DC will shift the flux generated by the AC signal above the horizontal axis of the BH curve,
DC might bias the AC flux swing into a non linear part of the BH curve if the core can not handle it,
DC in a wire wrapped around a steel core will generate a force equal to F= .4Pi * current * turns
most 600 ohm transformers have about 600 turns, you have 5 ma of DC current,
F=1.2566 * ,005 * 600 = 3.77 Gilberts?
we can distribute that magnetic force along the closed loop path of a a core, how big is that loop?
a typical 600:600 uses a core along the lines of 375 EI (3/8" tongue)
but for DC flux calculations we want the magnetic path length, not the cross section,
the MPL for 375 EI is 7.3 cm.
so or H field in Oersteds is 3.77 Gilberts/7.3 cm = .516
now what?
well, we have a formula that relates core permeability to flux density (B) and H.
it goes like this: B=uH, where u is relative permeability, which is just the perm of the steel compared to the perm of air,
so B=u * .516
80 Ni (supermalloy) takes 5000 Gauss and has a perm of 100,000 in clear weather.
so if we have a 80 Ni core in this Bogen, then B(dc)= .516 * 100,000= 51,600 Gauss which is big trouble.
if we have 50 Ni, then max flux = 12,000 and perm is about 20,000, so
B (dc) = .516 * 20,000 = 10,320 Gauss,
we are under the 12 KG limit, but not by much,
what about 4% Si? takes 18 KG and has a perm of 10,000, so B=.516 * 10,000 = 5.16 KG
conclusion?
80 Ni is out of the question, 50 Ni will work if you put a small gap in there or use F lams which has a natural gap, Silicon will work with no gap, but the core will have to be big to get minimum inductance.
here is a graph showing the minor loop due to incremental perm being shifted by DC,
the formula also shows that the perm can be altered with a gap to keep DC flux down.
cores have a natural gap, yours would need to be Bdc= F/Gap, B=3.77 (from above)/Gap, Gap= 3.77/12000 (50 Ni) = .000314" just to get you below saturation and leaving no room for AC flux (signal)
