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Components and technical solutions for power electronics

Closed loop Hall effect current sensors

ES, ESM, CS, MP and EL sensors

Formulas: | Abbreviations | |

N_{P} x I_{P} = N_{S} x I_{S} V_{A} = e + V_{S} + V_{M}V _{S} = R_{S} x I_{S}V _{M} = R_{M} x I_{S} | N_{P} : turn number of the primary windingI _{P} : primary currentIPN : nominal primary currentN _{S} : turn number of the secondary windingI _{S }: output secondary currentV _{A} : supply voltagee : voltage drop across output transistors (and in the protection diodes, if relevant) V _{S} : voltage drop across secondary windingV _{M} : measuring voltageRS : resistance of the secondary windingR _{M} : measuring resistance |

Values of "e" with a bipolar sensor supply

Sensor | ES100 | ES300...ES2000 | ESM1000 | CS300...CS1000 | CS2000 | MP or EL |

Voltage "e" | 2,5 V | 1 V | 2 V | 2,5 V | 1,5 V | 3 V |

Reminder of the sensor electrical connection

N_{P}/N_{S} | =1/1200 |

I_{PN} | =300A |

R_{S} | =33Ω (at +70°C) |

I_{S} | = 0,15A (at IP_{N}) |

e | =1V |

I_{S} = (N_{P} / N_{S}) x I_{P} | = (1 / 2000) x 520 | = 0,26A peak |

R_{M} = V_{M} / I_{S} | = 8 / 0,26 | = 30,77Ω |

We must check that the sensor can measure these 520A peak, i.e.

:V_{A} ≥ e + V_{S} + V_{M}

If V_{A} = ±15V (±5%), then we must check that

15 x 0,95 ≥ 1 + (33 x 0,26) + 8 which is false since 14,25V< 17,58V

Therefore a supply greater than or equal to 17.58V must be selected. Select a ±24V (±5%) supply.We verify that 24 x 0.95 ≥ 17.58V.__Conclusion:__An ES300C sensor can measure a peak of 520A in the following conditions:

V

R

to obtain an 8V signal at a peak of 520A

R

We must check that the sensor can measure these 520A peak.

V

If V

15 x 0.95 ≥ 1 + (33 x 0.26) + 5 which is false since 14.25V< 14.58V

Therefore a supply greater than or equal to 14.58V must be selected. Select a ±24V (±5%) supply or a ±15V supply with a tighter tolerance, for example ±15V (±2%).

(since 15V x 0.98 > 14.58V)

An ES300C sensor can measure a peak of 520A in the following conditions:

V

R

to obtain a 5V signal at a peak of 520A.

In general, the larger the measuring signal required, the larger the load resistance and the higher the sensor supply voltage should be. The thermal aspect of the sensor should be considered.

For example, the conditions are:

V

R

From the base formulas, we obtain the following formula:

I

Now calculate the equivalent primary current:

I

An ES300C sensor can measure a peak of 552A in the following conditions:

V

R

Note: the 552A peak current must not be a continuous current. For specific requirements, contact your distributor.

R

I

Now calculate the equivalent primary current:

I

V

R

Max. operating temperature = +50°CNote: the 582A peak current must not be a continuous current. For specific requirements, contact your distributor. In general, the lower the ambient temperature, the more important the sensor measurable current. The thermal aspect of the sensor should be considered.

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 300A sensor), then the elements are determined as follows:

I

Now calculate the voltage obtained at the terminals of the measuring resistance:

● for a turn ratio of 1/2000:

V

● for a turn ratio of 1/1500:

V

V

R

V

V

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.

Taking the conditions in point 2.3 again. The calculations were based on a supply voltage of ±15V (±5%).Reworking the calculations with a supply of ±24V (±5%).

From the base formulas, we obtain the following formula:

I

Now calculate the equivalent primary current:

I

VA = ±24V (±5%)

RM = 15Ω

Note: the 908A 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.