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SA4301A Datasheet(PDF) 18 Page - Sames

Part # SA4301A
Description  Polyphase Energy Metering IC with Pulse Outputs
PDF  23 Pages
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Manufacturer  SAMES [Sames]
Direct Link  http://www.sames.co.za
Logo SAMES - Sames

SA4301A Datasheet(HTML) 18 Page - Sames

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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����٠= ��������
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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