Midterm2 solutionsFall 2022

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1 EE143 Microfabrication Technology Fall 2022 Prof. J. Bokor Midterm Exam 2 Name: _____________________________________________ Signature: __________________________________________ SID: __________________________ CLOSED BOOK. TWO 8 1/2” X 11” SHEETS OF NOTES, AND SCIENTIFIC POCKET CALCULATOR PERMITTED. MAKE SURE THE EXAM PAPER HAS 13 PAGES. DO ALL WORK ON THE EXAM PAGES. USE THE BACK OF PAGES IF NECESSARY. TIME ALLOTTED: 80 MINUTES Fundamental constants you might need: Boltzmann’s constant, k = 1.38 x 10 -23 J/K Permittivity of free space, ε ο = 8.85 x 10 -12 F/m Permeability of free space, µ ο = 1.26 x 10 -6 H/m Speed of light in vacuum, c = 2.998 x 10 8 m/s Electron charge, e = 1.6 x 10 -19 C Free electron mass, m o = 9.1 x 10 -31 kg Electron volt, 1 eV = 1.6 x 10 -19 J Thermal voltage, kT/q = 0.0258 V (at 300K) Relative dielectric constant of silicon, K s = 11.8 Relative dielectric constant of silicon dioxide, K o = 3.9 Effective masses in silicon at 300K. Electrons: m n * = 1.18 m o ; Holes: m p * = 0.81 m o Silicon band gap at 300K, E g = 1.12 eV Silicon intrinsic carrier concentration at 300K, n i = 10 10 cm -3 Avogadro’s number, N A = 6.02 x 10 23 Silicon conduction band effective density of states, N C = 2.8 x 10 19 cm -3 Silicon valence band effective density of states, N V = 1.04 x 10 19 cm -3

2 Ion Implantation

3 Diffusion

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4 Solid Solubility

5 Silicon resistivity vs. doping

6 1. Ion Implantation and Diffusion [20 points] We have implanted Boron (B+) into n-type Si <100> with a background doping concentration of N D = 10 16 cm -3 (uniformly doped) The wafer is covered with 200 nm of SiO 2 . The acceleration voltage is 60 kV. The dose is 2x10 13 cm -2 . Assume that the implantation behavior in SiO 2 is identical to that of Si. a. Find the projected range and straggle. [4 points] From the graph, projected range R p = 200 nm. Straggle, ∆ R p ≈ 55 nm 2 points each b. Find the peak B concentration. [5 points] 𝑄𝑄 = √ 2 𝜋𝜋 𝑁𝑁 𝑝𝑝 ∆𝑅𝑅 𝑝𝑝 𝑁𝑁 𝑝𝑝 = 𝑄𝑄 √ 2 𝜋𝜋 ∆𝑅𝑅 𝑝𝑝 = 2 × 10 13 𝑐𝑐𝑐𝑐 −2 √ 2 𝜋𝜋 × 55 × 10 −7 𝑐𝑐𝑐𝑐 = 1.45 × 10 18 𝑐𝑐𝑐𝑐 −3 3 points for equation. 2 points for correct number (within factor 3) c. Calculate the percentage of the Boron dose that ended up in the silicon. [2 points] The peak is right at the SiO 2 – Si interface. Therefore 50% of the Boron is in the silicon, and 50% is in the SiO 2 . 1 point for correct approach. 1 point for correct answer.

