I am having difficulty in identifying whether feedback is negative or positive (i'm not talking about feedback topology)

My teacher told that there is a method in which, after deactivating independent sources, we break the loop and move from that breaking point around the loop, then and return to it after traversing some path. This allows us to determine the loop gain

He says that we can break it at any point that we wish but he always breaks it on Opamp's output. He said that is for convenience without giving reason; in his demonstrations he always moves from output to input to output.

If the calculated loop gain is negative then the circuit has negative feedback. If the calculated loop gain is zero zero then the circuit has zero feedback. If the calculated loop gain is positive then the circuit has positive feedback.

After that he gave some examples (in all examples there was at least one op amp present). But I want to know: does this method only work for op amps, and if it not works for all circuits then why not?

Please give me insight into intuition behind such method. Sorry for the long question

thank you for answering

  1. It will work with nearly all circuits, not just op-amp circuits.
    1. And I'm only saying "nearly all" because there's always exceptions -- just assume that it always works.
    2. It can get tricky if you break it at a point where the resulting circuit's output impedance is close to the resulting circuit's input impedance. Then you have to account for how that "input" loads that "output".
    3. Note, too that for some circuits and for some break points, you may want to consider the feedback to be a current, not a voltage -- there are some transistor amplifiers like this.
  2. Your instructor was probably breaking the circuit at the op-amp output because the output impedance is low, and it "looks into" a relatively high input impedance. As an exercise, you can always take some of his examples and try breaking the circuit at some other point (the next most convenient point in an op-amp circuit would be at the \$V_-\$ pin). Work out the math, and see if you get an answer that's the same, or substantially so.
  • \$\begingroup\$ yes it work as well for negative terminal too........okay so output and -ve terminal(since current is zero in op-amp terminals) nodes approximately sees an open circuit i.e, high impedence.that is reason for choosing them as convenient points .But if i take another point(other than output ,-ve terminal) then i've to take into account resistance looked by that node , then where will i connect that resistance (i guess between that node and ground). i just want to ask how to take loading in account?. \$\endgroup\$ – user141515 Feb 24 '19 at 18:43
  • \$\begingroup\$ Yes, that's right. \$\endgroup\$ – TimWescott Feb 25 '19 at 0:45
  • \$\begingroup\$ I usually discourage students to use the \$V_-\$ pin as it assumes that it is part of the feedback loop, which is not necessarily the case. If the opamp uses feedback, the output is always a part of it. \$\endgroup\$ – Sven B Feb 25 '19 at 9:04
  • \$\begingroup\$ @SvenB That's a good point. When I teach I find that there is a balancing act between guiding people to getting enough success to learn what they need to now, without accidentally teaching them right into a rut that they'll have trouble getting out of later. \$\endgroup\$ – TimWescott Feb 25 '19 at 17:26

Logic can be used with transistors as every collector or drain inverts (= 180 deg) from the input.

Because Op Amps have many transistors each with 45 deg phase shift at the unity gain frequency, it becomes inherently unstable unless a cap is included inside to act as an integrator thus a breakpoint hear 10Hz with open loop gain and 90-degree phase shift.

So when negative feedback is applied this phase margin reduces from 90 degrees towards 60 degrees (typical ideal design) and less if there are external phase shifts from RC or RLC components filters that reduce this phase margin. Since a 1st order RC filter adds 45-degree phase shift at the breakpoint and 90-degree several decades up, at some frequency in between, if the unity gain phase margin becomes zero or negative ( positive feedback) the step response turns from slighting ringing to steady oscillation ( read Barkhausen criteria).

For example, if an open collector with pullup is added to the output, you need to reverse the +/- inputs to use negative feedback topology. This boosts the output current in one direction but also reduces the phase margin overall.

Thus we use the method your Prof described to examine the open loop gain vs phase vs frequency and decide what the minimum gain is possible without inadequate gain or phase margin before the Barhausen Criteria for Oscillation occurs. Some IC's and the system cannot be stable unless a custom lead-lag RC/C compensation is added.

It does not always apply to Current mode drivers or higher order systems than 1st order internally compensated Op Amps such as machine control systems. Then other analytical tools like Root Locus are used instead of Bode Plots on gain/phase margin..

  • \$\begingroup\$ i haven't read oscillators and compensators yet .so, that phase margin thing is going over my head but thank you for answering :) \$\endgroup\$ – user141515 Feb 24 '19 at 18:47

It is convenient to break a loop where Rout=0 or Rin=infinity, as the analysis will become easier. So, as the output of the 'ideal' opamp has zero impedance, so it is easier to do the analysis by breaking the loop at output of opamp


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