Grouping bipartite test method for testing alternating current withstand voltage by wire test instrument

Abstract

The invention relates to the technical field of wire testers, in particular to a grouping and two-way test method for testing alternating current withstand voltage by a wire tester, which comprises the following steps: S1, averagely dividing a wire harness with N test wires into m groups, numbering each group by N1 = N2 =... = Nm, and setting the number of leads in each group to be N / m; s2, alternating current withstand voltage dichotomy measurement is carried out on the m groups of wire harnesses in sequence; and S3, after each group is subjected to the two-way test, carrying out a pair of other tests among the groups. The grouping dichotomy method is adopted in the grouping dichotomy testing method for testing the alternating current withstand voltage of the wire rod testing instrument, the dichotomy testing method is a testing method combining a dichotomy testing method with a pair of other testing methods, the testing efficiency of testing a large-scale wire harness group can be greatly improved, the testing time can be shortened, and meanwhile the grouping dichotomy testing method conforms to the limiting value of testing current.

Description

TECHNICAL FIELD

[0001] The application relates to the technical field of wire testing instruments, and particularly relates to a grouping and bisection testing method for testing AC voltage resistance of a wire testing instrument. BACKGROUND

[0002] Generally, large-scale wire harness testing, such as integrated wire harnesses in the fields of aviation and rail transit, must use wire testing instruments (such as a wire testing instrument shown in FIG. 1) to confirm whether the wire harnesses meet the limit values specified in safety specifications and to avoid major safety accidents caused by the wire harnesses in use. Figure 1 The wire testing instrument includes a CPU processing module, a high-voltage generating module, a channel switching module and the like.

[0003] Therefore, to ensure that each wire is a good product, each wire harness group must be subjected to AC voltage resistance testing (as shown in FIG. 2) before leaving the factory. Figure 2 For example, in the field of aviation, each wire harness group tested at a time contains several hundred or even thousands of wires (as shown in FIG. 3). Figure 3 For example, 1000 wires are tested for 1000V AC voltage resistance, and the AC voltage resistance between every two wires needs to be tested for 60 seconds. Figure 4 If other testing methods are used, that is, one wire is set to low voltage, and all other wires are set to the same high voltage (as shown in FIG. 4), then the AC voltage resistance of each wire to other wires is tested in turn (as shown in FIG. 5 and FIG. 6). Figure 5 Figure 6 Therefore, 1000 wires need to be tested 1000 times * 60 seconds, that is, 16.7 hours. Figure 7 Although the test current is only about 50uA, which meets the testing requirements, the test speed is obviously very slow. Figure 8 If the bisection method is used, that is, half of the wires are set to low voltage, and the other half of the wires are set to high voltage (as shown in FIG. 7), then the bisection testing of the already bisectioned pins is performed (as shown in FIG. 8 and FIG. 9). Figure 9 Therefore, the already bisectioned groups are tested again by bisection every time, and the test speed can be log2 1000 ​*60 seconds, or 10 minutes. However, since the AC withstand voltage test uses high AC voltage, the parasitic capacitance between the wires will exhibit resistivity with respect to the high AC voltage, which is called capacitive reactance. The formula for capacitive reactance is Xc=1 / (2*π*f*C), where f is the frequency of the high AC voltage, which is 50Hz in this case, and C is the parasitic capacitance between the wires, which is approximately 50pF in this case. The calculated capacitive reactance between the two wires is approximately 63.7MΩ. Under a 1000V AC voltage, the current between the two wires is 1000V / 63.7MΩ=15.7uA. Therefore, after 500 equivalent resistors are connected in parallel, the current becomes 500*15.7uA=7.85mA, which is much greater than the 2mA current required for the test, resulting in a poor test result and failure to pass the test.

[0004] Therefore, in order to ensure that the test results meet the current limit, we have to choose another test scheme and conduct long-term tests, which seriously affects the production capacity. Summary of the Invention

[0005] The purpose of this invention is to overcome the deficiencies in the prior art and provide a grouping and binary testing method for testing AC withstand voltage of wire testing instruments that can improve the testing efficiency of large-scale wire harness groups, shorten the testing time, and at the same time meet the test current limit.

[0006] To achieve the above objectives, the present invention provides a grouping and binary testing method for testing AC withstand voltage using a wire testing instrument, comprising the following steps: S1 First, divide a wire harness with N test wires into m groups on average, and number each group N1=N2=…=Nm. The number of wires in each group is N / m. S2 sequentially performs AC withstand voltage bi-tests on these m groups of wire harnesses; S3 After each group has completed the binary test, another pair of tests is performed between the groups.

[0007] In a preferred embodiment of the present invention, the AC withstand voltage bisection measurement method in step S2 is as follows: First, perform a binary search test on N1. Initially, divide N1 into two groups for AC withstand voltage testing, leaving the other groups' wires suspended. Each test lasts for T seconds. After each test, continue binary searching for each already divided pin, and so on, until all pins of N1 have undergone the binary search test. The total time required is T*log2. (N1) Then, the binary search method for N1 is used to perform binary search on N2, and so on, until the binary search for Nm is completed. The final time required for the binary search of Nm is T*log2. (Nm) The binary search test time for all test lines is log2. (N / m) *m*T.

