For Most Practical Waveguides the Loss Associated with Dielectric Loss

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For Most Practical Waveguides the Loss Associated with Dielectric Loss Adapted from notes by ECE 5317-6351 Prof. Jeffery T. Williams Microwave Engineering Fall 2019 Prof. David R. Jackson Dept. of ECE Notes 7 Waveguiding Structures Part 2: Attenuation ε,, µσ 1 Attenuation on Waveguiding Structures For most practical waveguides and transmission lines the loss associated with dielectric loss and conductor loss is relatively small. To account for these losses we will make this approximation: αα≈+cd α Attenuation constant Attenuation constant due due to conductor loss to dielectric loss (ignore dielectric loss) (ignore conductor loss) 2 Attenuation due to Dielectric Loss: αd Lossy dielectric ⇒ complex permittivity ⇒ complex wavenumber k σ εε= − j c ω σ k=ω µε = kk µε = µε(1 − j tan δ ) =−−εε′jj ′′ c00 rrc rr d ω =εε′ − ′′ k= k′ − jk ′′ ccj εc′′ =εc′ 1 − j εc′ =εδ′ (1 − j tan ) Note: kk′ = Re{ } cd =εε0 r (1 − j tan δ ) Note : εεrc′ is denoted as r . (e.g., ε r = 2.1 for Teflon). 3 Dielectric Attenuation for TEM Mode TEM mode: kkz = where kk= 0 µεrr (1− j tan δd ) Assume a small dielectric loss in medium: tanδ << 1 Use 1−zz ≈ 1 − /2 for z<< 1 ⇒≈kk0 µεrr(1 − j( tan δd) / 2) kk′ ≈ 0 µεrr 1 kk′′ ≈ µεtan δ 2 0 rr d 4 Summary of Dielectric Attenuation TEM mode kzd=−=βα jk β =Re(kk) = ′ αd =−=Im(kk) ′′ β≈ k0 µεrr 1 α≈ k µεtan δ d 2 0 rr d k=−= k′ jk ′′ k0 µεrr 1 −j tan δd 5 Dielectric Attenuation for Waveguide Mode An exact general expression for the dielectric attenuation: 22 kzd=−=−βα j kk c k=−= k′ jk ′′ k0 µεrr 1 −j tan δd Remember: The value kc is always real, regardless of whether the waveguide filling material is lossy or not. Note: The radical sign denotes the principal square root: −<ππarg( z) < 6 Dielectric Attenuation for Waveguide Mode (cont.) Approximation for the wavenumber of a waveguide mode : 22 kzd=−=−βα j kk c 22 =k0 µεrr(1 −− jk tan δd ) c 2 22 =(k00µεrr −− k c) jk µεrrtan δ d Assume a small dielectric loss: 2 22 k00µεrrtan δ d<< kk µε rr − c Then use: 11zz a− z = a1/ −( za) ≈ a 1 − =a − for z<< a 22a a 7 Approximate Dielectric Attenuation (cont.) Hence, we have: 2µε δ 221jk0 rr tan d We assume here that we kkz≈( 0 µεrr −− k c) 2 kk22µε − are above cutoff. ( 0 rr c) This gives us: 22 β≈−(kk0 µεrr c) k 2µεtan δ α ≈ 0 rr d d 2β 8 Summary of Dielectric Attenuation Waveguide mode (TMz or TEz) 22 kzd=−=−βα j kk c 22 β =Re kk − c 22 αdc=−−Im kk 22 β≈−kk0 µεrr c k 2µεtan δ α ≈ 0 rr d d 2β k=−= k′ jk ′′ k0 µεrr 1 −j tan δd 9 Attenuation due to Conductor Loss Assuming a small amount of conductor loss: We can assume that the fields of the lossy guide are approximately the same as those for lossless guide, except with a small amount of attenuation. ⇒ We can use a “perturbation” method to determine αc . Notes: . Dielectric loss does not change the shape of the fields at all in a waveguide or transmission line, since the boundary conditions remain the same (PEC). Conductor loss does disturb the fields slightly. 10 Surface Resistance This is a very important concept for calculating loss at a metal surface. x εµ, ε,, µσ z Note: The tangential fields determine the power going into εµ, the metal. Plane wave in a good conductor Note: In this figure, z is the direction normal to the metal surface, not the axis of the waveguide. Also, the electric field is assumed to be in the x direction for simplicity. 11 Surface Resistance (cont.) σ Assume 1 (for the metal) ωε The we have (for the metal): 1/2 1/2 σ σ µσ1− j ωµσ kj=ω µε − ≈− ωµ j = ω =(1 −j ) ω ωω2 2 Hence ωµσ k=−≈ k′ jk ′′ (1 −j) 2 Therefore ωµσ kk′≈≈ ′′ 2 12 Surface Resistance (cont.) ωµσ kk′≈≈ ′′ 2 1 Denote δ ≡=d “skin depth” = “depth of penetration” p k′′ e−1 ≈ 0.37 (For z = δ, fields are down to 37% Note : of their values at the surface.) Then we have 2 δ =d = At 3 GHz, the p ωµσ skin depth for copper is about 1 1.2 microns. kk′≈≈ ′′ δ 13 Surface Resistance (cont.) Frequency δ 2 1 [Hz] 6.6 [cm] δ = 10 [Hz] 2.1 [cm] ωµσ 100 [Hz] 6.6 [mm] 1 [kHz] 2.1 [mm] 10 [kHz] 0.66 [mm] Example: copper (pure) 100 [kHz] 0.21 [mm] −7 µµ=0 =4 π × 10 [H/m] 1 [MHz] 66 [µm] σ =5.8 × 107 [S/m] 10 [MHz] 20.1 [µm] 100 [MHz] 6.6 [µm] 1 [GHz] 2.1 [µm] Note: 10 [GHz] 0.66 [µm] × 7 A value of 3.0 10 [S/m] is often 100 [GHz] 0.21 [µm] assumed for “practical” copper. 