3.155J6.152J Microelectronic Processing Technology Fall Term. 2003 Bob HAndley Martin schmidt Problem set 6 Out Nov, 12. 2003 Due no,26,2003 Sputter deposition: Read Plummer Chap 9, Sections 9.2.2.2 to 9.3.10. Consider reading Ohring 1. You need to deposit a high quality (low electrical resistivity) Al film at a very high rate(v> I micron/min) and achieve good step coveage using sputter deposition. Referring to information in the class notes and text, answer the following three questions. (Grade will depend more on how you justify your design, rather than on its correctness-which is harder to determine. a) Design, and justify your design, for a sputtering system to deposit this al with attention paid to configuration of the anode(s), cathode(s)and substrate placement (be creative here), as well as the use of dC or rF power source, possible give some conditions on power requirements and critical dimensions in b)What sputtering conditions would you use?(type of gas, gas pressure, sputtering voltage, bias voltage, substrate temperature) c)Assuming a sticking coefficient of unity for both Al and oxygen, what oxygen partial pressure could you tolerate in the chamber to keep the oxygen content in the film less than 1%o for your chosen conditions? Some possibly useful information is shown below Q 8 ○○○OC SOLUTION 1. a) Text suggests DC(RF is acceptable), magnetron(for high deposition rate) sputtering; biasing will need to be justified in terms of film stress and density Here is one way to configure magnets and electrodes for more grazing incidence of ions on target(0 just less than 90) and greater sputter yield
1 3.155J/6.152J Microelectronic Processing Technology Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 6 Out Nov. 12, 2003 Due Nov. 26, 2003 Sputter deposition: Read Plummer Chap. 9, Sections 9.2.2.2 to 9.3.10. Consider reading Ohring 1. You need to deposit a high quality (low electrical resistivity) Al film at a very high rate (v > 1 micron/min) and achieve good step coveage using sputter deposition. Referring to information in the class notes and text, answer the following three questions. (Grade will depend more on how you justify your design, rather than on its correctness – which is harder to determine.) a) Design, and justify your design, for a sputtering system to deposit this Al with attention paid to configuration of the anode(s), cathode(s) and substrate placement (be creative here), as well as the use of DC or RF power source, and biasing. If possible give some conditions on power requirements and critical dimensions in the chamber. b) What sputtering conditions would you use? (type of gas, gas pressure, sputtering voltage, bias voltage, substrate temperature). c) Assuming a sticking coefficient of unity for both Al and oxygen, what oxygen partial pressure could you tolerate in the chamber to keep the oxygen content in the film less than 1% for your chosen conditions? Some possibly useful information is shown below: SOLUTION 1. a) Text suggests DC (RF is acceptable), magnetron (for high deposition rate) sputtering; biasing will need to be justified in terms of film stress and density. Here is one way to configure magnets and electrodes for more grazing incidence of ions on target (q just less than 900 ) and greater sputter yield: q S 90°
Target(s) ubstrate(s) Either planar or cylindrica symmetry: the latter has advantage of very large ratio of target area to substrate area. Other geometries acceptable but should be justified Magnets are arranged so their field lines run nearly parallel to target surface They could be oriented in opposition to each other(so overlapping fields cancel)to minimize field at substrates. Target/substrate spacing, a, will be governed by radius of curvature of electrons in B field. If the radius is comparable to a, then many Ar+ ions may curve away from the target and strike the substrate. If r<<a, then the ar+ ions will spiral about the b field with many possible strikes on the target: F=q(vx B)and F=mvIr indicate r= mv/(qB). The velocity may be governed by the power: P a (12)mv2≈eV≥1kev( class notes slides10.11) b) Either Nezo or Arao could be used to sputter Alz7 Use large negative bias perhaps -200 V on substrate 15). Chose pressure and temperature on basis of ideas in slide 13: low pressure and high temperature for good step coverage, but to avoid porous film, keep temperature quite high, text recommends 150 to 300C c)The given film growth rate of vilm= I micron/min a 17 nm/s=1. 7x 10 cm/s 1.7x 10 /s. So the oxygen flux should be less than about 104/s. Using / =p om s and assuming an al film density of 10-cm, the al flux on the substrate is J=nvr √2mnkT we get a pressure of 2.6x 10 Pa(1 atm= 10Pa= 760 T), so P(O,)210-Torr 2. It is very difficult to evaporate stoichiometric SiO; you often get SiO, with 1 <x<2, which implies a mixture of SiO and SiO2. But you still need to get as close as possible to What partial pressure of O2 must you have in your vacuum chamber so that the flux of O, on the substrate is the same as that of Si? Your evaporation source has a surface area of l cm, the substrate is at a planetary radius of 20 cm, and your crucible is heated to 1500c What is the mean free path of O2 in this partial pressure(assume the diameter of an o2 molecule is 0.3 nm) c)What does your measurement in b)mean for your process?
