Hello,

Probably you could use ellipsometry to get both real and imaginary parts of the dielectric constant of the semiconductor.

I hope this answers your question.

Hi Punam,

In addition to the solution proposed by Christophe, you could assemble a capacitor device with parallel plates and your semiconductor thin film between them. After you should perform a capacitance vs frequency characterization. With the values of capacitance (C), semiconductor film thickness (t) and area of contacts (A) you could obtain the semiconductor dielectric constant (k) through the following simple equation:

C = k . A / t

Hope it may help you,

Dear Punam,

Real part of permittivity of a semiconductor can be accurately measured

if it is larger (or of the same order) as the imaginary part. The imaginary part of permittivity for semiconductor at a given frequency depends on conductivity according to formulae Im(eps)=sigma/(omega*eps0)

where: sigma - conductivity , omega - angular frequency eps0- permittivity of vacuum. As you can easily find the imaginary part of permittivity decreases

with frequency. For high resistivity (low conductivity) semiconductors having resistivity > 1 kOhm*cm the imaginary part of permittivity becomes smaller than the real part at microwave frequencies. Therfore accurate

measurements of the real part of permittivity are typically measured at

microwave frequencies or higher. For more details see e.g.

IEEE TRANSACTIONS ON MICROWAVE THEORY AND TECHNIQUES, VOL. 54, NO. 11, NOVEMBER 2006 p.3995

Dear Dr. Punam Tiwary

Alsalamno Alicom wa rahmato Allah wa barakatoh

Have a nice time and good day.

Herein find an attached file that I had previously prepared and supported by references, it may help you and contains answer of your question " How to find relative permitivitty of any composite semiconductor? ". Please check the attached file.

Waiting your feedback.

yours

A. Eldenglawey