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)