CALCULATE CFDEN VALUE FOR DESIGN
WRITE CFDEN VALUE TO CFDEN REGISTER
ADDR. 0x15 = CFDEN
SET ITEST = Ib, VTEST = VNOM, PF = 1
SET HALF LINE CYCLES FOR ACCUMULATION
IN LINECYC REGISTER ADDR. 0x1C
SET MODE FOR LINE CYCLE
ACCUMULATION ADDR. 0x09 = 0x0080
ENABLE LINE CYCLE ACCUMULATION
INTERRUPT ADDR. 0x0A = 0x04
RESET THE INTERRUPT STATUS
READ REGISTER ADDR. 0x0C
NO
INTERRUPT?
YES
RESET THE INTERRUPT STATUS
READ REGISTER ADDR. 0x0C
NO
INTERRUPT?
YES
READ LINE ACCUMULATION ENERGY
ADDR. 0x04
CALCULATE WGAIN. SEE EQUATION 43.
WRITE WGAIN VALUE TO THE WGAIN
REGISTER: ADDR. 0x12
Figure 72. Calibrating Watt Gain Using an Accurate Source
Equation 43 describes the relationship between the expected
LAENERGY value and the LAENERGY measured in the test
condition:
WGAIN
=
INT
⎜⎛
⎜
⎝
⎜⎛
⎜⎝
LAENERGYIB(expected )
LAENERGYIB (nominal )
−
1⎟⎟⎠⎞
×
212
⎟⎞
⎟
⎠
(43)
The nominal LAENERGY reading, LAENERGYIB(nominal), is the
LAENERGY reading with the test current applied. The expected
LAENERGY reading is calculated from the following equation:
ADE7763
LAENERGYIB(expected) =
⎜⎛
⎟⎞
INT
⎜
⎜
⎜⎜⎝
CFIB(expected) × Accumulation Time(s)
CFNUM + 1
× WDIV
CFDEN + 1
⎟
⎟
⎟⎟⎠
(44)
where CFIB(expected) (Hz) is calculated from Equation 30,
accumulation time is calculated from Equation 33, and the line
period is determined from the period register according to
Equation 34.
For this example:
Meter Constant:
MeterConstant(imp/Wh) = 3.2
Test Current:
Ib = 10 A
Line Voltage:
Vnominal = 220 V
Line Frequency:
fl = 50 Hz
Half Line Cycles:
LINECYCIB = 2000
CF Numerator:
CFNUM = 0
CF Denominator:
CFDEN = 489
Energy Reading at Base Current:
LAENERGYIB (nominal) = 17174
Period Register Reading: PERIOD = 8959
Clock Frequency:
CLKIN = 3.579545 MHz
CFexpected is calculated to be 1.9556 Hz according to Equation 30.
LAENERGYexpected is calculated to be 19186 using Equation 44.
CFIB(expected)(Hz) =
3.200 imp/Wh × 220 V × 10 A
× cos(ϕ) = 1.9556 Hz
3600 s/h
LAENERGYIB(expected) =
INT
⎜⎛
⎟⎞
⎜ CFIB(expected) × LINECYC IB / 2 × PERIOD × 8 / CLKIN ⎟
⎜
⎜⎜⎝
CFNUM + 1
× WDIV
CFDEN + 1
⎟
⎟⎟⎠
LAENERGYIB(expected) =
INT
⎜⎛
⎜
⎜
1.9556
×
2000
/
2
×
8959
1
×
8
/(3.579545
×
10
6
)
⎟⎞
⎟⎟1
=
⎜
⎝
489 + 1
⎟
⎠
INT(19186.4) = 19186
Rev. A | Page 37 of 56