(Guanella 1944) described a 1:4 balun, of a type often known as a current balun.
An ideal current balun delivers currents that are equal in magnitude and opposite in phase.
A good current balun will approach the ideal condition. It will deliver approximately equal currents with approximately opposite phase, irrespective of the load impedance (including symmetry).
Common mode current will be small.
If the load impedance is not symmetric, then the voltages at each output terminal will not be equal in magnitude and opposite in phase. (Note that for a truly ‘isolated’ load, one well represented as a two terminal load, the currents MUST be equal in magnitude and opposite in phase, but the voltages may not be equal in magnitude and opposite in phase.)
A simplified model
Above is a schematic of the Guanella 1:4 balun as often presented, this is an edited graphic from the ARRL manual, so may be familiar to readers.
It represents the things as comprising two transformers T4 and T5, a very naive model, but one that may be understood by readers with quite basic knowledge.
Note that the schematic shows only two load terminals, the astute reader will realise that a lumped component analysis is that the current into one load terminal MUST be equal in magnitude and phase to the current out of the other load terminal. It is an oversimplification and if it did apply, there would be no need for a current balun, the currents are already balanced.
With that in mind, let’s discuss the Guanella 1:4 balun using the above model and assuming ideal 1:1 component transformers.
An ideal 1:1 transformer obeys the following rules:
- Rule1: magnetising current is approximately zero;
- Rule2: the voltage across each winding is equal to the other;
- Rule3: the current through each winding is equal to the other;
- Rule4: the phase and magnitude of current in a winding is uniform; and
- Rule5: zero loss (ie no core loss, no conductor loss).
These rules apply and may be referenced in the analysis below.
The ‘dot notation’ is used to marking phase relationships of windings:
- voltages at the dot terminals are in-phase; and
- current flowing into the primary winding dot gives rise to current flowing out of the secondary winding dot.
Let’s rearrange and relabel the circuit for purpose of discussion and clarity in understanding.
Above, L1 and L2 are coupled coils (flux coupling factor k is 1, all flux due to current in one winding cuts the other winding) form one transformer (T4 from the ARRL’s schematic), and L3 and L4 are coupled coils (flux coupling factor k is 1, all flux due to current in one winding cuts the other winding) forming the other transformer (T5 from the ARRL’s schematic).
The load is a simple three terminal asymmetric load.
To analyse this circuit, lets work from right to left.
Readers might print the schematic above and annotate it as they work through the steps.
- Assume V1=4V, polarity as marked;
- Assume current of 1A flows downwards in R1, 1Ω, dropping 1V (+ to high end).
- The voltage difference across L2 is V1-1=3V +ve at the dot end.
- By virtue of 3. and Rule2, the voltage difference across L1 is 3V +ve at the dot end.
- By virtue of 2., 1A flows downward in L2, outwards from source V1 + terminal .
- By virtue of 5, Rule3 and Ruke4, 1A flows upwards in L1.
- L3 is in series with L1, so by virtue of Kirchoff’s Current Law, 1A flows upwards in L3.
- The current in L1 & L3 is outwards from the source V1 + terminal.
- Total current at source V1 + terminal is 2A outwards from the terminal.
- By virtue of 7 and Rule3, 1A flows downwards in L4.
- By virtue of 10, 1A flows downwards in R7 and the voltage difference across R7 is 7V -ve down.
- By virtue of 11, the voltage difference across L4 is 7V +ve at the dot end.
- By virtue of 12. and Rule2, the voltage difference across L3 is 7V +ve at the dot end.
- By virtue of Kirchoffs Voltage Law, the voltage across L3 and L1 in series is -3+7=4V +ve to the bottom.
- The voltage across L1+L3 reconciles with the given value of source V1, 4V.
The key points are that:
- the voltages and currents calculated obey Kirchoff’s Laws;
- the voltages and currents calculate reconcile with the given data;
- The current in conductors feeding R1 and R2 are equal in magnitude and opposite in direction (ie those currents balance);
- the total load voltage is twice the source voltage, and the load current is half the total source current confirming that it is a 1:4 impedance transformation.
Issues in applying this model
In this model using ideal components, if the connection between R1 and R2 is not grounded, there is not a determined solution, it is impossible to calculate a division of source voltage across L1 and L3. R1 or R2 can be zero, but there must be a ground connection. More on this later.
Rework the above for some other cases, eg V1=2, R1=1, R2=3.
Print and annotate the diagram step by step.
The real RF world
Significant factors include:
- conductor lengths in the transformer are significant length in terms of wavelength, there is current phase change along the winding;
- there is signficant flux leakage;
- conductor loss is not zero, though usually quite small;
- magnetising current is significant;
- if powdered iron or ferrite cores are used, core loss is not zero and can be quite large depending on core material, frequency, flux density.
These ‘imperfections’ mean NONE of the Rules Rule1 – Rule5 apply exactly. For this reason, this is probably not a good model for designing or making inferences about most RF Guanella 1:4 baluns.
Using this model, for a Guanella 1:4 balun to have very good current balance, the magnetising current of the transformers must be insignificant, much much smaller than the load current (a hundredth or less) which broadly means that magnetising impedance needs to be much much greater than load impedance.
This model is not a good model for a Guanella 1:4 balun at radio frequencies.
- Guanella, G. Sep 1944. New methods of impedance matching in radio frequency circuits. The Brown Boveri Review.