In general, impedance has an upper limit that falls as frequency rises. You can make a 23kohm winding at 10kHz, you'll probably struggle at 100k, and you'll probably fail above 1MHz.
So you did...what... that binoc core is probably not much more than 1 uH/t^2, so 350 ohms at 1MHz is at least 50uH, or 7, call it 8 turns -- primary. And the secondary is then 8-10 times or at least 64 turns.
What do you suppose the Zo of those windings is? That is, considering the wire diameter and spacing, what's the transmission line impedance of a given turn with respect to its surroundings (assumed ground for an instantaneous wave at a point on the turn in question)?
Now, real multilayer windings aren't straight transmission lines, they're a jumble of lines stacked in series, both in the sense of the wire going around for multiple turns but also in that there are layers of transmission lines all the way down. (An equivalent circuit might use ideal single transmission lines, stacked so that all the port 1's are in series, and the port 2's are skew connected to the port 1's because that's how turns work.) Point is, Zo can be much higher for a multilayer winding, indeed for a wave of the hand it's around N_layers * Zo. But also dispersive, because such a structure doesn't have a constant velocity or impedance.
So, if the 64 turns are arranged in a square of 8x8 turns, they might be around 50 ohms for a given turn, or 8*50 = 400 ohms for the whole thing. That's a long ways away from 23k.
Which means equivalent capacitance will dominate, a proportional distance away from cutoff.
Cutoff being approximately the 1/4-wave electrical length of the winding (wire length).
If one turn is, oh I don't know, maybe 2cm, then 64 turns is 1.28m, or a 1/4 wave of 60MHz (at say 0.7c). 23kohm is 57 times above 400, so the cutoff at that impedance we should expect to be more like 1.04MHz.
Probably you're very near the peak, or just past it. Or maybe pretty far into it, I don't know. The -j impedance says it's capacitive, in any case.
Also, note that the above design exercise assumes nearly zero bandwidth -- the SRF Q would turn out pretty modest after all (probably < 1), but that wasn't an assumption I used! To cover the whole AM BCB, you need at least an octave more inductance (40% more turns), which will drop the HF cutoff by about 40% (40% more winding length!).
How to fix it?
Reduce winding length, apparently by at least 50%.
- Use a higher A_L core. Something with a thick cross section, maximizing A_e and minimizing turn length. Binoc cores are usually quite good at this to begin with, but you may consider a pot core, or maybe nanocrystalline (which is rolling off pretty well by these frequencies, but may still offer higher impedance than ferrite).
- Consider a smaller core, and finer wire. Winding length goes as the linear dimension of the component. This does reduce power handling, however.
- There are fortunately very few reasons to have tremendously high impedances, for RF purposes; you've likely made a design decision, which has turned out very poorly as it happens, and now you've, in part, discovered why no one does it that way. Redesign for a lower Zo. This will also shorten the winding, extending bandwidth even further; Zo doesn't need to be decreased by much.
- Or ditch the transformer altogether. For AM, just use a JFET preamp or something? It's not like you need a low noise floor anyway, the band is chock full of atmospheric noise. (Unless this is like some physics apparatus or something, and you're just using AM BCB as a reference point, not the actual goal.)
Tim