RF transformer - past resonance
jwL || 1/jwC = jwL/(1-LCw^2)
which means as frequency increases the effective inductance rises, until at the resonant frequency 1/sqrt(LC) its impedance is infinite, and past the resonant frequency there is an effective negative inductance, i.e. a capacitance.
I wonder if the transformer can still be used at frequencies higher than the resonant frequency. Sure, the capacitance dominates so each coil is effectively a capacitor not an inductor. But nevertheless, the inductive part is still conducting current and creating magnetic fields, so it should induce voltage in the other coil, right? If so, the transformer should still work? Will it however be somehow "less effective" because of the capacitance which is present, and if so, how? I have not succeeded in figuring that out yet, if anyone can, please help me.
After thinking for a while, I came to the conclusion that this would probably work in theory. But in practice, the losses would be too high and the Q too low to be useful.
Normally, neither the main (parallel) inductance of a transformer nor the parallel capacitance and the respective parallel resonance is determining the usable frequency range. As the main inductance of an ideal transformer is near to inifinity, it's parallel resonance frequency is rather low and may overlap the used frequency range without being noticed.
The critical parameter however is the leakage inductance, that forms a low pass with the winding capacitances. If the capacitance can't be reduced, a tighter coupling of the windings (lower leakage inductance) is an appropriate means to increase the transformer bandwidth.