Task 4: Doping Density of the Silicon Substrate

There are still some problems about Task 2 and 3; therefore, we decide to introduce Task 4 before them.

After studying for several minutes, we found a complex equation about doping density as shown below. 

A	Area of capacitor;
k	Boltzmann’s constant: 1.38*10-23J/K;
T	Temperature: normally choose 300K;
q	The minimum charge quantity: 1.6*10-19C
e0	Permittivity of vacuum: 8.85*10-12F/m;
es	Relative permittivity: In this case we choose 13.9 (Si);
C	Capacitance: 48pF ~ 3nF;
ND	Intrinsic carrier concentration: We normally use 1016m-3 at 300K.

Table 1: The Meanings of Variables 

After simplifying the equation, we got N_D/(2.14*10²⁰)=ln(N_D/10¹⁶). Let’s just recognize N_D/(2.14*10²⁰) as part one and the right hand side as part two. As always do, we used two methods to calculate.

First Method

We just applied formula in Matlab to solve the equation and the answer is shown below. It seems that the equation has two solutions, however, one answer exceed the limit. The solution 10¹⁶ indicates that the substrate was not doped and it did not fit the condition we got. Therefore this answer was abandoned.

Figure 1: Answers to equation

Second Method

Although we have already known that the solution 10¹⁶ m^-3 was not we needed, we still plotted part one and two in one graph between 10¹⁵ and 10¹⁷. From Figure 2 we know that the calculation for first answer was right.

Figure 2: The First Answer to Equation

Then we did some calculations and found that another solution was from 10²¹ to 10²². We plotted the images between 10²¹ and 10²² and the solution that we want is approximately 2.7*10²¹ m^-3.

Figure 3: The Second Answer to Equation