Page 105 - Current & voltage sensors catalog
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Calculation guide
         Closed loop Hall effect current sensors

         ES, CS, MP and EL sensors


                                 2.5 - What influence does the turn ratio have on the sensor's performance?
                                 Taking the conditions of point 2.3 again. The calculations were based on a turn ratio of 1/2000. If this
                                 ratio is 1/1500 (non standard ratio for a 300 A sensor), then the elements are determined as follows:
                          1SBC146057V0014  I S = (N P / N S) x I P   = (1 / 1500) x 552  = 0.368 peak   (I P = 522 A from 2.3 above)
                                 Now calculate the voltage obtained at the terminals of the measuring resistance:
                                 ●  for a turn ratio of 1/2000:
                                  V M = R M x I S   = 15 x 0.276   = 4.14 V
                                 ●  for a turn ratio of 1/1500:
         ES300C
                                  V M = R M x I S   = 15 x 0.368   = 5.52 V
                                 Conclusion
                                                                                                                    4
                                 An ES300C sensor can measure a peak of 552 A in the following conditions
                                 V A = ±15 V (±5%)
                                 R M = 15 Ω
                                 V M = 4.14 V with a turn ratio of 1/2000
                                 V M = 5.52 V with a turn ratio of 1/1500
                                 In general, the lower the turn ratio, the more important the output current and the higher the measuring
                                 voltage. The thermal aspect of the sensor should be considered.
                                 2.6 - What influence does the supply voltage have on the sensor's performance?
                                 Taking the conditions in point 2.3 again. The calculations were based on a supply voltage of ±15 V (±5%).
                                 Reworking the calculations with a supply of ±24 V (±5%).
                                 From the base formulas, we obtain the following formula:
                                 I SMAX = (V AMIN - e) / (R S + R M)   = [(24 x 0.95) – 1] / (33 + 15)  = 0.454 A peak
                                 Now calculate the equivalent primary current:
                                 I P = (N S / N P) x I S   = (2000 / 1) x 0.454 = 908 A peak
                                 Conclusion
                                 An ES300C sensor can measure a peak of 908 A in the following conditions:
                                 V A = ±24 V (±5%)
                                 R M = 15 Ω
                                 Note: the 908 A peak current must not be a continuous current.
                                 In general, the higher the supply voltage, the more important the measuring current and the higher the
                                 measuring voltage. The thermal aspect of the sensor should be considered.
                                 NB: for calculations with unipolar supply (e.g. 0…+24 V), contact your distributor.


























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