This is a reference buffer circuit that make a single-end audio input to be the differential inputs for ADC.

Circuit-1: ToBeAnalyzedCircuit

It does not like typical Op-Amp circuit of the same purpose like below.

Circuit-2: TypicalOpAmpCircuit

I understand how the circuit-2 works, but I can not figure out how the circuit-1 works. How the input signal be transformed into a pair of inverted signals?

What is the OP-Amp configuration of IC1 and IC2?

Dose the IC1 use a non-inverting amplifier configuration with negative feed-back?

If so, for audio signal the C1 will short-cut the output and negative-input, then why we need the R2 and and R1 resistor?

Which configuration of IC2 is used, the non-inverting amplifier or the inverting amplifier configuration?

Could you explain the work of R1,R2,R3,R4,R5 and C1,C2,C3,C4?


  • \$\begingroup\$ Fig 1 is a standard '2-amplifier Instrumentation amplifier' circuit. Now you know what to search for, you should find plenty of resources on line. \$\endgroup\$ – Neil_UK May 23 '19 at 15:52
  • \$\begingroup\$ @Neil_UK thanks for the clue, it's very useful! \$\endgroup\$ – Bob Yang May 23 '19 at 16:03

What is the OP-Amp configuration of IC1 and IC2?

IC1 is a voltage follower, its kind of hard to see but this question has more info.

IC2 is simply an inverter that is inverting the output of IC1 because the ADC takes a differential input. Essentially this circuit takes an AC signal and buffers it for the ADC and adds Vcc/2 of offset to the DC portion of the signal presumably to get it to a Vcc to 0V range of the ADC.

The difference between the pass band of IC1 and a voltage follower is this:

enter image description here

R1 really only kind of buffers IC1 from the ADC at high frequencies, it changes the impedance at high frequencies that the ADC sees. The ADC sees less of C1, for some ADC's this makes a difference in sampling error.

Could you explain the work of R1,R2,R3,R4,R5 and C1,C2,C3,C4?

C3 is a high frequency filter for the inputs of the ADC C4 is to provide a very high low pass filter with the resistor bridge formed by the 3.3k resistors. A better (but more expensive way) is to use a high precision voltage reference instead of a voltage divider.

R3 is for a DC pathway to the reference voltage, which the 1uf cap will need because IC1's input impedance is high. It also fixes IC1's output to Vcc/2.


The overall circuit is a 2-amplifier Instrumentation amplifier, as Neil describes.

The role of \$R_1,R_2,C_1\$ is to maintain the stability of amplifier IC1 across all frequencies where IC1 has gain (well above the audio band).

When IC1 is driving a high-impedance load it operates as a unity-gain non-inverting buffer amplifier with feedback from the output taken through \$R_2\$.

When IC1 is driving a low-impedance or heavy capacitive load IC1 sees only the isolation resistor \$R_1\$ and feedback is taken through \$C_1\$.

  • \$\begingroup\$ Thanks @sstobbe. According to the datasheet of CS5381, the AIN+ and AIN- input impedance is 2.5K, so can we say IC1 is driving a high-impedance load? \$\endgroup\$ – Bob Yang May 23 '19 at 17:00
  • 1
    \$\begingroup\$ Yes, 2.5K is typically a reasonably light load for an op-amp. It is capacitor C3 which is too large for a typical op-amp to drive directly. \$\endgroup\$ – sstobbe May 23 '19 at 17:05

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