Question 2. Determine the oxide relative permittivity

    The oxide relative permittivity could be found by calculating the capacitance of the oxide. For a p‑type MOS capacitor, the oxide capacitance is measured in the strong accumulation region. At that region, the capacitance could be represented by equation:

                                 (1)

    For the reason that in the accumulation region the capacitance for C_acc is pretty large, which can be ignored, Thus, the capacitance measured in the accumulation region could be represented as:

                                              (2)

    Where:

                                                        (3)

                                                     

                 

    C_max equals to 2.919 nF, as indicated in the C-V plot and A_c is the area of the MOS capacitors with the diameter of 0.55 mm. The value of C_ox is 0.0123 Fm^-2.

    For the reason that high-k material will form two different layers of oxide, which are HfSiO and SiO2. The cross-sectional photo for the MOS capacitor is also shown in below:

Taken from [1]

    (The picture is only used to indicate the structure of the MOS capacitor, in other words, the figures for the thickness of the oxide should be ignored.)

    Thus the capacitance for oxide should be:

   (4)

    And by using formulas:

                                               (5)

    The equation (4) can be simplified to:

    (6)

    Replacing all the figures the results for the oxide relative permittivity is: 10.643.

Reference:

[1]  M, Wang et.al. (2004) "Electrical Performance Improvement in SiO/HfSiO High-k Gate Stack for

Advanced Low Power Device Application " 2004 IEEE International Conference on Integrated Circuit Design and Technology.