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Slide 1 Show where all Units come from Show all new work If using a previously calculated value, cite your source 1 Problem XXXXXXXXXXpoints) Demonstrate the fcc and bcc crystal systems are...

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Slide 1
Show where all Units come from
Show all new work
If using a previously calculated value, cite your source
1
Problem XXXXXXXXXXpoints)
Demonstrate the fcc and bcc crystal systems are reciprocals of each other by:
 
Start from fcc real space lattice vectors and show that the fcc reciprocal space lattice vectors are equivalent to a bcc real space system.
Start from bcc real space lattice vectors and show that the bcc reciprocal space lattice vectors are equivalent to an fcc real space system.
2
Problem XXXXXXXXXXpoints)
A Silicon substrate (wafer) is doped n-type with neutral phosphorous atoms. The device it will be part of must operate at 253K.
Evaluate by reading it from the standard chart reproduced in Sze/Ng as Fig. 10. Make an assumption: that will not change significantly at 253K although will.
3
Problem 2. continued (40 points)
A Silicon substrate (wafer) is doped n-type with neutral phosphorous atoms. The device it will be part of must operate at 253K
Evaluate by googling Ioffe Semiconductor:
Si/ Band structure and ca
ier concentration / Temperature dependences
Evaluate graphically from Sze/Ng Fig. 9
4
Problem 2. continued (40 points)
A Silicon substrate (wafer) is doped n-type with neutral phosphorous atoms. The device it will be part of must operate at 253K
Find the ratio of neutral to ionized donors at 253K
    Boltzman constant
5
Problem XXXXXXXXXXpoints)
You will build a a p+n junction by depositing a silicon plus boron epitaxial layer on your (plasma cleaned) Problem 2 silicon wafer surface using molecular beam epitaxy. This is expensive and used when precise device fa
ication matters.
Since precision matters for your device, first check that your silicon wafer is in fact doped by running (several) C-V experiments that consistently produce the following result at room temperature:
6
Problem 3 continued. (35 points)
You will build a a p+n junction by depositing a silicon plus boron epitaxial layer on your (plasma cleaned) Problem 2 silicon wafer surface using molecular beam epitaxy. This is expensive and used when precise device fa
ication matters.

What is the value for based on the C-V results? You may assume that you have full ionization during the room temperature measurement:
YES or NO: Is your wafer doped
Since you also need precise wiring, you want to order a custom n-type silicon wafer that provides an minimal neutral region but enough material to avoid punch-through during operation of your p+n device (punch-through: when maximum > n-side).
Assume device operation range. Evaluate the maximum value of on the n-side of the p+n silicon wafer.
Should you order the thickness wafer or the thickness wafer? State your logic.
7
Problem XXXXXXXXXXpoints)
Your p+n junction in silicon is now being initially tested at room temperature for ON/OFF operation, with emphasis on achieving a good OFF. Unfortunately, the oxygen donor defect closest to midgap is causing problems.
What is the value of for the oxygen donor defect closest to midgap?
(Note: and  both defects. Use is: for useful dopants and for un-useful traps)
8
Problem 4 continued. (40 points)
Your p+n junction in silicon is now being initially tested at room temperature for ON/OFF operation, with emphasis on achieving a good OFF. Unfortunately, the oxygen donor defect closest to midgap is causing problems.
Suppose an oxygen donor defect is located physically close to a phosphorus dopant.
Plot the energy levels for both the oxygen defect and phosphorus dopant on the graph below, similar to HW02, Pr. 4 Solutions, slide 21:
Do you expect the donor function of the phosphorus to be adversely impacted by a nea
y oxygen defect or not?
9
Problem 4 continued. (40 points)
Your p+n junction in silicon is now being initially tested at room temperature for ON/OFF operation, with emphasis on achieving a good OFF. Unfortunately, the oxygen donor defect closest to midgap is causing problems.
Evaluate the maximum depletion region width for your device operation bias:
Evaluate the generation cu
ent density using standard assumptions for the oxygen defect as a single-level trap in silicon at room temperature:
    
