3.155J/6.152J Microelectronic Processing Technology Fall Term, 2003 Bob O'Handley Martin Schmidt Quiz # 1 Out: Oct. 22, 2003 Due: Oct. 22, 2003 Gas kinetics, vacuum technology 1. a) Write a compact sentence that describes the molecular origin of gas pressure, including its temperature and mass dependence. [5] A: P is due to change in momentum Dp = m Dv of molecules striking a surface; velocity increases as T so and accounting for the flux of m Dp ~ mT nv atoms hitting the surface Dp µ J mk T µ x mk T , where n is the B 2 B number density of molecules in chamber. b) In a 50 cm diameter vacuum chamber at room temperature and a pressure of 10 † mT, mostly N2 ( assume a nitrogen molecular diameter of 2 Å), what is the mean free path of the N2 molecules and describe their transport as either viscous or ballistic. [6] A: l = 1.75 cm, viscous. c) You want to reduce the pressure in your diffusion-pumped vacuum chamber from 10-5 Torr to 10-6 Torr. Comment on the efficacy of the following options. i) There is a good diffusion pump 5 feet from your chamber; it is known to produce a vacuum of 10-7 Torr in its chamber. You run a 1/2” tygon tube from that diffusion pump to an inlet valve on your chamber. [3] ii) Bake out your system at about 150oC overnight using heating tape. [3] iii) Pump liquid nitrogen through an unused substrate holder in your chamber. [3] A: i) Conductance of a long thin tube is too poor to allow much pumping; 1/2” diameter ≈ l, so flow is viscous. ii) Bake-out would help by desorbing water vapor from chamber walls which could then be pumped out of the chamber. iii) This would also help by condensing many residual gas molecules on the cold substrate holder, effectively removing them from the gaseous state. † † 1 Chemical vapor deposition 2. You are using a CVD system for which the data shown below have been proven reliable for the growth of poly Si. a) Write the chemical reaction for Silane pyrolysis to grow Si at 650°C. [3] b) Is this film growth reaction-rate limited or gas transport limited? . [3] c) How would you calculate the activation energy for this reaction? . [3] d) Calculate the activation energy from the figure. [3] e) You want to grow your Si film more smoothly and with higher density. To achieve this, would you do the following (explain in 1 sentence) i) Increase or decrease processing temperature, T? . [2] ii) Increase or decrease gas flow rate? . [3] iii) Increase or decrease Silane concentration, Cg . [3] A a) SiH4 (g) => Si(s) + 2H2(g). b) From figure above, it is reaction-rate limited (strong linear ln(v) vs 1/T). c) DG is given by the slope of these ln(v) vs. 1/T curves in the reaction-limited regime. d) Using the two points at about 650 C, 0.1 µm/min (1.67 ¥ 10-7 cm/s) and 750 C, 0.01 µm/min, the change in ln(v) is 2.3 and the change in 1/T is 1.06 ¥ 10-4. Include Boltzman constant to get DG ≈ 1.87 eV (3 ¥ 10-19 J). e) i) Increase T, it gives more surface mobility to reacting species. Also correct would be to say decrease T to slow down the reaction-limited growth rate. ii) Reaction-limited process is not affected by gas flow rate. Also accepted was: decrease gas flow rate to have less burial of surface-diffusing species, which is more appropriate for transport-limited case. iii)Decrease SiH4 concentration to slow down all fluxes. 2 Oxidation, diffusion 3. a) Explain why SiO2 grows on Si not at the outer surface (gas/SiO2 interface), but at the Si / SiOx interface. Say why this is beneficial. [4] A: Diffusion of O in SiO2 > diffusion of Si in SiO2. This leads to a clean Si / SiOx interface. b) Briefly describe 3 other beneficial physical or chemical qualities of SiO2. [3x2] A: (any three) Diffusion barrier for P, B, As; therefore, can be used as a mask. Etched by HF which does not attack Si; allows creation of a mask from SiO2. High electrical resistivity; useful as insulator. Epitaxial, clean interface; reproducible electrical properties. High dielectric breakdown; stable under high voltage. Stable to 900°C or 10-9 Torr; unaffected by subsequent processing. c) There are two general types of solutions to the diffusion equation that are relevant to processing, one Gaussian and the other a complementary error function. i)List the two boundary conditions and one initial condition for the