RF射频电路设计英文课件Lecture10_Manufacturabilityofaproductdesign

1、lecture 10richard li, 200911. introduction2. implication of 6 design o 6 and yield rate o 6 design for a circuit block3. approaching 6 design o by changing parts value o by replacing a single part with multiple parts 4. monte carlo analysis o a bpf (band pass filter) o simulation with monte carlo an

2、alysis o sensitivity of parts on the parameter of performanceappendixes o fundamentals of random process o table of the normal distribution lecture 10 : manufacturability of a product designrichard chi-hsi li 李缉熙李缉熙cellular phone:(prc)email : lecture 10richard li, 20092 1. introductiont

3、he unique criterion to measure success or failure of product development is the manufacturability of a product. what is the manufacturability of a product? 1) high yield rate, including * satisfication of specifications; * good repetition or identity of product; * high reliability. 2) low cost, incl

4、uding the cost of * material and parts, * manpower, * manufactory maintenance and equipment. lecture 10richard li, 20093specificationssimulationsfabrication of samples by handtestingalttoppilot productionmassive productiontolerance analysisfigure 1 typical design procedures of a new product with acc

5、eptable manufacturability3rd stage : production2nd stage : pre-production 1st stage : r & dlecture 10richard li, 20094o 6 and yield rate(z)figure 2 definition of design tolerance toldesign and process capability 6usllsldesign tolerance = toldesignprocess capability = 6defects = 6.68%defects = 6.68%

6、-4 -3 -2 -1.5 -1 0 +1 +1.5 +2 +3 +4 z =(c-m)/mczlslusltoldesign)()(1lslzfuslzfdefects)()(2210222zzeerfdxzfx2.implications of 6 design lecture 10richard li, 20095figure 3 relationship between yield rate f(z) and design tolerance toldesign . 0 1 2 3 4 5 6 7 100%90%80%70%60%50%40%30%20%10% 0%yield rate

7、 f(z), %design tolerance toldesign , 99.74%95.44%68.26%38.30%86.64%98.76%0%99.98% table 1 relationship between the yield rate f(z) and the design tolerance toldesign design tolerance yield rate toldesign f(z) 0 0% 1 38.30% 2 68.26% 3 38.30% 4 86.64% 5 98.76% 6 99.74% 7 99.98%lecture 10richard li, 20

8、096o 6 design for a circuit blocka 6 circuit design is a design such that the yield rate of this circuit block in mass-production reaches 99.74% when 6 tolerance of all the parts applied in the circuit block is allowed. actually, the expected yield rate is 100%, although 99.74% is close to 100%.lect

9、ure 10richard li, 200973. approaching 6 design o by changing parts 6 value(z)usllsldesign tolerance process capability = 6defects = 6.68%defects = 6.68%usllsldefects = 0.13%defects = 0.13%process capability = 6 -4 -3 -2-1 0 +1 +2 +3 +4 -4 -3 -2 -1 0 +1 +2 +3 +4 design tolerance = toldesignz =(c-m)/z

10、 =(c-m)/design tolerance = toldesignusllslfigure 4 convert 3 design to 6 design by 50%reduction of standard deviation from to , that is, = /2 = original = /2lecture 10richard li, 20098o by replacing a single part with multiple parts 2 k6 k(a) one resistor of 2 k is replaced by three 6 k resistors in

11、 parallelm = 2k = 200 m = 6 k/3 = 120 4 pf16pf(b) one capacitor of 5 pf is replaced by four 16 pf capacitors in seriesm = 4 pf = 0.5 pf m = 16 pf/4 = 0.2 pf 7 nh(c) one inductor of 7 nh is replaced by two 14 nh inductors in parallel.m = 7 nh = 0.2 nh m = 14 nh/2 = 0.15 nh figure 5 one part is replac

12、ed by multi-parts.14 nh14 nh6 k6 k16pf16pf16pflecture 10richard li, 20099table 1 replacement of original 1 resistor, 1 capacitor, and 1 inductor by 3 resistors, 4 capacitors, and 2 inductors.original partmulti-partsm m .resistor 2 k200 6 k/3120 (1 resistor)(3 resistors in parallel)capacitor4 pf0.5 p

