Executive Summary
Assuming:
Then for round coax, make the inside diameter of the outside conductor 2.302 times larger than the diameter of the inside conductor.
If the shield is square, and the inner conductor is still round, make the inside length of one side of the shield 2.134 times larger than the diameter of the inside conductor.
Explanation, for Round Coax
A coaxial transmission line has some inductance per unit length $L$, and some capacitance per unit length $C$. The conductors also have some (very low) resistance $R$, and the dielectric has some (very low) conductance $G$. Though it's important to note that these properties are distributed throughout the entire length of the transmission line, lumping them all together we can model a transmission line as:
The characteristic impedance is then:
$$ Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}} $$
where:
- $R$ is the resistance per unit length, considering the two conductors to be in series,
- $L$ is the inductance per unit length,
- $G$ is the conductance of the dielectric per unit length,
- $C$ is the capacitance per unit length,
- $j$ is the imaginary unit, and
- $\omega$ is the angular frequency.
If we assume a good conductor and dielectric, then $R$ and $G$ are negligible1. The equation then simplifies to:
$$ Z_0 \approx \sqrt{\frac{L}{C}} $$
For round coaxial transmission lines, we can calculate $C$ and $L$ as:
$$ C = {2 \pi \epsilon_0 \epsilon_r \over \ln(D/d)} $$
$$ L = {\mu_0 \mu_r \over 2 \pi} \ln(D/d) $$
where:
Putting these together and simplifying we get:
$$ \begin{align} Z_0
&\approx \sqrt{\frac
{{\mu_0 \mu_r \over 2 \pi} \ln(D/d)}
{{2 \pi \epsilon_0 \epsilon_r \over \ln(D/d)}}
} \\
&\approx \sqrt{\frac
{\mu_0 \mu_r \ln(D/d)^2}
{(2 \pi)^2 \epsilon_0 \epsilon_r}
} \\
&\approx \sqrt{\frac
{\mu_r}
{\epsilon_r}
} \sqrt{\frac
{\mu_0}
{(2 \pi)^2 \epsilon_0}
} \ln(D/d) \\
&\approx \sqrt{\frac
{\mu_r}
{\epsilon_r}
} 59.96 \ln(D/d) \\
\end{align} $$
For an air dielectric, $\epsilon_r$ and $\mu_r$ are so close to 1 than you can further simplify:
$$ Z_0 \approx 59.96 \ln(D/d) $$
You can also see that only the ratio of the diameters of the conductors are relevant, so it doesn't matter what units of length are used, as long as they are the same.
We can solve for $D/d$, and set $Z_0 = 50 \Omega$:
$$ \begin{align}
56.40 \ln(D/d) &\approx Z_0 \\
\ln(D/d) &\approx \frac{Z_0}{59.96} \\
D/d &\approx e^{Z_0/59.96} \\
D/d &\approx e^{50/59.96} \\
D/d &\approx 2.302
\end{align} $$
Explanation, for Square Shield with a Round Center Conductor
This is a rather less common situation. Kevin Schmidt, W9CF has a full explanation. The equations are rather more complex, owing to the more complex geometry of the conductors, and are most easily solved by computer. However, a good approximation for impedances around 50Ω, with an air dielectric, is:
$$ Z_0 \approx 59.96 \ln(1.079 D/d) $$
or solved for $D/d$:
$$ D/d \approx {e^{Z_0/59.96} \over 1.079} $$
Where:
- $d$ is the diameter of the inner conductor, and
- $D$ is the inside length of one side of the shield.
1: this is a valid assumption for all HF frequencies, and VHF with low-loss dielectrics. At higher frequencies, skin effect and dielectric loss make $R$ and $G$ more significant, but it's likely this will manifest as an impracticably lossy line before the impedance is significantly affected.
