How to measure the dielectric constant of a PCB. Any paper ?

A PCB is basically a capacitor. Take a double plate PCB, unetched, and measure its capacitance. C=E*(A/d) , C-capacity (F), A is the plate area (square meters), d-thickness of dielectric (m). E is absolute permittivity. Divide it by absolute permittivity of vacuum , 8,85 * 10(-14) F/cm (take care of units), and you'll have what you need.

/pisoiu

an diference:
etch a resonator such as microstrip patch and lower copling, mesure the resonator frequency,and then can manupulation the permittivity.

Of course the permittivty is a complex number and thus has magnitude and phase, so one should be careful in extracting this value. Of course a piece of equipment called the Thermal Mechanical Analyzer (TMA) with the electrical option will measure a sample's complex permittivty as a function of frequency directly, but unfortunately this piece of equipment is very expensive.

MoonShine

Moonshine
You have said that "permitivity is a complex number"
Can you explain this point in detail

Thank you very much in advance

Hi,

there is a so called "Ring Resonator Method" for measuring the dielectric constant and loss tangent of the RF base materials. Look for the following post:

http://www.elektroda.pl/eboard/ftopic72539.html

and for the following papers:

hxxp://www.imaps.org/presentations/ceramic2001/PBarnwell-S5.pdf

hxxp://www.imapsk.or.kr/html/publication/9-3/2.pdf

rgds, Al

assuming you have network analyzer. with time domain capability
make microstrip line with two discontinuities in it located L mm
between them. From measuring the time domain response. you will get the time difference between the two points as it propagate in air. Now calculate the distance in air D = dt x c. c=light vel.
L / D or D / L will give the square root of dielectric constant

To get accurate results make different L-s.

the imaginary part of a complex permittivity corresponds to the loss.