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Components and technical solutions for power electronics

	Formulas:	Abbreviations
	NP x IP = NS x IS  VA = e + VS + VM VS = RS x IS VM = RM x IS	NP : turn number of the primary winding IP  : primary currentIPN : nominal primary current NS : turn number of the secondary winding IS : output secondary current VA : supply voltage e : voltage drop across output transistors   (and in the protection diodes, if relevant) VS : voltage drop across secondary winding VM : measuring voltageRS : resistance of the secondary winding RM : measuring resistance

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_A = ±24V (±5%)
R_M = 30,77Ω
to obtain an 8V signal at a peak of 520A

2.2 - What are the consequences, if the required signal is only 5V?

R_M = V_M / I_S                    = 5 / 0,26  =                                        19,23Ω
We must check that the sensor can measure these 520A peak.
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) + 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)

Conclusion :
An ES300C sensor can measure a peak of 520A in the following conditions:
V_A = ±15V (±2%)
R_M = 19,23Ω
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.

2.3 - What is the maximum current measurable by an ES300C in specific conditions?

For example, the conditions are:
V_A = ±15V (±5%)
R_M = 15Ω
From the base formulas, we obtain the following formula:
I_SMAX = (V_AMIN - e) / (R_S + R_M)                     = [(15 x 0,95) - 1] / (33 + 15)           = 0,276A peak
Now calculate the equivalent primary current:
I_P = (N_S / N_P) x I_S                                        = (2000 / 1) x 0,276                         = 552A peak

Conclusion :
An ES300C sensor can measure a peak of 552A in the following conditions:
V_A = ±15V (±5%)
R_M = 15Ω
Note: the 552A peak current must not be a continuous current. For specific requirements, contact your distributor.

2.4 - What influence does the ambient temperature have on the sensor's performance?
Taking the conditions from point 2.3 (preceding example). The calculations were made using a maximum default operating temperature of +70°C. If this maximum temperature is +50°C, then the measuringrange can be increased as follows:

R_S = 33Ω at +70°C                           At +50°C, R_S = 30,5Ω                         then,
I_SMAX = (V_AMIN - e) / (R_S + R_M)         = [(15 x 0,95) - 1] / (30,5 + 15)           = 0,291A peak
Now calculate the equivalent primary current:
I_P = (N_S / N_P) x I_S                            = (2000 / 1) x 0,291                           = 582A peak

Conclusion :
An ES300C sensor can measure a peak of 582A in the following conditions:
V_A = ±15V (±5%)
R_M = 15Ω
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.


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 300A sensor), then the elements are determined as follows:
I_S = (N_P / N_S) x I_P           = (1 / 1500) x 552            = 0,368 peak             (I_P = 522A 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                     = 4,14V
●  for a turn ratio of 1/1500:
V_M = R_M x I_S                  = 15 x 0,368                     = 5,52V

Conclusion :
An ES300C sensor can measure a peak of 552A in the following conditions
V_A = ±15V (±5%)
R_M = 15Ω
V_M = 4.14V with a turn ratio of 1/2000
V_M = 5.52V 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 ±15V (±5%).Reworking the calculations with a supply of ±24V (±5%).

From the base formulas, we obtain the following formula:
I_SMAX = (V_{A MIN} - e) / (R_S + R_M)          = [(24 x 0,95) - 1] / (33 + 15)         = 0,454A peak
Now calculate the equivalent primary current:
I_P = (N_S / N_P) x I_S                              = (2000 / 1) x 0,454                       = 908A peak

Conclusion :
An ES300C sensor can measure a peak of 908A in the following conditions:
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.