[ §6 · calculators · current transformer ]

Current transformer

Ratio, currents, burden voltage and the lead the magnetising current gives, plus what wiring inductance in the sense loop costs you, which becomes critical once the burden is small.

first stage turnssecond stage turnsprimary current, peak · Aburden · Ωfrequency · kHzcore A_L · nH/turn²wiring L in the loop · µHtarget inductor L1 · µH

ratio 160:1CT secondary current 1.00 A at 160 A primaryburden voltage 3.90 Vburden power 3.90 W peakL of second stage 920 µHlead from magnetising 0.08° (subtracted, see CT timing) About 1 A: the loneoceans and WaskaLabs rule.Magnetising error is negligible. Wiring of 0.30 µH between the sense point and the burdenadds straight to L1: on its own it gives 13.48° = 28 % of the target angle fit a physical 1.09 µH instead of 1.39 Over 15 % of the angle comes from the stray part. Keep the burden and CT leads as short as possible, and set L1 ONLY on the scope.

formula

N = N₁·N₂     I₂ = I_pri/N     U = I₂·R     L₂ = A_L·N₂²
φ_mag ≈ arctan( R / (ω·L₂) )     (falls as 1/N₂²)
anything in series in the measured loop ADDS to L1

limits

The CT's own leakage is negligible, settled. This page used to carry an open question here, with one model putting the leakage at 13° and another at almost nothing. It is resolved by where the signal is sensed: across L1 + R, which it must be, or there would be no lead at all. That puts the leakage inside the magnetising term, where 920 µH of L_m swamps it. Moving the leakage from 0 to 1 µH shifts the total lead by 0.000085°. The old 13° belonged to a topology that is not on the board.

Wiring inductance is a different animal and still bites. Whatever sits between the sense point and the burden adds straight to L1. At 51 Ω that was a couple of per cent; at 3.9 Ω, 0.3 µH is 28 % of the target angle. Never set L1 by its nominal value, only on the scope, and keep the burden and CT leads extremely short.

The "optimum turns count" stays removed. It rested on the stray phase going as N², which the same model that closed the question above disproves.

The lead from magnetising printed above is the same quantity the CT timing page reports, and it enters the phase-lead budget with a minus.