ADE7761ARS-REF View Datasheet(PDF) - Analog Devices

Part Name

Description

Manufacturer

ADE7761ARS-REF

Energy Metering IC with On-Chip Fault and Missing Neutral Detection

Analog Devices 

ADE7761

The low frequency output of the ADE7761 is generated by

accumulating this active power information. This low frequency

inherently means a long accumulation time between output

pulses. The output frequency is, therefore, proportional to the

average active power. This average active power information can

in turn be accumulated (for example, by a counter) to generate

active energy information. Because of its high output frequency

and therefore shorter integration time, the CF output is propor-

tional to the instantaneous active power. This is useful for

system calibration purposes that would take place under steady

load conditions.

INSTANTANEOUS

POWER SIGNAL

INSTANTANEOUS

ACTIVE POWER SIGNAL

V× I

2

0V

CURRENT

VOLTAGE

INSTANTANEOUS INSTANTANEOUS

POWER SIGNAL ACTIVE POWER SIGNAL

CH1

E CH2

ADC

HPF

MULTIPLIER

ADC

DIGITAL-TO-

FREQUENCY

F1

F2

LPF

DIGITAL-TO-

FREQUENCY

CF

INSTANTANEOUS

POWER SIGNAL –p(t)

INSTANTANEOUS

ACTIVE POWER SIGNAL

T V× I

E TIME

p(t) = i(t).v(t)

WHERE:

v(t) = V × cos(ϖt)

V×I

i(t) = I × cos(ϖt)

2

p(t) = V × I {1 + cos (2ϖt)}

2

L Figure 20. Signal Processing Block Diagram

Power Factor Considerations

The method used to extract the active power information from

O the instantaneous power signal (by low-pass filtering) is still

valid even when the voltage and current signals are not in

phase. Figure 21 displays the unity power factor condition and

a displacement power factor (DPF = 0.5), that is, current signal

S lagging the voltage by 60°. If one assumes that the voltage and

current waveforms are sinusoidal, the active power component

of the instantaneous power signal (dc term) is given by

B (V × I/2) × cos(60°)

O This is the correct active power calculation.

V× I

× cos(60°)

2

0V

VOLTAGE

CURRENT

60°

Figure 21. Active Power Calculation over PF

Nonsinusoidal Voltage and Current

The active power calculation method also holds true for

nonsinusoidal current and voltage waveforms. All voltage and

current waveforms in practical applications have some

harmonic content. Using the Fourier transform, instantaneous

voltage and current waveforms can be expressed in terms of

their harmonic content:

∞

v(t) = VO + 2 × ∑Vh ×sin(hωt + αh )

(1)

h≠0

where:

v(t) is the instantaneous voltage.

VO is the average value.

Vh is the rms value of voltage harmonic h.

αh is the phase angle of the voltage harmonic.

∞

i(t) = IO + 2 × ∑ Ih ×sin(hωt + βh )

(2)

h≠0

where:

i(t) is the instantaneous current.

IO is the dc component.

Ih is the rms value of current harmonic h.

βh is the phase angle of the current harmonic.

Rev. A | Page 16 of 28

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