝐶𝑛 are constants – 𝐶𝐶𝑛 needs to be zero, why? – 𝐶𝐶𝑛 is the value of 𝐸𝐸𝑧𝑧 at the conductor surface, 𝑥𝑥 = 0

• Using the expansion of 𝜏𝜏,

𝐸𝐸𝑧𝑧 = 𝐸𝐸0𝑒𝑒−𝑛+𝑗𝑗𝛿𝛿 𝜏𝜏 → 𝐸𝐸𝑧𝑧 = 𝐸𝐸0𝑒𝑒

−𝜏𝜏𝛿𝛿𝑒𝑒−𝑗𝑗𝜏𝜏𝛿𝛿

• Similarly for 𝑯𝑯 and 𝑱𝑱, 𝐻𝐻𝑦𝑦 = 𝐻𝐻0𝑒𝑒

−𝜏𝜏𝛿𝛿𝑒𝑒−𝑗𝑗𝜏𝜏𝛿𝛿 , 𝐽𝐽𝑧𝑧 = 𝐽𝐽0𝑒𝑒

−𝜏𝜏𝛿𝛿𝑒𝑒−𝑗𝑗𝜏𝜏𝛿𝛿

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Field Penetration

• Note: Assumed time dependence 𝐸𝐸𝑧𝑧 𝑥𝑥, 𝜕𝜕 = Re 𝐸𝐸𝑧𝑧 𝑥𝑥 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡

𝐸𝐸𝑧𝑧 𝑥𝑥, 𝜕𝜕 = Re 𝐸𝐸0𝑒𝑒−𝜏𝜏𝛿𝛿𝑒𝑒−

𝑗𝑗𝜏𝜏𝛿𝛿 𝑒𝑒𝑗𝑗𝑗𝑗𝑡𝑡 = Re 𝐸𝐸0𝑒𝑒

−𝜏𝜏𝛿𝛿𝑒𝑒𝑗𝑗 𝑗𝑗𝑡𝑡−𝜏𝜏𝛿𝛿

𝐸𝐸𝑧𝑧 𝑥𝑥, 𝜕𝜕 = 𝐸𝐸0𝑒𝑒−𝜏𝜏𝛿𝛿 cos 𝑗𝑗𝜕𝜕 −

𝑥𝑥𝛿𝛿

– Amplitude decreases exponentially as 𝑥𝑥 increases – Same behavior for 𝐻𝐻𝑦𝑦 and 𝐽𝐽𝑧𝑧

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Field Penetration

29 Ramo. Fields and Waves in Communication Electronics. 3rd ed. p. 152.

Skin Depth

• Depth at which the fields have decreased to 1 𝑒𝑒⁄ (36.9%) of the values at the conductor surface

• Skin depth, 𝛿𝛿

𝛿𝛿 =1𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎

– Decreases with 𝑗𝑗, 𝜎𝜎. Value for ideal conductor?

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𝜎𝜎 𝛿𝛿𝑠𝑠 @ 10 GHz

Aluminum 3.816 × 107S 8.14 × 10−7m

Gold 4.098 × 107S 7.86 × 10−7m

Copper 5.813 × 107S 6.60 × 10−7m

Silver 6.173 × 107S 6.40 × 10−7m

Internal Impedance

• Total current (per unit width) in the semi-infinite conductor

𝐽𝐽𝑠𝑠𝑧𝑧 = � 𝐽𝐽𝑧𝑧𝑑𝑑𝑥𝑥∞

𝜏𝜏=0= � 𝐽𝐽0𝑒𝑒

−𝑛+𝑗𝑗𝛿𝛿 𝜏𝜏𝑑𝑑𝑥𝑥∞

𝜏𝜏=0=

𝐽𝐽0𝛿𝛿1 + 𝑗𝑗

• Using Ohm’s law at the surface,

𝐸𝐸𝑧𝑧0 =𝐽𝐽0𝜎𝜎

• Impedance per unit length and unit width

𝑍𝑍𝑠𝑠 =𝐸𝐸𝑧𝑧0𝐽𝐽𝑠𝑠𝑧𝑧

=𝐽𝐽0 𝜎𝜎⁄

𝐽𝐽0𝛿𝛿 1 + 𝑗𝑗⁄ =1 + 𝑗𝑗𝜎𝜎𝛿𝛿

– 𝐸𝐸𝑧𝑧0 is applied at the surface, 𝐽𝐽𝑠𝑠𝑧𝑧 is the resulting current

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Internal Impedance

• Impedance for a unit length and unit width

𝑍𝑍𝑠𝑠 =1 + 𝑗𝑗𝜎𝜎𝛿𝛿

• Internal resistance (surface resistivity)

𝑅𝑅𝑠𝑠 =1𝜎𝜎𝛿𝛿

=𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎

Ω

• Internal reactance

𝑗𝑗𝐿𝐿𝑖𝑖 =1𝜎𝜎𝛿𝛿

= 𝑅𝑅𝑠𝑠

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Internal Impedance

• Internal resistance (surface resistivity)

𝑅𝑅𝑠𝑠 =1𝜎𝜎𝛿𝛿

=𝑗𝑗𝑗𝑗𝑗𝑗𝜎𝜎

Ω

– Note: DC resistance, 𝑅𝑅 = 𝑛𝜎𝜎𝑙𝑙𝐴𝐴

– Same DC resistance per unit length and width as a plane conductor of depth 𝛿𝛿

33 http://www.caplinq.com/blog/linqstat-volume-resistivity-vs-volume-conductivity-vs-surface-resistivity_267/