This article explains the operation of a simple nominally 1:4 impedance transformer using transmission line (TL) elements.

Above is a diagram of the device. The currents shown are differential currents in the coax (ie wholly inside the coax), the current on the outside of the shield is not shown on the diagram.

At very low frequencies it may be intuitive that \(V_1\approx V_2\) and \(I_1\approx I_2\), but as frequency increases, a more exact solution is needed.

We can write an expression for input impedance of each transmission line section in terms of its terminating impedance \(Z_t=V_2/I_2\), the line length l and frequency dependent complex propagation coefficient \(\gamma=\alpha+ \jmath \beta\\\).

\(Z_{i}=Z_0 \frac{Z_t/Z_0+tanh((\alpha+\jmath \beta)l)}{1+(Z_t/Z_0)tanh((\alpha+\jmath \beta)l)}\)

We can see intuitively that because the TL sections are of equal length and that the voltages at the left hand side are equal (by virtue of the parallel connection) and that the currents at the right hand side are equal (by virtue of the series connection) that the voltages and currents in both TL sections are equal. That means that \(Z_t=Z_{load}/2\), and knowing \(Z_0, \: \gamma\) and \( l\) we can calculate the combined input impedance \(Z_{input}=Z_i/2\).

Above is the calculated input impedance for a pair of 1m long sections of RG402 with 50+j0Ω load. It can be seen that at low frequencies, input impedance approaches 50/4Ω but as frequency increases the impedance transformation in each line section becomes more significant.

Note that the example and discussion assume no common mode TL current, and the explanation holds valid if the application does not have significant common mode current.

An upcoming article will use this device in a shielded loop antenna where the symmetry of the loop and the coupling transformer both serve to effectively minimise common mode current.