9/27/2017
Course Instructor Dr. Raymond C. Rumpf Office: A‐337 Phone: (915) 747‐6958 E‐Mail: [email protected]
EE 4347 Applied Electromagnetics
Topic 4a Transmission Lines
Transmission Lines These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited Slide 1
Lecture Outline • Introduction • Transmission Line Equations • Transmission Line Wave Equations • Transmission Line Parameters – and
– Characteristic Impedance, Z0 • Special Cases of Transmission Lines – General transmission lines – Lossless lines – Weakly absorbing lines – Distortionless lines • Examples – RG‐59 coaxial cable – Microstrip design
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Introduction
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Map of Waveguides (LI Media)
• Confines and transports waves. Waveguides • Supports higher‐order modes. • Has one or less conductors. • Usually what is implied by Transmission Lines “Pipes” the label “waveguide.” • Contains two or more conductors. • No low‐frequency cutoff. • Thought of more as a circuit clement Metal Shell Pipes Dielectric Pipes Homogeneous Inhomogeneous • Enclosed by metal. • Composed of a core and a cladding. • Has TEM mode. • Supports only • Does not support TEM mode. • Symmetric waveguides have no • Has TE and TM quasi‐(TEM, TE, • Has a low frequency cutoff. low‐frequency cutoff. modes. & TM) modes. Homogeneous Channel Waveguides • Supports TE and TM modes • Confinement along two axes. coaxial microstrip • TE & TM modes only supported in circularly symmetric guides.
rectangular circular dual‐ridge Single‐Ended stripline coplanar optical Fiberphotonic crystal rib Inhomogeneous • Supports TE and TM modes Slab Waveguides buried parallel coplanar strips only if one axis is uniform. plate • Otherwise supports quasi‐TM • Confinement only along one axis. and quasi‐TE modes. • Supports TE and TM modes. • Interfaces can support surface waves. Differential
shielded pair slotline no uniform axis uniform axis dielectric Slab large‐area interface (no TE or TM) (has TE and TM) parallel plate Transmission Lines Slide 4
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Transmission Line Parameters RLGC
We can think transmission lines as being composed of millions of tiny little circuit elements that are distributed along the length of the line.
In fact, these circuit element are not discrete, but continuous along the length of the transmission line.
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RLGC Circuit Model
It is not technically correct to represent a transmission line with discrete circuit elements like this. However, if the size of the circuit z is very small compared to the wavelength of the signal on the transmission line, it becomes an accurate and effective way to model the transmission line.
z
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L‐Type Equivalent Circuit Model
Distributed Circuit Parameters There are many possible circuit models for R (/m) Resistance per unit length. transmission lines, but most produce the same Arises due to resistivity in the equations after analysis. conductors. L (H/m) Inductance per unit length. Arises due to stored magnetic R z L z energy around the line. G (1/m) Conductance per unit length. Arises due to conductivity in the dielectric separating the conductors. Gz Cz 1 G R C (F/m) Capacitance per unit length. Arises due to stored electric energy between the conductors. z zz
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Relation to Electromagnetic Parameters
Every transmission line with a homogeneous fill has: , , LC G C
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Fundamental Vs. Intuitive Parameters
Fundamental Parameters Intuitive Parameters Electromagnetics Electromagnetics , , n, , , , tan
Transmission Lines Transmission Lines
RLGC, , , Z0 , , , VSWR
The fundamental parameters are the The intuitive parameters provide intuitive most basic parameters needed to solve a insight about how signals behave on a transmission line problem. transmission line. However, it is difficult to be intuitive They isolate specific information to a about how they affect signals on the line. single parameter. An electromagnetic analysis is needed to determine R, L, G, and C from the The intuitive parameters are calculated geometry of the transmission line. from R, L, G, and C .
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Example RLGC Parameters
RG‐59 Coax CAT5 Twisted Pair Microstrip
R 36 mΩ m R 176 mΩ m R 150 mΩ m L 430 nH m L 490 nH m L 364 nH m G 10 m G 2 m G 3 m C 69 pF m C 49 pF m C 107 pF m
Z0 75 Z0 100 Z0 50
Surprisingly, almost all transmission lines have parameters very close to these same values.
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Transmission Line Equations
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E & H V and I
Fundamentally, all circuit problems are electromagnetic problems and can be solved as such.
All two‐conductor transmission lines either support a TEM wave or a wave very closely approximated as TEM.