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7 d. Now we anneal the wafer in nitrogen for 30 mins at 1000C to activate the implanted B atoms. Find the junction depth. (Hint: you can use limited source Gaussian diffusion profile approximation. For simplicity, you can assume that the diffusion constant in the SiO 2 is the same as in the Si.) [9 points] 𝐷𝐷 ~ 2 × 10 −14 𝑐𝑐𝑐𝑐 2 / 𝑠𝑠 𝑁𝑁 ( 𝑥𝑥 , 𝑡𝑡 ) = 𝑄𝑄 � 2 𝜋𝜋 ( ∆𝑅𝑅 𝑝𝑝 2 + 2 𝐷𝐷𝑡𝑡 ) 𝑒𝑒 − 𝑥𝑥 2 2 ( ∆𝑅𝑅 𝑝𝑝 2 +2𝐷𝐷𝐷𝐷 ) set x = 0, at the implant peak at the SiO 2 – Si interface. 𝑁𝑁 0 = 2 × 10 13 𝑐𝑐𝑐𝑐 −2 � 2 𝜋𝜋 × [(55 × 10 −7 𝑐𝑐𝑐𝑐 ) 2 + (2 × 2 × 10 −14 𝑐𝑐𝑐𝑐 2 𝑠𝑠 × 1800 𝑠𝑠 )] 𝑁𝑁 0 = 7.89 × 10 17 𝑐𝑐𝑐𝑐 −3 Assuming constant Q, x j is formed from a new Gaussian centered at the projected range: 𝑥𝑥 𝑗𝑗 = � 2 �∆𝑅𝑅 𝑝𝑝 2 + 2Dt � ln � 𝑁𝑁 0 𝑁𝑁 𝐵𝐵 � = � 2((55 × 10 −7 𝑐𝑐𝑐𝑐 ) 2 + 2 × 2 × 10 −14 𝑐𝑐𝑐𝑐 2 / 𝑠𝑠 × 1800 𝑠𝑠 ) × 𝑙𝑙 n( 7.89 × 10 17 10 16 ) 𝑥𝑥 𝑗𝑗 = 298.9 𝑛𝑛𝑐𝑐 2 points for correct D (within factor 2). 5 points for correct equation (3 points if close). 1 point if correctly plugged previous numbers into correct equation. 1 point if final number correct.

8 2. Thin Film Deposition [30 points] a. Suppose aluminum is deposited by e-beam evaporation on the structure pictured above, and the deposition occurs at 10nm/min for 20 minutes. Draw the result. Assume the evaporator tool has been engineered to have perfectly vertical deposition (i.e., no sidewall coverage). [4 points] 3 points for correct profile. 1 point for correct thickness. b. It is decided a more conformal coating would be better, so you decide to use a CVD process instead of e-beam evaporation. Draw the result, assuming the final film is perfectly conformal and the same thickness as in part a. [4 points] 3 points for correct profile. 1 point for correct thickness. 200 nm 200 nm 200 nm

9 c. What happens if the CVD deposition from part b. is continued for three times as long? Draw the result and give a qualitative explanation. [8 points] The opening will get closed after ~400 nm of deposition leaving a ‘void’. Deposition will continue on top of the structure. 2 points for plugged gap. 2 points for void. 1 point for thickness in cavity. 1 point for thickness on top. 2 points for explanation d. Qualitatively plot the deposition rate as a function of 1/temperature for CVD. Label each regime on the curve in terms of the rate limiting step that applies in that regime. [6 points] 3 point for correct curve shape, axes. 3 points for correct regime labels 600 nm 400 nm 400 nm

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10 e. Suppose we maintain all CVD conditions the same except the gas flow velocity is reduced. Add a sketch of the new deposition rate curve versus 1/T to your figure above and give a brief explanation. [4 points] Mass transport is reduced due to reduced gas flow. Surface reaction rate is not affected. 2 points for correct curve. 2 points for explanation f. Calculate the impingement rate and mean free path for gold (M = 197 g/mol; atomic radius = 0.146nm) at 350K and a pressure of 7.5 x 10 -6 Torr. [4 points] ϕ = 3.5 ∗ 10 22 𝑃𝑃 √𝑀𝑀𝑀𝑀 = 10 15 / 𝑐𝑐𝑐𝑐 2 𝑠𝑠 λ = 𝑘𝑘𝑀𝑀 √ 2 π𝑑𝑑 2 𝑃𝑃 = 12.75 𝑐𝑐 1 point for each equation. 1 point for each correct number (with units)

