validity of the lumped element impedance

[StoppTidigare]

impedance inverter

Hi all, could somebody discuss with me the validity of the lumped element impedance from a Smith-chart view point ?

According to P. Vizmuller "RF Design Guide" page 165, a pi-section, for instance 2 cap and 1 inductor, can be viewed as a lowpass approximation to a lambda/4 -line.

The component values should be choosed as:
L=Zo/(2pi*f)
C=1/(2pi*f*Zo),
Zo=characteristic impedance.

Kindest regards,
StoppTidigare

 

[sergio mariotti]

impedance inverter filter

what's exactly the topic you want to discuss?
I' don't have a copy of Vizmuller's book.
I'have no doubt about the Smith-Chart (S-C) validity in every condition.
But some cases or application may be difficult to plot.
Are you writing about a filter or the infinitesimal lenght section of a transission line?
For example, if you are writing about a filter, the "rotations" due to each Cap and Ind are easy plottable but... at fixed frequency and omitting dissipative losses.
Also i agree with you that lumped components should be shorter than Lambda/4.

If you are writing about the infinitesimal transmisison line lenght section, modelled as C-shunt, L-series, ladder network, you may plot on S-C as a Cshunt of a given susceptance, followed by Lseries of a reactance that the end point come back to the center. This operation should be re-itered "infinitely" until you'll reach the load. Note that you are everytime on center. This mean that your signal is flowing on matched line.

Finally if you are a novice i suggest you Agilent AN-95

Bye

 

[flatulent]

Re: Impedance inverter

Yes. You are approximating the series inductance and shunt capacitance of a line with lumped values. This will give you the same impedance and phase shift at low frequencies as a physical transmission line. As you go higher the approximation does not work well and eventually you get no throughput. (low pass filter).

You should be able to plot these elements on a smith chart and end up at the starting point.

 

[StoppTidigare]

Re: Impedance inverter

Hi all, think I have it now, case somebody is interested:

Say that I have a short circuit that I want to make look as an open circuit.
Then add a lowpass pi-network, and

starting at the far left in the ZY-chart, the first shunt C will still let me be at the shortcircuit point. Adding then series L with the value as in previous post will take you along constant resistance circle 0 to 0 +j1 .0

Then adding a shunt C, takes you along the constant g-circle = infinity to
the point r=infinity, x=0

Kindest regards,
StoppTidigare