An important property of TEM waves is that E is uniquely related to V and H and uniquely related to E. VEd I Hd L L
This let’s us analyze transmission lines in terms of just V and I. This makes analysis much simpler because these are scalar quantities!
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Transmission Line Equations
The transmission line equations do for transmission lines the same thing as Maxwell’s curl equations do for unguided waves.
Maxwell’s Equations Transmission Line Equations H VI E RI L t zt E I V H GV C t zt
Like Maxwell’s equations, the transmission line equations are rarely directly useful. Instead, we will derive all of the useful equations from them.
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Derivation of First TL Equation (1 of 2) R z L z
+ 2 3 + I zt, Vzt , 1 Gz Cz 4 Vz zt,
‐ ‐
z zz
Apply Kirchoff’s voltage law (KVL) to the outer loop of the equivalent circuit: Izt, Vzt , IztRz , Lz Vz zt, 0 t 1 2 4 3
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Derivation of First TL Equation (2 of 2)
We rearrange the equation by bringing all of the voltage terms to the left‐hand side of the equation, bringing all of the current terms to the right‐hand side of the equation, and then dividing both sides by z. Izt, Vzt,, IztRz Lz Vz zt ,0 t Vz zt,, Vzt Izt , RI z, t L zt
In the limit as z 0, the expression on the left‐hand side becomes a derivative with respect to z.
Vzt,, Izt RI z, t L zt
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Derivation of Second TL Equation (1 of 2)
R z L z 1 2 + I zt, I zzt , + 3 4 Vzt , Gz Cz Vz zt,
‐ ‐
z zz
Apply Kirchoff’s current law (KCL) to the main node the equivalent circuit: Vz zt, Izt , Iz zt, GzVz zt, Cz 0 t 1 2 3 4
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Derivation of Second TL Equation (2 of 2)
We rearrange the equation by bringing all of the current terms to the left‐hand side of the equation, bringing all of the voltage terms to the right‐hand side of the equation, and then dividing both sides by z. Vz zt, IztIzztGzVzztCz,, , 0 t Iz zt,, Izt Vz zt , GV z z, t C zt
In the limit as z 0, the expression on the left‐hand side becomes a derivative with respect to z.
Izt,, Vzt GV z, t C zt
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Transmission Line Wave Equations
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Starting Point – Telegrapher Equations
We start with the transmission line equations derived in the previous section.
Vzt ,, Izt Izt ,, Vzt RI z, t L GV z, t C time‐domain ztzt
For time‐harmonic (i.e. frequency‐domain) analysis, we Fourier transform the equations above.
dV z dI z R jLIz GjCVz frequency‐domain dz dz
Note: Our derivative d/dz became an ordinary derivative because z is the only independent variable left. These last equations are commonly referred to as the telegrapher equations.
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Wave Equation in Terms of V(z)
dV z dI z R jLIz Eq. (1)GjCVz Eq. (2) dz dz To derive a wave equation in terms of V(z), we first differentiate Eq. (1) with respect to z. dV2 z dIz RjL Eq. (3) dz2 dz
Second, we substitute Eq. (2) into the right‐hand side of Eq. (3) to eliminate I(z) from the equation. dV2 z R jL G jCVz dz 2 Last, we rearrange the terms to arrive at the final form of the wave equation. dV2 z RjLGjCVz 0 dz 2
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Wave Equation in Terms of I(z)
dV z dI z R jLIz Eq. (1)GjCVz Eq. (2) dz dz To derive a wave equation in terms of just I(z), we first differentiate Eq. (2) with respect to z. dI2 z dVz GjC Eq. (3) dz2 dz
Second, we substitute Eq. (1) into the right‐hand side of Eq. (3) to eliminate V(z) from the equation. dI2 z GjCRjLIz dz 2 Last, we rearrange the terms to arrive at the final form of the wave equation. dI2 z GjCRjLIz 0 dz 2
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Propagation Constant,
In our wave equations, we have a common term GjCRjL . Define the propagation constant to be j GjCRjL
Given this definition, the transmission line equations are written as dV2 z 2Vz 0 dz2 dI2 z 2 Iz 0 dz2
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Solution to the Wave Equations
If we hand the wave equations off to a mathematician, they will return with the following solutions.
2 dV z zz 2Vz 0 Vz Ve Ve dz2 00
2 dI z zz 2 Iz 0 I zIeIe dz2 00
Forward wave Backward wave
Both V(z) and I(z) have the same differential equation so it makes sense they have the same solution.
