3.155J/6.152J Microelectronic Processing Fall Term. 2003 Bob handley Martin schmid Problem set s solutions Out Oct 29, 2003 Due Nov 5. 2002 1.(Plummer 10.3)In a certain process, it is desired that the pitch of metal lines be equal to or less than 1.Omm(the pitch equals one metal linewidth plus one spacing between metal lines, measured at top of features). Assume that the metal linewidth and spacing are equal(that is, 0.5um each). The height of such structures is also 0. 5um, and the minimum lithographic dimension is 0. 25um a. What minimum degree of anisotropy is needed in an etch process in order to produce such a structure? b. What minimum pitch could be obtained for such a structure with wet etching?(Again with minimum lithograph dimension of 0. 25um, thickness of 0. 5um and equal metal width and spacing) Answer(a) Consider the anisotropic etching effect to obtain 0.5um equal linewidth and spacing, suppose the mask width is 1-x. 0.5um the anisotropy of etching would be A =1 .5+x 2×0.5 min(A)=05+mn(x)=0.5+0.25=075 (b). Metals are usually polycrystalline structure and thus are subjected to isotropical etch in wet etching. Therefore, we can determine the anisotropy of wet etching is A, =0 Similar to part(a), we will have this following equation between the pitch p and the mask (p-x) P r =1-Wmask-WmetaL =1 2=1-(P-x)→p=2(+x) →min(P)=2(1+min(x)=2×(1+0.25)=2.51m3.155J/6.152J Microelectronic Processing Fall Term, 2003 Bob O'Handley Martin Schmidt Problem set 5 Solutions Out Oct. 29, 2003 Due Nov. 5, 2002 1. (Plummer 10.3) In a certain process, it is desired that the pitch of metal lines be equal to or less than 1.0mm (the pitch equals one metal linewidth plus one spacing between metal lines, measured at top of features). Assume that the metal linewidth and spacing are equal (that is, 0.5µm each). The height of such structures is also 0.5µm, and the minimum lithographic dimension is 0.25µm. a. What minimum degree of anisotropy is needed in an etch process in order to produce such a structure? b. What minimum pitch could be obtained for such a structure with wet etching? (Again with minimum lithograph dimension of 0.25µm, thickness of 0.5µm and equal metal width and spacing.) Answer (a). Consider the anisotropic etching effect, to obtain 0.5µm equal linewidth and spacing, suppose the mask width is 1-x, the anisotropy of etching would be: Af = 1 − wmask − wmetal / x 2 f 1 − x − 0 .5 = 1 − = 0 .5 + x 2 × 0.5 min( Af ) = 0 .5 + min( x) = 0 .5 + 0 .25 = 0 .75 (b). Metals are usually polycrystalline structure and thus are subjected to isotropical etch in wet etching. Therefore, we can determine the anisotropy of wet etching is A = 0 . f Similar to part (a), we will have this following equation between the pitch p and the mask spacing x: 0 = Af = 1 − wmask − wmetal = 1 − ( p − x) − 2 p = 1 − ( p − x) Ω p = 2 (1+ x) 2x f 2 × 0 .5 2 Ω min( p) = 2 (1+ min( x)) = 2 × ( 1+ 0 .25 ) = 2 .5µm 1