@Medved, your calculations are really strange. First of course, bipolar transistor inverter with a commutating ferrite ring is never run at 140 kHz, I'd say some 50kHz is a practical maximum.

Anyway, if you drive 3mH/2200pF series circuit at 140kHz you will never reach 0.45A current.

Let's do the math carefully. Right, you have approximately +- 150V squarewave output after a half-bridge. First harmonic amplitude will be 4/pi*150V=191V. RMS = 191/SQR(2) = 135V. Next, coil impedance will be 2×π×140000×3×10⁻³= j2638 Ohms. Capacitor will cancel some of this impedance by 1÷(2×π×140000×2200×10⁻¹²) = -j517 Ohms. So resulting impedance will be j2121 Ohms. Ignoring 3rd and higher harmonics, first harmonic current will be just 64mA even not taking electrode impedances into account.

My bad, publishing without check (managed to screw up quadratic root formula...). The "220V" should have tell me there is something wrong...

The correct frequency is indeed 71kHz and the voltage gets into 450V range (Xl=1346, Xc=1013), which is about what really to expect...


140kHz is not that rare with the cheepeese "USB chargers" at low load (use the same transistors as small CFL's, but the base drive is reduced by the feedbeck to get that high frequency/low magnetizing current, so the Ts is minimized there), so that did not sound to me as that strange, mainly for part of the sequence where it remains for less than a second or so (so the elevated losses are not a problem)...