# Lossless planar line geometry

A customer requested a $50 ohm$ loss-less planar line, with the restriction that the width of the strip cannot exceed $3 mm$. What should the distance, $d$ be? However due to error during manufacturing, if the final characteristic impedance value turns out to be $52 ohm$, how much would the $d$ differ?

The impedance of a loss-less line is

$Z0 = \sqrt{(L/C)}$.

The inductance of a line parallel to a conducting wall is

$L = \frac{\mu*l}{2\pi}*ln(d/a)$ where $l$ is length, $d$ is distance to wall and $a$ is wire radius

The capacitance of line parallel to a conducting wall is

$C = \frac{(2\pi\epsilon*l)} {ln(d/a)}$

These are found in electrodynamics, by writing the Maxwell equations for the propagation of electromagnetic waves in a lose-less medium. Therefore:

$Z^2 = L/C = [\mu/(4\pi\epsilon)]\ln^2(d/a)$

for $Z =50 ohm$, $a = 3 mm$, $\mu =4\pi*10^{-7} (H/m)$ and $1/(4\pi\epsilon) =9*10^9$

one obtains

$ln^2(d/a) = \frac{50^2}{(4*10^{-7}*9*10^9} =0.221$

$d/a =1.6$

$d = 1.6*3 =4.8 mm$

For $Z = 52 ohms$

$ln^2(d/a) = 0.239$

$d/a =1.63$

$d =4.892 mm$

The difference in $d$ is $\Delta d=4.892-4.8 =0.092 mm$