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STPM10 数据表(PDF) 38 Page - STMicroelectronics |
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STPM10 数据表(HTML) 38 Page - STMicroelectronics |
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38 / 47 page ![]() Theory of operation STPM10 38/47 DocID17728 Rev 5 Equation 11: ����1(����) = �������� �������� ⋅ ∫ i(t) ⋅ dt = − V ⋅ I ⋅ cosϕ 2 − V ⋅ I ⋅ cos(2wt + ϕ) 2 See figure above (9) Equation 12: ����2(����) = ����(����) ⋅ i(t) = − V ⋅ I ⋅ cosϕ 2 − V ⋅ I ⋅ cos(2wt + ϕ) 2 See Figure 26: "Active energy computation diagram" (10) After these two operations, another stage performs the subtraction between the results p2 and p1 and a division by 2, obtaining the active power: Equation 13: p(����) = (����2(t)−����1(t)) 2 = V⋅I⋅cosφ 2 See Figure 26: "Active energy computation diagram" (11) In this way, the AC part V ⋅ I⋅ cos(2ωt + ϕ)/2 has been removed from the instantaneous power. The absence of any AC component allows a very fast calibration procedure. It only requires the setting of (using the internal device programming registers) the voltage and current sensor conversion constants, using the effective voltage and current (Vrms, Irms) readings provided by the device built-in communication port, avoiding the time-averaged readings of the active power or the need for line synchronization. 7.23.2 Reactive power The reactive power is produced using the previously-computed signals. In case of shunt sensor the voltage signal is derived while the current signal is not. A first computation is to multiply the DS value of the integrated voltage channel with the value of the integrated current channel, which yields: Eqaution 14: ����1(����) = ∫ ����(����)�������� ⋅ l(t) = ����(����) ⋅ I(t) = (Vsinωt) ⋅ (− 1 ���� cos(�������� + ����)) = VI 2���� ∙ (���������������� − sin(2�������� + ����)) The second is to multiply the filtered DS value of the voltage channel with the value of the filtered current channel: Equation 15: ����2(����) = ����(����) ⋅ I(t) = Vωcosωt ⋅ Isin(ωt + ����) = �������� 2 ∙ ���� ∙ ((���������������� + sin(2�������� + ����)) From the above results, Q1(t) is proportional to 1/ω, while Q2(t) is proportional to ω. The correct reactive power would result from the following formula: Equation 16: ���� = 1 2 ⋅ ����1(����) ⋅ ω + ����2(t) ⋅ 1 ���� = �������� 2 ���������������� Since the above computation needs a significant additional circuitry, the reactive power in the STPM10 is calculated using only the Q1(t) multipli ed by ω, which means: Equation 17: ����3(����) = 1 2 ⋅ ����1(����) ⋅ ω = �������� 2 ⋅ (���������������� − sin(2�������� + ����)) |
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