Background

 

The object of this project is thin high-k gate stacks (HfSiO, 70%), which is bind up with MOSFET, while silicon-based integrated circuits have developed rapidly in the past years. High-k dielectric materials are one of the most popular research topics in the micro electronic industry. By using some special properties of some high-k dielectric materials, some new devices may be conducted and the limitation of the thin gate-oxide faced in semiconductor industry may be solved.

Electronics based on these improve the life quality and make our life more comfortable, like internet and mobile phones. On the basis of Moore’s law, the number of transistors on the integrated circuits doubled every two years, so that the gate stacks become the concern issue. Scientists and engineers made great efforts to incorporate the high-k dielectric into a transistor. According to SIA [1], in order to maintain the development speed of the semiconductor, researchers should develop constraints like new models or new materials, as the silicon-based semiconductor industry is rising so fast, and it is almost reached the limits if we want to make the size smaller. The only way is to increase the dielectric constant while the thickness of the dielectric medium is constant, to maintain the same capacitance and reduce the capacitance area. So, in this project, all the teammates used the high-frequency capacitance voltage characteristic data and identified some properties like EOT and doping density, in order to have a better understanding of the thin high-k gate stacks.

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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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