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AN2288 Scheda tecnica(PDF) 7 Page - STMicroelectronics |
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AN2288 Scheda tecnica(HTML) 7 Page - STMicroelectronics |
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7 / 29 page ![]() AN2288 Theory 7/29 So, the average active energy over a time period with N samples is: There is delay ( δ) between each voltage and current sample, due to this the real phase shift angle ( α) between v(t) and i(t) is different than theoretical phase shift angle (φ). So, there is a discrete error ( ε d): The average ε d will be 0. So, Ed = E. There is also error due to non-simultaneous sampling, which is shift error ( ε ns): The total active energy including not simultaneous sampling will be: 3.1 Analog front end The ADC used in this application is only one 10-bit SAR ADC for current as well as voltage sampling. The voltage is reduced by potential divider and lifted up by Vdd/2 (e.g. 2.5V) and then given to one of the ADC channels of ST7Lite20 microcontroller to get the digital converted value after every 200us. The Lite2 timer RTC2 interrupt is used to get 200us. The current value is measured using a current transformer with a 1:5000 turn ratio and with 36 Ohm shunt resistance. The voltage induced across the current transformer is lifted up by 3VDD/4 (e.g. 3.75V) to make the induced voltage unipolar (because the microcontroller works in one direction only). Two analog switches are used for tampering detection phenomena. The output of both switches are connected, but the inputs are different, one from phase line and the other one from neutral line. The output goes to gain multipliers e.g. operational amplifiers. There are four gain multipliers implemented by three operation amplifiers and with one analog switch. The gain factors are x2, x8, x32 and x128. The x32 and x128 are multiplexed using one op-amp and one analog switch. The output of the gain multipliers goes to the ADC channels of the microcontroller. The active channel for current is selected based on current range. E d t sampling vk ik × K=0 N-1 ∑ = ε d VI × 2N × ------------- 2 α 2Kωt φ – + () cos K=0 N-1 ∑ = P d P ε d + = ε ns VI × 2 ----------- δφ () sin δ N ---- 2 α 2Kωt φ – + () cos K0 = N1 – ∑ + = P ns P ε ns ε d ++ = P ns P ε ns + = E ns E d ε d ∑ + = |
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