Funtional role of op amps in a ADC filter circuit

I am a student trying to reverse engineer a rather interesting PCB, trying to learn from it but have come across some circuits that I am struggling to identify the roles of.

1. This is the first part.

"ref" that I have circled inputs a reference signal, which at first I thought was the GND. The negative input pin is connected to other op amps, serving as a ground, and also to the Ref pin of an instrumentation amplifier. The role is pretty clear that it is providing some kind of a stable reference for all the other amps, but I don't know exactly what it is doing.

2.

This is the second part of the circuit which takes in a signal filter from an instrumentation amp. OA3's positive input is connected to OA4's negative terminal. I am pretty sure that it is a rectifier of some sort, judging by the function of the board, but I have never seen one like this. The simulation results show that OA3 has no effect on the signal, the entire this just inverts the signal. Maybe I have my schematics wrong..?

Q3.

This is the last part of the question. OA5 takes in the signal from OA2(shown above) but weirdly it halves the signal, then amplifies it by 2. The negative termainal is connected to the negative terminal of OA4 from Q1. Why on earth would this kind of a circuit be here? Impedance matching doesn't seem probable because it is between cascade of op amps.

• Are you sure the first schematic is correct, the op amp has no feedback? – Colin Jan 24 at 9:21
• @Colin yes I have triple checked, there seems to be no feedback with the amp – Jung Haemin Jan 24 at 11:55
• It is behaving as a comparitor then, and either outputting Vcc or Vss – Colin Jan 24 at 11:57

Usually Vin- is connected to output Vref for unity gain and current limit but load regulated errors corrected.

1) is a standard Vcc/2 reference for single supply OpAmps using bipolar signals.

So you can call this as an "AC gnd" if you AC couple the outputs or offset after ADC = 0V with bipolar levels. "Gnd" can be anything you decide as 0V, even if floating above earth ground.

2) is a standard precision rectifier which reduces the diode drop by the open loop gain (to 0)

3) is a standard balanced differential amp with unity gain

When $$\\dfrac{R_f}{R_1}=\dfrac{R_g}{R_2}= \dfrac{V_{out}}{(V_2-V_1)} ~~, ~~ V_{CM12}{_{gain}} \$$=0, gain = ideally , actual = CMRR - 20 log R tol.

• Thank you so much for clarifying things. I have figured out/understood how the reference circuit works, does. I am doing some research on the other two, probably grasping the rectifier circuit. But would you care to explain a bit more about the last circuit? I cannot understand why the input if halved, then amplified. Cannot see why it is there. – Jung Haemin Jan 24 at 11:59