    
    
Evaluate the diode (Sze: ideal) cu
ent density expected for this p+n junction
Evaluate the total cu
ent density
10
Problem 4 continued. (40 points)
Your p+n junction in silicon is now being initially tested at room temperature for ON/OFF operation, with emphasis on achieving a good OFF. Unfortunately, the oxygen donor defect closest to midgap is causing problems.
YES or NO: is total cu
ent density in the range?
Look for a way to reduce from the oxygen traps:
    
    In the depletion region during OFF for the standard assumptions given:
Evaluate the trap concentration for
What does the trap concentration have to be to make
     ?
            
11
Answered 1 days After Mar 17, 2023

Solution

Amar Kumar answered on Mar 18 2023
32 Votes
Q1.
(i)
To demonstrate that the FCC and BCC crystal systems are reciprocals of each other, we can start by showing that the FCC reciprocal space lattice vectors are equivalent to a BCC real space system.
The FCC crystal system has a cubic unit cell with lattice vectors ? ̅, ? ̅, ? ̅, where each vector has a length of ?. The coordinates of the atoms in the FCC unit cell are (0,0,0), (0,0,1/2), (0,1/2,0), (1/2,0,0), (1/2,1/2,0), and (1/2,0,1/2), where the numbers represent fractional coordinates of the unit cell.
The reciprocal lattice vectors for the FCC system can be found using the formula:
? ̅^∗ = 2π/? (1, -1, 1)
? ̅^∗ = 2π/? (1, 1, -1)
? ̅^∗ = 2π/? (-1, 1, 1)
Substituting the value of ? for the length of the FCC lattice vectors, we get:
? ̅^∗ = 2π/a (1, -1, 1)
? ̅^∗ = 2π/a (1, 1, -1)
? ̅^∗ = 2π/a (-1, 1, 1)
Now, to show that these reciprocal lattice vectors are equivalent to a BCC real space system, we can use the formula for the BCC lattice vectors:
? ̅ = a/2 (1, 1, 1)
? ̅ = a/2 (-1, 1, 1)
? ̅ = a/2 (1, -1, 1)
Substituting the value of ? for the length of the BCC lattice vectors, we get:
? ̅ = a/2 (1, 1, 1)
? ̅ = a/2 (-1, 1, 1)
? ̅ = a/2 (1, -1, 1)
To find the reciprocal lattice vectors for the BCC system, we can use the formula:
? ̅^∗ = 2π/? (b ̅ x c ̅)
? ̅^∗ = 2π/? (c ̅ x a ̅)
? ̅^∗ = 2π/? (a ̅ x b ̅)
where V is the volume of the unit cell and x represents the vector cross product.
Substituting the BCC lattice vectors into this formula, we get:
? ̅^∗ = 2π/a (1, -1, 1)
? ̅^∗ = 2π/a (1, 1, -1)
? ̅^∗ = 2π/a (-1, 1, 1)
This is the same set of reciprocal lattice vectors that we obtained for the FCC system. Therefore, we have shown that the FCC and BCC crystal systems are reciprocals of each other.
(ii)
To demonstrate that the BCC and FCC crystal systems are reciprocals of each other, we can start by showing that the BCC reciprocal space lattice vectors are equivalent to an FCC real space system.
The BCC crystal system has a cubic unit cell with lattice vectors ? ̅, ? ̅, ? ̅, where each vector has a length of ? * √2. The coordinates of the atoms in the BCC unit cell are (0,0,0) and (1/2,1/2,1/2), where the numbers represent fractional coordinates of the unit cell.
The reciprocal lattice vectors for the BCC system can be found using the formula:
? ̅^∗ = 2π/? (1, 1, -1)
? ̅^∗ = 2π/? (-1, 1, 1)
? ̅^∗ = 2π/? (1, -1, 1)
Substituting the value of ? for the length of the BCC lattice vectors, we get:
? ̅^∗ = 2π/(a√2) (1, 1, -1)
