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Can someone explain to me exactly how a quarter-wave transmission line transformer is put into action? I mean, how does it operate and what values are usually taken into consideration, calculation wise?

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  • $\begingroup$ This question is too broad. Please include context and your research so far. $\endgroup$ – Juancho May 13 '15 at 14:13
  • $\begingroup$ I have included this is in my answer to your other question on Stub Impedance Matching $\endgroup$ – Harry Weston May 16 '15 at 15:00
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A quarter-wave transmission line makes some load appear to the source on the other end look like the load's dual impedance. The most common case is that the impedance at the end is a short or an open, and the quarter-wave transmission line makes it look to the source (V1 or V2) as the opposite:

schematic

simulate this circuit – Schematic created using CircuitLab

This is pretty weird. In the case of V1, it looks like it's connected to a short: a huge (theoretically infinite) current flows. But if you stick an ohmmeter on the same terminals where V1 is connected, it will tell you there's no connection between them.

In the case of V2, it looks like it's connected to nothing: no current flows. Yet if you measure with your meter again, it will tell you there's a dead short.

Since it seems like this is logically impossible, it bears mentioning that a short "looks like" an open (or vice versa) only at equilibrium. If you were able to slow time such that the speed of light were observable, you'd see that in both cases, which V1 is initially connected, some current does flow. Once that initial wave has reached the end of the transmission line, and then reflected back (maybe several times), does the system reach equilibrium.

In the more general case, the load on the right end doesn't have to be a short or an open: it can be any impedance:

Quarter wave impedance transformer.svg
"Quarter wave impedance transformer" by Courtesy Spinningspark at Wikipedia. Licensed under CC BY-SA 3.0 via Wikipedia.

In this case:

  • $Z_\text{in}$ is the impedance something will see, looking into the transmission line,
  • $Z_0$ is the characteristic impedance of the transmission line, which is commonly $50\:\Omega$, and
  • $Z_L$ is the impedance of the load at the end.

These three parameters are then related by the equation:

$$ {Z_\text{in} \over Z_0} = {Z_0 \over Z_L} $$

Since these are complex numbers the math can be a little unintuitive, so some consequences, in plain English:

  • inductors look like capacitors
  • capacitors look like inductors
  • small resistors look like big resistors
  • big resistors look like small resistors
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