《Microelectronics Process》Problem set 4 solutions

3.155J6,152J Microelectronic processing Fall Term. 2003 Bob O'Handley Martin Schmidt Problem set 4 solutions Out oct 8. 2003 Due Oct 15, 2002 I. Plot resolution and depth of field as a function of exposure wavelength for a projection aligner with 100nm< A <500nm. Assume NA=0.26. Recalculate on the same plot for NA=0.41. Discuss the implication of these plots for the technologist that must manufacture transistors with 0.5 um features R k10.6 NA 0.5 DOF=± (MA)2(NA)2 To resolve 0.5 um feature size with NA=0. 26, one can choose exposure wavelength less than 216 nm (e.g. 193 nm ArF DUV) and also carefully control the resist layers topography to keep the pattern in focus Resolution and Depth of Focus Vs Exposure wavelength 4000 2000 -1000 DOF (NA0. 26) Exposure wavelength(nm) If NA=0.4, one can choose exposure wavelength less than 340 nm(e.g. ArF 193nm or KrF 248nm)

1 3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 4 Solutions Out Oct. 8, 2003 Due Oct.15, 2002 1. Plot resolution and depth of field as a function of exposure wavelength for a projection aligner with 100nm < λ < 500nm. Assume NA = 0.26. Recalculate on the same plot for NA =0.41. Discuss the implication of these plots for the technologist that must manufacture transistors with 0.5 µm features. NA NA k R 1λ 0.6λ = = 2 2 2 ( ) 0.5 (NA) NA k DOF λ λ = ± = ± To resolve 0.5 µm feature size with NA=0.26, one can choose exposure wavelength less than 216 nm (e.g. 193 nm ArF DUV) and also carefully control the resist layer’s topography to keep the pattern in focus. -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 100 150 200 250 300 350 400 450 500 Resolution and Depth of Focus Vs Exposure wavelength R (NA=0.26) DOF (NA=0.26) DOF (NA=0.26) Resolution and Depth of Focus (nm) Exposure wavelength (nm) ArF 193nm If NA = 0.41, one can choose exposure wavelength less than 340 nm (e.g. ArF 193nm or KrF 248nm)

Resolution and Depth of Focus Vs Exposure wavelength 2000 1500 ArF 193nm KrF 248nm -500 一DoF(NA-0.41) -1500 2)A0.6 um thick layer of resist has Q0=40 mJ/cm" and Qf= 160 mJ/cm. Calculate the resist contrast and CMtF. If the resist thickness is cut in half, Qf reduces to 70 mJ/cm while Qo is unchanged. Assuming NA=0.4, use the figure below to determine the minimum linewidthfor an aligner with S=1.0 using both resist thicknesses with a source of 365 nm. The figure below plots mtf of the aligner for a set of lines and spaces. The lines and spaces are of equal width(W), and the spatial frequency is normalized by the Rayleigh criteria, R. In other words, a normalized spatial frequency of 0.5, corresponds to a linewidth W equal tor(since the equivalent source spacing of the lines is 2W) 0.9 s=0.5 0.8 0.7 s=0 0.6 0.5 0.4 0.3 0.2 0.1 S=025 010203040.50.6070.809 Normalized spatial frequency(linesunit length) /(cutoff frequency)

2 -2000 -1500 -1000 -500 0 500 1000 1500 2000 100 150 200 250 300 350 400 450 500 Resolution and Depth of Focus Vs. Exposure wavelength R (NA=0.41) DOF (NA=0.41) DOF (NA=0.41) Resolution and Depth of Focus (nm) Exposure wavelength (nm) ArF 193nm KrF 248nm 2) A 0.6 µm thick layer of resist has Q0 = 40 mJ/cm2 and Qf = 160 mJ/cm2 . Calculate the resist contrast and CMTF. If the resist thickness is cut in half, Qf reduces to 70 mJ/cm2 while Q0 is unchanged. Assuming NA = 0.4, use the figure below to determine the minimum linewidthfor an aligner with S = 1.0 using both resist thicknesses with a source of 365 nm. The figure below plots MTF of the aligner for a set of lines and spaces. The lines and spaces are of equal width (W), and the spatial frequency is normalized by the Rayleigh criteria, R. In other words, a normalized spatial frequency of 0.5, corresponds to a linewidth W equal to R (since the equivalent source spacing of the lines is 2W)

0.6 um resist 1.66 logo(2 /20) log1o (160/40) Q1-Q107-1 CMTF= =0.6 Q,+Q107+1160+40 0.3 um resist: y1ogo(Q/90)logm0(70/40) 4.11 CMTF= Q0107-170-40 =0.273 Qr+Qo10+170+40 Normalized spatial frequency: f=(line/unit length)/(cut off frequency =(1/2W)/(1/R)=R/2W= W=R/2f R=0612 =557nm NA To resolve the feature. we need mtf> cMtf. The minimum linewidth. i. e the maximum spatial frequency, can be directly read from the graph as the intersections between MTF=Cmtf and s=l 0.9 s=0.5 0.8 s=0 0.6 um resist 0.7 CMTF 05 0.4 Maximum frequencies 0.3 esist 03R2W CMTE 02 0.1 s=0.25 010203040.50.60.70.8091 Normalized spatial frequency(lines/unit length) /cutoff frequency) The maximum frequencies from the graph are 0.33 and 0.63 with respect to 0.6 um and 0.3 um resists, corresponding to a minimum linewidth of 844nm and 442nm. 3) Estimate the diffraction-limited resolution for an X-Ray exposure system using photons with an energy of l ke V and a mask to wafer separation of 20 microns hc6.62×10-34×2.998×103 → =1.24×10-m=1.24nm E 103×1.6×10 Minimum resolvable feature size due to the X-ray exposure system's Fresnel diffraction Proximity printing) is: (ag) 2 =(1. 24x10-x20)2=0.157um

3 0.6 µm resist: 1.66 log (160 / 40) 1 log ( / ) 1 10 0 10 = = = Qf Q γ 0.6 160 40 160 40 10 1 10 1 1/ 1/ 0 0 = + − = + − = + − = γ γ Q Q Q Q CMTF f f 0.3 µm resist: 4.11 log (70 / 40) 1 log ( / ) 1 10 0 10 = = = Qf Q γ 0.273 70 40 70 40 10 1 10 1 1/ 1/ 0 0 = + − = + − = + − = γ γ Q Q Q Q CMTF f f Normalized spatial frequency: f=(line/unit length)/(cut off frequency)=(1/2W)/(1/R)=R/2WŸW=R/2f. nm NA R 557 0.61 = = λ To resolve the feature, we need MTF≥CMTF. The minimum linewidth, i.e. the maximum spatial frequency, can be directly read from the graph as the intersections between MTF=CMTF and S=1. The maximum frequencies from the graph are 0.33 and 0.63 with respect to 0.6 µm and 0.3 µm resists, corresponding to a minimum linewidth of 844nm and 442nm. 3) Estimate the diffraction-limited resolution for an X-Ray exposure system using photons with an energy of 1 keV and a mask to wafer separation of 20 microns. m nm E hc hc E 1.24 10 1.24 10 1.6 10 6.62 10 2.998 10 9 3 19 34 8 = × = × × × × × = Ÿ = = − − − λ λ Minimum resolvable feature size due to the X-ray exposure system’s Fresnel diffraction (Proximity printing) is: (λg) (1.24 10 20) 0.157µm 1/ 2 3 1/ 2 = × × = − 0.6 µm resist CMTF 0.3 µm resist CMTF Maximum frequencies R/2W