13、f16 pf/40.2 pf(1 capacitor)(4 capacitors in series)inductor7 nh0.2 nh14 nh/20.15 nh(1 inductor)(2 inductors in parallel)lecture 10richard li, 2009104. monte carlo analysiscvr1r1c2l3l2l1l1inl1l1outl2c3c4vr1(c1)vr1(c1)vr1(c1)vr1(c1)l1 : inductor 4.5 nh / 27 k / 0.2 pfl2 : inductor 8.8 nh / 27 k / 0.2

14、pfl3 : inductor 40.5 nh / 27 k / 0.2 pfc1 : capacitance of varactor 13.75 pf when cv=5vc2 : capacitor 2.4 pf c3 : capacitor 100 pfc4 : capacitor 10000 pfr1 : resistor 20 k cv : 5 vfigure 6 schematic of a uhf tunable filter o bptf (band pass tunable filter)lecture 10richard li, 200911 tunable fltlnam

15、ixerfltfigure 7 a tunable filter is added in the front end of the receiver for better selectivitylecture 10richard li, 200912frequencyfigure 8 frequency response of a tunable filter moved from low end to high end as the control voltage is increased(a) popular cases : bandwidth is changed as the freq

16、uency response is movedfrequency(b) ideal cases : bandwidth is kept unchanged as the frequency response is movedlecture 10richard li, 200913* main coupling : inductive coupling - an improper coupling unable a tunable filter tuned over a wide frequency tuning range . where q = quality factor of filte

17、r ;o = tuned central frequency ;bw = bandwidth o ;l = inductance of coupling inductor ;r = equivalent resonant resistance of tank circuit o .oorcbw40obwoorcbwq21rlbwqoo222orcbwlrbw2- in the case of inductor coupling,- in the case of capacitor coupling,lecture 10richard li, 200914from the above expre

18、ssions , it is found that1) in the case of capacitor-coupling, the bandwidth of the filter is dependent of the tuning frequency. it is increased as the tuning frequency is increased. 2) in the case of inductor-coupling, the bandwidth of the filter is an independent of the tuning frequency. the bandw

19、idth could be kept in an almost constant for a wide tuning frequency range. it is therefore concluded that the correct coupling element between the 2 tank circuits must be an inductor, but not a capacitor nor others. lecture 10richard li, 200915* second coupling : capacitive coupling - the capacitor

20、, c2, is the second coupling component between the two tank circuits; - it forms a “zero” at the imaginary frequency; - this “zero” traces the central frequency, o , over a wide frequency tuning range. lecture 10richard li, 200916* monte carlo analysisfigure 7.15 simulation page for monte-carlo anal

21、ysis (strictly speaking, it is not a gaussian but a normal distribution here.)fltport 1port 250 terminator50 terminatorc1 = cac2 = cbl1 = lal2 = lbl3 = lcequation ca = randvar gaussian, 13.75 pf 5%equation cb = randvar gaussian, 2.4 pf 5%equation la = randvar gaussian, 4.5 nh 7%equation lb = randvar

22、 gaussian, 8.8 nh 7%equation lc = randvar gaussian, 40.5 nh 7%figure 9 simulation page for monte-carlo analysis (strictly speaking, it is not a gaussian but a normal distribution here.) lecture 10richard li, 200917* frequency response without tolerance200 300 400 fo 500 600 700 -10s21, db0-20-40-60

23、10-50-70-80-90-30f , mhzil=1.76 dbimag. rej. =83.1 dbf o =435.43 mhzfigure 10 typical frequency response of tunable filter without tolerance lecture 10richard li, 200918* case #1 : bandwidth is too wide 200 300 400 fo 500 600 700 -10s21, db0-20-40 10-50-90-30 f , mhzil=1.76 dbimag. rej. =80.1 dbf o