2 Target(s) Substrate(s) Either planar or cylindrical symmetry; the latter has advantage of very large ratio of target area to substrate area. Other geometries acceptable but should be justified. Magnets are arranged so their field lines run nearly parallel to target surface. They could be oriented in opposition to each other (so overlapping fields cancel) to minimize field at substrates. Target/substrate spacing, a, will be governed by radius of curvature of electrons in B field. If the radius is comparable to a, then many Ar+ ions may curve away from the target and strike the substrate. If r << a, then the Ar+ ions will spiral about the B field with many possible strikes on the target: F = q(v x B) and F = mv2 /r indicate r = mv/(qB). The velocity may be governed by the power: P ≈ (1/2)mv2 ≈ eV > 1 keV (class notes slides 10,11). b) Either Ne20 or Ar40 could be used to sputter Al27. Use large negative bias, perhaps -200 V on substrate (slide 15). Chose pressure and temperature on basis of ideas in slide 13: low pressure and high temperature for good step coverage, but to avoid porous film, keep temperature quite high, text recommends 150 to 300C. c) The given film growth rate of vfilm = 1 micron /min ≈ 17 nm/s = 1.7 x 10-6 cm/s, and assuming an Al film density of 1023 cm-3, the Al flux on the substrate is J = nvfilm = 1.7 x 1017/s. So the oxygen flux should be less than about 1014/s. Using † J = P 2pmkT , we get a pressure of 2.6 x 10-9 Pa (1 atm = 105 Pa = 760 T), so P(O2) ≈ 10-11 Torr. 2. It is very difficult to evaporate stoichiometric SiO2; you often get SiOx with 1 < x < 2, which implies a mixture of SiO and SiO2. But you still need to get as close as possible to SiO2. a) What partial pressure of O2 must you have in your vacuum chamber so that the flux of O2 on the substrate is the same as that of Si? Your evaporation source has a surface area of 1 cm2 , the substrate is at a planetary radius of 20 cm, and your crucible is heated to 1500o C. b) What is the mean free path of O2 in this partial pressure (assume the diameter of an O2 molecule is 0.3 nm). c) What does your measurement in b) mean for your process?
SOLUTION 2.a)Aahs=1cm2,r=20cm,Ts=1500C(1773K) P 28 甲27m4m from table. pl= 2x10-Torr 2×103x[3( (Pa/Torr)]Icm2√293×32 =23×103Pa=1.7×10-Tomr 4π×202cm2√1773×28 b)λ= 万dP 23 cm(using T, P of Si) c) There are virtually no collisions between O, molecules in the chamber, for that matter between O2 and Si vapor molecules. Both species do collide with walls of chamber. Without collisions, the two gases(Si, O,)are not in equilibrium Also Si vapor follows a straight line path to substrate. It is only on the substrate that Si has a chance to interact with O
3 SOLUTION: 2. a) Acrucible = 1 cm2 , r = 20 cm, TSi =1500 C (1773K) † JO2 JSi =1 = PO2 Peq.vap Si ¥ 2pkTSi mSi 2pkTH2OmO2 Acru 4pr 2 , mSi = 28 amu. † mO2 = 32 amu from table, † PSi 1500 C @ 2 ¥10-3 Torr † PO2 = 2 ¥10-3 ¥[133(Pa Torr)]1cm 2 293 ¥ 32 4p ¥202 cm2 1773¥ 28 = 2.3 ¥10-5 Pa =1.7 ¥10-7 Torr b) † l = kT 2pd2 PTot = 23 cm (using T, P of Si) c) There are virtually no collisions between O2 molecules in the chamber, for that matter between O2 and Si vapor molecules. Both species do collide with walls of chamber. Without collisions, the two gases (Si, O2) are not in equilibrium. Also Si vapor follows a straight line path to substrate. It is only on the substrate that Si has a chance to interact with O2