? ̅^∗ = 2π/(a√2) (-1, 1, 1)
? ̅^∗ = 2π/(a√2) (1, -1, 1)
Now, to show that these reciprocal lattice vectors are equivalent to an FCC real space system, we can use the formula for the FCC lattice vectors:
? ̅ = a/2 (1, 1, 0)
? ̅ = a/2 (-1, 1, 0)
? ̅ = a/2 (0, 0, 1)
Substituting the value of ? for the length of the FCC lattice vectors, we get:
? ̅ = a/2 (1, 1, 0)
? ̅ = a/2 (-1, 1, 0)
? ̅ = a/2 (0, 0, 1)
To find the reciprocal lattice vectors for the FCC system, we can use the formula:
? ̅^∗ = 2π/? (b ̅ x c ̅)
? ̅^∗ = 2π/? (c ̅ x a ̅)
? ̅^∗ = 2π/? (a ̅ x b ̅)
where V is the volume of the unit cell and x represents the vector cross product.
Substituting the FCC lattice vectors into this formula, we get:
? ̅^∗ = 2π/(a√2) (1, 1, -1)
? ̅^∗ = 2π/(a√2) (-1, 1, 1)
? ̅^∗ = 2π/(a√2) (1, -1, 1)
This is the same set of reciprocal lattice vectors that we obtained for the BCC system. Therefore, we have shown that the BCC and FCC crystal systems are reciprocals of each other.
Q2.
(i)
To evaluate ?_?−?_? for a Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorus atoms at 253K, we can refer to the standard chart in Sze/Ng as Fig. 10.
Assuming that ?_?−?_? will not change significantly at 253K, we can read the value of ?_?−?_? for phosphorus from the chart. From the chart, we can see that the value of ?_?−?_? for phosphorus in Silicon is approximately 0.56 eV.
Therefore, ?_?−?_? for the given n-type Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorus atoms at 253K is approximately 0.56 eV.
(ii)
To evaluate the concentration of electrons (?_?) in the given n-type Silicon substrate doped with ?_?=1 ? [10]^15 [??]^(−3) neutral phosphorous atoms at 253K, we can refer to Fig. 9 of the Ioffe Semiconductor website.
From Fig. 9, we can see that the intrinsic ca
ier concentration (?_i) of Silicon at 253K is approximately 1.0 x 10^10 cm^(-3). The website also provides the following formula for calculating the concentration of electrons in n-type Silicon:
?_?=?_?+?_i^2/?_?
Substituting the given values, we get:
?_?=1?10^15+ (1.0?10^10)^2/1?10^15
?_?=1.01?10^15 cm^(-3)
Therefore, the concentration of electrons (?_?) in the given n-type Silicon substrate doped with ?_?=1 ? [10 ]^15 [??]^(−3) neutral phosphorous atoms at 253K is approximately 1.01?10^15 cm^(-3).
(iii)
To evaluate the intrinsic ca
ier concentration (?_i) of Silicon at 253K graphically from Fig. 9 in Sze/Ng, we can follow these steps:
Locate the temperature axis on the bottom of the graph and find the point co
esponding to 253K.
From the point on the temperature axis, draw a vertical line to the curve representing Silicon (Si).
From the intersection of the vertical line and the curve for Si, draw a horizontal line to the left until it intersects the ?_? axis.
Read the value of ?_? from the ?_? axis at the point where the horizontal line intersects.
Using these steps, we can see that the ?_? of Silicon at 253K is approximately 1.0 x 10^10 cm^(-3).
(iv)
To find the ratio of neutral to ionized donors at 253K, we can use the equation:
?_D = ?_D^+ + ?_D^0
where ?_D is the total donor concentration, ?_D^+ is the ionized donor concentration, and ?_D^0 is the neutral donor...
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