I am trying to build a line-following robot with the schematics I found below. I have just started to understand the use of feedback in op-amp. Ie. The continuous balancing mechanism creates a very stable gain determined by resistors connected to the op-amp.

I have tried to trace the current path marked in green and orange lines for the respective op-amp. Both lead to the same inverting and non-inverting input terminal of different op-amp.

I am kind of puzzled by this.

1) What is the point to have feedback to the different input terminals?

2) Because of the different input terminals used, how do I know which is the negative feedback? And is there even a thing like "positive feedback" happening here?

3) Or could it be a schematic drawn wrongly because it is just not technically practical nor possible to have the feedback wired in this manner?

Many thanks in advance.

enter image description here

  • 5
    \$\begingroup\$ The devices in your schematics are not OpAmps: they are comparators with open collector outputs. And also, even if comparators are used with positive feedback in several circuit, in this one is not so: here the devices are used as open loop, therefore (ideally) there is no feedback at all. If you want to understand how feedback (positive / negative) works by analyzing a circuit, this one is not suitable. \$\endgroup\$ – Daniele Tampieri Jun 5 '19 at 6:01
  • 1
    \$\begingroup\$ Together those two 500K potentiometers are adjusted, for good/accurate line following. \$\endgroup\$ – analogsystemsrf Jun 5 '19 at 8:06
  • 1
    \$\begingroup\$ @DanieleTampieri thanks for the advice. \$\endgroup\$ – Jack Oat Jun 5 '19 at 11:36

The schematic you show can be a bit confusing since they've drawn the entire +5V signal as the upper line, and the ground as the lower line. In this case there is no positive feedback or negative feedback - you've drawn loops but these are not negative/positive feedback loops. The top wire of the schematic is +5V meaning that resistors R3/R9 are just a voltage divider for the output of IC1A, and R14,R10 are a voltage divider for IC1B.

| improve this answer | |
  • \$\begingroup\$ I was confused by the voltage divider string used together with the op-amp look-alike. May I know the use of the voltage divider at the output of IC1A in this case? \$\endgroup\$ – Jack Oat Jun 5 '19 at 6:09
  • 2
    \$\begingroup\$ New questions should usually be asked as a new question instead of in a comment. However, here it's simple: It's really a voltage divider between the output voltage of IC1A and the +VCC rail, with the purpose of dividing the voltage that reaches the base of Q1. \$\endgroup\$ – Marcus Müller Jun 5 '19 at 6:12
  • 5
    \$\begingroup\$ @JackOat and this is not a "opamp look-alike". It's a comparator. Same symbol. also, see the answer to your previous question, which explains the difference between opamps and comparators very nicely. You're faster at asking questions than understanding the answers! \$\endgroup\$ – Marcus Müller Jun 5 '19 at 6:14
  • \$\begingroup\$ @MarcusMüller When you say "Same symbol", isn't it also valid that the opamp and comparator are "look-alike" as far as the schematic is concerned? After more understanding, the way to differentiate them is by checking the presence of negative feedback(closed loop) to the triangular symbol in the schematic. \$\endgroup\$ – Jack Oat Jun 5 '19 at 11:23
  • 1
    \$\begingroup\$ @JackOat, you should get used to the triangle being used as a general symbol for any kind of amplifier, not just op-amps and comparators. It is also used for rf amplifiers, digital buffers, etc, etc. \$\endgroup\$ – The Photon Jun 5 '19 at 14:39
