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SA4301A Datasheet(PDF) 18 Page - Sames |
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SA4301A Datasheet(HTML) 18 Page - Sames |
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18 / 23 page ![]() SPEC-1538 (REV. 7) 18/23 29-09-2017 SA4301A TYPICAL APPLICATION The following description outlines the basic process required to design a typical three phase energy meter using the SA4301A. The meter is a 3-phase 4-wire meter capable of measuring 3x220V/60A/50Hz with a precision better than Class 1. The meter uses a stepper motor counter with 100imp/kWh and the calibration LED has a constant of 800imp/kWh. The most important external circuits required for the SA4301A are the current input networks, the voltage input networks as well as the bias resistor. All resistors should be 1% metal film resistors of the same type to minimize temperature effects. Bias Resistor A bias resistor of R34 = 47k sets optimum bias and reference currents on chip. Calibration of the meter should be done using the voltage inputs and not by means of the bias resistor. Current Input Networks Three current transformers are used to measure the three line currents. The output of each current transformer is terminated with a low impedance resistor split into two equal parts to obtain purely differential current input signals. The voltage across the termination resistors is converted to the required differential input currents through the current input resistors. Anti-alias filters are incorporated on these input resistors to filter any high frequency signal components that could affect the performance of the SA4301A. The voltage drop across the current transformer termination resistors at maximum rated current should be in the order of 100mVRMS. The current transformers have a low phase shift and a turns ratio of 1:2500. The value of the termination resistors R1, R2 is therefore ����1 = ����2 = 100�������� × ������������ ���������������� × 1 2 ≈ 2Ω = �������� where NCT is the current transformer ratio (2500) and IMAX is the maximum input current (60A). The four current input resistors (R3, R4, R5, R6) should be of equal size to optimize the input networks low pass filtering characteristics, so the values can be calculated as follows: ����3 = ����4 = ����5 = ����6 = ���������������� ������������ × �������� 2 × 16�������� = 1.5����Ω = �������� For optimum performance the cut-off frequency of the anti- alias filter should be between 10kHz and 20kHz. The equivalent resistance associated with each capacitor is RC/2 so the capacitor values should be in the order of ����1 = ����2 = 1 ������������������������ = 1 ���� × 10������������ × 1.5����Ω ≈ 22�������� = �������� where fCI is the cut-off frequency of the anti-alias filter of the current input network. The current input networks for channel 2 and channel 3 are identical. Voltage Input Networks The voltage sense inputs require an input current of 11μARMS at VNOM (220V) according to Table 1. The mains voltage is divided by means of a voltage divider to a lower voltage that is converted to the required input current by means of the input resistor. Once again an anti-alias filter is required to remove any high frequency signals that could affect the performance of the SA4301A. The phase shift of the current transformers is compensated by means of this anti-alias filter as well, by purposefully increasing the cut-off frequency. The input resistor R22 sets the current input into the device. This resistor should not be too large else the capacitor for the anti-alias filter will be quite small which could cause inaccurate phase shift due to parasitic capacitances. Therefore R22 = 100k is chosen and the voltage at the centre of the trimpot should be 1.1V (11μA x 100k ). The calibration range of the voltage input network should be about ±15% to ensure that all component tolerances can be catered for, so the total tuning range can be set to ±0.17V. Therefore the voltage across the trimpot and R23 is 1.27V. Choosing a 1k trimpot results in ����23 = 1����Ω (2 × 0.17) × (1.27 − 2 × 0.17) ≈ 2.7����Ω The effect of R22 can be ignored in the above equation, given the fact that R22 is significantly larger than P1 and R23. Now let RA = R19 + R20 + R21 and �������� = (����1 + ����23) × ( 220���� 1.27���� − 1) ≈ 637����Ω so choose R19 = 240k , R20 = 220k and R21 = 180k. |
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