[0008] In a preferred embodiment of the present invention, in step S3, a pair of other test methods are as follows: All wires in group N1 are set to low voltage, and all wires in other groups are set to high voltage. The test time is T. After one test, all wires in group N2 are set to low voltage, and all wires in other groups are set to high voltage. The test time is T, and so on. The time to complete the test of all m groups of wires is m*T. At this time, the test current of each group to the other groups is N / m*(V / (1 / (2*π*f*C))). The total time after all tests are completed is log2. (N / m) *m*T+m*T.

[0009] In summary, the present invention has the following beneficial effects: The grouping binary test method for testing AC withstand voltage of wire testing instruments of the present invention adopts the grouping binary method, which is a test method that combines binary test with a pair of other test methods, which can greatly improve the test efficiency of large-scale wire harness groups, shorten the test time, and at the same time comply with the test current limit value. Attached Figure Description

[0010] Figure 1 This is a schematic diagram of the structure of existing wire testing instruments; Figure 2 This is a schematic diagram of the principle of AC withstand voltage testing using existing wire testing instruments; Figure 3 This is a schematic diagram of the test harness structure when performing AC withstand voltage tests using existing wire testing instruments; Figure 4 This is a schematic diagram of the structure of a test harness in an existing pair of other test methods; Figure 5 It is the principle of an existing pair of other testing methods. Figure 1 ; Figure 6 It is the principle of an existing pair of other testing methods. Figure 2 ; Figure 7 This is a schematic diagram of the test harness structure in the existing bisection test method; Figure 8 The principle of existing bisection testing methods Figure 1 ; Figure 9 The principle of existing bisection testing methods Figure 2 ; Figure 10 This is a flowchart of the grouping and binary testing method for testing AC withstand voltage using wire testing instruments according to the present invention; Figure 11 This is a schematic diagram of the harness grouping structure in the grouped binary test method of the present invention; Figure 12This invention relates to the AC withstand voltage binary measurement principle of the group binary test method. Figure 1 ; Figure 13 This invention relates to the AC withstand voltage binary measurement principle of the group binary test method. Figure 2 ; Figure 14 This invention relates to the AC withstand voltage binary measurement principle of the group binary test method. Figure 3 ; Figure 15 This is a pair of other test principles of the grouping and binary testing method of the present invention. Figure 1 ; Figure 16 This is a pair of other test principles of the grouping and binary testing method of the present invention. Figure 2 . Detailed Implementation

[0011] The preferred embodiments of the present invention will now be described in detail with reference to the accompanying drawings, so that the advantages and features of the present invention can be more easily understood by those skilled in the art, thereby providing a clearer and more explicit definition of the scope of protection of the present invention.

[0012] like Figure 10 The method for grouping and dividing AC withstand voltage testing in a wire testing instrument includes the following steps: S1 First, a wire harness with N test wires is divided into m groups, and each group is numbered N1=N2=…=Nm, with the number of wires in each group being N / m; S2 The AC withstand voltage is measured in two groups sequentially; S3 After each group has completed the two-division test, another test is performed between the groups.

[0013] The present invention will now be described in detail.

[0014] S1 harness grouping: Since testing AC withstand voltage with too many wire harnesses in parallel can lead to excessive current, a wire harness with N test leads can be divided into several groups, i.e., m groups. Each group is numbered, with the first group being N1 (e.g., N1). Figure 11 As shown), that is, N1=N2=…=Nm=N / m, In other words, the number of wires in each group is N / m, and when each group performs an AC withstand voltage test individually, the current limit will not be exceeded due to the resistive characteristics of parasitic capacitance. That is, N / (2*m)*(V / (1 / (2*π*f*C)))<2mA.

[0015] S2 AC withstand voltage bi-point measurement: First, perform a bisection test on N1. Initially, divide N1 into two groups for AC withstand voltage testing, while leaving the wires of the other groups suspended (e.g., ...). Figure 12 As shown), this involves performing a binary search test on N / m wires in group N1, with each test lasting T seconds. After the test is completed, the binary search is then performed again on each of the already binary-split pins (e.g., ...). Figure 13 and Figure 14 As shown), and so on, the time required to complete all binary search tests on N1 is T*log2. (N1) ; Then, following the same binary search method as N1, binary search is performed on N2. The total time required to complete the binary search for N2 is T*log2. (N2) ; Following this logic, the final time required for the Nm binary search test is T*log2. (Nm) ; Since the N wires are divided into m groups, N1 = N2 = ... = Nm, meaning the binary search test time for all test lines is log2. (N / m) *m*T.

[0016] S3 and another pair of tests: After each group has completed the binary test, another test is performed between the groups: all wires in group N1 are set to low voltage, and all wires in other groups are set to high voltage. Figure 15 As shown, the test time is T. After one test, all wires in group N2 are set to low voltage, and all wires in other groups are set to high voltage. Figure 16 As shown, the test time is T, and so on, the time to complete the test of all m groups of wires is m*T; At this point, the test current of each group to other groups is N / m*(V / (1 / (2*π*f*C))). The total time after all tests are completed is log2. (N / m) *m*T+m*T.

