Copyright
1996 
2016
Webmaster:
Kirt
Blattenberger,
BSEE  KB3UON
RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ...
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Two Conductors in Parallel (Unbalanced) Above Ground Plane For D << d, h Z_{0}= (69/ε^{½}) log_{10}{(4h/d)[1+(2h/D)^{2}]^{½}} 
Single Conductor Above Ground Plane For d << h Z_{0}= (138/ε^{½}) log_{10}(4h/d) 
Two Conductors in Parallel (Balanced) Above Ground Plane For D << d, h_{1}, h_{2} Z_{0}= (276/ε^{½}) log_{10}{(2D/d)[1+(D/2h)^{2}]^{½}} 
Two Conductors in Parallel (Balanced) Different Heights Above Ground Plane For D << d, h_{1}, h_{2} Z_{0}= (276/ε^{½})log_{10}{(2D/d)[1+(D^{2}/4h_{1}h_{2})]^{½}} 
Single Conductor Between Parallel Ground Planes For d/h << 0.75 Z_{0}= (138/ε^{½}) log_{10}(4h/πd) 
Two Conductors in Parallel (Balanced) Between Parallel Ground Planes For d << D, h Z_{0}= (276/ε^{½}) log_{10}{[4h tanh(πD/2h)]/πd} 
Balanced Conductors Between Parallel Ground Planes For d << h Z_{0}= (276/ε^{½}) log_{10}(2h/πd) 
Two Conductors in Parallel (Balanced) of Unequal Diameters Z_{0}= (60/ε^{½}) cosh^{1} (N) N = ½[(4D^{2}/d_{1}d_{2})  (d_{1}/d_{2})  (d_{2}/d_{1})] 
Balanced 4Wire Array For d << D_{1}, D_{2} Z_{0}= (138/ε^{½}) log_{10}{(2D_{2}/d)[1+(D_{2}/D_{1})^{2}]^{½}} 
Two Conductors in Open Air Z_{0}= 276 log_{10}(2D/d) 
5Wire Array For d << D Z_{0}= (173/ε^{½}) log_{10}(D/0.933d) 
Single Conductor in Square Conductive Enclosure For d << D Z_{0}≈ [138 log_{10}(ρ) +6.482.34A0.48B0.12C]/ε^{½} A = (1+0.405ρ^{4})/(10.405ρ^{4}) B = (1+0.163ρ^{8})/(10.163ρ^{8}) C = (1+0.067ρ^{12})/(10.067ρ^{12}) ρ= D/d 
Air Coaxial Cable with Dielectric Supporting Wedge For d << D Z_{0}≈ [138 log_{10}(D/d)]/[1+(ε1)(θ/360)]^{½}) ε = wedge dielectric constant θ= wedge angle in degrees 
Two Conductors Inside Shield (sheath return) For d << D, h Z_{0}= (69/ε^{½}) log_{10}[(ν/2σ^{2})(1σ^{4})] ν = h/d σ = h/D 
Balanced Shielded Line For D>>d, h>>d Z_{0}= (276/ε^{½}) log_{10}{2ν[(1σ^{2})/(1+σ^{2})]} ν = h/d σ = h/D 

Two Conductors in Parallel (Unbalanced) Inside Rectangular Enclosure For d << D, h, w ∞ Z_{0}= (276/ε^{½}) {log_{10}[(4h tanh(πD/2h)/πd) ∑ log_{10}[(1+μ_{m}^{2})/(1ν_{m}^{2})]} ^{m=1} μ_{m}=sinh(πD/2h)/cosh(mπw/2h) ν_{m}=sinh(πD/2h)/sinh(mπw/2h) 