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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?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ That's not a balun. \$\endgroup\$ – Olin Lathrop Jun 18 '13 at 22:45
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    \$\begingroup\$ @OlinLathrop There is more than one way to make a balun, and OP's circuit is widely recognized to be a balun. See, for example, minicircuits.com/pages/BalunApplicationNote.htm \$\endgroup\$ – The Photon Jun 19 '13 at 3:59
  • \$\begingroup\$ @olin lathrop please have a look at this cn-william.com/e2v/ad-ap/… \$\endgroup\$ – harish Jun 19 '13 at 5:15
  • \$\begingroup\$ That symbol is used for a "Classical Transformer type Balun", it isn't an unusual representation. \$\endgroup\$ – Anindo Ghosh Jun 19 '13 at 12:15
  • \$\begingroup\$ @Theph: That appeared to be some Atmel app note about impedance-matching transformer coupled signals, having little relevance to baluns. At least the first couple pages didn't. If you are referring to something specific, point it out. See addition to my answer. \$\endgroup\$ – Olin Lathrop Jun 19 '13 at 13:24
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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.

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  • \$\begingroup\$ so what can be the solution, to take a transformer with 1:1.5 turns ratio? If not what could be the possible solution? \$\endgroup\$ – harish Jun 19 '13 at 7:28
  • \$\begingroup\$ @harish see edits. \$\endgroup\$ – Phil Frost Jun 19 '13 at 11:20
  • \$\begingroup\$ Is 1:1.5 fine, because 1:1.4 transformers are not available to my knowledge \$\endgroup\$ – harish Jun 19 '13 at 13:19
  • \$\begingroup\$ @harish it depends on how much mismatch you can tolerate. Read about the reflection coefficient to calculate how much power would be reflected. How much of that reflected power is eventually lost depends on how much is absorbed before it's reflected back again. At HF, a reflection coefficient of \$\lvert \Gamma \rvert < 0.25\$ is probably fine; at higher frequencies, transmission line losses become greater, so that extra bouncing back and forth means more lost power. \$\endgroup\$ – Phil Frost Jun 19 '13 at 14:05
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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.

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    \$\begingroup\$ A balun is by definition a balanced-to-unbalanced transformer, nothing prevents a conventional transformer of the right parameters being used as one. That symbol and representation have been used in several publications found on a quick scan, including some seminal patents (e.g. C L Ruthroff, US 3037175). \$\endgroup\$ – Anindo Ghosh Jun 19 '13 at 12:19
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    \$\begingroup\$ I think you could make an argument that you can make a better balun with the connection you describe, but saying this other configuration isn't a balun at all does not seem accurate. Further, you can't use that configuration to match impedances, which seems to be half the problem here. See vk5ajl.com/projects/baluns.php#voltage_tfmr for some discussion. \$\endgroup\$ – Phil Frost Jun 19 '13 at 13:48
  • \$\begingroup\$ @Phil: That page uses "balun" rather liberally. It is wrong to call a simple transformer, as in my bottom schematic, a balun. A classic balun is not for impedance matching. I realize the OP is asking about impedance matching, but he is incorrectly applying the term "balun" to what is just a straight transformer. A transformer can change impedance, and that's what he's asking about, but that doesn't make it a balun. \$\endgroup\$ – Olin Lathrop Jun 19 '13 at 16:02
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    \$\begingroup\$ OlinLathrop, I agree with @PhilFrost --- a balun is defined by its function, not by some specific circuit design. Your second diagram doesn't show a balun. But as soon as you ground one of 4 nets connected to the transformer it does. \$\endgroup\$ – The Photon Jun 19 '13 at 16:13

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