# FMCC EUROPE Components and technical solutions for power electronics

Calculation guide
Closed loop Hall effect voltage sensors
EM010 sensors

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

 Formulas: Abbreviations NP x IP = NS x IS VA = e + VS + VMVS = RS x ISVM = RM x IS R = RE + RP R = UP / IP NP : turn number of the primary windingUP : primary voltageIP  : primary currentIPN : nominal primary currentNS : turn number of the secondary windingIS : output secondary currentVA : supply voltagee : voltage drop across output transistors   (and in the protection diodes, if relevant)VS : voltage drop across secondary windingVM : measuring voltageRS : resistance of the secondary windingRM : measuring resistanceRE :  external resistance in series with the primary circuit of the voltage sensorRP : 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

2 - Measurement circuit calculation (secondary part of the sensor)

Example with an EM010 sensor without primary resistance supplied with the sensor (EM010BBFHP1N)

 NP/NS =10000/2000 IPN =10mA RP = 1500Ω (at +25°C) RS =60Ω (at +70°C) IS = 50mA (at IPN) e =1,5V

2.1 -  What load resistance (RM) is required to obtain a 10V measuring signal (VM = 10V) when the IP current = 12mA peak?

 IS = (NP / NS) x IP = (10000 / 2000) x 0,012 = 0,060A peak RM = VM / IS = 10 / 0,060 = 166,67Ω

We must check that the sensor can measure these IP = 12mA peak, i.e.
:VA ≥ e + VS + VM

If VA = ±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:
VA = ±24V (±5%)
RM = 166,67Ω
to obtain an 10V signal at a peak of 12mA (VM = 10V for IP = 12 mA)
Note: the 12mA peak current must not be a continuous current. For specific requirements, contact your distributor.

2.2 - What are the consequences, if the required signal is only 5V(VM = 5V)?
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

2.3 - What is the maximum measurable current by an EM010BBFHP1N in these specific conditions?

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.

2.4 - What influence does the ambient temperature have on the sensor's performance?

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.

2.5 - What influence does the turn ratio have on the sensor's performance?

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.

2.6 - What influence does the supply voltage have on the sensor's performance?

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.

3 - Sensor primary circuit calculation

Example with an EM010 sensor without primary resistance supplied with the sensor (EM010BBFHP1N)

 NP/NS =1000/2000 IPN =10mA RP = 1500Ω (at +25°C) RS = 60Ω (at +70°C) IS = 50mA (at IPN) e = 1,5V

3.1 - What primary resistance RE is required in series with the sensor to obtain a primary current IP = 12mA when the primary voltage UP = 1500V?

 R = RE + RP and R = UP / IP therefore RE = (UP / IP) - RP RE = (1500 / 0,012) - 1500 i.e. RE = 123,50kΩ

3.2 - What power is required for the primary resistance RE added in series with the sensor?

Taking the same conditions as point 3.1 above.

PRE is the power dissipated in the resistance RE.
PRE = RE x IP2 = 123 500 x 0,0122 = 17,8W
For obvious reliability reasons, select a resistance with a nominal power of at least 5 times this calculated power, i.e. approx 90W.

3.3 - What influence does the temperature have on the determination of the primary resistance RE to be connected in series with the sensor?

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:
RP = 1500Ω at +25°C gives a resistance of 1642Ω at +50°C.
By redoing the calculations with RP = 1642Ω, we obtain RE = 123.36kΩ, i.e. a difference of 0.1%.The ambient temperature has only a very little influence on the calculation of primary resistance.