CT timing
How much time the current transformer itself costs, and with which sign. The answer is counter-intuitive: a CT gives lead, not delay, so the lead asked of the inductor goes down by this much rather than up. The phase-lead budget on this site had it as a plus until this calculation settled it.
I_burden = (I_pri/N) · jωL_m / (jωL_m + R) phase = 90° − arctan(ωL_m/R) = arctan( R / (ω·L_m) ) > 0 ⇒ LEAD t = φ / (360°·f) L_m = A_L · N₂² the first stage is loaded by the second's input, i.e. R/N₂², so its share is nil
Measure instead of computing where you can. Open the primary side of the second stage and measure the inductance of its secondary: that is L_m. The measured field takes priority over A_L.
Measuring the time directly is possible but needs a reference channel: a second CT with a high ratio and a low burden, so its own error is small, or a current clamp. Then compare its zeros against the voltage on the working burden. Going through L_m is more practical.
The magnetising term is the whole of it. The CT's leakage sits inside this same term rather than beside it. The signal is sensed across L1 + R, which puts the leakage upstream of the sense point and under 920 µH of L_m. Moving it from 0 to 1 µH shifts the total lead by 0.000085°. This is therefore the figure, not a lower bound; the 13° that used to be quoted belonged to a topology that is not on the board.
What is not in here is the wiring between the sense point and the burden. That adds straight to L1 and is worth 28 % of the angle at a 3.9 Ω burden. See the current transformer page.