I'm having little difficulties understanding the following schematic:

enter image description here

I understand the part regarding current distribution. I understand why this schematic suppresses common mode current. Also I realize that if Z1 = V/2I was 50 Ohm than the balun converts the impedance to Z2 = 2V/I, 200 Ohm.

What I don't understand is the voltage part on this schematic. Why the voltage is +3V/2 and -V/2 in the right part (2V in total), where V/2 came from in the middle part and what does it mean?


1 Answer 1


It's a confusing diagram.

Some of the voltages are differences between points you might intuitively assume. For example, the $2V$ at the right is the voltage across the entire load (which also confusingly, is modeled by two identically named series resistors).

But the other voltages are implicitly relative to the ground symbol attached to the connector shield.

Lemma: the voltage across any winding must be the same since we are considering only the differential mode and the transformers are the same. Hopefully that's intuitive enough.

The voltage across any winding must be $V/2$ (though the polarity may differ). Consider this circuit:

enter image description here

By the KVL, the voltages around this circuit must add to zero. Since the voltage across the connector is $V$ by definition, the windings must each be half that.

Now start at the point labeled $-V/2$ and follow back to ground:

enter image description here

Having gone through one winding (minding polarity) the voltage is one times the winding voltage, or $-V/2$

And the other point labeled $3V/2$:

enter image description here

This way goes through a winding and the connector, so the voltage is $V + V/2 = 3V/2$.

The difference between these points is:

$$ 3V/2 - (-V/2) = 2V $$

which is the output voltage you'd expect considering conservation of energy or just what a 4:1 balun is supposed to do.

  • $\begingroup$ Thank you for a great explanation! There is one thing though that although seem to be logical is not absolutely obvious. Let's say we know that the voltage drop on a given transformer is V/2 and we determined it's polarity. Knowing that how we determine the voltage change for the second winding? Does it depend on the current direction or only on polarity? If it's not too much trouble could you please add a little more details on that to your answer? $\endgroup$ Commented Aug 21, 2019 at 21:33
  • 1
    $\begingroup$ @AleksanderAlekseev-R2AUK As long as these are 1:1 ideal transformers (assumed by this basic analysis) the voltage on the two windings are always the same, with the same polarity. (That is, if the dot side is positive on one winding, the dot side is also positive on the other winding.) It follows from Faraday's law of induction, and the assumption of no leakage flux in an ideal transformer. $\endgroup$ Commented Aug 21, 2019 at 22:57

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