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current value measured under 0 - 5 A (ac) with a current transformer

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francisco6329

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Hello friends, I found a signal conditioning circuit for a power transformer to be applied to the analog input of a microcontroller.

the output transformer is an AC voltage with an amplitude in millivolts, but does not understand the circuit to convert that signal into a DC voltage proportional to load current. And there's a part where mixed-voltage AC voltage DC, search the Internet and the circuit is an attenuator.

In the circuit using an operational amplifier LMC6484AIM and datasheet appears in an application where the same situation happens, please attach a picture of the datasheet.

Please help with technical information and documents to help me understand this application of operational amplifiers
 
here are the imagens
 

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The current (proportional to load current) through current transformer secondary winding L1 develops an AC voltage Vmv across burden resistor R4. IC1B is an inverting unity-gain amp which subtracts Vmv from a 0.61V DC offset. IC1C and IC1D are each inverting amplifiers with a gain of 10 and a reference input of 0.61V. The output from IC1D is an AC signal of 0.61-100*Vmv. There is no rectification.
I don't understand what you mean by 'the same situation' in the schematic with LMC6484AIM (Fig 14). In Fig 14 the LM185BX is a 2.048 voltage reference. A fraction of 5-2.048V is tapped off by resistors and applied to the non-inverting input of unity-gain amp LMC6484 to derive a refernce 2.5VDC for the COM pin of the ADC. The top opamp is a unity-gain non-inverting buffer for the input signal Vin, and has an input bias voltage of 2.5VDC.
 
The output from an AC signal is IC1D of 0.61 to 100 * VMV. There is no rectification.

then, is an AC signal that is connected to the analog input of the microcontroller. I have worked in processing the input DC voltage, but never with an AC voltage as the ADC of the microcontroller responds to an AC voltage at the input.
 
0.61 to 100 * VMV
No, not "to". The equation meant "0.61 minus 100*Vmv". So, provided 100*Vmv is less than 0.61 the input to the micro will be positive. But maybe the circuit shown is intended to feed a rectifying stage (not shown) if required?
 
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