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ADE7762
TRANSFER FUNCTION
FREQUENCY OUTPUTS F1 AND F2
The ADE7762 calculates the product of six voltage signals (on
current channel and voltage channel) and then low-pass filters
this product to extract active power information. This active
power information is then converted to a frequency. The
Example 1
In this example, with ac voltages of ±500 mV peak applied to
the voltage channels and current channels, the expected output
frequency is calculated as follows:
f 1 to 7 = 0 . 56 Hz , SCF = S0 = S1 = 1
= 500 mV peak ac =
V REF = 2 . 4 V ( nominal reference value )
frequency information is output on F1 and F2 in the form of
active high pulses. The pulse rate at these outputs is relatively
low, for example, 2.09 Hz maximum for ac signals with SCF =
S0 = 0; S1 = 1 (see Table 6). This means that the frequency at
these outputs is generated from active power information
V AN = V BN = V CN = IA = IB = IC
0 . 5
V rms
2
(13)
6 . 313 × ( V AN × I A + V BN × I B + V CN × I C ) × f 1 to 7
Freq =
V REF
accumulated over a relatively long period. The result is an
output frequency that is proportional to the average active
power. The averaging of the active power signal is implicit to
the digital-to-frequency conversion. The output frequency or
pulse rate is related to the input voltage signals by the following
equation:
2
(12)
where:
Freq is the output frequency on F1 and F2 (Hz).
V AN , V BN , and V CN are the differential rms voltage signal on
voltage channels (V).
I A , I B , and I C are the differential rms voltage signal on current
channels (V).
V REF is the reference voltage (2.4 V ± 8%) (V).
f 1 to 7 is one of seven possible frequencies selected by using the
logic inputs SCF, S0, and S1 (see Table 5).
Note that if the on-chip reference is used, actual output fre-
quencies can vary from device to device due to a reference
tolerance of ±8%.
6 . 313 × 0 . 5 × 0 . 5 × 0 . 58
Freq = 3 × = 0 . 230 Hz (14)
2 × 2 × 2 . 4 2
As can be seen from these two example calculations, the maximum
output frequency for ac inputs is always half of that for dc input
signals. The maximum frequency also depends on the number
of phases connected to the ADE7762. In a 3-phase, 3-wire delta
service, the maximum output frequency is different from the maxi-
mum output frequency in a 3-phase, 4-wire Wye service. The
reason is that there are only two phases connected to the analog
inputs, but also that in a delta service, the current channel input
and voltage channel input of the same phase are not in phase in
normal operation.
Example 2
In this example, the ADE7762 is connected to a 3-phase, 3-wire
delta service as shown in Figure 18. The total active energy
calculation processed in the ADE7762 can be expressed as
Table 5. f 1 to 7 Frequency Selection 1
Total Active Power = ( V A ? V C ) × I A + ( V B ? V C ) × I B
(15)
SCF
0
1
0
1
0
1
0
S1
0
0
0
0
1
1
1
S0
0
0
1
1
0
0
1
f 1 to 7 (Hz)
2.24
4.49
1.12
4.49
5.09
1.12
0.56
where:
V A , V B , and V C represent the voltage on Phase A, Phase B, and
Phase C, respectively.
I A and I B represent the current on Phase A and Phase B,
respectively.
With respect to the voltage and current inputs in Equation 7
and Equation 8, the total active power (P) is
P = ( V A ? V C ) × ( I AP ? I AN ) + ( V B ? V C ) × ( I BP ? I BN )
P = ? 2 × V A × cos ( ω l t ) ? 2 × V C × cos ? ? ω l t +
? ? ×
1
1 1 1 0.56
f 1 to 7 is a fraction of the master clock and therefore varies if the specified
CLKIN frequency is altered.
?
2 × I A × cos ( ω l t ) +
? ?
4 π ? ?
3 ? ?
? 2 × V B × cos ? ω l t +
? ? v 2 × V C × cos ? ω l t +
4 π ? ?
? ? ×
? ?
? ?
2 π ? ?
3 ? ?
3 ? ?
2 × I B × cos ? ? ω l t +
2 π ?
3 ?
?
?
(16)
Rev. 0 | Page 23 of 28
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