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Sometimes designs call for quarter-wave sections of transmission lines. There are some situations when using actual transmission lines would not be practical:

  • The necessary characteristic impedance is something not easily obtained
  • A quarter wavelength of transmission line would be too physically large

Can I accomplish the same thing with lumped elements like inductors and capacitors? If so, how?

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1 Answer 1

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It can be done! Remember that a transmission line consists of some self-inductance per unit length, and some capacitance per unit length, and the ratio of these determines the line's characteristic impedance:

$$ Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}} $$

Intuitively then, a lumped element implementation might look like some series inductance and some parallel capacitance. We need something that's symmetrical, so a pi network1 like this should do it:

schematic

simulate this circuit – Schematic created using CircuitLab

All we have to do is determine the appropriate values. Since transmission lines work in both directions this arrangement is going to be symmetrical, so C1 = C2. And as it turns out, the values are very elegant: the reactance of the components is equal to the characteristic impedance. Put mathematically:

$$ X_{L1} = Z_0 \\ X_{C1} = Z_0 $$

And while we're at it, let's review the formulae for reactance:

$$ X_L = 2 \pi f L \\ X_C = 1 / (2 \pi f C) $$

Putting those together, with a bit of algebra, you get:

$$ L = {Z_0 \over 2 \pi f} $$

$$ C = {1 \over 2 \pi f Z_0} $$

Let's say we want a characteristic impedance of 86.6 ohms so we can realize a 3-way Wilkinson power splitter. We'll use the frequency 435 MHz for the middle of the 70cm band.

$$ L = { 86.6\:\Omega \over 2\pi \cdot 435\:\mathrm{MHz} } = 31.5\:\mathrm{nH} $$

$$ C = {1 \over 2\pi \cdot 435\:\mathrm{MHz} \cdot 86.6\:\Omega } = 4.22 \:\mathrm{pF} $$

So now the circuit with real values, terminated with an 86 ohm load:

schematic

simulate this circuit

If this is indeed a quarter-wave transmission line, we should see R1 90 degrees out of phase with the input. Let's run a time-domain simulation:

voltage

current

Bingo! The impedance transforming properties of a quarter wave transmission line are also preserved: if the output is open, the source will see a short, and so on. So this circuit can be used in place anywhere that calls for a quarter-wave transmission line.


1 This pi network makes a low pass filter, which can be a nice side-effect since it reduces harmonic distortion.

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  • $\begingroup$ As a bonus, this circuit works as a Low Pass filter with a -3 dB cut off at about 620 MHz. -10 dB at 880 MHz. -20 dB at 1.3 GHz. $\endgroup$ May 18, 2016 at 20:56
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    $\begingroup$ That looks suspiciously like the symbol Pi. :) $\endgroup$
    – David Hoelzer
    May 19, 2016 at 17:10
  • $\begingroup$ How do we know that this provides Zo ohms impedance? $\endgroup$ Aug 1, 2017 at 14:40
  • $\begingroup$ @NourhanElsayed Not sure what you mean. It doesn't provide any impedance: it provides an impedance transformation equivalent to a quarter-wave section of some transmission line of some impedance. If you are wondering how we know what "some impedance" is, the math is in the answer. $\endgroup$ Aug 1, 2017 at 16:04

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