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Transmission Line Parameters:
Attenuation Coefficient, Phase Constant,
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Derivation and (1 of 7)
Step 1 – Start with our expression for .
j GjCRjL
Square this expression to get rid of square‐root on right‐hand side.
j2 GjCRjL
Expand this expression.
22j2 RG j RC j LG 2 LC
Collect real and imaginary parts on the left‐hand and right‐hand sides. 22 j2 RG 2 LC j RC LG
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Derivation and (2 of 7)
Step 2 – Generate two equations by equating real and imaginary parts.
2 RCLG
22 j2 RG 2 LC j RC LG
22RGLC 2
We now have two equations and two unknowns.
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Derivation and (3 of 7)
Step 3 – Derive a quadratic equation for 2.
2 RC LG Eq. (1a) 22RG 2 LC Eq. (1b)
Solve Eq. (1a) for . RC LG Eq. (2) 2
Substitute Eq. (2) into Eq. (1b) and simplify.
2 22RC LG RG LC 2 2 RC LG 2 22RG LC 4 2 44442RCLG2 2 RG 22 LC 2 422LC RG RC LG 0 2 Transmission Lines Slide 27
Derivation and (4 of 7)
Step 4 – Solve for 2 using the quadratic formula.
bb2 4 ac Recall the quadratic formula: ax2 bx c0 x 2a Our equation for is in the form of a quadratic equation where a 1 2 2 bLCRG 422 2 LC RG RC LG 0 2 cRCLG 2 x 2 The solution is
222 LC RG LC RG4 RC LG 2 2 2 RG2222222 LC R L G C 2
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Derivation and (5 of 7)
Step 5 – Resolve the sign of the square‐root.
RGLCR2222222 LGC 2 2
In order for this expression to always give a real value for , the sign of the square‐root must be positive.
The final expression is
RGLCR2222222 LGC 2 2
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Derivation and (6 of 7)
Step 6 – Solve for 2 using our expression for 2.
Recall Eq. (1b): 22RGLC 2
RG2222222 LC R L G C 2 2
We obtain an equation for 2 by substituting our expression for 2 into Eq. (1b).
RG2222222 LC R L G C 22RGLC 2 RG2222222 LC R L G C 2 2
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Derivation and (7 of 7)
Step 7 – We arrive at our final expressions for and in terms of the fundamental parameters R, L, G, and C by taking the square‐root of our latest expressions for 2 and 2.
RG2222222 LC R L G C 2 RGLCR2222222 LGC 2 Both and must be positive quantities for passive materials. This means we take the positive sign for the square‐root.
RG2222222 LC R L G C 2
RG2222222 LC R L G C 2
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Transmission Line Parameters:
Characteristic Impedance, Z0
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Characteristic Impedance, Z0 ()
The characteristic impedance Z0 of a transmission line is defined as the ratio of the voltage to the current at any point of a forward travelling wave. VV00 Z0 I00I
Definition for a forward Definition for a backward travelling travelling wave. wave. Notice the negative sign!
Most characteristic impedance values fall in the 50 to 100 range. The specific value of impedance is not usually of importance. What is important is when the impedance changes because this causes reflections, standing waves, and more.
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Derivation of Z0 (1 of 5)
Step 1 – Substitute our solution into the transmission line equations.
z z Vz Ve00 Ve z z Iz Ie00 Ie
dV z dI z R jLIz GjCVz dz dz
d zz d zz Ve00 Ve Ie00 Ie dz dz z z z z RjL Ie00 Ie GjC Ve00 Ve
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Derivation of Z0 (2 of 5)
Step 2 – Expand the equations and calculate the derivatives.
d zz d zz Ve00 Ve Ie00 Ie dz dz z z z z R jL Ie00 Ie GjCVe 00 Ve
zz zz Ve00 Ve Ie00 Ie z z z z R jLIe00 R jLIe GjCVe00 GjCVe
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Derivation of Z0 (3 of 5)
Step 3 – Equate the expressions multiplying the common exponential terms. VRjLI00
Ve z Ve zzz R jLIe R jLIe 00 0 0
VRjLI00
IGjCV00
Ie z Ie zzz G jCVe G jCVe 00 0 0
IGjCV00
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Derivation of Z0 (4 of 5)
Step 4 – Solve each of our four equations for V0/I0 to derive expressions for Z0.
V RjL VRjLI 0 00 Z0 I0
V0 RjL VRjLI00 Z0 I0
V 0 Z IGjCV00 0 IGjC0
V 0 Z IGjCV00 0 IGjC0
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Derivation of Z0 (5 of 5)
Step 5 – Put Z0 in terms of just R, L, G, and C.
