Page 136 - Six Sigma for electronics design and manufacturing

Determining the Manufacturing Yield and Test Strategy
105
x
p(x, ) = e ( /x!)
–
0
(4.3)
–
P (at least 1 defect) = 1 – P (no defects or X = 0) = 1 – e
( /0!)
–
–
First time yield = FTY = 1 – P (at least 1 defect) = 1 – (1 – e ) = e
(4.4)
Since = np = DPU
FTY = e
–DPU
(4.5)
When an assembly is made from similar parts or operations, such as
the transistors in an IC or soldering in a PCB, then the FTY for the
assembly can be derived from the total DPUs of the individual opera-
tions. Sometimes, this yield is referred to as total yield (Y T ) or assem-
bly estimated yield (Y A ) to distinguish it from FTY. It can be derived a
follows:
Y T = Y A = e – DPU (4.6)
In six sigma quality, the DPUs are very small, and approximations
can be performed without sacrificing the accuracy of the yield esti-
mates. In this case, the general equation for yield can be further sim-
plified by the power series expansion of exponential functions:
3
FTY = e –DPU = 1 – DPU/1! + DPU /2! – DPU /3! + DPU /4! + . . .
4
2
+ (–1) n+1 DPU /n! (4.7)
n
Since the DPU is small in six sigma quality (0.000034), we can ig-
nore all the terms beyond the first two:
FTY = 1 – DPU = 1 – (# of defects/# of opportunities) (4.8)
and
Y T = Y A = (1 – DPU) n
where n is the number of operations to be analyzed for defects.
4.2.1 Example of calculating yield in a part with
multiple operations
In Figure 4.1, the wire bonding of an IC is shown. The chip is centered
in the middle of the IC package frame, and wires are bonded from the
chip to the frame. There are two bonds per IC termination. If there
are 256 connections in the IC frame, and the bonding operation DPU
is 100 PPM, what is the FTY for the bonding of an IC?
There are three methods of calculating the FTY, either by using the
Poisson distribution [Equation (4.6)], the first two terms of the expo-