Conclusion

 

  According to the experiment and tasks we finished, it is a summing-up to whole project.

  The relative permittivity of HFSiO4 is 10.6 and the relative permittivity of SiO2 is 3.9, thus the relative permittivity of the oxide layer increased with the concentration of Hf. EOT is 2.8nm, which thinner than the thickness of the gate stack 4.9nm.

  The doping density of the silicon substrate is 4.8*1021m-3, this value will be used in later calculations. The work function difference assuming a gold (Au) gate is equal to 0.08eV, this work function can also be used to calculate the VG at mid-gap condition. The results of problem 6 and problem 7 are the base of calculate oxide charge density and the corresponded charges number.

  The charge number of flatband condition is less than the mid-gap condition. This is the influence of deltaV, deltaV equals VG subtracting Vmidgap, and it will cause the shifting of C-V curve as figure 6. Ideal C-V curve needs more voltage to achieve the position as experimental C-V curve.



                                Figure6. C-V plot comparison between ideal and experimental

  Generally, CMOS with high-k gate can effectively alleviate the problem of electronic leakage on the micro scale, and the electrical properties are better than the normal CMOS in electrical area.


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Task 2 Determine the Oxide Relative Permittivity (εr or k)

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  The oxide relative permittivity can be calculated using the mathematical characteristics of capacitors under the accumulation region (which is the highest region of the C-V plot). Under the accumulation condition, the MOS capacitor can be seen as the series connection of the C ox and C acc .            As C acc is exponentially dependent on and therefore very large.  (C acc =dC acc /d Φ S )  So, in above equation, the term of accumulation capacitance can be ignored and the total capacitance can be seen equal to oxide capacitance. Then we can obtain the oxide capacitance by reading the max value from the C-V plot, from given data, the value is 2.92E+03pF. Then we can calculate the oxide relative permittivity through function below: With given data, we have: By substituting equations 2.2 to 2.5 into equation 2.1, we obtained  ε ox =6.8.
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Task5. Work Function Difference (Calculate the work function difference assuming a gold (Au) gate.）

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  The work function Ф M is defined as the minimum energy required for an electron with an energy equal to Fermi level to escape from the metal into vacuum. It is shown in figure 2. For metals, the work function Ф M = Ф VA - Ф FM is constant. Thus,  Ф M (Au)=5.1eV. For semiconductors, the work function is: which is shown in figure 2. According to known conditions, the electron affinity of Si is and the Bandgap energy of silicon is The Fermi potential Ф F is equal to and is about 25mV at 300k. Thus, Fermi potential: and the work function of semiconductors: The definition of Mental-Semiconductors work function difference is: Thus, the work function difference: Figure 2: Energy band diagram explaining work function
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Task 6. Flatband voltage (Calculate the flatband voltage）

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  Flatband voltage is defined as the gate voltage when the band of semiconductor is not bent as shown in figure 3. In ideal case, V G =0   In actual case, V G ≠ 0. There are two reasons. The first one is that there is a work function difference between the metal and the semiconductor. The second one is that the net charge density in the oxide layer is actually non-zero, such as Fixed charge of oxide. Figure 3: Band diagram of flat band As V G is gradually reduced, the accumulation capacitance becomes smaller and has the effect of reducing the measured capacitance. When V G is equal to zero volts. The semiconductor capacitance is now defined by the so-called Debye length: thus: Figure 4: The a.c. equivalent circuit at flatband. thus: The area: Then, The corresponding Flatband voltage V FB is close to 0.93 V.
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