Task 8 and 9: Oxide Charge Density and Comparison

We finished Task 8 and 9 this Friday. We know that Q=C_ox*ΔV, whereΔV is the shift of CV plot. ΔV can be recognized as the difference between gate voltage(V_G) and oxide voltage(V_o). The equation shown below can be used to calculate gate voltage.

Flatband Condition (work function difference needed)

Under this circumstance, the gate voltage is 0V and we don’t need the foregoing equation. The equation of Q can be simplified as Q=C_ox*(-V₀). Since V₀=V_FB-V_MS, use the results of Task 5 and 6, it’s easy to find out that V₀=0.455V. Finally, total charge and charge density are equal to -5.17*10^-7C/cm² and -3.23*10¹²cm^-2 respectively.

Midgap Condition (work function difference ignored)

In this condition, surface potential is equivalent to Fermi energy. Plug all data in the equation and gate voltage is 0.327V. Task 7 provides midgap voltage which is 1.23V andΔV should be -0.902V. Eventually, total charge in this area is -6.4*10¹²cm^-2 and its density is -1.0242*10^-6C/cm².

Comparison

We just shared and compared the data with the group who did the project slightly different from us. This task asks us to explain the reason why relative permittivity drops with Hafnium percentage in oxide. Two groups were all agree with the explanation shown below. High-k dielectrics have much higher relative permittivity than silicon dioxide and since the amount of Hafnium decreases (equivalent to increasing of silicon dioxide), whole permittivity of diode inevitable drop.