1. Applied Electromagnetic Waves
ECE 3317
Applied Electromagnetic Waves
Prof. David R. Jackson
Fall 2018
Notes 16
Plane Waves in
Good Conductors

2. Good Conductor Good conductor: Hence Recall Use Therefore
Example: copper
Hence
Recall
Use
Therefore

3. Skin Depth
Denote
“skin depth”
Then we have

4. Skin Depth (cont.)
Hence

5. Skin Depth (cont.) Example: copper Frequency  1 [Hz] 6.6 [cm] 10 [Hz]
6.6 [mm]
1 [kHz]
2.1 [mm]
10 [kHz]
0.66 [mm]
100 [kHz]
0.21 [mm]
1 [MHz]
66 [m]
10 [MHz]
21 [m]
100 [MHz]
6.6 [m]
1 [GHz]
2.1 [m]
10 [GHz]
0.66 [m]
100 [GHz]
0.21 [m]
Example: copper

6. Skin Depth (cont.)
The same penetration principle holds for curved conductors, as long as the radius of curvature is large compared with the skin depth.
The distance z is measured from the boundary of the conductor.
Coax
Penetration into inner conductor
Penetration into outer conductor

7. Regions of strong currents
Skin Depth (cont.)
Coax
Regions of strong currents
The fields are confined inside the coax if

8. Equivalent surface current
Surface Impedance
Equivalent surface current

9. Surface Impedance (cont.)
Actual current
Surface current model
Hence

10. Surface Impedance (cont.)
Define the surface impedance:

11. Surface Impedance (cont.)
Hence
We then have

12. Surface Impedance (cont.)
Define “surface resistance” and “surface reactance” from
We then have:

13. Surface Impedance (cont.)
Frequency Rs
1 [Hz]
2.610-7 []
10 [Hz]
8.310-7 []
100 [Hz]
2.610-6 []
1 [kHz]
8.310-6 []
10 [kHz]
2.610-5 []
100 [kHz]
8.310-5[]
1 [MHz]
2.610-4 []
10 [MHz]
8.310-4 []
100 [MHz]
[]
1 [GHz]
[]
10 [GHz]
0.026 []
100 [GHz]
0.083[]
Example: copper

14. Impedance of Wire Find the high-frequency impedance for a solid wire.
Z = impedance seen by source + -
Note: The current mainly flows on the outside surface of the wire!

15. Impedance of Wire (cont.)
Surface-current model:
Z = R + j X = impedance
Hence
Therefore, we have

16. Impedance of Wire (cont.)
Equivalent circuit: R jX

17. Impedance of Wire (cont.)
Example: copper wire a
R = X
10 [m]
6.57 []
0.1 [mm]
0.657 []
1 [mm]
[]
10 [mm]
[]
Assume:

18. Impedance of Wire (cont.)
Compare with the same wire at DC: DC
1.0 GHz a R
10 [m]
2.7 []
0.1 [mm]
0.027 []
1 [mm]
2.710-4 []
10 [mm]
2.710-6 [] a
R = X
10 [m]
6.57 []
0.1 [mm]
0.657 []
1 [mm]
[]
10 [mm]
[]

19. Coax We use the surface resistance concept to calculate the
resistance per unit length R of coax.
For a length l :
Resistance per unit length:

20. Coax (cont.)
Coax Formulas
(R, L, G, C)
Recall:

21. Coax (cont.) Approximate attenuation in dB/m for RG59 coax: Frequency
(from Wikipedia)
Frequency
RG59 Coax
1 [MHz]
0.01
10 [MHz]
0.03
100 [MHz]
0.11
1 [GHz]
0.40
5 [GHz]
1.0
10 [GHz]
1.5
20 [GHz]
2.3

22. Coax (cont.)
Fiber-optic guides give a much lower attenuation than coax:
dB/m
Frequency
RG59 Coax
1 [MHz]
0.01
10 [MHz]
0.03
100 [MHz]
0.11
1 [GHz]
0.40
5 [GHz]
1.0
10 [GHz]
1.5
20 [GHz]
2.3
Fiber optic cable
Typical single-mode fiber optic cable: 0.3 dB/km
Typical multimode fiber optic cable: 3 dB/km

23. Coax (cont.) Internal reactance per unit length:
The skin effect will also contribute to an extra inductance per unit length, called the “internal inductance” per unit length.
This is due to magnetic energy stored inside the conductor.)
Internal reactance per unit length:
Internal inductance per unit length:
The internal inductance is usually neglected in practice.
(It is usually small compared with the external inductance, which is calculated assuming perfect conductors.)