HW2%20Solution

HW2%20Solution - EEL5225 Principles of MEMS Transducers HW2...

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Unformatted text preview: EEL5225 Principles of MEMS Transducers HW2 Fall 2010 Semester Assigned: Friday, 9/3 Due: Monday, 9/13 1. A 0.5 µm oxide is required as a dielectric layer for a MEMS device, and the furnace in your cleanroom is limited to 1000°C. Using the Deal-Grove model, compute the wet and dry oxidation times needed to grow the oxide on a bare (100) silicon wafer. Determine how much silicon is consumed. For bare (100) Si wafer: Wet oxidation at 1000 deg.C : Find the appropriate constants (Campbell pg.72 or slides): Deal Grove Equations: (No initial oxide, and no "fudge factor" for wet oxidation) Dry oxidation at 1000 deg.C : Deal Grove Equations: Principles of MEMS Transducers Prepared by D. Arnold Page 1 of 4 September 8, 2010 Silicon consumed : 2. To achieve the 0.5 µm oxide thickness from problem 1 in a shorter amount of time while maintaining a high-quality dielectric, you decide to use a dry/wet/dry oxide, where the initial and final dry steps are 15 min. each. a. Compute the time required for the wet step to achieve a 0.5 µm total thickness. b. Compute the thickness of each of the three layers. Sketch and label the layers. c. Discuss how much longer the dry/wet/dry process takes compared to a pure wet oxidation, but how much time is saved compared to a pure dry oxidation. Dry / Wet / Dry oxidation: 1st step: Dry1: From equation: …… First, a dry oxide is grown on a Bare wafer having native oxide. 3rd step: Dry2: Now : …… Since we know the time for the 3rd step (dry2) and final ox. Thickness, calculate the new tau and initial ox. Thickness. 2nd step: Wet1: hence Constants : and …… For the 2nd step(wet) the initial ox. Thickness is the ox. Grown in 1st step (dry1) and the final ox. Thickness is the previously calculated ox thickness in 3rd step (dry2) Principles of MEMS Transducers Prepared by D. Arnold Page 2 of 4 September 8, 2010 a. Hence: b. Thickness of oxide layers: c. Total oxidation time: Extra time compared to wet: Time saved compared to dry: 3. Explain why some silicon is consumed during thermal oxidation, and derive the equation xSi consumed = 0.46 xox. Hint: Examine the chemical reaction equation. Material parameters (Reference: CRC Handbook of Chemistry and Physics) Principles of MEMS Transducers Prepared by D. Arnold Page 3 of 4 September 8, 2010 From the chemical reaction, Si + O2 → SiO2, we know that one mole of Si is consumed to form one mole of SiO2. The volume occupied by one mole of Si is given by the atomic weight of Si/ mass density of Si. Similarly, the volume occupied by one mole of SiO2 is given by the molecular weight of SiO2 divided by the mass density of SiO. Assuming that the area of the Si is the same as the area of the SiO2, then we can calculate the ratio of the thickness of Si consumed to the thickness of SiO2 formed as follows: Hence 4. For a lift-off metallization, explain whether an evaporator or sputterer would be preferred for depositing the metal and why. Use pictures if necessary. Evaporation results in poor step coverage while sputtering leads to uniform step coverage. For lift-off metallization, it is better to use “Evaporation” because after the removal of the photoresist, a good edge profile will be achieved. Principles of MEMS Transducers Prepared by D. Arnold Page 4 of 4 September 8, 2010 ...
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