MIL-HDBK-217电路实例计算MTBF

1、MIL-HDBK-217电路实例计算MTBFMIL HDBK 217 實例計算簡介(一) 參考電路：(二)零件計數法可靠度預估(PARTS COUNT)1. 用於設計初期估計裝備總失效率，適於招標或欲簽訂單時的可靠度預估。2.必須資料a. 零件屬性與型式b. 該零件使用數目c. 零件品質水準d. 裝置所在的環境3. 零件失效率計算：I=nepuip=ni(g ) I=1g=零件基本的失效率(f/106 hrs)=零件的品質因數Ni=第i 類零件的數量4. 例題計算：Gf=40(Ta)a.r =5個(0.018 f/106 hrs 3)=0.27 f/106hrsb.ce =4個(0.290 f

2、/106 hrs 3)=3.48 f/106 hrsc.tr =2個(0.016 f/106 hrs 10)=0.32 f/106 hrsd.diode =2個(0.0031 f/106 hrs 6)=0.031 f/106 hrse.led =1個(0.033 f/106 hrs 10)=0.33 f/106 hrsequip=r+ce+tr+diode led=4.431f/106 hrs(三)零件應力分析法可靠度預估(PART STRESS ANALYSIS)1. TRANSISTOR 類a. 求 Q1 之 Imax = V / R3 = 10V / 39K = 0.256 mAPw =

3、 V I = 0.5V 0.256 = 0.128 mWI 取 RMS 值時 ( 0.256 / 1.414 = 0.18 mW )Pw = 0.18 ( 10- 0.18 39) = 0.54mWST = 0.54mW / 400 mW = 0.0014 ( C. F. )C.F = ( Tmax 25 ) / 150 = 100 25 / 150 = 0.5T = 40 0C + ( 175 100 ) = 115 0C查T = 115 0C, ST = 0.1 得b = 0.0019 f / 106hrs查e = 5.8 ( Gf ) a = 1.5 ( Linear )q = 12 (

4、 PLASTIC ) r = 1 ( s2 =Applied Vce / Vceo = 10V / 60V = 20%得 0.3c = 1.0 ( Single transistor ) p = b ( e a q r s2 c) p = 0.0019 ( 5.8 1.5 12 1 0.3 1.0 ) = 0.05951 f / 106hrsb. 求 Q2 之 Imax = ( Vdc Vf ) / 1K = 8 mAPw = V I = 8 0.5 = 4 mWI 取 RMS 值時 ( 8 / 1.414 = 5.65 mA )Pw = 5.65 ( 10- 2 5.65 1) = 13.3

5、mWST = 13.3mW / 400 mW = 0.033 ( C. F. )C. F. 同 Q1 所以 T = 115 0C查 T = 115 0C，ST = 0.1 得 b = 0.0019f / 106hrs查 e = 5.8 ( Gf ) a = 1.5 ( Linear ) q = 12 ( PLASTIC ) r = 1 ( s2 = Applied Vce / Vceo = 10V / 60V = 20% 得 0.3c = 1.0 ( Single transistor ) p q2 = b ( e a q r s2 c) p q2 = 0.0019 ( 5.8 1.5 12

6、1 0.3 1.0 ) = 0.05951 f / 106hrsC. 求晶體之總失效率p( Q )=pq1+pq 2 = 0.05951 + 0.05951= 0.11902f / 106hrs2.LED p(led) = be(環境) t(溫度) q (品質)查表得 b = 0.00065 f / 106hrs ( Single LED )t=LED=Ta+300C=Tj=700C 查表得 430e = 2.4 ( Gf )q = 0.5 ( LOWER ) ， 1.0 (PLASTIC) p ( led ) = 0.00065 430 1.0 2.4= 0.6708 f / 106hrs3

