References

 

[1]: International Technology Roadmap for Semiconductors, [online] Available: http://www.itrs.net/mtrs/publntrs.nsf

[2] S. K. Kundu et al,, "A study on high-κ gate stack for MOS-FET," 2015 International Conference and Workshop on Computing and Communication (IEMCON), pp. 1-5, Oct. 2015.

[3] Thomas Y. Hoffmann. (2010, March) Integrating high-k /metal gates: gate-first or gate-last? [Online]. Available: https://sst.semiconductor-digest.com/2010/03/integrating-high-k/

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Task 3: Find the Equivalent Oxide Thickness (EOT)

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 The equivalent oxide thickness is stated as a distance, which imply the thickness of silicon oxide that is required to get the similar effect as using high-k materials. The term is commonly used to describe the field effect transistors and is often given in nanometers. These formulas have been used: so that, and we can figure out that ε hk=10.6506 while ， we can get that EOT=2.8084nm.
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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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Task8. Find the oxide charge density (Qox in Ccm-2) at the Flat-band condition (taking into account the work function difference) and at the Mid-gap condition? How many charges does this correspond to (Nox cm-2)?

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  This problem should be discussed in two situations. First, the oxide charge density at the flat-band condition. By using this formula: The value of Qox is -1.045* 10 6 C/cm 2 Then, we can use this formula to calculate the count of charges. We know the value of q, this is a common value, 1.602*10 -19 Then the value of N OX is 6.52*10 12 Second, at mid-gap situation, we need different formulas to calculate the value of Q OX , and there are the formulas: After calculate the value of VG , deltaV is 0.422-1.35=-0.928V, and we also knew the value  of area, so Q OX is  -1.14*10 -6 C/cm 2 In the end, the value of N OX at mid-gap condition is 7.116*10 12
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