Task 1 The Plot of the C-V Curve

 

The data was allocated in txt format and we put them into excel and obtained this C-V plot.

Figure 1: C-V plot C-V plot (HfSiO, 70% Hf).

According to the plot drawn from given data, the substrate involved in this project is p type.

This conclusion is made from the characteristic of surface field effect. There are three conditions of the energy band diagram, accumulation, depletion and inversion.

Under the accumulation condition of p type Si MOS system, when negative voltage is applied on the gate, positive holes in the semiconductor will be attracted to the region close to the oxide and accumulate there.

When voltage applied on the gate increases to positive, the positive holes accumulated before will be repelled and leave behind some negative acceptances. Then this condition is called depletion condition.

Finally, when the gate voltage is high enough, the fermi level of semiconductor in the region near to the oxide will be bend higher than the intrinsic level, and thus create inversion layer. Although the width of depletion region will not increase anymore, the inversion layer will growth because of the build-up of minority carriers.

In addition, the plot can tell us that this MOS capacitor is tested under high frequency.

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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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