24、=435.43 mhzfigure 11 frequency response of tunable filter with tolerance - bandwidth is too wide-70-80-60lecture 10richard li, 200919* case #3 : bandwidth is too narrows21, db 200 300 400 fo 500 600 700 -100-20-40-60 10-50-70-80-90-30f , mhzil=1.76 dbimag. rej. =80.1 dbf o =435.43 mhzfigure 12 frequ

25、ency response of tunable filter with tolerance - bandwidth is too narrowil=5.10 dblecture 10richard li, 200920* case #2 : bandwidth is appropriate 200 300 400 fo 500 600 700 -10s21, db0-20-40 10-50-70-80-90-30f , mhzil=1.76 dbimag. rej. =80.1 dbf o =435.43 mhz-60figure 13 frequency response of tunab

26、le filter with tolerance -bandwidth is appropriatelecture 10richard li, 200921* yield rate & its histogram of il less than 2.5 db ( for case #2 )yield rate, % 100.00-2.16 -1.71il, db350%figure 14 display of insertion loss histogram and yield rate for il60 db lecture 10richard li, 200924* image rejec

27、tion performance of parts with tolerance ( for case #2 )-64-8413.2 142imag.rej., dbc1 , pf-64-842.3 2.5c2 , pfimag.rej., db-64-848.3 9.3l2 , nhimag.rej., dbl1 , nh-64-844.2 4.8imag.rej., db-64-8438.1 42.6l3 , nhimag.rej., dbnote : capacitors are more sensitive to the image rejection than inductors !

28、figure 17 the effect of individual parts value on the image rejection lecture 10richard li, 200925o a bpf (band pass filter)c1 l c3 l c4 l c3 l c1port 1inc2 c2 c2 c2 c5 c2 c2 c2figure 18 bpf (band pass filter), 403 470 mhz. l = 20 nh c1 = 3.9 pf c2 = 4.3 pf c3 = 7.5 pf c4 = 9.1 pf c5 = 3.9 pfport 2o

29、utlecture 10richard li, 200926figure 19 frequency response of band pass filter f = 403 470 mhz 200 300 400 fo 500 600 700 -10s21, db-20-400 0-50-30f , mhzil= -1.05 dbf o =435.21 mhz-70-80-60-90-100lecture 10richard li, 200927vnswoeufigure 20 s11 of band pass filter f = 403 470 mhz 200mhz403mhz470mhz

30、700mhzlecture 10richard li, 200928* simulation with monte carlo analysis bpf(band pass filter)403 = 470 mhzport 1port 250 terminator50 terminatorl = lac1 = cac2 = cbc3 = ccc4 = cdc5 = ceequation la = randvar gaussian, 20 nh 7%equation ca = randvar gaussian, 3.9 pf 5%equation cb = randvar gaussian, 4

31、.3 pf 5%equation cc = randvar gaussian, 7.5 pf5%equation cd = randvar gaussian, 9.1 pf 5%equation ce = randvar gaussian, 3.9 pf 5%figure 21 simulation page for monte-carlo analysis. number of iteration = 50.lecture 10richard li, 200929figure 22 frequency response of band pass filter, 403 470 mhz, nu

32、mber of iteration = 50 200 300 400 fo 500 600 700 -10s21, db-20-400 0-50-30il= -2.30 db-70-80-60-90-100f , mhzf o =435.21 mhzlecture 10richard li, 200930swoevnu200mhz403mhz470mhz700mhzfigure 23 s11 of band pass filter, 403 470 mhz number of iteration = 50lecture 10richard li, 200931yield rate, % 100

33、.00-2.3 -0.8il, db0figure 24 display of insertion loss histogram and yield rate for |il| 2.5 db (iteration number = 50) number in iteration2010 0 table 2 distribution of iteration number vs. insertion loss insertion loss, db -2.30 to -2.05 -2.05 to -1.80 -1.80 to -1.55 -1.55 to -1.30 -1.30 to -1.05

