2Transmission LinesA device used to transfer energy from one point to another point “efficiently”Efficiently = minimum loss, reflection and close to a perfect match as possible (VSWR = 1:1)Important to be efficient at RF and microwave frequencies, since freq used are higher than DC and low freq applicationsas freq gets higher, any energy loss is in transmission lines are more difficult and costly to be retrieved
3Some Types of Transmission Lines Some types of transmission lines are listed as follow:Coaxial transmission linetransmission line which a conductor completely surrounds the otherBoth shares the same axis, separated by a continuous solid dielectric or dielectric spacersFlexible – able to be bent without breaking
4Some Types of Transmission Lines WaveguidesHollow-pipe structure, in which two distinct conductor are not presentOpen space of the waveguide is where electromagnetic energy finds the path of least resistance to propagateDo not need any dielectric medium as it uses air as medium of energy transferPlanar transmission linesPlanar – looks like a 3D line that have been run over and flattenedUsually made up of a layer of dielectric, and one or several ground (metallic planes)Four types of planar lines discussed in this chapter; (1) Stripline, (2) microstrip (3) dielectric waveguide (4) SlotlineOur focus in this chapter
5But Why Planar? Waveguides Coaxial lines High power handling capabilitylow lossbulkyexpensiveCoaxial lineshigh bandwidth,convenient for test applicationsdifficult to fabricate complex microwave components in the medium
6But Why Planar? Planar Transmission Lines Compact Low cost Capability for integration with active devices such as diodes, transistors, etc
7STRIPLINEFigure 3.1: Stripline transmission line (a) Geometry (b) Electric and magnetic field lines.
8STRIPLINEAlso known as “sandwich line” – evolved from “flattened” coaxial transmission lineThe geometry of a stripline is shown in Figure 3.1.Consist of a; (1) top ground plane, (2) bottom ground plane and (3) a center conductorW is the width of thin conducting strip (centered between two wide conducting ground planes).b is the distance of ground planes separation.The region between the ground planes is filled with a dielectric.Practically, the centered conductor is constructed of thickness b/2.
9STRIPLINEFigure 3.2: Photograph of a stripline circuit assembly.
10STRIPLINE The phase velocity is given by: [3.1]Thus, the propagation constant of the stripline is:[3.2]
11STRIPLINEFrom equation [3.1], c = 3 x 108 m/sec is the speed of light in free-space.The characteristic impedance of a transmission line is given by:[3.3]L and C are the inductance and capacitance per unit length of the line. There is a solution as explained in [M. Pozar’ book].The resulting formula for the characteristic impedance is:[3.4]
12STRIPLINEWhere We is the effective width of the center conductor given by:[3.5]These formulas assume a zero strip thickness, and are quoted as being accurate to about 1 % of the exact results.It is seen from equation [3.4] and [3.5] that the characteristic impedance decreases as the strip width W increase.
13STRIPLINEWhen designing stripline circuits, one usually needs to find the strip width, W.By given characteristic impedance (and height b and permittivity εr), the value of W can be find by the inverse of the formulas in equation [3.4] and [3.5].The useful formulas is:[3.6]Where,[3.7]
14STRIPLINE The attenuation due to dielectric loss is: [3.8]The attenuation due to the conductor loss:[3.9]
15STRIPLINEWith:[3.10][3.11]Where t is the thickness of the strip
16STRIPLINE [EXAMPLE 2.1]Find the width for a 50 Ω copper stripline conductor, with b = 0.32 cm and εr = If the dielectric loss tangent (tan δ) is and the operating frequency is 10 GHz, calculate the attenuation in dB/λ. Assume the conductor thickness of t = 0.01 mm and surface resistance, Rs of Ω
17SOLUTION [EXAMPLE 2.1] Since and Eq [3.6] gives the width as W = bx = (0.32)(0.830) = cm. At 10 GHz, the wave number is
18SOLUTION [EXAMPLE 2.1] since A = 4.74 The dielectric attenuation is Surface resistance of copper at 10 GHz is Rs = Ω. Then from eq [3.9]since A = 4.74The total attenuation constant is
19SOLUTION [EXAMPLE 2.1] In dB; At 10 GHz, the wavelength on the stripline is;So in terms of the wavelength the attenuation is
20SOLUTION [EXAMPLE 2.1]But why do we need to convert Np/m to dB/m using this way?loss to the transmission line is reflected by the attenuation constant. The amplitude of the signal decays as e-α.The natural units of the attenuation constant are Nepers/meter, but we often convert to dB/meter in microwave engineering. To get loss in dB/length, multiply Nepers/length by
21STRIPLINE [EXAMPLE 2.2]Find the width for 50 Ω copper stripline conductor with b = 0.5 cm and εr = 3.0. The loss tangent is and the operating frequency is 8 GHz. Calculate the attenuation in dB/λ. Assume a conductor thickness of t = cm. The surface resistance is 0.03 Ω.
22SOLUTION [EXAMPLE 2.2]The wave number at f = 8 GHz
23SOLUTION (cont) [EXAMPLE 2.2] The dielectric attenuation isNp/mThe conductor attenuation is
24SOLUTION (cont) [EXAMPLE 2.2] Total loss:Np/mIn dB:dB/m
25SOLUTION (cont) [EXAMPLE 2.2] The guided wavelength:The attenuation in dB/λ:
26STRIPLINE DISCONTINUITY Figure 3.3: geometry of enclosed stripline
27STRIPLINE DISCONTINUITY By derivation found in M.Pozar’s book (page 141), the surface charge density on the strip at y = b/2 is:[3.12]The charge density on the strip line by uniform distribution:[3.13]
28STRIPLINE DISCONTINUITY The capacitance per unit length of the stripline is:[3.14]The characteristic impedance is then found as:[3.15]
29MICROSTRIPFigure 3.3: Microstrip transmission line. (a) geometry. (b) Electric and magnetic field lines.
