balun transformer design

Question

The schematic shows a balun (unbalanced i/p to balanced output).In the design one condition is input impedance must be equal to output impedance since its 1:1 transformer and it has to follow impedance and turns ratio theory. My query is, on the output side if I have 50 ohm trace impedance instead of 25 ohm, what will be the effect. Is it going to further reduce my output power?

What else can be the effect if the impedance (i/p and o/p) is not matched for 1:1 transformer?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

If the transformer is ideal, then the problem will be the same as if you attached any \$50\Omega\$ source to a \$25\Omega\$ load. The transformer is not relevant. That is, a portion of any power arriving at the transformer from either direction will be reflected. This could be power generated by your source, or power reflected by your load.

Note that the power reflection, in itself, doesn't reduce your output power. The reflected power will bounce around until it finds a place to go, either radiated as EM energy, or converted to heat by resistive components. However, your voltage source might not like it very much, depending on what it is. The reflected power might result in voltages outside of its specifications, or if it operates on feedback, it may become unstable. The ideal voltage source in your schematic, however, won't care.

If the transformer isn't ideal, then you are introducing additional impedances into the circuit. The windings of the transformer have some capacitance. There is some magnetic flux not shared by the primary and secondary which appears as a series inductor. And of course, the windings have some resistance. The extent to which these are significant depend on many parameters, like the construction of the transformer and the operating frequency.

If you can't alter your source or load to match, then you could use a transformer with a turns ratio of:

$$ T = \sqrt{\frac{R_S}{R_L}} = \sqrt{\frac{50\Omega}{25\Omega}} = \sqrt{2} \approx 1.4 $$

Another possibility is an L-section of a capacitor and inductor to achieve a narrow-band match, just as you would if the transformer were not there.

Answer 2

No, the schematic does NOT show a balun. It shows a ordinary transformer. A balun is essentialy two magnetically-coupled chokes. They are arranged so that common mode currents add to the magnetic field created by the combined windings and differential mode currents subtract to cancel and not make a magnetic field. The net effect is the choke has high impedance for common mode signals and in the ideal case is a dead short for differential mode signals. Another name for a balun is "common mode choke".

Added:

I see there is some misundertanding about what I said. Yes, a ordinary transormer can be used as a balun if it has a 1:1 ratio, preferably bifilar wound. In fact that's how baluns are often made. What makes it a balun or a transformer is how it is used.

In a tranformer, signal or power is put in one winding and taken out of the other. The magnetic field is integral in transferring this signal from the input to the output. In a balun, the signal goes thru both windings in such a way that the device as a whole resists (looks like a inductor) common mode signals and passes differential mode signals.

The circuit above is using the device as a transformer, not a balun. Clearly the signal is meant to be transferred between the two windings by the magnetic field. The same transformer could possibly be used as a balun, but that is not how it is being used in this case.

Here is a example of a balun:

Note that this passes DC and that if the balun is ideal, it has no effect on the differential mode component of the signal. It does, however, add series inductance to the common mode comonent of the signal.

And here is the same device used in a ordinary transformer role:

Note how the circuit in question has the second topology, not the first.