34、-1.05 to -0.8number in iteration 4 12 1 20 12 1lecture 10richard li, 200932-0.8-2.318.86 20 20.14il, dbl=la , nh(il vs. l) tolerance of l : 7%-0.8-2.34.085 4.3 4.515il, dbc2=cb , pf(il vs. c2) tolerance of c2 : 5%-0.8-2.33.705 3.9 4.095 il, dbc1=ca , pf(il vs. c1) tolerance of c1 : 5%figure 25 sensi

35、tivity of individual parts value on the insertion loss-0.8-2.3il, dbc5=ce , pf3.705 3.9 4.095(il vs. c5) tolerance of c5 : 5%-0.8-2.38.645 9.1 9.555il, dbc4=cd , pf(il vs. c4) tolerance of c4 : 5%-0.8-2.37.125 7.5 7.875 il, dbc3=cc , pf(il vs. c3 ) tolerance of c3 : 5%lecture 10richard li, 200933-0.

36、8-2.38.645 9.1 9.555il, dbc4=cd , pffigure 26 sensitivity of c4 on the insertion loss after its tolerance of 5% is replaced by 7%(il vs. c4) tolerance of c4 : 7%lecture 10richard li, 200934a.1 fundamentals of random processappendixesrelative number of resistorsfigure a.1 histogram of relative number

37、 of resistors versus the value of resistor 0.800k 0.850k 0.900k 0.950k 1.000k 1.050k 1.100k 1.105k 1.120k r, ohmlecture 10richard li, 200935 ohms , 1000m%52remmrtoliilativeiir2 ohms .mrz2222)(xexdxdxxzfzezx020222)()(* gaussian probability function , * average value :* relative tolerance :* variance

38、or standard deviation ri , * sample value :where* gaussion distribution :lecture 10richard li, 200936 -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohm(z)figure a.2 distribution of the random variable, r, is a normal probability function or a gaussian distribution0.39890.24

39、200.05400.00440.0001lecture 10richard li, 200937* tolerance and normal distribution (z)figure a.3 integral of (z) from 0 to z or from m to m+z . -4 -3 -2 -1 0 +1 z +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+z m+2 m+3 m+4 r, ohmlecture 10richard li, 200938* six sigma and 100% yield rategaussian distri

40、bution :dxzfzemx022222)(where m = average of the variable x ; = square root of the variable x.normal probability function is a gaussian distribution function when :m = 0 , and = 1 .)()(2210222zzeerfdxzfxdyexerfxy022)(lecture 10richard li, 200939figure a.4 at each interval, z, the appearing percentag

41、e of a random variable with a normal distribution when the interval is-1 z +1 ,then the area isf(z) = 68.26%,when the interval is-2 z +2 ,then the area isf(z) = 95.44%,when the interval is-3 z +3 ,then the area isf(z) = 99.74%,when the interval isz +3,then the area isf(z) = 0.26%, (z)0.13%0.13%34.13

42、%34.13%13.59%13.59%2.15%2.15% -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmlecture 10richard li, 200940* 6, cp, and cpkfigure a.5 the various terminologies of a process with normal distribution musllsldesign tolerance process capability = 6 defectsdefects -4 -3 -2 -1 0

43、+1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmusl = upper specification limit, , lsl = lower specification limit, .lecture 10richard li, 200941lslusltolerance)()(1lslzfuslzfdefects6_pr_lsluslcapabilityocesstolerancedesigncpcapability index, cp , is defined as)1(kccppk2/lsluslmmkwhere

44、m = nominal process mean, , m = actual process mean, .adjusted capability index, cpk, is defined aswhere usl = upper specification limit, , lsl = lower specification limit, .lecture 10richard li, 200942example #102/lsluslmmk%56. 5%72.47*21)2(21zfdefects6667. 0646lsluslcp6667. 0)01 (6667. 01kccppkfig

45、ure a.6 a normal distribution with m=10, =0.01, usl-lsl=4, m=m .specification mean, m process mean, m usllsl2.28%2.28% -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmlecture 10richard li, 200943example #202/lsluslmmk%26. 0%87.49*21)3(21zfdefects0000.1666lsluslcp0000. 1)01

