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