Much is written about antenna system balance, this article looks at balance issues with the very common ATU configuration that uses a Ruthroff 4:1 voltage balun to adapt coax transmitter output to two wire open feed line. This type of balun is employed in most ham market ATUs that contain an integral balun.
Above is Ruthroff’s equivalent circuit, Fig 3 from his paper (Ruthroff 1959).
If one looks carefully at the transmission line form, there is effectively a two wire line wound into a helix (usually on a magnetic core) and connected from the unbalanced source to one half of the load inverting the connection for the necessary phase reversal.
Ideally, Vout of this line is equal to Vin, ie Vout/Vin should be 1∠0°. That is unlikely as it implies a zero length transmission line which provides the decoupling of the phase inverting line.
This article looks at the Ruthroff 4:1 balun balance using the very popular MFJ949E as an example.
Above is a pic of the MFJ949E Ruthroff 4:1 balun. The transmission line is not uniform, but let’s make an approximation to predict its behavior with a centre tapped 100Ω load, the centre of which is connected to the ground terminal.
A theoretical analysis
The following is a model using TWLLC of a similar transmission line with 50+j0Ω load (one half of the load is connected to the transmission line). We will assume that Zcm of the TLT is very high.
Approximate MFJ949E Ruthroff balun TL 

Parameters  
Conductivity  5.800e+7 S/m 
Rel permeability  1.000 
Diameter  0.000600 m 
Spacing  0.002000 m 
Velocity factor  0.750 
Loss tangent  0.000e+0 
Frequency  7.000 MHz 
Twist rate  0 t/m 
Length  0.600 m 
Zload  50.00+j0.00 Ω 
Yload  0.020000+j0.000000 S 
Results  
Zo  170.52j2.02 Ω 
Velocity Factor  0.7500 
Twist factor  1.0000 
Rel permittivity  1.778 
Length  6.725 °, 0.018680 λ, 0.600000 m, 2.669e+3 ps 
Line Loss (matched)  1.22e2 dB 
R, L, G, C  7.976649e1, 7.666990e7, 0.000000e+0, 2.637228e11 
Line Loss  4.12e2 dB 
Efficiency  99.06 % 
Zin  5.113e+1+j1.854e+1 Ω 
Yin  0.01728488j0.00626760 S 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (source end)  5.288e1+j1.322e1, 0.545∠166.0°, 5.271 dB, 3.40, 1.531 dB 
Γ, ρ∠θ, RL, VSWR, MismatchLoss (load end)  5.466e1+j4.146e3, 0.547∠179.6°, 5.247 dB, 3.41, 1.541 dB 
Vout/Vin  8.582e1j3.457e1, 9.252e1∠21.9° 
Iout/Iin  1.006e+0j3.533e2, 1.006e+0∠2.0° 
S11, S21  4.334e2+j1.754e1, 9.561e1j2.102e1 
Y11, Y21  1.168e3j4.917e2, 1.168e3+j4.951e2 
NEC NT  NT t s t s 1.168e3 4.917e2 1.168e3 4.951e2 1.168e3 4.917e2 ‘ 0.600 m, 7.000 MHz 
k1, k2  7.679e6, 0.000e+0 
C1, C2  2.428e1, 0.000e+0 
Mhf1, Mhf2  2.340e1, 0.000e+0 
dB/m @1MHz: cond, diel  0.007679, 0.000000 
γ  2.339e3+j1.978e1 
Among all that detail is an estimate of Vout/Vin, it is 9.252e1∠21.9°. That is to say that if there is 1.0V on the ‘direct’ balun terminal, the other terminal via the transmission line will be 0.9252∠21.9° V.
Since the current in each of our 50Ω 1% load resistors is V/50, we can calculate the common mode current by summing them, and it will not be zero because of the lower voltage and particularly the significant phase delay of the ‘indirect’ terminal.
Let’s measure the real thing
Connecting a scope to the terminals and measuring the voltage applied to each of the 50Ω resistors, we can see that the orange trace is lower in amplitude (about 90%) and delayed by about 8ns (20.2°)
We can also use the scope to add the waveforms to find the common mode component.
The red trace is CH1+CH2.
Lets put those numbers into a calculator as unscaled divisions from the scope.
Above, the calculated phase difference is 22° which reconciles well with the estimate from the scope trace delay above. It also reconciles well with the theoretical prediction earlier.
The ratio 2Ic/Id is 40%, this is pretty awful current balance on a symmetric load.
Now this balun is worse at higher frequencies because the attenuation in the balun transmission line increases, and the phase delay increases almost proportional to frequency.
This is quite a small balun physically, larger baluns will have longer transmission line and greater phase delay, so even worse performance on a perfectly symmetric load.
So, whilst ideal voltage baluns deliver equal but opposite currents into a perfectly symmetric load, practical Ruthroff 4:1 voltage baluns might not be all that good.
Conclusions
Measured behavior of the Ruthroff 4:1 voltage balun accords with prediction using Ruthroff’s transmission line model of the device.
Voltage symmetry is poor, predictable, and it is load dependent, and it gives rise to significant common mode current, even in a symmetric antenna system.
If you think your balun is working well… it is probably because you have not measured common mode current.
References / links
 Duffy, O. 2007. A model of a practical Ruthroff 1:4 balun.
 Ruthroff C. 1959. Some broad band transformers. Proceedings of the IRE.