money wrote:
K7ITM wrote:

money wrote:
I have a circuit in which there is a series capacitance of value 680
pF. The characteristic impedance of the line is 50 ohms. Can any
suggest how do i plot the capacitance in the smith chart.

If suppoe it was a parallel capacitance, how do i do it?

Owen's right. A picture is worth, generally, about a kiloword.

I'd also recommend you download one of the free Smith chart programs.
You may also even find a decent web aplet that will plot directly in
your browser window.

For a quick more direct answer: for a series capacitance, plot on the
impedance grid. The capacitance must be converted first to a
capacitive reactance: Xc = -1/(2*pi*f*C). That is the length you need
to move along an arc of a circle passes through your starting point and
is tangent to the unit circle at infinite impedance. You will see some
of these plotted on an impedance grid; your starting point may already
lie on one of the plotted ones. For example, if the chart is
normalized to 50 ohms, the 50+j0 point lies at the center of the chart.
Let's say that the starting point is 25 ohms, and the frequency is
9.36MHz. Then the capacitive reactance will move you counterclockwise
(in the normal presentation of the chart) along the arc that passes
through the 25 ohm point (or the 0.5 point on a chart normalized to 1
ohm), about 35 degrees, to 25-j25 ohms. That should not be a surprise:
the reactance of the capacitor at 7MHz is 25 ohms capacitive; an
impedance of -j25 ohms. Series impedances simply add.

For a parallel capacitance, find the suseptance of the capacitor at the
frequency of interest, and follow a arc of constant conductance on the
admittance overlay for the Smith chart. Remember, admittances of
parallel components add, just as impedances of series components add.
For the case in point, again at 9.36 MHz, you should go clockwise about
75 degrees along the arc of a circle which is tangent to the unit
circle at impedance = 0 (infinite admittance) and end up at an
admittance of 0.04+j0.04 mhos, or an impedance of 12.5-j12.5 ohms.

If you go back to the first example, with a series 680pF capacitance,
and then add a shunt inductance, you will travel from the 25-j25 ohm
point, counterclockwise along a constant conductance circle. If you
make the inductance 850nH, you will find that you end up at an
impedance of 50 ohms: you have matched the original 25 ohms to a 50
ohm system, at that frequency, with an "L" network. The Smith chart
lets you visualize the match very quickly, once you are comfortable
with it.

A computerized Smith chart made the numbers above much easier than the
words!

Cheers,
Tom



The circuit is actually an impedance transformation circuit to match it
with line impedance 50 ohms. the frequency of operation is 900 MHz.

With a software i matched an amplfier circuit with line impedence 50
ohms. 680pF(series capacitance) is part of the impedance transformation
circuit. I do get the point of traversal in a smith chart. For series
capacitance i need to move along the constant resistance circle
anti-clockwise.

But if i use the formula capacitive reactance = 1 / C W then normalise
it to 50 ohms i get negligible reactance of 0.05.... Is it correct?

Yes, right. It is a very small reactance at 900MHz. In other words, I
assume it is just a DC blocking capacitor, if it is part of a properly
designed network. It must be other parts which are doing the actual
matching job.

Do you know the impedance of the port which you are matching to 50
ohms? Or do you have the network already, and you want to
"reverse-engineer" the circuit to discover what the impedance is that
is matched to 50 ohms?

I would expect that parts used in a 900MHz matching network would be in
the range from perhaps 0.5pF to 10pF--maybe a little beyond, and also
inductances in the low nanohenry range. And such small capacitances
and inductances can be even small copper areas and short, narrow traces
on printed circuit boards. Often it is better to analyze everything in
terms of transmission lines, instead of discrete capacitances and
inductances, when you are operating at such high frequencies.

Cheers,
Tom

"Frank's" wrote in message
news:YOhdh.17285$YV4.7275@edtnps89...

"money" wrote in message
ps.com...
I have a circuit in which there is a series capacitance of value 680
pF. The characteristic impedance of the line is 50 ohms. Can any
suggest how do i plot the capacitance in the smith chart.

If suppoe it was a parallel capacitance, how do i do it?

Try the free Smith Chart demo at: http://fritz.dellsperger.net/.
Probably the best electronic version I have seen.