Recall our expression for : j GjCRjL
We can substitute this into either of our expressions for Z0. RjL Z 0 GjC
Proceed with the first expression.
2 R jL RjL R jL Z0 GjCRjLGjCRjL GjC
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Final Expression for Z0 ()
We have derived a general expression for the characteristic impedance
Z0 of a transmission line in terms of the fundamental parameters R, L, G, and C. VV Definition: 00 Z0 I00I R jL R jL Expression: Z 0 GjC GjC
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Dissecting the Characteristic Impedance, Z0
The characteristic impedance describes the amplitude and phase relation between voltage and current along a transmission line. With this picture in mind, the characteristic impedance can be written as
Vz Ve z ZZ 0 00Z0 j zzV0 zZ0 Iz Ie000 e ZVe e Z0 The characteristic impedance can also be written in terms of its real and imaginary parts.
Z00RjX 0
Reactive part of Z0. This is not equal to jL or 1/jC.
Resistive part of Z0. This is not equal to R or G.
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Special Cases of Transmission Lines:
General Transmission Line
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Parameters for General TLs
Propagation Constant, jGjCRjL
Attenuation Coefficient, RG2222222 LC R L G C 2 Phase Constant, RG2222222 LC R L G C 2
Characteristic Impedance, Z0 RjL ZRjX 00 0 GjC
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Special Cases of Transmission Lines:
Lossless Lines
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Definition of Lossless TL
When we think about transmission lines, we tend to think of the special case of the lossless line because the equations simplify considerably.
For a transmission line to be lossless, it must have RG0
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Parameters for Lossless TLs
Propagation Constant, jjLC
Attenuation Coefficient,
0
Phase Constant,
LC
Characteristic Impedance, Z0 L ZRjX 00 0 C L RX 0 00C
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Special Cases of Transmission Lines:
Weakly Absorbing Line
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Definition of Weakly Absorbing TL
Most practical transmission lines have loss, but very low loss making them weakly absorbing.
We will define a weakly absorbing line as
RL and G C
Ensures very little conduction between the lines through the dielectric.
Ensures low ohmic loss for signals propagating through the line.
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Parameters for Weakly Absorbing TLs
Attenuation Coefficient, 1 R GZ0 2 Z0 Conductance through the dielectric dominates attenuation in high‐impedance transmission lines.
Resistivity in the conductors dominates attenuation in low‐impedance transmission lines.
In weakly absorbing transmission lines, there usually exists a “sweep spot” for the impedance where attenuation is minimized.
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Special Cases of Transmission Lines:
Distortionless Lines
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Definition of Distortionless TL
In a real transmission line, different frequencies will be attenuated differently because is a function of . This causes distortion in the signals carried by the line.
RG2222222 LC R L G C 2 To be distortionless, there must be a choice of R, L, G, and C that eliminates from the expression of , effectively making independent of frequency .
The necessary condition to be distortionless is
RG L C
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Parameters for Distortionless TLs
Propagation Constant,
jRGjLC
Attenuation Coefficient,
RG
Phase Constant, To be distortionless, we must have . is a measure of how quickly a signal accumulates phase. Different frequencies have LC different wavelengths and therefore must accumulate different phase through the same length of line.
Characteristic Impedance, Z0 RL ZRjX 00 0 GC RL RX 0 00GC
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Example:
Properties of RG-59 Coax
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The Lossless Circular Coax
Fundamental Parameters (derived in EE 3321) 2 C Fm ln ba 1 b L ln H m 24 a a Attenuation Coefficient, 0 b Phase Constant, and
Characteristic Impedance, Z0 b Z00RjX 0 ln ab 2 a b RX00ln 0 2 a
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Typical RLGC for RG‐59 Coax at 2 GHz
The typical RG‐59 coaxial cable operating at 2.0 GHz has the following RLGC parameters:
R 36 mΩ m L 430 nH m G 10 m C 69 pF m
Calculate the transmission line parameters , , , and Z0.
Classify the line as lossless, weakly absorbing, distortionless, etc.
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Solution (1 of 3)
Our equations mostly utilize the angular frequency instead of the ordinary frequency f. 2f 2 2.0 1091 s 12.5664 10 9 rad s
The characteristic impedance Z0 is RjL Z 0 GjC 36 mΩ mj 12.5664 109 rad s 430 nH m 10 mj 12.5664 109 rad s 69 pF m 78.94j 1.92 104
Note the imaginary part of Z0 is very small indicating that our line is very low loss.
