Conclusion

    The project was accomplished ahead of schedule and all the tasks were fulfilled. The comparison between Project 27 (50% Hf) indicates that a higher oxide permittivity will show some of the positive effect on MOS capacitor, but negative side should not be ignored. Further research need to target on the fixed oxide charge density and try to decrease it.

    Apart from the general calculation, a MatLab program was written to calculate the electrical properties of the high-k gate stacks, The output result is shown below:

    The program needs to input the Capacitance, Voltage and Resistance data for calculation. Therefore, it can be used to calculate other gate materials with input data. If you are interested in the source code, please contact us.

Question 9. Compare the Results with Project 27 (50% Hf)

    Finally the results were compared with Project 27 (50% Hf). All the results between two groups are summarized below:

	Project 26 (70% Hf)	Project 27 (50% Hf)
Type of Substrate	P-type	P-type
Oxide Relative Permittivity	10.64	8.67
EOT (nm)	2.81	3.04
Doping Density (m-3)	4.244x1021	2.7x1021
Work Function Difference (eV)	0.065	0.08
Flatband Voltage (V)	0.932	0.53
Midgap Voltage (V)	1.367	1.21
Oxide Charge Density (cm-3)
Flatband Condition	-6.660x1016	-3.23x1016
Mdigap Condition	-7.841x1016	-6.4x1016

    It can be noticed that a higher percentage of Hf will lead to a higher oxide relative permittivity. It could be explained by the fact that the relative permittivity of Hf is higher than SiO2. Also, the equivalent oxide thickness will be decreased. Nevertheless the EOT is still too large for commercial purpose (which should be less than 2 nm). Higher percentage of Hf will also lead to a higher doping density, which will then increase the cost of manufacture.

    Another thing should be considered is that the fixed oxide charge density almost doubled in flatband condition. According to our supervisor, the charge density should be controlled less than 1x10¹⁰ in order to get a better quality of CMOS. Thus further research should be carried out to decrease the amount of fixed oxide charge density.

Question 10. Write program for plotting the ideal high-frequency capacitance voltage plot.

    As mentioned above, there is a small number of negative charges in oxide, which need extra voltage to neutralize to reach the same capacitance according to the equation:

    According to the previous task, the differences between ideal value and real value can be calculated in flat-band and mid-gap conditions.

    As shown above, the two voltage difference are almost same. In fact, there do exist a constant voltage difference between ideal and real values at same capacitance.

Question 7. Calculate the mid-gap voltage

When gate voltage of this p-type gate is continuously increased to be positive(decreased to be negative for n type), majority carriers (holes in this case) are forced away from the surface of the semiconductor and form a depletion region, which is called depletion or mid-gap. 

Similarly to the flat-band c=voltage, the mid-gap voltage can be calculated indirectly by matching the capacitance at that moment with the datum given in 'HFCV_70%Hf.txt'. The total capacitance will be the oxide capacitance in series with the depletion capacitance.

    (1)

(2)

Therefore,

 (3)

To improve the accuracy, the slop k at the matched point can be calculated, and the mid-gap voltage will be:

  (4)

Finally, the value of flat-band voltage is found to be 1.366 V.

Question 6. Calculate the flat-band voltage

When gate voltage of this p-type gate is increased (decreased for n type), the accumulation capacitor is decreased which will in turn reduce the measured capacitance. When the gate voltage is zero for ideal case, the condition is thus called flat-band. However, in real cases, because of the exist of oxide charges, the gate voltage will be a complex function of oxide charges instead of zero.

Fortunately, the flat-band voltage can be calculated indirectly by matching the capacitance at that moment with the datum given in 'HFCV_70%Hf.txt'. The total capacitance will be the oxide capacitance in series with the Debye capacitance.

    (1)

        (2)

    (3)

According to the calculation, the result of C_FB is equal to 345.27pF. 

However, it is obviously inaccurate to just find the nearest capacitance point in datum and use its voltage as the flat-band voltage. To improve the accuracy, the slop k at the matched point can be calculated, and the flat-band voltage will be:

  (4)

Finally, the value of flat-band voltage is found to be 0.932 V.

Question 5. Calculate the Work Function Difference Assuming a Gold (Au) Gate.

    The following figure shows the energy-band diagram of three separated components form the MOS capacitor.[1]

    For the case of using a gold (Au) gate, ΦM is the work function of metal and ΦS is the work    function of semiconductor. Thus, ΦMS is the work function difference between the gate and semiconductor. 

         (1)

    where 

 (2)

    Xs is the Electron Affinity of the Si, the value of which is 4.14 eV.

    Eg is the Bandgap energy, the value of which is 1.12 eV.

    Φfp is the Fermi potential for p-type silicon.

    According to the formula:

(3)

    Thus,

   (4)

    Therefore

  (5)

Reference:

[1]	N. Arora, MOSFET Modeling For VLSI Simulation, Singapore: World Scientific Printers, 2007.

Question 8. Finding the Oxide Charge Density at the Flatband Condition and at the Midgap Condition

    Recall from the former discussion that the flatband voltage could be calculated by the following equation:

                                  (1)   

    Where 

      (2)

    Therefore, replacing equation 2 to equation 1, Q can be found to be:

  (3)

    i.e.

 (4)

    In mid gap condition, the ideal case for the mid gap voltage should be:

   (5)

    By replacing all the figures, the result for mid gap voltage is 0.454 V. Then the total charge can be calculated by:

(6)

    Thus, the charge density is

  (7)