73,

Frank (VE6CB)



That is a good Smith Chart program. I have been looking for one.

"money" wrote in message
oups.com...

Jerry Martes wrote:

"money" wrote in message
ps.com...
I have a circuit in which there is a series capacitance of value 680
pF. The characteristic impedance of the line is 50 ohms. Can any
suggest how do i plot the capacitance in the smith chart.

If suppoe it was a parallel capacitance, how do i do it?


Hi Money

A purely reactive load, plots on the circumference (or perimeter) of
the
Smith Chart. It plots on the lower half (the minus impedance half) if it
is
capacitive, as in your case.

If you pick a frequency around 10 MHz, the capacitor will have close to
75
ohms X sub C.
Since you are working with a 50 ohm line, 75 is plotted on the lower
half
of the chart, at its perimeter and with a value of 75/50, or, 1.5 on the
chart.

I really like getting help thru Wikipedia. If you read a few of their
sites on "Smith Chart" and still have questions, I'd be happy to tell you
what little I know about Smith Charts.

Jerry


Yeah Jerry..... I have seen Wikipedia. :-) But from where did you get
75 ohms X sub C.... Wat does it mean?

Converting 680pF series capacitance to capacitive reactance we need to
use Xc = 1 / CW.... Is it not??

Hi money

I got 75 by making a mistake. The 680 pF capacitor will be about 24
ohms capacitive reactance at about 10 MHz., ?right?

When 24 ohms of capacitive reactance terminates the 50 ohm line, that
impedance is shown as R0-J24/50, which is found on the perimiter of the
chart. All impedances with R = zero are located on the outer perimiter of
the Smith Chart. Find the point on the chart where -J 0.48 is shown.
That point is in the lower left of the chart, when the chart is positioned
so the purely resistive axis is reading left to right.
So, that answers your original question "how do I plot xxx". I asummed
that you were aware that you cant actually plot capacitance on a Smith
Chart. The Smith Chart identifys only Impedance. The Smith Chart
identifies *all* impedances that have a real R value.

In 1966 I wrote a short article on matching with a Smith Chart for
Electronic Design. It isnt worth much by today's standards. But, I'd be
happy to send you a copy if you are interested in using a Smith Chart for
impedance matching.

Jerry

Try the free Smith Chart demo at: http://fritz.dellsperger.net/.
Probably the best electronic version I have seen.

73,

Frank (VE6CB)


Yeah frank i just downloaded the smith chart demo..... I am looking
into it. It is great & is surely helping me a lot. Thanks a ton
frank... :-)

Glad you found it interesting. If you need to know more of the
theory behind the Smith Chart I would recommend
"RF Circuit Design" by Christopher Bowick. It is available
on www.amazon.com for $26.37. The treatment is fairly non-
mathematical, but it does give great insight into how the Smith
Chart works.

Frank

On Tue, 05 Dec 2006 17:15:36 GMT, "Frank's"
wrote:


"money" wrote in message
ups.com...
I have a circuit in which there is a series capacitance of value 680
pF. The characteristic impedance of the line is 50 ohms. Can any
suggest how do i plot the capacitance in the smith chart.

If suppoe it was a parallel capacitance, how do i do it?

Try the free Smith Chart demo at: http://fritz.dellsperger.net/.
Probably the best electronic version I have seen.

Hi Frank,

Why is it that when people develop electronic implementations of the
Smith Chart (like this one) that they don't implement things like
complex Zo and attenuation in transmission line elements, complex
reference impedance (incorrectly termed Zo in this prog)?

Owen
--

Hi Frank,

Why is it that when people develop electronic implementations of the
Smith Chart (like this one) that they don't implement things like
complex Zo and attenuation in transmission line elements, complex
reference impedance (incorrectly termed Zo in this prog)?

Owen

Hi Owen,

Yes that has always bugged me. Not so much a complex
Zo, but transmission line losses would be nice.

Frank

Jerry Martes wrote:
In 1966 I wrote a short article on matching with a Smith Chart for
Electronic Design.

Aha, that's where I ran across your name. I still
have a copy of that article somewhere.
--
73, Cecil http://www.w5dxp.com

Yeah sure....

Please do send it...