In our class we encountered several opamp circuits, for example, Non-inverting integrator:


simulate this circuit – Schematic created using CircuitLab

or we can consider the sallen-key type high pass filter:


simulate this circuit

Question: In such circuits, are these considered as positive feedback or negative feedback, that is, can we say the input terminals of opamp can be virtual grounded or not?

Question: Also what is the use of this kind of positive feedback?

My intuition is that it depends on capacitor time constant, for large time constant the negative feedback will dominate. But for small time constant, the capacitor will get quickly charged and discharged, so positive feedback will dominate.

  • 2
    \$\begingroup\$ Neither input is at ground so no virtual ground. The opamp still tries to drives its output such that both inputs are equal though. Virtual ground is just a special case of that. \$\endgroup\$
    – DKNguyen
    Commented Nov 10, 2019 at 19:54
  • 1
    \$\begingroup\$ I'd just use KCL and work out the transfer function for each and let that speak for itself. And if, for example, you are talking about a high pass Sallen Key (with gain), then why not just say that? And, at most, show the transfer function development. Why do you need to call it NFB or PFB? How does that help you or anyone you are talking to? \$\endgroup\$
    – jonk
    Commented Nov 10, 2019 at 21:07
  • \$\begingroup\$ learn to do the math, following online examples to get started. \$\endgroup\$ Commented Nov 10, 2019 at 21:31
  • \$\begingroup\$ @jonk I was trying to intuitively predict opamp output in that case, but I agree with you, best is to use KCL and analyse \$\endgroup\$
    – jeea
    Commented Nov 11, 2019 at 8:03
  • \$\begingroup\$ electronics.stackexchange.com/questions/444983/… \$\endgroup\$
    – G36
    Commented Nov 11, 2019 at 14:27

2 Answers 2


Regarding positive/negative feedback, the answer is not so simple. Remember that each "negative" feedback will turn into positive feedback for rising frequencies due to unavoidable phase shift of the amplifier. More than that, positive feedback allows stable operation as long as the (positive) loop gain is below unity.

Therefore, based on the classical feedback model the following definition gives the distinction between positive and negative feedback:

(1) The denominator of the closed-loop function is D(s)=[(1-LG(s)] with LG(s)=loop gain

(2) Negative feedback for |1/D(s)|<1 and positive feedback for |1/D(s)>1|

Hence, pos. fedback will enhance the closed-loop gain if compared with the gain without feedback (and vice versa for negative feedback).

In the given example (active 2nd-order high-pass stage) we have a fixed negative feedback loop and a positive frequency-dependent feedback loop. However, the net feedback will always be negative. It is the task of the positive loop to decrease the overall net negative feedback within a certain frequency range (around the pole frequency of the C-R network).

Comment (EDIT)

The example shown (highpass stage) allows to define three different feedback loops (because we can define three different openings) and, hence, three different loop gain expressions.

1.) The above considerations apply to the case where the "naked" opamp is considered as active unit. Hence, both feedback loops are to be opened at the same time directly at the opamp output node.

In the two following cases , only one of the two remaining loops are to be opened.

2.) As an alternative, we can consider the opamp with the resistive negative feedback as a "gain-of-two" amplifier and open only the frequency-dependent feedback loop with the two R-C sections. A visual inspection reveals that the "gain-of-two" amplfier now has positive feedback only. This is confirmed by investigation (simulation) of the expression |1/D(s)| which is always larger than unity.

3.) As another alternative, we can define a frequency-dependent block (opamp together with the positive R-C feedback path). In this case, which is a theoretical case only, we open the resistive feedback path only and investigate (simulate) again the expression |1/D(s)|.

As a result, we will see that there is negative feedback - except a small frequency region around the pole frequency of the filter (app. 2 kHz) where the negative feedback turns into positive feedback. See the attached graph which shows the expression |1/D(s)| .

enter image description here


To analyze ideal op amp circuits, (that have some feedback that yields non-infinite solutions), we say the two input terminals of the op amp are at the same potential, not necessarily 0v.

The “virtual ground” comes into play if one of the terminals is forced to 0v, perhaps by connecting the noninverting input to ground. Then the inverting input, since it at the same potential as the non-inverting input, is a “virtual ground”.


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