30MICROSTRIPMicrostrip line is one of the most popular types of planar transmission line.Easy fabrication processes.Easily integrated with other passive and active microwave devices.The geometry of a microstrip line is shown in Figure 3.3W is the width of printed thin conductor.d is the thickness of the substrate.εr is the relative permittivity of the substrate.
31MICROSTRIPThe microstrip structure does not have dielectric above the strip (as in stripline).So, microstrip has some (usually most) of its field lines in the dielectric region, concentrated between the strip conductor and the ground plane.Some of the fraction in the air region above the substrate.In most practical applications, the dielectric substrate is electrically very thin (d << λ).The fields are quasi-TEM (the fields are essentially same as those of the static case.
32MICROSTRIP The phase velocity and the propagation constant: [3.16][3.17]Where εe is the effective dielectric constant of the microstrip line used to compensate difference between the top and bottom of the circuit lineThe effective dielectric constant satisfies the relation:and is dependent on the substrate thickness, d and conductor width, W
33MICROSTRIPThe effective dielectric constant of a microstrip line is given by:[3.18]The effective dielectric constant can be interpreted as the dielectric constant of a homogeneous medium that replaces the air and dielectric regions of the microstrip, as shown in Figure 3.4.Figure 3.4: equivalent geometry of quasi-TEM microstrip line.
34MICROSTRIP The characteristic impedance can be calculated as: [3.19]For a given characteristic impedance Z0 and the dielectric constant Єr, the W/d ratio can be found as:[3.20]
35MICROSTRIPWhere:Considering microstrip as quasi-TEM line, the attenuation due to dielectric loss can be determined as[3.21]Where tan δ is the loss tangent of the dielectric.
36MICROSTRIPThis result is derived from Equation [2.37] by multiplying by a “filling factor”:Which accounts for the fact that the fields around the microstrip line are partly in air (lossless) and partly in the dielectric.The attenuation due to conductor loss is given approximately by:[3.22]Where Rs = √(ωμ0/2σ) is the surface resistivity of the conductor.
37MICROSTRIP [EXAMPLE 2.3]Calculate the width and length of a microstrip line for a 50 Ω characteristic impedance and a 90° phase shift at 2.5 GHz. The substrate thickness is d = cm, with εr = 2.20.Solution: M.Pozar’ book (page: 146).
38SOLUTION [EXAMPLE 2.3] εe = 1.87 First we need to find W/d for Z0=50 Ω, and initially guess that W/d > 2. From eq [3.20];B = and W/d = 3.081Otherwise, we would use the expression for W/d<2Then W = (3.081.d) = cm. From eq [3.18];εe = 1.87
39SOLUTION [EXAMPLE 2.3]The line length, l, for a 90o phase shift is found as;
40MICROSTRIP [EXAMPLE 2.4]Design a microstrip transmission line for 70 Ω characteristic impedance. The substrate thickness is 1.0 cm, with εr = What is the guide wavelength on this transmission line if the frequency is 3.0 GHz?
41SOLUTION [EXAMPLE 2.4]Initially, it is guessed that W/d < 2
42SOLUTION cont [EXAMPLE 2.4] Since the W/d < 2 assumption is valid;We proceed to calculate εe
43SOLUTION cont [EXAMPLE 2.4] Thus the guided wavelength is given by;
44MICROSTRIP [EXAMPLE 2.5]Design a quarter wavelength microstrip impedance transformer to match a patch antenna of 80 Ω with a 50 Ω line. The system is fabricated on a 1.6 mm substrate thickness with εr = 2.3, that operates at 2 GHz.
45MICROSTRIP [EXAMPLE 2.5] From the quarter wave transformer equation; Which is also Z0 of lineZSZ0RL
46SOLUTION [EXAMPLE 2.5]Since W is not known, guess that W/d < 2
47SOLUTION cont [EXAMPLE 2.5] From the calculation, the initial assumption of W/d < 2 is incorrect. The next formula (where W/d > 2) is used
48SOLUTION cont [EXAMPLE 2.5] The next step is to find the effective dielectric constant (εe)To determine the quarter wavelength of the line, the guided wavelength λg need to be determined
49SOLUTION cont [EXAMPLE 2.5] Thus the quarter wave length of line is determined by dividing the full wavelength by 4;
50MICROSTRIPFigure 3.5: Geometry of a microstrip line with conducting sidewalls.
51MICROSTRIP DISCONTINUITY By derivation found in M.Pozar’s book (page 147), the surface charge density on the strip at y = d is:[3.23]The charge density on the microstrip line by uniform distribution:[3.24]
52MICROSTRIP DISCONTINUITY The capacitance per unit length of the stripline is:[3.25]The characteristic impedance is then found as:[3.26]
54DIELECTRIC WAVEGUIDE The dielectric waveguide is shown in Figure 3.6. εr2 is the dielectric constant of the ridge.εr1 is the dielectric constant of the substrate.Usually εr1 < εr2The fields are thus mostly confined to the area around the dielectric ridge.Convenient for integration with active devices.Very lossy at bends or junctions in the ridge line.Many variations in basic geometry are possible.
55SLOTLINEFigure 3.7: Geometry of a printed slotline.
56SLOTLINE The geometry of a slotline is shown in Figure 3.7. Consists of a thin slot in the ground plane on one side of a dielectric substrate.The characteristic impedance of the line can be change by changing the width of the slot.