1
\$\begingroup\$

I came across the following slide on power combiner circuit and I have questions about it.

  1. Why traverse across the upper half circle ?

  2. Why "ZT-line is the geometric mean of Zout and Zin." ?

Smith chart of λ/4 Transmission Line

\$\endgroup\$
1
  • 1
    \$\begingroup\$ have you read the math (the derivations) behind the Smith Chart? \$\endgroup\$ Commented Mar 18, 2019 at 14:32

2 Answers 2

2
\$\begingroup\$

Why traverse across the upper half circle ?

When translating an impedance towards the generator we move clockwise around the chart.

When translating an impedance towards the load, we move counter-clockwise around the chart.

Why "ZT-line is the geometric mean of Zout and Zin." ?

This is a design decision that was made to get the desired result. If you choose a different \$Z_T\$, you won't get the equivalent load (\$Z_T\$ line plus actual load) to match your input impedance.

Notice how they overlaid the Smith chart scaled for \$Z_T\$ on top of the \$Z_0\$ Smith chart. You need to be able to rotate the load around from the 0-ohm point to the \$\infty\$-ohm point on the \$Z_T\$ Smith chart and have that end up being the correct place on the \$Z_0\$ Smith chart. That only happens if you choose \$Z_T=\sqrt{Z_{out} Z_{in}}\$

\$\endgroup\$
2
  • \$\begingroup\$ Why overlay the Smith chart scaled for ZT on top of the Z0 Smith chart ? \$\endgroup\$
    – kevin998x
    Commented Mar 19, 2019 at 1:08
  • \$\begingroup\$ why is Zin located at the right side of Zout ? \$\endgroup\$
    – kevin
    Commented Mar 19, 2019 at 4:55
1
\$\begingroup\$

Why traverse across the upper half circle ?

Traverse is across the upper half of a circle is because of adding physical line with length θ=90° delays reflection from Г(Zout) by 2*θ=180° (90° before reflection and 90° after reflection). On a Smith chart phase delay corresponds to clockwise rotation around some center depending on Zline(phase decreases clockwise). Cutting physical line length will result in counter-clockwise direction. Rotation go through upper or lower half circle depending on particular values of Zout and quarterwave transformer line impedance Zt.

Why "ZT-line is the geometric mean of Zout and Zin." ?

Maybe for emphasizing that it is not an arithmetic mean (Zout+Zin)/2.

Formula for Zt is derived without using Smith chart. Smith chart is a graphical aid.

Why overlay the Smith chart scaled for ZT on top of the Z0 Smith chart?

Maybe it is more convenient to use Z0=Zt Smith chart for obtaining impedance transformation graphically without formulas (Smith chart center at Zt, not at 50 Ohm)

For deeper understanding of Smith chart I recommend to calculate impedance change for different line lengths using formulas, then calculate corresponding reflection coefficient Г (vector) for each case and draw it on a Smith chart.

\$\endgroup\$
4
  • \$\begingroup\$ "it is more convenient to use Z0=Zt Smith chart for obtaining impedance transformation graphically without formulas" <-- could you explain more on this reason of putting two smith charts together on top of each other ? \$\endgroup\$
    – kevin998x
    Commented Mar 20, 2019 at 15:53
  • \$\begingroup\$ What do you exactly mean by "Smith chart center at Zt, not at 50 Ohm" ? \$\endgroup\$
    – kevin998x
    Commented Mar 20, 2019 at 16:01
  • \$\begingroup\$ @kevin998x , you may draw Smith chart for any impedance. Usually Z0=50 Ohm is chosen. But you may use any other Z0 in this formula: Г=(Z-Z0)/(Z+Z0). If Z0=Zt, then Zt will be in the center of Smith chart, and quarter-wave transforming will go through circle around this center. \$\endgroup\$ Commented Mar 21, 2019 at 2:24
  • \$\begingroup\$ "use Z0=Zt Smith chart for obtaining impedance transformation graphically without formulas" <-- what do you exactly mean by impedance transformation ? And how is normalizing Z0 with respect to Zt related to the use of two smith chart overlaid together ? \$\endgroup\$
    – kevin998x
    Commented Mar 22, 2019 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.