11 3. Device physics [25 points] A bar of Si is doped with 10 16 cm -3 of Sb on one side, and with Ga on the other, so that the resulting band diagram at room temperature is as shown below with (E C, L – E F ) = (E F – E V, R ). Assume uniform doping, step junction and complete ionization. (L and R stand for left and right) a. Mark on the diagram which side has been doped by Sb and which by Ga. Also mark the Si on each side as p-type or n-type. Assume complete ionization. Then calculate the electron and hole concentrations on the left and the right (n L , p L , n R , p R ). [10 points] n L = N D = 10 16 cm -3 p L = n i 2 /n L = 10 4 cm -3 𝑛𝑛 𝐿𝐿 = 𝑁𝑁 𝐶𝐶 𝑒𝑒 − 𝑞𝑞 ( 𝐸𝐸 𝐶𝐶 , 𝐿𝐿 −𝐸𝐸 𝐹𝐹 ) 𝑘𝑘𝑘𝑘 �𝐸𝐸 𝐶𝐶 , 𝐿𝐿 − 𝐸𝐸 𝐹𝐹 � = 𝑘𝑘𝑀𝑀 𝑞𝑞 ln � 𝑁𝑁 𝐶𝐶 𝑛𝑛 𝐿𝐿 � = 0.206 𝑒𝑒𝑒𝑒 �𝐸𝐸 𝐹𝐹 − 𝐸𝐸 𝑉𝑉 , 𝑅𝑅 � = �𝐸𝐸 𝐶𝐶 , 𝐿𝐿 − 𝐸𝐸 𝐹𝐹 � = 0.206 𝑒𝑒𝑒𝑒 𝒑𝒑 𝑹𝑹 = 𝑵𝑵 𝑽𝑽 𝒆𝒆 − 𝒒𝒒 ( 𝑬𝑬 𝑭𝑭 −𝑬𝑬 𝑽𝑽 , 𝑹𝑹 ) 𝒌𝒌𝒌𝒌 = 𝟑𝟑 . 𝟕𝟕𝟕𝟕 × 𝟕𝟕𝟏𝟏 𝟕𝟕𝟏𝟏 𝒄𝒄𝒄𝒄 −𝟑𝟑 = N A n R = n i 2 /p R = 2.69 x 10 4 cm -3 1 point for each correct labels. 1 point each for n L and p L . 3 points for E C,V -E F. 2 points for p R . 1 point for n R (if correctly calculated from p R then give credit) Sb, n Ga, p E C, L – E F E F – E V,R

12 b. An MOS gate stack is built on the right side of this bar of Si at room temperature by depositing 100 nm of gate oxide, followed by p+ polysilicon gate. Calculate the following for the resulting MOS capacitor. (Assume E F of p+ polysilicon gate coincides with E V , i.e. degenerately doped gate, and assume the gate stack is built far away from the Si on the right) i. Gate oxide capacitance per unit area [2 points] C ox = ε ox /t ox = 34.5 nF/cm 2 1 point for formula. 1 point for number (incl units). ii. Flat-band voltage [4 points] V FB = Φ gate – Φ body = [E F – E V, R ]/q = 0.206 V 3 points for formula. 1 point for number (give credit if uses wrong Ec-E F from part a. iii. Threshold voltage. (Hint, be careful to convert all numbers in the formula to MKS units!) [5 points] For n-Si body MOS capacitor 𝑒𝑒 𝑘𝑘 = 𝑒𝑒 𝐹𝐹𝐵𝐵 + |2 𝜑𝜑 𝐹𝐹 | + 1 𝐶𝐶 𝑂𝑂𝑂𝑂 � 2 𝑞𝑞𝜀𝜀 𝑠𝑠𝑠𝑠 𝑁𝑁 𝐴𝐴 |2 𝜑𝜑 𝐹𝐹 | V FB = 0.206 V, |2 𝜑𝜑 𝐹𝐹 | = 0.412 𝑒𝑒 So, 𝑒𝑒 𝑘𝑘 = 𝑒𝑒 𝐹𝐹𝐵𝐵 + |2 𝜑𝜑 𝐹𝐹 | + 1 𝐶𝐶 𝑂𝑂𝑂𝑂 � 2 𝑞𝑞𝜀𝜀 𝑠𝑠𝑠𝑠 𝑁𝑁 𝐴𝐴 |2 𝜑𝜑 𝐹𝐹 | = 0.625 V 3 points for formula. 2 points for number (1 point for sign). Give credit if used wrong input numbers from previous parts.

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13 c. Calculate the sheet resistance of a 100 nm thick Aluminum electrode with a resistivity of 2.7 x10 -8 Ohm·µm. [2 points] R s = ρ / t = 2.7 x 10 -8 Ohm · µm / 100 x 10 -3 µm = 2.7 x 10 -7 Ohm /□ 1 point for formula. 1 point for number (including units) d. Calculate the resistance of Aluminum metal line with dimensions 100 µm long and 0.2 µm wide. [2 points] R = R s x (100 / 0.2) = 1.35 x 10 -4 Ohm 1 point for formula 1 point for number (including units)