Page 142 - Six Sigma for electronics design and manufacturing

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Determining the Manufacturing Yield and Test Strategy
Solution method 1. Calculating total yield using nDPU
–{(solder DPU · n · 2.5) + assembly DPU · n + component DPU · n}
FTY {total} = e
–nDPU
# Parts
Solder
Assembly Component Total nDPU FTY = e
defects
defects
yield
n
defects
defects
100
0.03
0.105
0.05
0.025
90%
0.525
59%
0.15
500
0.25
0.125
0.25
35%
0.3
1.05
0.5
1000
Solution method 2. Calculating the total yield by multiplying
individual process yields
Solder Assembly Component Total yield
# Parts yield yield yield Y(solder) · Y(assembly)
n e –ndpu e –ndpu e –ndpu · Y(component)
100 0.975 0.951 0.97 90%
500 0.882 0.779 0.861 59%
1000 0.779 0.606 0.741 35%
Solution method 3. Calculating the total yield using power se-
ries expansion. In this method, the solution is derived by calculat-
ing the total yield by multiplying individual process yields based on
1 – DPU component expansion, where DPU is the process defect rate for
one component. Note that the defect rate for the PCB operations
should not be used, because some of the values are too high (i.e., the
DPU for total assembly defects is 0.5) to ignore the higher-order
terms in the power expansion.
Solder Assembly Component Total yield
# Parts yield yield yield Y(solder) · Y(assembly)
n (1 – DPU) n (1 – DPU) n (1 – DPU) n · Y(component)
100 0.975 0.951 0.97 90%
500 0.882 0.779 0.861 59%
1000 0.779 0.606 0.741 35%
The total yield results using all three methods of calculations men-
tioned above were approximately equal in values.
It can be shown that as the number of components increases in the
PCBs, first-time yields decrease significantly, assuming that the qual-
ity level of the assembly process remains the same. In order to achieve
higher first-time yields for complex PCBs of more than 500 parts, the
quality level of the assembly process steps has to be improved from
hundreds of PPM defects to tens of PPM defects.