Circuit on Figure1 shows a linear transfer of AC voltage to isolated output as AC voltage and not as DC equivalent. All normal spikes on AC mains will pass to output and can be measured with A/D in any manner you vant as shown on Figure2.
Do not be confused with ground symbol on Figure1. All what is on right side of optocoupler is galvanically isolated from mains. Only first OPA and left side of optocoupler together with transformerless supply are connected to mains.
RS485 line can be additionally isolated with optocouplers.
As you wish. IL300 itself is nonlinear but it has first photodiode connected as negative feedback in first OPA which linearizes transfer characteristic of optocoupler. LED diode current in optocoupler has such shape that Ip2 current has exact shape uf mains voltage. Nonlinear distortion is 0.08%. Last paragraph of the article describes overall parameters of this circuit:
The main specifications of the circuit are 5300V-ac-rms galvanic isolation, 0.08% linearity, 470-ppm/°C thermal shifts in VOUT, 2° phase shift at 50 Hz, and dc to 1-kHz bandwidth at –3 dB. If you connect the output to a 10-bit A/D converter, one LSB is equivalent to 0.5V in the 110V power line.
hope to be closer to 0.1V resolution
1st you must be given an accuracy class.
I assume class that requires 0.2% max error under all conditions.
Dielectric strength : IEC 688 2.5KV AC rms 1 minute between input terminals & case.
This can be achieved with dual opto isolators to linearize one with the other using a MUX'd DAC output to result in zero difference for each polarity and phase. 0.1% resistor values are required.
In case this idea has not been suggested, consider making an expanded scale.
For instance, if voltage will never drop below 90 V, then set up your measuring system to respond only to volt levels from 90 to 120V.
You're right. You can use the 3 phases to a Buck supply for micro in case of loss of any phase or better have battery backup.
Then use BiPhase code thru a 3kV signal transformer so there is no DC signal and it is self clocking.
Does anyone know what the capacitors accomplish in post # 9?
Probably a safety factor, to provide galvanic isolation (sort of) from mains. I add 'sort of' because if they fail by shorting, you lose the safety factor.
They need to be rated to withstand peak mains voltage (and add another 50 percent for greater safety).
They are 100 uF. A large enough value, so that they pass the low frequencies of mains AC. In that sense it is still possible for your devices to be exposed to full mains voltage, if your project is not hooked up properly to mains wiring.
AC coupling to shift the signal to the voltage range of the single supply amplifier. The 100 µF, resulting in high-pass cut-off frequency of 0.05 Hz has been apparently chosen for very low phase error, I wonder if it's really necessary. If so, you'll better go for a DC coupled amplifier.Does anyone know what the capacitors accomplish in post # 9?
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