WEEK 2 

TASK 2-DETERMINE THE OXIDE RELATIVE PERMITTIVITY

 The oxide relative permittivity  was calculated when the transistor was in the accumulation region because the capacitance is at its peak in this region. The capacitor of the MOS can be represented by circuit comprising of the series combination of the oxide capacitor and the accumulation layer capacitor.

the series combination of the capacitor was represented by the expression below.

The capacitor of the accumulation layer depends on the majority carriers which are holes for the p type substrate. The capacitor of the accumulation region is large because the majority carriers which are holes that accumulates depends exponentially  on the surface charge thereby making it very large for a negative voltage applied to the gate. this made the capacitance to be of the accumulation region to be ignored  
the value of Cmax given was 2.99×10^-9 

The area of the capacitor was found by using the this formula

The Area of silicon wafer calculated was equal to =2.6430×10^-7

The oxide capacitance consists of two layers which is represented in an electrical circuit as the parallel combination of the high k capacitor and the silicon oxide capacitor. which was represented in the formula below.


                                                                          


note: the capacitor of the accumulation region was ignored 
for the oxide relativity permittivity to be calculated the, the capacitance of the silicon oxide was initially calculated 

the value gotten for silicon oxide capacitance was 

with the capacitance value of the silicon oxide known the capacitance of the high-k material was gotten with the value of Cmax given by the our supervisor

the value gotten was

 this value gotten was used to calculate the oxide relative permittivity by using this formula

WEEK 2 

TASK 3-FIND THE EQUIVALENT OXIDE THICKNESS

The equivalent oxide thickness was by using the formula used to find the capacitance of the oxide which is the oxide capacitance given was 2.99×10^-9.

The thickness of the oxide was expressed in this form

The equivalent oxide thickness(EOT) is therefore equal to 3.03nm

This is the the thickness of the silicon oxide required to make the capacitance of the silicon oxide equivalent to the capacitance of the high k gate material.