[0017] If 1000 wires are divided into 10 groups that meet the test current limit, and each test lasts for 60 seconds, then N, m, and T are substituted into the formula. log2 (N / m) *m*T+m*T log2 (1000 / 10) *10*60+10*60=7*10*60+10*60=4200 seconds+600 seconds=1.33 hours Substitute C=50pF, f=50Hz, N=1000, m=10, V=1000V into the equation. N / (*m)*(V / (1 / (2*π*f*C))) =1000 / 10*(1000 / (1 / (2*3.14159*50*50pF))))=100*15.7uA=1.57mA<2mA, which meets the test requirements; The 1.33 hours is far less than the 16.7 hours required for a pair of other methods. Although the time is longer than the 10 minutes of the traditional binary method, the test current can meet the limit value, making it an efficient and feasible test method.

[0018] The following examples will illustrate this point.

[0019] There is a wire harness consisting of 320 test leads. The AC withstand voltage test current is required to be less than 1mA, each test lasts 60 seconds, the AC voltage is 750V, and the capacitive reactance of the wires is calculated to be 63.7MΩ. Using one pair of other testing methods, the test current is approximately 0.01mA per test, and the total test time is 60 * 320 = 19200 seconds = 5.3 hours. Using the binary divider method, the test current is approximately... 320 / 2*(750 / 63.7MΩ) = 1.8mA > 1mA, which does not meet the test requirements; use the grouped binary test method as follows: First, divide the 320 test lines into 10 test groups consisting of 32 test lines each, which is N=320, m=10, N1=N2=…=N10=32, T=60 seconds as described in the above invention.

[0020] First, a binary test is performed on N1, while N2, N3, ... N10 are all set to floating. The current of N1 in each test is approximately 32 / 2*(750 / 63.MΩ) = 0.188mA, which meets the requirement of 1mA test current. The time to complete the binary test of N1 is 60*log2. (32) =60*4=240 seconds.

[0021] Then, following the test procedure for N1, tests were performed on other test groups such as N2, N3, etc. The current for each test was approximately 0.2mA, all meeting the requirement of a 1mA test current. The time to complete the binary test for each group was 60*log2. (32) =60*4=240 seconds.

[0022] After testing all the test groups, the total testing time was 240 * 10 = 2400 seconds = 0.7 hours, while all of them met the requirement of 1mA test current.

[0023] After all groups completed the binary test, another pair of tests were conducted between the groups. The test current for each test was approximately 32 / (750 / 63.MΩ) = 0.38mA, which also meets the requirement of 1mA test current. Each test lasted 60 seconds, and a total of 10 tests were conducted between the groups, which is 60*10=600 seconds.

[0024] Therefore, the total comprehensive test time is 2400 + 600 = 3000 seconds, or 0.83 hours, which is much shorter than the traditional 5.3-hour test time for another pair. The maximum test current is also within 0.4mA and does not exceed 1mA, which meets the test requirements.

[0025] The above are all preferred embodiments of this application, and are not intended to limit the scope of protection of this application. Therefore, all equivalent changes made in accordance with the structure, shape, principle and application direction of this application should be covered within the scope of protection of this application.

Claims

1. A group two division test method for wire testing instrument to test AC voltage resistance, characterized in that, It comprises the following steps: S1 first divide a harness with N test lines into m groups, number each group N1=N2=…=Nm, and the number of lines in each group is N / m; S2 perform AC voltage withstand two-division measurement on the m groups of harnesses in turn; S3 after two-division measurement is performed on each group, perform one pair of other tests between groups.

2. The group two division test method for AC voltage withstanding test of wire testing instrument according to claim 1, characterized in that, In step S2, the AC voltage withstand two-division measurement method is as follows: First, the N1 is tested by dichotomy, the first time, the N1 is divided into two groups, the other group of conductors are suspended, each test time is T seconds, after the test, each group is tested by dichotomy, and so on, until the N1 is tested by dichotomy, the time required is T*log2 (N1) Then, the N2 is tested by dichotomy according to the dichotomy test method of N1, and so on, until the Nm is tested by dichotomy, the time required for the final Nm dichotomy test is T*log2 (Nm) The dichotomy test time of all test lines is log2 (N / m) *m*T.

3. The group two-part test method for AC voltage withstanding test of a wire testing instrument according to claim 2, wherein, In step S3, the one pair of other test method is as follows: The N1 group of all leads is set to low voltage, and the other group of leads is set to high voltage, and the test time is T. After testing once, the N2 group of all leads is set to low voltage, and the other group of leads is set to high voltage, and the test time is T. In this way, the m group of leads is tested completely, and the time is m*T. At this time, the test current of each group to the other group is N / m*(V / (1 / (2*pi*f*C)), and the total time after all the tests are completed is log2 (N / m) *m*T+m*T.

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