I have a doubt in smith chart
 

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?

I have a doubt in smith chart
 

On 4 Dec 2006 14:13:15 -0800, "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?

I doubt this is a very effective communications channel for an
introduction to the Smith Chart, you need pictures.

Google for "smith chart tutorial", you will find plenty, and amongst
them you should find a suitable intro.

Owen
--

I have a doubt in smith chart
 

Owen Duffy wrote:
On 4 Dec 2006 14:13:15 -0800, "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?

I doubt this is a very effective communications channel for an
introduction to the Smith Chart, you need pictures.

Google for "smith chart tutorial", you will find plenty, and amongst
them you should find a suitable intro.

Owen
--

Yep thanx

I have a doubt in smith chart
 

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

I have a doubt in smith chart
 

"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

I have a doubt in smith chart
 

"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)

I have a doubt in smith chart
 

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?

I have a doubt in smith chart
 

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??

I have a doubt in smith chart
 

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

Yeah Tom..... U r right.... i get it. If its a series capacitance we
have to move anti-clockwise along the constant r cirlce. So first all
we got to do is convert 680pF capacitance value to reactance. As the
frequency used is 900Mhz( RF range) & the line impedence is 50 ohms how
do we proceed....?

Do tell if i am wrong.....
we know that Xc = 1/Cw .So we get Xc=0.26 . we need to normalise this
reactance value. so divide this value by 50. we get 0.005 . But this is
a negligible value, is it not??

I have a doubt in smith chart
 

Frank's 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?

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... :-)