Microchip Technology ARD00330 データシート
PIC18F87J72 Single-Phase Energy Meter Calibration User’s Guide
DS51964A-page 16
© 2011 Microchip Technology Inc.
EQUATION 2-4:
ENERGY WITHOUT PHASE ERROR AT PF = 0.5 LAG
EQUATION 2-5:
PHASE COMPENSATION ANGLE CALCULATION
EQUATION 2-6:
PHASE COMPENSATION FACTOR CALCULATION
2.3.2.3.2
Phase Compensation between Voltage and Current
To determine the state of the signal between the two successive samples, use the
interpolation method.
If x(Θ-N) and x(Θ) are known, where, “Θ” is the
interpolation method.
If x(Θ-N) and x(Θ) are known, where, “Θ” is the
phase of signal that can vary between
0
o
to 360
o
and “N” is the angle between two successive samples = (360 * frequency of
signal/sampling rate), then it is possible to determine the state of the signal at any inter-
val between ‘Θ-N’ and ‘Θ’ with an assumption that signal varies linearly in the
intermediate region as shown in Equation 2-7.
To apply this method to a line signal, the sampling rate has to be selected such that the
interval between successive samples is small. This ensures that the above stated
assumption holds true.
val between ‘Θ-N’ and ‘Θ’ with an assumption that signal varies linearly in the
intermediate region as shown in Equation 2-7.
To apply this method to a line signal, the sampling rate has to be selected such that the
interval between successive samples is small. This ensures that the above stated
assumption holds true.
Note 1: The Sign convention of Phase Compensation angle is:
• Positive, if the current leads voltage
• Negative, if the current lags voltage
• Negative, if the current lags voltage
Actual Energy at PF = 0.5 lag ( Φ = 60
o
):
E
PF0.5
= V
RMS
x I
RMS
x Cos( ø ) x Period
= V
RMS
x I
RMS
x Cos( 60 ) x Period
Where:
Φ = Input Phase between V and I
Error = (E
PF0.5
- E
PF0.5_error
) * 100 / E
PF0.5
Error/100 = 1 – (E
PF0.5_error
/ E
PF0.5)
Substituting equations and simplify:
Error/100 = 1 – (Cos(60 + β) / Cos(60))
= 1 – (Cos(60 + β) / 0.5)
Cos(60 + β) = 0.5 x (1 + Error/100) = n
β = [COS
-1
(n) – 60] in degree is the phase compensation angle
PHASE_COMPENSATION = β x 128 / Angle between samples
= β x 128 / 5.529
o
Angle between samples = 360
o
/ Samples per sine cycle
= 360
o
/ 65.1
= 5.529
o
Samples per sine cycle = Sampling rate/Frequency of Signal
= 3906 / 60Hz
= 65.1 samples per cycle
Magnitude of Phase compensation factor:
PHASE_COMPENSATION (magnitude) = |PHASE_COMPENSATION |
= PHASE_COMPENSATION x 2