Page 638 - Design for Six Sigma a Roadmap for Product Development

Tolerance Design 591

100
y
R (2 fL ) 2
2
Assume that f 50 Hz, the nominal value for the resistor T R 9.5 , the tol-
erance R 1.0 , the nominal value for the inductor T L 0.01 H, and the
tolerance L 0.006 H. Assume that for R and L, C p 1.33. Finally, assume
that the tolerance R can be reduced by 0.5 with an additional cost of $0.15
and that L can be reduced by 0.003 H with an additional cost of $0.20.
The customer’s tolerance for the circuit current is y 10.0 1.0 A, and
we would like to satisfy this requirement with C p 2.0. So the required
standard deviation for the current y is

1.0
0
req 0.1667
3C p 3
2.0
The actual variance of circuit current is approximately
2 2 2 2
∂f
∂f
2
L
R
∂L
∂R
where
R 1.0
R 0.25
3C p 3
1.333
0.006
L
L 0.0015
3C p 3
1.333
2
2
2
(100R)
(100(2 f) L)
2 , 2
∂f
∂f

2 3
2
2 3
2
∂R (R (2 fL) ) ∂L (R (2 fL) )
When R 9.5, L 0.01 and
2 0.899, 2 9708.2
∂f
∂f

∂L
∂R
So
2 2 2 2 2 2
∂f
∂f
2
0.899
0.25 9708.2
0.0015 0.078
R
L
∂L
∂R
0.078 0.27
Because req , the current circuit will not be able to satisfy the customer
requirement with C p 2.
We can compute
0.15
"C R
p R 0.316
("f) 2 R 0.474