46、 (0000. 11kccppkfigure a.7 a normal distribution with m=10, =0.01, usl-lsl=6, m=m .usllsl0.13%0.13%specification mean, m process mean, m -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmlecture 10richard li, 200944example #3 5000. 02/412/lsluslmmk%16%87.49%13.341)()3(1zfzfd

47、efects6667.0646lsluslcp3334. 0)5 . 01 (6667. 01kccppkfigure a.8 a normal distribution with m=10, =0.01, usl-lsl=4, m=9.99 .specification mean, m process mean, m usllsl0.13%15.87% -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmlecture 10richard li, 200945example #412/422/l

48、sluslmmk%50%50%01)4()0(1zfzfdefects6667.0646lsluslcp0) 11 (0000. 11kccppkfigure a.9 a normal distribution with m=10, =0.01, usl-lsl=4, m=10.02 .usllsl50%specification mean, m process mean, m -4 -3 -2 -1 0 +1 +2 +3 +4 z =(r-m)/m-4 m-3 m-2 m-1 m m+1 m+2 m+3 m+4 r, ohmlecture 10richard li, 200946* yiel

49、d rate and dpu pdnndpuwhere nd = number of defects found at all acceptance points,np = number of units processed.lecture 10richard li, 200947 !xexpx* poisson distributionpoisson distribution, the occurrence of random defects in manufactured product can be statistically predicted. poisson distributio

50、n is a mathematic model to describe the probability distribution of the arrived entities. for instance, in one hour lunch time, say, statistically from 12.00 to 13.00 oclock, the number of customers getting into a restaurant for lunch. assuming that the average number of customers is , and the rando

51、m number of arrivals at the same time interval is x, then the probability of the random arrival number, px, islecture 10richard li, 200948 dpudpueedpup!000dpuefty !xedpuxpdpux and, in generalthis is the relationship between the defect-free probability and the average dpu. the first time yield, fty,

52、can be approximated by the formula:lecture 10richard li, 200949countpartsdpupartppmratedefectpartppm_/_/ nftydftycftybftyaftyftyrolled*.*regardless of process flow or order, the rolled yield can be calculated from the summation of the dpu values in all the steps.if a process with n steps, 1,2,3,4.n,

53、 and the value of dpu in each step is a, b, c, d, n respectively. then, in terms of the formula, the first time yield, fty, is e-a, e-b, e-c, e-de-n correspondingly. the fist time rolled yield for the process isndcbarolledefty.lecture 10richard li, 200950example : rolled yield for a printed circuit

54、board. assuming that there are totally 3 steps : parts placed, parts soldered, and parts assembly. a) part placement = 300 ppm, it means 300 parts failed in 1million parts.b) part assembly defect rate = 800 ppm, it means 800 assembly-components failed in 1 million parts.c) solder defect rate = 200 p

55、pm, it means 200 connected points failed in 1million parts.*) on average, there are 2.3 connections for each part. table a.1 calculated fty for a printed circuit board assembly by 3 process stepsparts parts parts # of parts placementassemblysoldered totalrolled ftyabc a+b+ce-(a+b+c) 1000.0300.0800.0

56、46 0.15685.6%5000.1500.4000.230 0.78045.8%10000.3000.8000.460 1.56021.0% lecture 10richard li, 200951a.2 table of the normal distriution z 01 234 5 6 7 8 90.0.0000.0040.0080.0120.0160.0199.0239.0279.0319.03590.1.0398.0438.0478.0517.0557.0596.0636.0675.0714.07540.2.0793.0832.0871.0910.0948.0987.1026.1064.1103.11410.3.1179.1217.1255.1293.1331.1368.1406.1443.1480.15170.4.1554.1591.1628.1664.1700.1736.1772.1808.1844.18790.5.1915.1950.1985.2019.2054.2088.2123.2157.2190.22240.6.2258.229