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ADE7760ARSRL 数据表(PDF) 14 Page - Analog Devices |
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ADE7760ARSRL 数据表(HTML) 14 Page - Analog Devices |
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14 / 24 page ![]() ADE7760 Rev. 0 | Page 14 of 24 Power Factor Considerations The method used to extract the active power information from the instantaneous power signal (by low-pass filtering) is still valid even when the voltage and current signals are not in phase. Figure 18 displays the unity power factor condition and a displacement power factor (DPF = 0.5), that is, current signal lagging the voltage by 60°. If one assumes the voltage and current waveforms are sinusoidal, the active power component of the instantaneous power signal (dc term) is given by (V × I/2) × cos(60°). This is the correct active power calculation. INSTANTANEOUS POWER SIGNAL INSTANTANEOUS ACTIVE POWER SIGNAL INSTANTANEOUS POWER SIGNAL INSTANTANEOUS ACTIVE POWER SIGNAL 60° CURRENT CURRENT VOLTAGE 0V 0V VOLTAGE V× I 2 V× I 2 × cos(60°) Figure 18. 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 har- monic content. Using the Fourier transform, instantaneous voltage and current waveforms can be expressed in terms of their harmonic content: ) sin( 2 ) ( 0 h h h o t h V V t V α + ω × × + = ∑ ∞ ≠ (1) where: v(t) is the instantaneous voltage. Vh is the rms value of voltage harmonic h. αh is the phase angle of the voltage harmonic. ) sin( 2 ) ( 0 h h h o t h I I t i β + ω × × + = ∑ ∞ ≠ (2) 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. Using Equations 1 and 2, the active power P can be expressed in terms of its fundamental active power (P1) and harmonic active power (PH): H 1 P P P + = where: 1 1 1 1 1 1 1 I V P β − α = Φ Φ × = ) cos( (3) and h h h h h h h H I V P β − α = Φ Φ × × = ∑ ∞ = ) cos( 2 (4) As can be seen from Equation 4, a harmonic active power component is generated for every harmonic, provided that harmonic is present in both the voltage and current waveforms. The power factor calculation has previously been shown to be accurate in the case of a pure sinusoid; therefore, the harmonic active power must also correctly account for power factor, because it is made up of a series of pure sinusoids. Note that the input bandwidth of the analog inputs is 7 kHz with the internal oscillator frequency of 450 kHz. HPF and Offset Effects Equation 5 shows the effect of offset on the active power calculation. Figure 19 shows the effect of offsets on the active power calculation in the frequency domain. ) cos( ) cos( 2 )) cos( ( )) cos( ( ) ( ) ( 0 0 1 1 0 t I V t I V I V I V t I I t V V t I t V 1 1 1 0 1 1 0 ω × × + ω × × + × + × = ω × + × ω × + = × (5) As can be seen from Equation 5 and Figure 19, an offset on Channel 1 and Channel 2 contributes a dc component after multiplication. Because this dc component is extracted by the LPF and used to generate the active power information, the offsets contribute a constant error to the active power calcula- tion. This problem is easily avoided in the ADE7760 with the HPF in Channel 1. By removing the offset from at least one channel, no error component can be generated at dc by the multiplication. Error terms at cos(ωt) are removed by the LPF and the digital-to-frequency conversion (see the Digital-to- Frequency Conversion section). |
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