Explanation for circuit with 2-stage op-amp, with diode in between first output and second input

I am a beginner in op-amp design and would like to get an explanation to a circuit that has a diode wired in between the output of op-amp-1 (stage 1) and the input of op-amp-2 (stage 2).

The inverting input of op-amp-1 takes a voltage from 0V <= x <= 5V, while the non-inverting input is wired to a 1kOhm potentiometer. The output has a 100kOhm||0.001uF for feedback.

Then, that output is connected to a 10kOhm in series with a 1N4149 diode. That signal is sent through a 0.1uF||100kOhm pot before being used as an input to the second stage (op-amp-2).

Furthermore, if someone could explain the the inclusion of the 0.1uF in parallel with the pot, that would be very beneficial to my understanding.

Thank you for looking this over.

• Can you tell us what this circuit is supposed to do? That would help someone explain the reason for the diode and other components. – Peter Bennett Nov 5 '18 at 17:04
• The output T1OUT carries a voltage, which is the processed voltage of saline, blood and air - each with their own expected resistances and voltages (1.5 for saline, 0.9 for blood, 0.0V for air). The green parts U17 and U18 stand in for this silicon tube configuration that allows these fluids and air to be present for detection. When saline or blood flow through the tube, the op-amps will process this and allow the T1OUT output to provide information to a computer program to process the voltage. I did not create this schematic, although my job is to interpret it. – Anthony the Kid Nov 5 '18 at 18:21
• That is where I am currently stuck..getting to the nitty-gritty of the reasoning behind the stages and particular components used along the way. – Anthony the Kid Nov 5 '18 at 18:21