Task 3 The equivalent oxide thickness (EOT)

(12 February, 2015) 

In recent years, some high-k materials are used in the MOSFET which take place of silicon dioxide. The equivalent oxide thickness is describe the thickness that the silicon dioxide need which to act as the same effect as the thickness of high-k material. There are two methods to solve the problem.

Method 1 

using the equation 

             

From the data we had, Cmax=2.92*10^-9 F 

Solving the equation we have   

EOT=2.808nm

Method 2

using the equaiton 

                            

 

simplify this equation, we get the equation about EOT 

The end we get the EOT=2.8nm