ADE9078ACPZ datasheet(57/108 Pages) AD

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ADE9078

Data Sheet

Rev. 0 | Page 56 of 107

QUICK START

There are a few important steps to note when using the

ADE9078 IC.

For most applications, ensure that the PM1 and PM0 pins are

low to enter normal measurement mode (PSM0).

The following initialization sequence is recommended:

1. Wait for the RSTDONE interrupt, indicated by the IRQ1 pin

going low.

2. Configure the xIGAIN, xVGAIN, and xPGAIN registers

via the SPI to calibrate the measurements.

3. If other calibration values are required, for example, to

improve rms performance at low input signal levels, write

these registers.

4. If the CFx pulse output is used, configure the CFxDEN and

xTHR registers.

5. Configure the expected fundamental frequency (50 Hz or

60 Hz network) in the SELFREQ bit and write VLEVEL =

0x117514.

6. If a Rogowski coil sensor is used, write the INTEN bit in

the CONFIG0 register to enable the digital integrator on

the IA, IB, and IC channels. To enable the digital integrator on

the neutral current, IN, channel, set the ININTEN bit.

Additionally, write DICOEF = 0xFFFFE000 to configure

the digital integrator. If current transformers are used,

INTEN and ININTEN in the CONFIG0 register must = 0.

7. If the service bring measured is something other than

4-wire wye, see Table 24 to determine how to configure

ICONSEL and VCONSEL in the ACCMODE register.

8. Write a 1 to the run register.

9. Write a 1 to the EP_CFG register.

The ADE9078 IC sampling capacitors vary device to device (see

Table 1). For this reason, gain calibration is required to be able to

accurately measure connected loads. If a current transformer

sensor is used, phase calibration is required to remove any device

to device variation in the phase error to accurately measure loads

over power factor.

Use the following example to determine if the ADE9078 IC is

correctly measuring the input voltage signal.

In this example, a 1 MΩ and 1 kΩ resistor divider network

measures the voltage between the Phase A voltage and the neutral.

If the input signal is 240 V rms, the expected voltage at the input to

the ADE9078 IC is 240 V rms ×1000/(1000 + 1,000,000) =

0.2397 V rms. The ADE9078 ADC full-scale input is ±1 V,

0.707 V rms. Thus, 0.2397 V rms/0.707 V rms = 33.9% of full scale.

It is recommended to scale the nominal voltage input to about ½ of

full scale to allow room for overvoltage events. As described in the

Filter-Based Total RMS section, the full-scale voltage rms register

output reading is given as 52,866,837d. Thus, with this 33.9% of

full-scale input, the expected VRMS register reading is 17,921,858.

Note that the actual xVRMS register reading varies based on the

external component gain error, combined with the ADE9078 IC

device to device gain error. Assume that 18,000,000d is read when

the 240 V rms load was applied. Thus, there are 18,000,000 output

codes per 240 V rms, which means there are 75,000 output codes

per volt. Take the xVRMS register reading and divide by 75,000 to

determine the voltage in volts.

Volts = xVRMS/75,000

A similar exercise can be performed to determine if the ADE9078

IC is correctly measuring the input power.

For example, if the same 240 V rms signal is applied along with

a 10 A load, the applied power is 240 V × 10 A = 2.4 kW. Assuming

that the 10 A load is connected to a current transformer with

1000:1 turn ratio on the secondary side, the current is 10 mA.

Assume a center tapped burden resistor is used so that there is

10 Ω total burden resistance. Thus, 10 mA × 10 Ω yields a 0.1 V

rms signal. The ADE9078 IC allows full-scale inputs of ±1 V,

0.707 V rms, which means that the 10 A input is 0.1 V

rms/0.707 V rms = 14.1% of full scale.

For the active power measurement, with a 240 V × 10 A load, the

ADE9078 IC sees 33.9% of full scale on the voltage side and 14.1%

on the current side, so 33.9%×14.1% = 4.8% of the full-scale output

power. As described in the Total Active Power section, the xWATT

applied, 4.8% × 20,823,646 = 999,535 is the expected register

reading. Assume that 1,000,000d is read when the 240 V rms, 10 A

load is applied. Thus, there are 1,000,000 output codes per 2.4 kW,

which means there are 416,667 output codes per kW. Read the

xWATT register and divide by 416,667 to determine the power

in watts.

Watts = xWATT/416,667

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