Analyze Test Confidence to Enhance Throughput
Christopher Jones, M/A-COM Division of AMP, Lowell, MA -- Test & Measurement World, 9/1/1999
| At M/A-COM’s Semiconductor Business Unit, we test more than 4 million RF ICs per week. A major factor in reaching this capacity has been our attention to improving test yield. We have observed that poor device-lead to test-board contact is a major cause of low yield and is a common issue for RF devices and plunge-to-board testing. To increase our yield, we needed to improve our “test confidence”—the probability of making a good connection. We determine test confidence by measuring and monitoring connection repeatability, and we use it to qualify new automated part handlers and test interfaces. The test-confidence measurement is based on the relationship between the yields of a prime test and a retest of devices that failed the prime test. Devices that pass the retest do so because their performance is borderline or because of poor connection repeatability. You can eliminate borderline devices from the retest yield by setting specification limits that account for tester variation.The retest yield will then depend on connection repeatability alone. Your test confidence will determine the value of retest. Consider the following example: Assume 100 parts that have an absolute yield of 90% are entered into a test in which your test confidence is 80%. The first-pass test results will equal the product of parts, absolute yield, and confidence, or Overall, you passed 86 parts out of a possible 90, and you performed 128 tests to measure 100 parts. From an operations point of view, you rejected four good parts and increased test time by 28%. This is the effect of low test confidence. This example is expanded in Figure 1, which shows the effects of varying test confidence on the cumulative yield for a three-pass test—an initial test and two retests. Low test confidence clearly leads to yield loss and excessive retesting. Our experience suggests these rules of thumb:
Deriving Test Confidence
As discussed in the previous example, the probability that a device passes the first test, P(T1P), is not equal to the absolute yield. The first-pass yield, Y1, equals the probability that a device is good times the test confidence, C,
and the number of devices to pass the first test, T1P, is equal to the test quantity times the yield:
As a result, the first-pass yield equals the absolute yield scaled down by the test confidence. Inverting this relationship, you can solve for the absolute yield:
Thus, given a known confidence rating you can forecast the absolute yield at the end of the first-pass test and determine if the value of the second-pass test is worth the effort of performing it. You can continue to solve for test confidence by looking at the second-pass test. Test 2 sample size, S2, is equal to the original quantity less the devices that passed the first test:
The probability that a device is good, P(S2G), is equal to the ratio of good devices to total devices:
Given these equations, you can solve for test confidence in measurable terms. Begin with the last equation in terms of test confidence:
and substitute for P(S2G) and then T1P:
Factor S1 and substitute for Y:
Then solve for test confidence:
Simplifying the Analysis
Christopher Jones is test engineering manager at AMP MA/COM’s Semiconductor Business Unit in Lowell, MA. He received his B.S.E.E. from the University of Massachusetts Amherst and an M.S. in Manufacturing Engineering from Boston University. | ||||
| Copyright 1999, Test & Measurement World. Published by Cahners Business Information, Newton, MA. |





















