1 00:00:18,516 --> 00:00:23,437 Estimating Fs and ALF for Various CTs 2 00:00:23,984 --> 00:00:33,203 In measurement CTs (Current Transformers), a parameter called Fs (Security Factor) exists 3 00:00:33,516 --> 00:00:38,515 which is used to describe the CT and its accuracy class 4 00:00:38,984 --> 00:00:51,406 According to the IEC60044-1 standard, Fs is defined such that at a current of Fs times the rated current 5 00:00:51,720 --> 00:01:04,687 the composite error in the secondary is 10%. For example, for a CT with a class of 0.5Fs10, at 10 times the rated current 6 00:01:04,922 --> 00:01:11,640 the composite error is 10%. On the CT nameplate provided in this example 7 00:01:11,797 --> 00:01:18,515 Core number one is a measurement type and its class is 0.2Fs10 8 00:01:19,375 --> 00:01:27,109 . This means that at 10 times the rated current, the composite error at the output is 10% 9 00:01:27,422 --> 00:01:35,000 The 0.2 figure before Fs represents the accuracy class of the CT at the rated current 10 00:01:35,468 --> 00:01:42,733 This means that at 100% and 120% of the rated current on the primary side 11 00:01:42,891 --> 00:01:48,906 the ratio error at the secondary terminal is less than 0.2%. 12 00:01:53,047 --> 00:02:00,630 For protective CTs, a similar concept exists known as ALF (Accuracy Limit Factor) 13 00:02:00,842 --> 00:02:11,411 To explain this parameter, consider the previous example. Core number 2 is a protective core with a class of 5P20 14 00:02:11,724 --> 00:02:24,224 In this example, ALF = 20, and according to the definition provided by IEC60044-1, at 20 times the rated current 15 00:02:24,693 --> 00:02:35,240 the composite error is 5%. For a more detailed explanation, consider a protective CT with a class of 10P5 16 00:02:35,709 --> 00:02:46,568 In this case, ALF = 5, and according to the definition, at 5 times the rated current, the composite error is 10%. 17 00:02:47,271 --> 00:02:58,443 One key point about these two parameters is that for measurement CTs, the actual Fs is less than the value indicated on the nameplate 18 00:02:58,834 --> 00:03:06,724 whereas for protective CTs, the actual ALF is greater than the value shown on the nameplate 19 00:03:07,428 --> 00:03:10,084 Ratio Error and Composite Error 20 00:03:10,240 --> 00:03:20,865 According to the definition provided in the IEC60044-1 standard, the ratio error is represented as follows: 21 00:03:21,021 --> 00:03:27,427 ALF or Fs is where the composite error is 5% or 10%. 22 00:03:28,053 --> 00:03:44,069 According to the definition, ALF or FS is where Ipse/Ipn equals 5% or 10% of Ip/N . 23 00:03:45,459 --> 00:04:01,161 In indirect method according to the definition of ALF or FS which are also show as ALFi and FSi 24 00:04:02,880 --> 00:04:14,677 is where of Iext/Iout=0.1 .The same method is also used in the software 25 00:04:14,912 --> 00:04:19,130 Software Method for Calculating ALF and Fs 26 00:04:19,365 --> 00:04:30,068 In the software, to estimate Fs and ALF, the saturation curve and the relationship between the current and voltage of the parallel branch are used. 27 00:04:30,693 --> 00:04:34,599 To calculate the composite error using the indirect method (disregarding the ratio error) 28 00:04:34,834 --> 00:04:42,255 it can be said that the parallel branch current is the cause of the composite error between the primary and secondary sides 29 00:04:42,490 --> 00:04:54,755 Therefore, to ensure that the composite error at Fs times the rated current is 10%, the parallel branch current must be 10% of the output current 30 00:04:55,414 --> 00:05:03,695 In such conditions, the parallel branch voltage is also equal to Fs times the parallel branch voltage 31 00:05:03,930 --> 00:05:09,945 under conditions where the rated current passes through the CT terminal at the nominal load: 32 00:05:10,257 --> 00:05:18,694 According to the provided relationship, for a specified ALF (Fs), there is a line with a constant slope 33 00:05:19,164 --> 00:05:27,132 The intersection of this line with the saturation curve indicates the point where the CT’s composite error is 10%. 34 00:05:28,695 --> 00:05:40,961 By dividing the obtained voltage by the nominal impedance, the output current is obtained, and its ratio to the rated current will correspond to ALF (Fs)