The article Antennas – disturbing the thing being measured – open wire lines #2 did a simple analysis of current flows in the model scenario using ideal voltage balun drive.
That begs the question, what difference would a good current balun make?
We can get a good approximation of what happens by inserting the current balun’s Zcm in series with Z3. Let’s take Zcm to be 1130+j1657Ω (11t on a FT240-43).
We can solve this simple network with 1V of differential drive with a little complex maths, here in Python.
import math import cmath zcm=1130+1657j z1=33.59-190.8j z2=38.09-28.43j z3=118.6-109.2j z3=z3+zcm i1=1/(z1+1/(1/z2+1/z3)) i2=-1/(z2+1/(1/z1+1/z3)) i3=i1+i2 id=(i1-i2)/2 ic=(i1+i2)/2 cmath.polar(i1) cmath.polar(i2) cmath.polar(i3) cmath.polar(id) cmath.polar(ic) 20*math.log10(cmath.polar(ic)[0]/cmath.polar(id)[0])
… and with the results…
>>> import math >>> import cmath >>> >>> zcm=1130+1657j >>> z1=33.59-190.8j >>> z2=38.09-28.43j >>> z3=118.6-109.2j >>> z3=z3+zcm >>> i1=1/(z1+1/(1/z2+1/z3)) >>> i2=-1/(z2+1/(1/z1+1/z3)) >>> i3=i1+i2 >>> id=(i1-i2)/2 >>> ic=(i1+i2)/2 >>> cmath.polar(i1) (0.004348895656197927, 1.2509382087514753) >>> cmath.polar(i2) (0.004077924130781668, -1.945047482489986) >>> cmath.polar(i3) (0.0003547983517372658, 1.9258977176478989) >>> cmath.polar(id) (0.0042118533752797515, 1.224616432471892) >>> cmath.polar(ic) (0.0001773991758686329, 1.9258977176478989) >>> 20*math.log10(cmath.polar(ic)[0]/cmath.polar(id)[0]) -27.51043292031913
So, the common mode component of current at this point in each conductor is 27.5dB less than the differential component, 17.1dB lower than the ideal voltage balun drive. The current balun has significantly reduced common mode current at this point.
The solution of the network is a first approximation of the system behaviour, but it is not as good as that produced directly by NEC.