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Problem 1 Sample Process Sequence (30 points total)
(a) (20 points) Using surface micromachining, a pin joint can be fabricated with the cross-section and top view show below. The pin joint has a stationary pin (the anchor) on the wafer surface and a free spinning rotor which slides on top of the wafer surface. Note that the top of the stationary pin has a dimension larger than that of the rotor inner hole to keep the rotor in place.
You find the following brief description of the process flow in the notebook of a former EE 143 student. Sketch the cross-sections and top views at the highlighted processing steps (marked by bold font).
Process Description Cross-Sections Top Views Starting Material – Pure Si wafer --------------------------------- Si substrate
Deposit 1st^ level Phosphosilicate Glass (PSG) by CVD Deposit of 1st^ level Poly-Si by CVD
Pattern 1st^ level poly-Si (Mask #1)
Deposit 2nd^ level PSG
Pattern opening for stationary pin(Mask #2)
Deposit 2nd^ level poly-Si by CVD Pattern 2nd^ level poly-Si
Selectively etch away 1st^ level And 2nd^ level PSG using HF acid Final Structure
Problem 1 continued
(b) The following qualitative questions are related to the process flow in part (a). No partial credit will be given without and explanation or discussion.
(I) (3 points) To reduce the inertia of the rotor, some former EE 143 students proposed to replace the 1st^ level poly-Si with photoresist. Will this replacement be compatible with the process sequence?
(II) (3 points) Instead of depositing the 2nd^ level PSG by CVD, can we use thermal oxidation to form the 2nd^ level oxide? Discuss why or why not?
(III) (4 points) The process flow in part (a) uses two separate poly-Si deposition. Can we fabricate the device with only one layer of poly-Si? Explain why or why not?
Problem 2 Thermal oxidation (23 points total)
(a) (9 points) An ion implanter with an accelerating voltage of 50kV is used to implant the following ions into Si to an ion does of 10^15 ions/cm^2. Use the full gaussian approximation to estimate the maximum concentration of the Boron profile [in B atoms/cm^3 ]. Show all calculation.
(1) B+^ (atomic Boron ion, singly charged)
(2) B2+^ (atomic Boron ion, doubly charged)
(3) B 2 +^ (diatomic Boron molecular ion, singly charged)
(b) (4 points) If the Si substrate is n-type (background concentration of 10^16 /cm^3 ), which ion in part (a) will give the deepest junction depth?
(c) (4 points) Estimate the sheet resistance of the implant profile in part (b) using Rs ~ 1/(q•μ•atomic does)
(d) (6 points) Discuss why a much higher implant does of Boron is required to create a surface amorphous Si layer as compared with Arsenic.
Problem 4 Diffusion and Sheet Resistance (24 points total)
(a) Sheet Resistance Calculations (i) (2 points) What is the sheet resistance of a 1-μm-thick copper thin film with a resistivity of 2 x 10-6^ ohm-cm?
(ii) (3 points) Diffusion Predeposition of Arsenic is used to form a shallow junction in p-type Si (NB = 1 x 10^15 /cm^3 ). The solid solubility of As in Si is known to be 1 x 10^21 /cm^3. if the allowed junction depth is less than 0.1 μm, what is the lowest sheet resistance which can be achieved?
(iii) (3 points) Phosphorus is implanted into p-type (NB = 10^16 /cm^3 ) Si to form a full gaussian depth profile. The peak concentration and the junction depths are show in the sketch below. Use the Irvins Curve to calculate the sheet resistance of the phosphorus implant layer.
- Rp = 51.051+32.60883 E –0.033837 E^2 + 3.758e-5 E^3 –1.433e-8 E
∆∆∆∆ Rp = 185.34201 +6.5308 E –0.01745 E^2 +2.098e-5 E^3 –8.884e-9 E^4
GENERAL IMFORMATION
Q = 1.6 x 10-19^ coulombs
εs = 1.036 x 10-12^ F/cm for Si
ni for Si = 3.69 x 10^16 x T3/2^ exp[0.605eV/kT] = 1.45 x 10^10 cm-3^ at 300K Eg of Si = 1.12 eV at 300K Electron Affinity of Si = 4.05 eV D/μ = kT/q = 0.0259 volts at 300K
