Can anyone please explain, with sufficient mathematics, why the virtual short concept is not applicable to an operational amplifier in positive feedback.

  • \$\begingroup\$ You have any source where it states so? \$\endgroup\$ Jul 27, 2015 at 6:58
  • \$\begingroup\$ Virtual ground do you mean? \$\endgroup\$
    – Andy aka
    Jul 27, 2015 at 9:00
  • \$\begingroup\$ I think, "virtual ground" for inverting operations (non-inv. input grounded) and "virtual short" for non-inverting applications (no input node grounded). \$\endgroup\$
    – LvW
    Jul 27, 2015 at 13:34

2 Answers 2


At first, I assume that you speak about operational amplifiers and the virtual short across the opamps input terminals, right? In this case, your statement - in this general form - is not correct. Let me explain:

The term "virtual short" applies to amplifier units with a very large open-loop gain which may be set to infinity (during mathematical manipulations/calculations). However, this assumption is true if the opamp is dynamically stable and operated in its linear region only. Normally, this is the case for negative feedback. However, there are some other applications which use negative and positive feedback at the same time. As long as the negative feedback is dominating (negative feedback factor larger than the pos. feedback factor) the circuit remains stable - and the "virtual short" principle continues to apply.

More than that, there are active filter circuits - Sallen-Key topologies, for example - which need positive feedback for Q enhancement. These circuits have negative feedback for DC (stable operating point) and in addition positive feedback for some specific frequencies (pole frequency region). Of course, in case of large Q values the circuit operates closer to the stability limit than simple amplifiers - but as long as negative feedback dominates the active filter circuit is stable and working as desired (and the virtual short principle applies).

In summary, the "virtual short" scenario does not apply for opamps with a feedback arrangement that causes dynamic instability. This is the case - for example - if we have a pure resistive feedback to the non-inv. terminal with a loop gain larger than unity (however, slight positive feedback leading to a loop gain below unity is stable!).

However, there may be an exception to this rule during switching: The classical Schmitt trigger is such a circuit with resistive positive feedback. When this circuit is used to build a squarewave oscillator, the opamp output is switched between both supply voltage limits - and during the switching phase the circuit crosses the region where we have a linear relationship between input and output. During this very short time period, the voltage between both opamp input nodes is negligible snmall (virtual short circuit).

  • \$\begingroup\$ than you very much.I am speaking about operational amplifiers.If you don't mind could please explain intuitively how an opamp maintains zero voltage between it's input taking a simple non inverting amplifier as an example. \$\endgroup\$ Jul 30, 2015 at 4:33
  • \$\begingroup\$ Yes - I can. However, because this is another question (not positive feedback) I would suggest to you to open a new question. OK? \$\endgroup\$
    – LvW
    Jul 30, 2015 at 8:39

The virtual-short principle, as someone calls it, relies on the fact that an high-gain amplifier operates with overall negative feedback. This is true for any amplifier with differential inputs, not only operational amplifiers. In relation with these latter, though, I'll provide the following analysis.

First of all let's consider an almost-ideal opamp, which is a differential amplifier with gain A>>1 and infinite input impedances looking into both inputs (an ideal opamp would have also A=∞).

That means that its output is an amplified replica of the difference signal between its inputs. Let's call \$v_-\$ and \$v_+\$ the signals at the inverting and non inverting output of the opamp, then the output is given by \$v_o = A(v_+ - v_-)\$.

Now think of creating a negative feedback loop, i.e. a return path that takes a fraction B of the output back to the inverting input, whereas you provide the signal to be amplified right to the non-inverting output:

$$ v_- = B v_o = B A (v_+ - v_-) = B A v_+ - B A v_- $$ Bringing all terms containing \$v_-\$ to the first member you get: $$ v_- + B A v_ - = B A v_+ \qquad\Rightarrow\qquad (1 + B A) v_- = B A v_+ \qquad\Rightarrow\qquad v_- = \dfrac{B A}{1 + B A} v_+ $$

Now you have a relationship between the voltages at the inputs of the opamp. If the opamp is ideal \$A \to \infty\$, therefore the product \$B A \to \infty\$, hence the 1 at numerator can be neglected and the numerator becomes, in the limit, equal to the denominator and the fraction becomes unity.

We conclude that for an ideal opamp with negative feedback \$v_- = v_+\$, i.e. there is no voltage across its inputs. Therefore the input terminals behave like a short circuit, but that short is called virtual because, unlike a real short, it draws no current since the inputs have infinite impedance.

