ADE7762ARWZ-RL(Rev0) View Datasheet(PDF) - Analog Devices
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ADE7762ARWZ-RL Datasheet PDF : 28 Pages
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TYPICAL CONNECTION DIAGRAMS
CURRENT CHANNEL CONNECTION
Figure 16 shows a typical connection diagram for the current
channel (IAN). A current transformer (CT) is the current trans-
ducer selected for this example. Notice that the common-mode
voltage for the current channel is AGND and is derived by
center-tapping the burden resistor to AGND. This provides the
complementary analog input signals for IAP and IAN. The CT
turns ratio and Burden Resistor Rb are selected to give a peak
differential voltage of ±500 mV at maximum load.
In theory, it is better to center-tap Rb; however, this requires
very careful attention to the layout and matching of the resistors
to ensure that the channels have the same resistance. A single
resistor may be more practical and is a valid design choice.
CT
Rb
Rf
±500mV
IAP
Cf IAN
IP
Rf
Cf
PHASE NEUTRAL
Figure 16. Typical Connection for Current Channels
VOLTAGE CHANNEL CONNECTION
Figure 17 shows two typical connections for the voltage chan-
nel. The first option uses a potential transformer (PT) to pro-
vide complete isolation from the main voltage. In the second
option, the ADE7762 is biased around the neutral wire, and a
resistor divider is used to provide a voltage signal proportional
to the line voltage. Adjusting the ratio of Ra, Rb, and VR is a
convenient way of carrying out a gain calibration on the meter.
VR can be implemented using either a potentiometer or a
binary weighted series of resistors. Either configuration works,
however, the potentiometer is subject to noise over time. Two
fixed value resistors can be used in place of VR to minimize
the noise.
PT
Rf
±500mV
Rf
VAP
Cf
VN
AGND
Cf
PHASE NEUTRAL
Ra*
Cf
Rb*
VR*
±500mV
VAP
Rf
VN
Cf
PHASE NEUTRAL
*Ra >> Rf + VR; *Rb + VR = Rf
Figure 17. Typical Connections for Voltage Channels
ADE7762
METER CONNECTIONS
In 3-phase service, two main power distribution services exist:
3-phase, 4-wire or 3-phase, 3-wire. The additional wire in the
3-phase, 4-wire arrangement is the neutral wire. The voltage
lines have a phase difference of ±120° (±2π/3 radians) between
each other (see Equation 7).
VA(t )= 2 × VA × cos (ωlt )
VB (t )=
2
× VB
× cos ⎜⎝⎛ ωlt
+
2π
3
⎟⎠⎞
(7)
V C (t )=
2
× VC
×
cos
⎜⎝⎛ ωlt
+
4π
3
⎟⎠⎞
where VA, VB, and VC represent the voltage rms values of the
different phases.
The current inputs are represented by
( ) I A (t ) = 2 I A × cos ωlt + φ A
I B (t ) =
2
IB
× cos
⎜⎝⎛ ωlt
+
2π
3
+
φB
⎟⎠⎞
(8)
IC (t ) =
2
IC
× cos
⎜⎝⎛ ωlt
+
4π
3
+
φC
⎟⎠⎞
where:
IA, IB, and IC represent the rms value of the current of each
phase.
φA, φB, and φC represent the phase difference of the current and
voltage channel of each phase.
The instantaneous powers can then be calculated as follows:
PA(t) = VA(t) × IA(t)
PB(t) = VB(t) × IB(t)
PC(t) = VC(t) × IC(t)
Then,
( ) PA (t ) = VA × I A × cos(φ A ) − VA × I A × cos 2ωlt + φ A
PB (t ) =
VB
×
IB
×
cos(φ B
)
−
VB
×
IB
×
cos
⎜⎝⎛ 2ωl t
+
4π
3
+
φB
⎟⎠⎞
(9)
PC
(t )
= VC
×
IC
×
cos(φC
)
−
VC
×
IC
×
cos
⎜⎝⎛ 2ωl t
+
8π
3
+
φC
⎟⎠⎞
As shown in Equation 9, the active power calculation per phase
is made when current and voltage inputs of one phase are
connected to the same channel (A, B, or C). Then the
summation of each individual active power calculation gives the
total active power information, P(t) = PA(t) + PB(t) + PC(t).
Rev. 0 | Page 15 of 28
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