7、.Diode p ( diode ) = b e r q a c s2 ST = Imax / 110 mA = 0.256 / 110 = 0.002C.F. = 1 所以查表得 b = 0.00025e = 3.9 ( Gf )r = 1.0 ( q = 7.5 ( Lower )a = 1.0 ( Analog circuit c = 1.0 ( metallurgically bonded )s2 = appliedVr/rated Vr 100% =10/35 = 30% = 0.7 p (diode)=0.000253.91.07.51.01.00.72個= 0.01024 f /

8、 106hrs4.Resiste ra. R1 Prl = I2 R1 =10V/( R1+R2) 2 680K=0.14 mWST = 0.14mW / 250mW = 0.0056 (ST = 0.1 T = 40 0C ) 查表得 b = 0.00088 f / 106hrs e = 2.4 ( Gf ) r = 1.1 ( 0.1 - 1 Mohm ) q = 15.0 ( Lower ) p ( r1 ) = b e r q= 0.00088 2.4 15 1.1= 0.03485 f / 106hrsb. R2 Pr2 = 10 / ( 680 + 43) 2 43 = 0.008

9、 mWST = 0.008 / 250mW = 0.00033查表得b = 0.00088 f / 106hrse = 2.4 ( Gf )r = 1.0 ( 0 100Kohm )q = 15.0 ( Lower )p (r2) = b e r q= 0.00088 2.4 15 1.0= 0.03168 f / 106hrsc. R3 同上方法求得p ( r3 ) = b e r q= 0.00088 2.4 15 1.0= 0.03168 f / 106hrsd. R4 同上方法求得p(r4)= b e r q= 0.00088 2.4 15 1.0= 0.03168 f /106hrs

10、e. R5 Pr5 = (10-2) / 1K2 1K = 64mWST = 64mW / 250 mW = 0.256 取 ST = 0.3 T = 40 o C 查表得 b = 0.0011 f / 106hrse = 2.4 ( Gf )r = 1.0 (0 100Kohm)q = 15.0 ( Lower)p (r5) = b e r q= 0.0011 2.4 15 1.0= 0.0396 f / 106hrsf.求電阻之總失效率p (r) =p(rl)+p (r2) + p(r3) + p (r4) + p (r5)= 0.03485 + 0.03168 + 0.03168 + 0

11、.03168 + 0.0396 = 0.1695 f / 106hrsCE(電解電容器) 以25V電解計算時- (括號內則以16V 電解計算時)a. C1 ST= Vapplied / Vrated = 5 / 25 = 0.2 - ( 5 /16 = 0.312 )b = 0.019 - ( 0.025 )e = 2.4 (Gf)q = 10 (lower)cv = 0.4 ( p (c1) = b e q cv= 0.019 2.4 10 0.4= 0.1824 f / 106hrs - ( 0.24 ) b. C2 ST = 0.02V / 25V = 0.001 - ( 0.001)b

12、 = 0.018 - ( 0.018 )e = 2.4 ( Gf )q = 10 ( lower)cv = 0.4 ( p (c1) = b e q cv= 0.018 2.4 10 0.4= 0.1728 f / 106hrs - ( 0.1728 ) c. C3 ST = 10 / 25 = 0.4 - ( 0.7 )b = 0.025 - ( 0.055 )e = 2.4 ( Gf )q = 10 ( lower)cv = 0.4 ( p (c1) = b e q cv= 0.025 2.4 10 0.4= 0.24 f / 106hrs - ( 0.528 )d. C4 ST = 5

13、/ 25 = 0.2 - - ( 0.4 )b =0.019 - ( 0.025 ) e = 2.4 ( Gf ) q = 10 ( lower) cv = 0.4 ( p (c1) = b e q cv= 0.019 2.4 10 0.4= 0.1824 f / 106hrs - ( 0.24 )e.求電解電容器總失效率 p (c)= p (cl)+p (c2) + p(c2) + p (c3) + p (c4)= 0.1824 + 0.1728 + 0.24 + 0.1824= 0. 7776 f / 106hrs - ( 1.1808)5.電路總失效率p(total)=p(c)+p(r) + p(diode) +p(led)+p(Q)= 0.7776 + 0.1695 + 0.010