All,

This is a follow up to a dialog several months back on impedance
measurements. The question relates to smith charts and an HP 8407a with
a polar display. I was under the impression that the polar display and
the smith chart were pretty closely related. In fact hp made smith chart
overlays for the screens.

The smith chart predicts a variable capacitor connected across a source
will be represented by an impedance semicircle in the capacitive domain,
one of the points on the semicircle passes through the 50 ohm center. I
would expect to be able to create this using the 8407.

The 8407 creates a circle around the outside edge of the display. It is
a circle, however it does not pass through the center.

Why are these different? What above is incorrect?

thanks - Dan

dansawyeror wrote in
:

All,

This is a follow up to a dialog several months back on impedance
measurements. The question relates to smith charts and an HP 8407a with
a polar display. I was under the impression that the polar display and
the smith chart were pretty closely related. In fact hp made smith
chart
overlays for the screens.

The smith chart predicts a variable capacitor connected across a source
will be represented by an impedance semicircle in the capacitive
domain,
one of the points on the semicircle passes through the 50 ohm center. I
would expect to be able to create this using the 8407.

The 50 ohm centre is on the R=50 circle and the X=0 arc (which is
infinitely large in radius, so it is the the diameter from R=0
R=infinity).

Sit down and plot a few values of Z00+jx on a Smith chart, aren't they
all on the outside edge of the chart (the R=0 circle).

Owen

Owen Duffy wrote in news:Xns98E18366671D3nonenowhere@
61.9.191.5:

Sit down and plot a few values of Z00+jx on a Smith chart, aren't they
all on the outside edge of the chart (the R=0 circle).


That should read:

Sit down and plot a few values of Z=0+jx on a Smith chart, aren't they
all on the outside edge of the chart (the R=0 circle).


Owen

dansawyeror wrote:
Why are these different? What above is incorrect?

The horizontal center line on a Smith Chart is resistive.
The outside circle on a Smith Chart is reactive. A
capacitor is almost purely reactive. The reactance is
1/(2*pi*f*c) That value will always fall on the
outside circle of the Smith Chart. If the capacitor
has a reactance of 0-j50 ohms, it will be located on
the outside circle at the very bottom of the Smith
Chart normalized to a 0-j1.0 value.
--
73, Cecil http://www.w5dxp.com

Thank you, that is consistent with the analyzer. The smith chart program
is in error.

The analyzer also shows the response of an open coax to follow the same
circle at the edge of the chart. Is this accurate? The analyzer does not
correct for length, is this a result of that?

- Dan

Cecil Moore wrote:
dansawyeror wrote:
Why are these different? What above is incorrect?

The horizontal center line on a Smith Chart is resistive.
The outside circle on a Smith Chart is reactive. A
capacitor is almost purely reactive. The reactance is
1/(2*pi*f*c) That value will always fall on the
outside circle of the Smith Chart. If the capacitor
has a reactance of 0-j50 ohms, it will be located on
the outside circle at the very bottom of the Smith
Chart normalized to a 0-j1.0 value.

dansawyeror wrote in news:45DFA702.9000204
@comcast.net:

The analyzer also shows the response of an open coax to follow the same
circle at the edge of the chart. Is this accurate? The analyzer does
not
correct for length, is this a result of that?

Dan,

Perhaps a better way to understand a Smith chart as a visualisation tool
is to understand the Smith chart as a problem solving tool.

If you want to follow that path, print of a page or three of Smith chart
and plot solve some typical problems (without the aid of a computer
program), think about what the arcs and lines that you draw mean, and see
if you don't get an understanding of what points on a Smith chart mean.
You will need to find a text book or authoritive web article to walk you
through the process... but if you don't plot points for yourself on a
blank chart, you probably won't learn the principles.

The case you are talking about of the Z of an o/c stub at a range of
frequencies is an easy one to solve with a Smith chart. You plot the Z at
the load end of the stub, you draw the arcs for constant VSWR for each of
the frequencies of interest (the arcs are of different length for each
frequency because they are plotted in electrical length), and you read
off the source end Z.

That describes the mechanics of solving that problem, but it helps to
understand that the points on a Smith chart are the sum of two phasors on
a complex plane, and what they represent. You don't really understand the
Smith chart unless you understand what happens on a transmission line.

Owen

dansawyeror wrote:
The analyzer also shows the response of an open coax to follow the same
circle at the edge of the chart. Is this accurate?

It is accurate for lossless lines. Actual coax
values will tend to spiral in to a Z0 limit.
200 feet of open-circuit RG-58 used on 440 MHz
gets pretty close to 50 ohms after hundreds of
spirals.
--
73, Cecil http://www.w5dxp.com

This is a follow up to a dialog several months back on impedance
measurements. The question relates to smith charts and an HP 8407a with a
polar display. I was under the impression that the polar display and the
smith chart were pretty closely related. In fact hp made smith chart
overlays for the screens.

The smith chart predicts a variable capacitor connected across a source
will be represented by an impedance semicircle in the capacitive domain,
one of the points on the semicircle passes through the 50 ohm center. I
would expect to be able to create this using the 8407.

The 8407 creates a circle around the outside edge of the display. It is a
circle, however it does not pass through the center.

Why are these different? What above is incorrect?

My first question would be: is the 8407 calibrated correctly?

When you apply a 50ohm load do you get close to a single spot in the centre
of the screen, and when you apply your calibrated Open & Shorts do you get
close to single points on the X axis on the outer ring of the Smith Chart?
If not then you need to re-calibrate your test set up.

It is a long time since I used an HP8407, but I do recall that they were a
pain to calibrate correctly, lacking the auto calibration routines of modern
VNA's.

73
Jeff