Autotransformer design -- how many turns?
If it's running from an oscillator then the frequency is fixed therefore you can tune the auto-transformer so that there is minimal reactive load to the oscillator. A 33pF capacitor and a 1uH coil are resonant at about 28MHz but do some tweaking so that the unloaded inductor and capacitor are near enough fully resonant at 28 MHz.

All that is left is to pick a "tap" that is 70.7% of the turns up from 0V - this will make the 25 ohm resistor like a 50 ohm resistor when seen at the oscillator terminals.

The ratio of impedance is proportional to turns squared so 1/0.707 squared = 2.







simulate this circuit &ndash; Schematic created using CircuitLab

Here's what a simulation looks like with and without tuning capacitor using an auto-transformer with 1uH end-to-end.



Fortunately the blue curves show virtually the same performance attenuation-wise i.e. 6dB down at "Vout" on the simulated circuit but note that the red curves (phase angle) tell the truer story at 28 MHz - it's subtle but the circuit with the resonating capacitor has a phase angle of precisely zero at 28MHz. Without the tuning capacitor, the frequency would have to be well-over 100 MHz before this was as precise.



Just a further note about the inductance values on the circuit I simulated. These are based on a "pretend" 10-turn auto-transformer having an inductance of 1uH - this implies an \$A_L\$ of 0.01uH i.e. 10 turns squared x 0.01 = 1uH. 7.071 turns would give 0.5uH and 2.929 turns yields 0.086 uH.

The formula is a bit tortuous but is based on total inductance (Ls) being L1 + L2 + 2\$\sqrt{L1\cdot L2}\$. If you know the total inductance you require and you have one of the other values i.e. L1 then L2 is: -

\$L_2 = L_1 + L_S - \sqrt{(L_S + L_1)^2 - (L_S - L_1)^2}\$