Gaussian solution and sketch the form of this solution for small time and a later time. [5] ii) List the two boundary conditions and one initial condition for the complementary error function solution and sketch the form of this solution for small time and a later time. [5] A: i) Gaussian: BC: dC(0,t)/dx = 0, C(∞,t) = 0; C(x,t) time IC: C(x> 0, 0) = 0 Some gave C(0,0) governed by Q. I took off points because this is a normalization condition that sets the vertical scale of the Gaussean; it does not affect the shape of the solution. t ii) Comp erf: BC: C(0,t) = C0, C(∞,t) = 0; C(x,t) time IC: C(x> 0, 0) = 0 t There was some confusion with the initial slope of the erfc solution; it is not zero. 3 Ion implantation, diffusion 4. Phosphorus is implanted at 100 keV to a dose of Q = 2.5 ¥ 1014 cm-2 in a p-type Si 18 -3 wafer, NA = 10 cm . (Use data below for implanting into Si). a. At what depth is the implanted P concentration a maximum? . [4] b. What is the peak phosphorus concentration? [4] c. What is the depth of the junction before diffusion? [4] d. What happens to the value of Cp and its location if the implant is subjected to a diffusion process at an elevated temperature? [4] e. With what mathematical form does the junction depth evolve with time upon diffusion at an elevated temperature? [4] A -5 a. From figure, Rp = 1250 Å (1.25 ¥ 10 cm). 19 -3 b. DRp = 350 Å, Cp = Q/[÷(2π) DRp] = 2.85 ¥ 10 cm . c. The implant concentration profile is equal to the background acceptor 2 Ê x - R ˆ Á ( p ) ˜ 18 concentration: C(x) = Cp exp - 2 =10 . Inverting to solve for x Á 2DR ˜ Ë p ¯ 2 gives: x = Rp + 2DRp ln(28.5) = 0.22 mm d. The value of Cp decreases but its position remains the same (to first order) upon †diffusion. If you took “its location” to mean that of the junction, the answer is “junction moves deeper into Si”. †e. The implant profile advances deeper into the wafer like ÷(Dt). 4 Lithography 5. a) Sketch the intensity profile incident on a substrate for exposure of a small circular feature through a projection lithography system and a contact/proximity lithography system. In the case of the projection system, quantitatively label the horizontal axis. [5] A. a) Proximity Lithography Projection Lithography 1.22lf d d is the lens diameter Circular W Aperture Contact Incident Lithography Light Light Intensity b) Explain in words how the Rayleigh criterion for a diffraction limited system is mathematically determined. [4] A b) Rayleigh criterion defines the resolving power of an optic lens, i.e. the size of the minimum resolvable feature. Two adjacent points are distinguishable when one point projection’s peak lies on the other point’s first minimum. c) Derive the Rayleigh criterion in terms of numerical aperture, NA, and l, using the expressions; R = 1.22lf/d and NA = nsina. Define f and d. [6] 5 A B’ d R B a A’ n =1 in air f 1.22lf 1.22lf 0.61l 0.61l A. c) The minimum resolution R = = = = where d n(2 f sina) nsina NA f is the focal length (which may not be the same as the distance to the imaging plane) and d is the diameter of the lens. You might also include the fudge factor, k1, in the final form of R, to account for the departure of a real lens from ideality. d) Why would this resolution limit improve if the optical projection could be done in an oil medium. Use the Rayleigh criterion equations to explain this. [2]. A. d) According to Rayleigh criterion, the resolution limit will improve (smaller R) in a high reflective index ( n > 1) medium, for example, oil. e) How does phase-shift lithography improve the limit of resolving two features in close proximity? [3] A, e) The figure below shows how phase-shift lithography improves the resolution limit in close proximity. The left side shows that the light intensity, I = |amplitude|2, in the region between two exposure features will be enhanced by addition of the amplitudes of the two features point-by-point. The photo resist corresponding to this region will also be exposed and poorly resolved. The right hand side uses a phase-shift lens to shift the phase of one feature by exactly 180o. The amplitudes cancel in the overlap region, reducing the intensity of light between the neighboring features and greatly improve the resolutioin of the resulting pattern on the photo resist. 6 7.