EDIT (replying to a request for clarification from a comment)

First of all, let's see what happens to the output in the case above.

\begin{align} v_o &= A(v_+ - v_-) = \\[1 em] & = A \left( v_+ - \dfrac{B A}{1 + B A} v_+ \right) = A \left( 1 - \dfrac{B A}{1 + B A} \right) v_+ = A \left( \dfrac{1 + B A - B A}{1 + B A} \right) v_+ = \\[1 em] &= \dfrac{A}{1 + B A} v_+ \quad \underset{A \to \infty}{\to} \quad \dfrac 1 B \cdot v_+ \end{align}

So the output becomes a fixed multiple of the input signal. What happens in all the above if we used positive feedback instead. In this case the input signal would be applied on the inverting input and the feedback loop would be closed on the non-inverting input, which would take a fraction B of the output:

$$ v_+ = B v_o = B A (v_+ - v_-) = B A v_+ - B A v_- $$

Bringing all terms containing \$v_+\$ to the first member you get:

$$ v_+ - B A v_+ = - B A v_- \qquad\Rightarrow\qquad (B A - 1) v_+ = B A v_- \qquad\Rightarrow\qquad v_+ = \dfrac{B A}{BA - 1} v_- $$

Now the output will be:

\begin{align} v_o &= A(v_+ - v_-) = \\[1 em] & = A \left( \dfrac{B A}{BA - 1} v_- - v_- \right) = A \left( \dfrac{B A}{BA - 1} - 1 \right) v_- = A \left( \dfrac{BA - BA + 1}{BA - 1} \right) v_- = \\[1 em] &= \dfrac{A}{BA - 1} v_+ \quad \underset{A \to \infty}{\to} \quad \dfrac 1 B \cdot v_+ \end{align}

So it seems that both approaches (negative versus positive feedback) give the same result! Sadly the previous analysis is somewhat flawed. The problem is that the analysis only determined that the feedback system has a point of equilibrium for any given input for which the other input is the same. It tells nothing about the stability of such equilibrium! It turns out that the negative feedback result is stable, whereas the positive feedback one is unstable (it is like a pen left standing on its tip: it is theoretically a possible situation, but any slight vibration will make the the pen tilt and fall flat on the table).

To understand why the stability is different we cannot model the amplifier as a simple constant gain A. This assumes that the amplifier reacts instantly to input variations (in that case the positive feedback would be stable), but this is physically impossible. To analyze stability we need a better model for the opamp. We must model it using a transfer function in the s-domain and the simplest one is the dominant pole approximation, where the opamp behaves like a low-pass system with a single real pole.

So let's say the transfer function of the opamp is:

$$ A(s) = \dfrac{ A_0 } { 1 + \dfrac {s} {2\pi f_c} } = \dfrac{ 2 \pi f_c A_0 } { s + 2\pi f_c } = \dfrac {K} {s - p} $$

Where I defined \$K = 2 \pi f_c A_0\$ and \$p = - 2\pi f_c\$ to simplify the following calculations. Note that \$p\$ is the open loop pole, which is real and negative, indicating that the open loop amplifier is stable.

Moreover, notice that \$A_0\$ is the DC open-loop voltage gain of the opamp (\$A_{VOL}\$ in the datasheets), whereas \$f_c\$ is the corner frequency of its frequency response, which is related to the gain-bandwidth product (GBW or, approximately, \$f_T\$ on the datasheet) through the relation \$f_c \cdot A_0 = GBW \$. For typical opamps you may have \$ GBW \approx 10MHz \$ and \$ A_0 \approx 100dB = 10^5\$, therefore \$ f_c \approx 100Hz \$. Keep these value in mind for the following discussion.

Now let's consider the closed loop transfer function of the negative feedback amplifier:

$$ G_-(s) = \dfrac{A(s)}{B A(s) + 1} = \dfrac{\dfrac {K} {s - p}}{B \dfrac {K} {s - p} + 1} = \dfrac{K}{B K + s - p } = \dfrac{K}{s - p_-} $$

You can see that the closed loop system is again a low-pass single-pole system with a pole \$p_-\$ defined as:

$$ p_- = p - B K = - 2\pi f_c - B \cdot 2 \pi f_c A_0 = - 2 \pi f_c (1 + B A_0) < 0 $$

Thus the pole is negative and real for whatever value of B (which is positive) and \$A_0\$ (positive too).

Let's repeat the calculations for the positive feedback system. Its transfer function is:

$$ G_+(s) = \dfrac{A(s)}{B A(s) - 1} = \dfrac{\dfrac {K} {s - p}}{B \dfrac {K} {s - p} - 1} = \dfrac{K}{B K - s + p } = \dfrac{-K}{s - p_+} $$

In this case the closed loop pole is \$p_+\$:

$$ p_+ = p + B K = -2\pi f_c + B \cdot 2 \pi f_c A_0 = 2 \pi f_c (B A_0 - 1) $$

Since \$ B A_0 > 1 \$ (unless \$B < 1 / A_0\$, but this is unrealistic, since it would mean almost no feedback) you can see that \$ p_+>0\$, hence the closed loop system turns out to be instable!

What does it mean in practice? That any slight perturbation (e.g. noise) will make the output rise exponentially, which leads to the saturation of the opamap. Of course, once saturated the behavior becomes non-linear and the preceding analysis doesn't describe the operation of the system any longer.

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    \$\begingroup\$ ,Thank you very much for your time and concern.if you don't mind Could you please explain how the above equation fails in the case of positive feedback I mean how to prove "V+" is not equal to "V-" in case of positive feedback \$\endgroup\$ Jul 30, 2015 at 5:47

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