This might be a silly question. I saw this from "Opamp For Everyone" and was a bit confused with this circuit. The book doesn't mention the signal source at all, I think it does have a signal input, whose low frequency is filtered by C1 at first. The non-inverting input of the first op-amp has the exact reference voltage buffered by the second op-amp. C3 provides a decoupling effect to suppress current from entering the first op-amp. The first op-amp has a gain of -RF/RG which causes the output to have a magnified input with a 180-degree phase shift if it is by itself with a grounded non-inverting input. Because now the non-inverting input of the first op-amp has the reference voltage, the first op-amp is actually a differential op-amp in this circuit? What is the point for ADC's input if the input is minus by reference voltage?
Best Answer
not at all
Yes, the signal is connected to C1 and this constitutes a high pass filter with RG. But its primary purpose is to AC-couple the input. This way the circuit can arrange its DC biasing without having to take into account some unknown DC voltage from the source.
There are only two opamps. It appears you’re calling the top one the 2nd, the bottom one the 3rd. The bottom opamp is buffering the Vref from the ADC.
No, the input current of opamps is essentially zero. C3 combines with R1 to form a low pass filter to remove noise from Vref. The bottom opamp was put in so the Vref pins wouldn’t have to drive the RC filter directly.
Yes
Not really. The voltage of the two opamp inputs will be equal, so the inverting input will be fixed at a DC value of Vref. Because the opamp input draws no current, and C1 blocks any DC current, there is no DC current through RF, so the DC voltage on the other side of RF will be Vref too. The AC signal will be superimposed on that (after it is inverted and multiplied by RF/RIN).
The ADC input is actually the (inverted and gained) input plus the reference voltage.