14 Surface Resistance (cont.) x S z Fields evaluated on this plane 2 Pd = time-average power dissipated / m on S 11* P =Re(E × H ) ⋅= zˆ Re( EH* ) d 22x yz=0 15 Surface Resistance (cont.) Inside conductor: EHxy=η where µ µ µ ωµ η ==≈=j εσσσ “Surface resistance (Ω)” c ε −−jj ωω ωµ 1+ j ωµ Rs ≡ = 2σ 2 σ ωµ =(1 + j ) “Surface impedance (Ω)” 2σ Zss=(1 + jR) =(1 + jR ) s = Zs Note: To be more general: Note: Ets= ZnH( ˆ × t) EE xx= HH nˆ = outward normal yyz=0, air plane wave in metal 16 Surface Resistance (cont.) eff nˆ J s Conductor Ets= ZnH( ˆ × t) Et = tangential electric field at surface Ht = tangential magnetic field at surface nˆ = outward unit normal to conductor surface “Effective surface current” Hence we have eff Jst≈×( nHˆ ) eff Et≈ ZJ ss For the “effective” surface current density we imagine the The surface impedance gives us the ratio of the actual volume current density to tangential electric field at the surface to the be collapsed into a planar effective surface current flowing on the object. surface current. 17 Surface Resistance (cont.) Summary for a Good Conductor Ets= ZnH( ˆ × t) (fields at the surface) eff Et≈ ZJ ss (effective surface current) Zss=(1 + jR ) ωµ 1 R = = s 2σ σδ 2 δ = ωµσ 18 Surface Resistance (cont.) Frequency Rs 1 [Hz] 2.61×10-7 [Ω] ωµ 1 10 [Hz] 8.25×10-7 [Ω] Rs = = 2σ σδ 100 [Hz] 2.61×10-6 [Ω] 1 [kHz] 8.25×10-6 [Ω] Example: copper (pure) 10 [kHz] 2.61×10-5 [Ω] 100 [kHz] 8.25×10-5 [Ω] µµ= =4 π × 10−7 [H/m] 0 1 [MHz] 2.61×10-4 [Ω] σ =5.8 × 107 [S/m] 10 [MHz] 8.25×10-4 [Ω] 100 [MHz] 0.00261 [Ω]6.6 1 [GHz] 0.00825 [Ω] Note: 10 [GHz] 0.0261 [Ω] A value of 3.0×107 [S/m] is often assumed for “practical” copper. 100 [GHz] 0.0825 [Ω] 19 Surface Resistance (cont.) Returning to the power calculation, we have: 11** 12 Pd = Re(EHx yz )=0= Re((ZHs y00) H y) = R s H y0 22 2 1 2 In general, P = RH d2 st0 eff For a good conductor, Jst≈× nHˆ 0 This gives us the power dissipated per 1 2 Hence eff square meter of conductor surface, if we Pd= RJ ss know the effective surface current density 2 flowing on the surface. eff PEC eff PEC PEC limit: JJss→ Perturbation method : Assume that JJss≈ 20 Perturbation Method for αc −2α z Power flow along the guide: Pz()= Pe0 Power P0 @ z = 0 is calculated from the lossless case. dP( z) Power loss (dissipated) per unit length: P = − (∆P = −∆ Pz( )) l dz abs flow −2αz ⇒=Pl () z 2αα Pe0 = 2 P () z Pz( ) P (0) ⇒ α =ll = Note: α α 2Pz () 2 P0 = c for conductor loss 21 Perturbation Method: Waveguide Mode There is a single Pl (0) αc = conducting 2P0 boundary. S C z 1 * P0 =Re E ×⋅ H zˆ dS ∫∫2 For these calculations, we S z=0 neglect loss when we determine the fields and currents. R 2 P(0) = s Jd ls2 ∫ C z=0 Js = nHˆ × On PEC conductor Surface resistance of ωµ R = metal conductors: s 2σ 22 Perturbation Method: TEM Mode There are two S Pl (0) αc = conducting 2P0 boundaries. z 1 * P0 =Re E ×⋅ H zˆ dS ∫∫2 S z=0 For these calculations, we 1 2 neglect loss when we determine = ZI 2 0 the fields and currents. lossless (ZZ00= ) 1122 P(0) = RJ d+ RJ d l 22∫∫ss11ss CC12zz=00= Js = nHˆ × On PEC conductor Surface resistance of ωµ R =(σσ = or σ) metal conductors: s 2σ 12 23 Wheeler Incremental Inductance Rule The Wheeler incremental inductance rule gives an S alternative method for calculating the conductor attenuation on a transmission line (TEM mode): z It is useful when you have a formula for Z0. RZs ∂ 0 The formula is applied for each conductor αc = and the conductor attenuation from each of 2Z0η ∂ the two conductors is then added. In this formula, (for a given conductor) is the distance by which the conducting boundary is receded away from the field region. The top plate of a PPW line E E is shown being receded. H. Wheeler, “Formulas for the skin-effect,” Proc. IRE, vol. 30, pp. 412-424, 1942. 24 Calculation of R for TEM Mode From αc we can calculate R (the resistance per unit length of the transmission line): i( zt, ) Rz∆ Lz∆ i( z+∆ zt, ) Pl (0) αc = 2P0 + + v( zt, ) Gz∆ Cz∆ v( z+∆ zt, ) - 1 2 - P= ZI z 02 00 1 2 P(0) = RI Assume conductor loss only.
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