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ADE7759

Active Energy Metering IC with di/dt Sensor Interface

Analog Devices 

ADE7759

V1P

V1

PGA1

V1N

V2P

V2

PGA2

V2N

V2

V1

ADC 1

HPF

20

LPF2

20

ADC 2

0.1؇

1

DELAY BLOCK

1.12␮s/LSB

7

0

11111100

PHCAL [7:0]

–110␮s TO +103␮s

CHANNEL 2 DELAY

REDUCED BY 4.48␮s

(0.1؇ LEAD AT 60Hz)

FCH IN PHCAL [7:0]

V2

V1

60Hz

60Hz

Figure 28. Phase Calibration

0.30

0.25

0.20

0.15

0.10

0.05

0.00

–0.05

–0.10

100 200 300 400 500 600 700 800 900 1000

FREQUENCY – Hz

Figure 29. Combined Phase Response of the HPF and

Phase Compensation (100 Hz to 1 kHz)

0.30

0.25

0.20

0.15

0.10

0.05

0.00

–0.05

–0.10

40

45

50

55

60

65

70

FREQUENCY – Hz

Figure 30. Combined Phase Response of the HPF and

Phase Compensation (40 Hz to 70 Hz)

0.4

0.3

0.2

0.1

0.0

–0.1

–0.2

–0.3

–0.4

54

56

58

60

62

64

66

FREQUENCY – Hz

Figure 31. Combined Gain Response of the HPF and Phase

Compensation (Deviation of Gain in % from Gain at 60 Hz)

ACTIVE POWER CALCULATION

Electrical power is defined as the rate of energy flow from source to

load. It is given by the product of the voltage and current wave-

forms. The resulting waveform is called the instantaneous power

signal, and it is equal to the rate of energy flow at every instant

of time. The unit of power is the watt or joules/second. Equa-

tion 3 gives an expression for the instantaneous power signal in

an ac system.

v(t) = 2 V(ωt)

(1)

i(t) = 2 I sin(ωt)

(2)

where:

V = rms voltage

I = rms current

p(t) = v(t) × i(t)

p(t) = VI – VI cos(2ωt)

(3)

The average power over an integral number of line cycles (n) is

given by the expression in Equation 4.

P

=

1

nT

nT

∫

0

p (t ) dt

= VI

(4)

where T is the line cycle period. P is referred to as the active or

real power. Note that the active power is equal to the dc compo-

nent of the instantaneous power signal p(t) in Equation 3, i.e.,

VI. This is the relationship used to calculate active power in the

ADE7759. The instantaneous power signal p(t) is generated by

multiplying the current and voltage signals. The dc component

of the instantaneous power signal is then extracted by LPF2 (low-

pass filter) to obtain the active power information. This process

is illustrated in Figure 32. Since LPF2 does not have an ideal

“brick wall” frequency response (see Figure 33), the active power

signal will have some ripple due to the instantaneous power

signal. This ripple is sinusoidal and has a frequency equal to twice

the line frequency. Since the ripple is sinusoidal in nature, it will

be removed when the active power signal is integrated to calcu-

late energy—see Energy Calculation section.

–20–

REV. A

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