125kHz RFID System Design/Troubleshooting
If you have an LC tank circuit and want to reduce the inductance at resonance:

The inductor you have has inductance of L = 2.8mH:
At 125kHz the inductive reactance is \$\mathsf{2 \pi \, f \, L}\$ = 2199 Ohm.

The inductor you would like to have has inductance of L = 1.337mH:
At 125kHz the inductive reactance is \$ \mathsf{2 \pi \,f \, L}\$ = 1050 Ohm.

So you need to reduce inductive reactance by 2199-1050=1149 Ohm.

At 125kHz the reactance of a 1.1nF capacitor is \$\mathsf{ \frac {1}{2\,\pi \, f \, C}}\$ = 1157 Ohm, which is within a few Ohms of the 1149 "inductive" Ohms you want to cancel out.

Here's a simple model of your driver, coil and the 1.1nF "canceling cap" C1 

In the simulation you can edit (by right-clicking) the value of C1 and observe the effect on the coil current.

Note (in the simulation) that the voltage across capacitor C1 is quite high - make sure you consider the voltage rating of the capacitor you use!



The 63 Ohms of loss resistance in the coil model is probably a low-ball estimate, as inductor losses quickly increase with frequency due to the skin effect.

Finally, you are likely better off (reduced coil loss resistance) by reducing the number of windings on the inductor by (roughly) a factor sqrt(1.337mH/2.8mH)