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Components and technical solutions for power electronics

Closed loop Hall effect voltage sensors

EM010 sensors

**1 - Reminder of the key elements (closed loop Hall effect)**

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} R = R _{E} + R_{P} R = U _{P} / I_{P} | N_{P} : turn number of the primary windingU _{P} : primary voltageI _{P} : primary currentI _{PN} : 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 resistanceR _{E }: external resistance in series with the primary circuit of the voltage sensorR _{P} : internal resistance of the primary winding |

Values of "e" with a bipolar sensor supply

Sensor | EM100 |

Voltage "e" | 1,5 V |

Reminder of the sensor electrical connection

N_{P}/N_{S} | =10000/2000 |

I_{PN} | =10mA |

R_{P} | = 1500Ω (at +25°C) |

R_{S} | =60Ω (at +70°C) |

I_{S} | = 50mA (at I_{PN}) |

e | =1,5V |

I_{S} = (N_{P} / N_{S}) x I_{P} | = (10000 / 2000) x 0,012 | = 0,060A peak |

R_{M} = V_{M} / I_{S} | = 10 / 0,060 | = 166,67Ω |

We must check that the sensor can measure these I_{P} = 12mA peak, i.e.

:V_{A} ≥ e + V_{S} + V_{M}

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

15 x 0,95 ≥ 1.5 + (60 x 0,060) + 10 which is false since 14,25V< 15,10V

Therefore a supply greater than or equal to 15.10V must be selected. Select a ±24V (±5%) supply.We verify that 24 x 0.95 ≥ 15.10V.__Conclusion:__An EM010BBFHP1N sensor can measure a peak of 12mA in the following conditions:

V

R

to obtain an 10V signal at a peak of 12mA (V

Note: the 12mA peak current must not be a continuous current. For specific requirements, contact your distributor.

In the same way as for closed loop Hall effect current sensors (see page 109), if the required measuring voltage is reduced, carefully check that the ±15V (±5%) supply used in this example is sufficient to obtain a 5V signal with the conditions used in the preceding point.

15 x 0.95 > 1.5 + (60 x 0.060) + 5 which is true since 14.25V > 10.10

A closed loop Hall effect sensor is extremely sensitive to thermal aspects.In general, a voltage sensor can withstand the following variations in primary current:

● Up to 110% of the nominal primary current: continuous overload possible

● Up to 125% of the nominal primary current: overload 3min/hr possible

● Up to 150% of the nominal primary current: overload 50sec/hr possible

In all these cases, we recommend that you contact your distributor in order to obtain detailed information on this subject.

In the same way as for closed loop Hall effect current sensors (see page 109), if the maximum operating temperature of the sensor is reduced, the measurable primary current (and therefore the primary voltage) of the voltage sensor increases. The thermal aspect of the sensor should be considered.

For closed loop Hall effect voltage sensors, the turn ratio has a significant influence on the sensor's operation:

● Output current value

● Thermal capacity

● Maximum frequency of the measuring voltage

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.

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...+24V), contact your distributor.

N_{P}/N_{S} | =1000/2000 |

I_{PN} | =10mA |

R_{P} | = 1500Ω (at +25°C) |

R_{S} | = 60Ω (at +70°C) |

I_{S} | = 50mA (at I_{PN}) |

e | = 1,5V |

**3.1 - What primary resistance R _{E} is required in series with the sensor to obtain a primary current IP = 12mA when the primary voltage UP = 1500V?**

R = R_{E }+ R_{P} and R = U_{P} / I_{P} | therefore R_{E} = (U_{P} / I_{P}) - R_{P} |

R_{E }= (1500 / 0,012) - 1500 | i.e. R_{E} = 123,50kΩ |

**3.2 - What power is required for the primary resistance R _{E} added in series with the sensor?**

Taking the same conditions as point 3.1 above.

P

P

For obvious reliability reasons, select a resistance with a nominal power of at least 5 times this calculated power, i.e. approx 90W.

Taking the same conditions as point 3.1 above.

The sensor's ambient temperature can vary the resistance of the primary winding, therefore if the sensor's operating temperature is 50°C, the difference will have to be treated as follows:

R

By redoing the calculations with R