Task 2 Determine the oxide relative permittivity

(12 February, 2015 )

The total capacitance is equal to the value of the oxide capacitance and  the accumulation layer in series in accumulation stage.
so we have the equation

   [1] and we know that the  [2]                

so we know that

 [3]

and   the oxide capacitance contains two parts which are high-k material and Silicon dioxide. 

[4]

and we know that   

                     [5] 

the relationship of capacitance with permittivity  

                      [6] 

From the data, we get the maximum value of the capacitance is Cmax=2.92*10^-9 F. We know the diameter of capacitor is 0.55 mm, so the area of the capacitor is A=(d/2)^2*pi, and the thickness  of high-k is t1=3.3nm and the thickness of silicon dioxide is t2=1.6nm. 

Solving the equation [1][2][3][4][5][6] 

we get the oxide relative permittivity k=10.65151148.