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Solution (2 of 3)
The complex propagation constant is RjLGjC 9 36 mΩ mj 12.5664 10 rad s 430 nH m 9 10 mj 12.5664 10 rad s 69 pF m 6.23 1041j 68.45 m
From this result, we read off and . jj 6.23 1041 68.45 m
6.23 104 Np m Np is Nepers
68.45 rad m rad is radians
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Solution (3 of 3)
Is the line lossless? NO
No because R ≠ 0 and G ≠ 0. Also, we can determine this because ≠ 0 .
Is the line weakly absorbing? YES
? ? RL GC ? ? 36 mΩ m 12.5664 109 rad s 430 nH m 10 m 12.5664 109 rad s 69 pF m ? ? 0.036 5403.5 10 106 0.8671 Yes Yes Is the line distortionless? NO, but close
? RC LG ? 36 mΩ m 69 pF m 430 nH m 10 m ? 2.48 1012 4.30 10 12 No, but close
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Cable Loss Vs. Characteristic Impedance
As we adjust the cable dimensions (i.e. b/a), we change both its impedance and its loss characteristics. This let’s us plot the cable loss vs. characteristic impedance for a coax with different dielectric fills.
For the air‐filled coax, we observe minimum loss at around 77 , where b/a 3.5.
A coaxial cable filled with
polyethelene (r = 2.2), the minimum loss occurs at 51.2 (b/a = 3.6).
https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms Transmission Lines Slide 58
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Power Handling Vs. Characteristic Impedance As we adjust the cable dimensions (i.e. b/a), we affect the peak voltage handling capability (breakdown) and its power handling capability (heat). We observe the lowest peak voltage at just over 50 which we interpret as the point of best voltage handling capability.
We observe the lowest peak current at around 30 which we interpret as the point of best power handling capability.
https://www.microwaves101.com/encyclopedias/why‐fifty‐ohms Transmission Lines Slide 59
Why 50 Impedance is Best?
Two researchers, Lloyd Espenscheid and Herman Affel, working at Bell Labs produced this graph in 1929. They needed to send 4 MHz signals hundreds. Transmission lines capable of handling high voltage and high power were needed in order to accomplish this.
Data to the right was generated for an air‐filled coaxial cable.
Best for High Voltage: Z0 = 60 Best for High Power: Z0 = 30 Best for Attenuation: Z0 = 75
50 seems like the best compromise.
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Why 75 Impedance Standard for Coax?
Nobody really knows!!
The ideal impedance is closer to 50 , however this requires a thicker center conductor. Maybe 75 is a compromise between low loss and mechanical flexibility?
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Example:
Microstrip Design
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The Lossless Microstrip
w Attenuation Coefficient, 0
h r Phase Constant, 11 rr r,eff 2 2112 hw
k0,effr
Characteristic Impedance, Z0
60 8hw ln wh 1 thin lines eff wh4 ZRjX00 0 1120 wh 1 wide lines wh1.393 0.667ln wh 1.444 eff
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Problem Description
Typically, the manufacturing process fixes the value of dielectric constant r. This means the impedance of microstrips is controlled solely through the ratio w/h.
For this example, design a 50 microstrip transmission line in FR‐4, which as a dielectric constant of 4.5, to operate at 2.4 GHz.
w ? h
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Design Equations
To solve this problem, we must first derive some design equations. To do this, we solve our microstrip equations for w/h. This gives
Z 1 1 0.11 A 0 rr0.23 60 2rr 1 60 2 B Z0 r 8e A 2 A wh 2 thin lines w e 2 h 20.61 r 1 BB1 ln 2 1 ln B 1 0.39 wh 2 wide lines 2 rr
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Design Solution (1 of 2)
Applying our design equations, we get
A 1.5438 B 5.5831 w 1.8799 wh 2 thin lines h 1.8812 wh 2 wide lines Since the above numbers for w/h are essentially the same, we conclude that w 1.88 h
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Design Solution (2 of 2)
We learn from our manufacturing engineer that a convenient choice for substrate thickness h is 0.5 mm. From this, to get 50 the width w of the microstrip should be
wh1.88 1.88 0.5 mm 0.94 mm The phase constant for this line will be
eff 3.3941 22.410 91 s 2 f 1 k0 50.3 m cc00299792458 m s 50.3 m11 3.3941 92.67 m
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