Coil self-resonance

Above what frequency does this choke stop behaving as an inductor?

C = \frac{K \cdot D \cdot N}{1000} \qquad F = \frac{1}{2\pi\sqrt{L \cdot C}}

Method

1. Find the aspect ratio — winding height divided by coil diameter — and pick Medhurst’s K from it: about 0.6 for a squat coil (ratio under 0.5), 0.45 for a long thin one (over 2), and 0.52 in between.
2. Multiply K by the coil diameter in millimetres and by the number of turns, then divide by 1000. That gives the distributed self-capacitance in picofarads.
3. Convert that capacitance to farads and the inductance to henries.
4. Apply the same series resonance formula as any LC pair: one over 2π times the square root of L times C. The coil resonates with itself.
5. Divide the self-resonant frequency by the intended working frequency. That ratio is the margin, and it is the number to judge the design by rather than the raw frequency.

Assumptions

• Medhurst’s K is an empirical fit for single-layer solenoids. A multilayer VIC choke has considerably more self-capacitance than this predicts, so the real self-resonance is lower — often much lower.
• The distributed capacitance is treated as a single lumped capacitor across the coil. It is not; the real behaviour has multiple resonances, and this finds only the first.
• Nothing else is connected. Wiring, a screen, or a nearby earthed surface all add capacitance and pull the self-resonance down further.
• Air core, no ferrite. A core raises the inductance and therefore lowers the self-resonance.
• A margin of ten is a working rule, not a law. Below it the coil’s apparent inductance is already higher than its nominal value, which shifts the circuit resonance somewhere the design did not intend.