小火龙果 |
2023-10-16 10:19 |
SYNOPSYS 课程四十四:为什么 SYNOPSYS 不使用坐标断点?
SYNOPSYS有六个用于描述倾斜和偏心(TDC)的方法,其中只有一个涉及虚拟表面。(如果您愿意,也可以在没有额外表面的情况下指定几何体。)让我们举一些例子。这是一个典型的相机镜头,有八个镜片: &eX^ll 打开C28L1.RLE文件,运行这个宏文件: "1$hfs )G9,5[ RLE Y3f2RdGl ID 8-ELEMENT TELEPHOTO 124 ^G(+sb[t FNAME 'C28L1.RLE ' "V7 &@3 LOG 124 ?s{Pp WAVL .6562700 .5875600 .4861300 80O[pf*? APS 4 km!jxs UNITS MM '[Ch8Yf\ OBB 0.0000000 5.0000000 25.4000000 -3.67701038746 0.0000000 0.0000000 25.4000000 ;z^C\=om MARGIN 1.270000 KZTT2KsYl BEVEL 0.254001 >PiEu->P, 0 AIR ;(9q, ) 1 RAD 90.4200490000000 TH 11.00000000 ucC'SS 1 N1 1.61726800 N2 1.62040602 N3 1.62755182 cH\.-5NQ 1 CTE 0.630000E-05 =wX(a 1 GTB S 'SK16 ' 5?4jD]Z 1 EFILE EX1 30.200000 30.200000 30.200000 0.000000 *.NVc 1 EFILE EX2 30.200000 30.200000 0.000000 1'[_J 2 RAD -193.5240600000000 TH 3.00000000 /=ro$@ 2 N1 1.69220502 N2 1.69894060 N3 1.71544645 9mH/xP:y 2 CTE 0.790000E-05 n8>(m, 2 GTB S 'SF15 ' q%GlS=o" 2 EFILE EX1 29.161700 29.415700 30.200000 0.000000 5J8U] :Y) 2 EFILE EX2 28.244500 29.161700 0.000000 @phb5 3 RAD 645.1795900000005 TH 25.00000000 AIR FQCz_z 3 EFILE EX1 28.244500 29.161700 30.200000 D[ (A`!) 4 CV 0.0000000000000 TH 25.00000000 AIR 41jx+
0\Z 5 RAD -75.8953820000000 TH 5.00000000 ;?0k> 5 N1 1.51981155 N2 1.52248493 N3 1.52859442 q~:k[@`. 5 CTE 0.820000E-05 (y!<^Q 5 GTB S 'K5 '
^iaG>rvA 5 EFILE EX1 21.047029 21.105107 21.359108 0.000000 r5N.Qt8 5 EFILE EX2 21.105107 21.105107 0.000000 u>o2lvy8 6 RAD -67.6909630000000 TH 3.00000000 AIR =7uxzg/%Tj 6 EFILE EX1 21.105107 21.105107 21.359108 $&iw (BIq 7 RAD -80.0000000000000 TH 3.00000000 F%pYnHr< 7 N1 1.61502503 N2 1.62003267 N3 1.63207204 Z-,'M tD 7 CTE 0.820000E-05 &]Q\@;]Aq 7 GTB S 'F2 ' [1{uK&$e 7 EFILE EX1 21.153005 21.153005 21.407006 0.000000 p$
%D 7 EFILE EX2 21.153005 21.153005 0.000000 8(c,b 8 RAD -112.8857000000000 TH 60.00000000 AIR *XZlnO 8 EFILE EX1 21.153005 21.153005 21.407006 #^fDKM 9 RAD 134.3623100000000 TH 6.00000000 qkN{l88 9 N1 1.61502503 N2 1.62003267 N3 1.63207204 k&PxhDf 9 CTE 0.820000E-05 RZV6\j 9 GTB S 'F2 ' Jx8?x#} 9 EFILE EX1 20.680300 20.680300 21.000000 0.000000 xr*hmp1 9 EFILE EX2 20.680300 20.680300 0.000000 o3~ecJ?k 10 RAD -89.1513450000000 TH 3.00000000 L^zF@n^5A 10 N1 1.51981155 N2 1.52248493 N3 1.52859442 Una7O] 10 CTE 0.820000E-05 hWujio/h 10 GTB S 'K5 ' MxO0# 10 EFILE EX1 20.524700 20.524700 21.000000 0.000000 5&\% 10 EFILE EX2 20.028900 20.524700 0.000000 8QN#PaY 11 RAD 175.6904000000000 TH 9.00000000 AIR as?~N/} 11 EFILE EX1 20.028900 20.524700 21.000000 $($26g 12 RAD -54.1687770000000 TH 3.00000000 ErNL^Se1 12 N1 1.61726800 N2 1.62040602 N3 1.62755182 se1\<YHDS 12 CTE 0.630000E-05 ' s6SKjZS 12 GTB S 'SK16 ' #PpmR_IX 12 EFILE EX1 19.707434 19.707434 19.961435 0.000000 xu _: 12 EFILE EX2 19.961435 19.961435 0.000000 < | |