In an opamp, feedback on the positive input places it in saturation mode and the output is of the same sign as V+ - V-; feedback on the negative input places it in "regulator mode" and ideally Vout is such that V+ = V-.

  1. How does the opamp change its behaviour depending on the feedback? Is it part of a more general "behavioral law"? [Edit: Isn't it something in the lines of the voltage added increases the error instead of reducing it in the case of + feedback?]
  2. How can we analyse circuits where both are present?

Whoever answers both at the same time in a coherent manner wins a pot of votes.

enter image description here

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    \$\begingroup\$ There is a theorem that describes a general method to analyze circuits with any kind of feedback, is it what you are looking for? \$\endgroup\$ Commented May 30, 2014 at 12:30
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    \$\begingroup\$ There is an OUTSTANDING explanation of basic op-amp operation on this site somewhere, I just can't find it. Some of the more veteran members of the site may link it here, so I'll just add this comment: Suffice to say that you're probably thinking of op-amps only in terms of their inputs trying to be equal. It's a bit more nuanced than that. \$\endgroup\$
    – scld
    Commented May 30, 2014 at 12:31
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    \$\begingroup\$ Yes to both of you, I think general analysis methods rely on a sound understanding of the behaviour of opamps so I want to address both of these. \$\endgroup\$ Commented May 30, 2014 at 12:42
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    \$\begingroup\$ To answer the question, it is necessary to know what is connected to the pos. terminal: An ideal voltage or current source ? Some additional resistors? \$\endgroup\$
    – LvW
    Commented May 30, 2014 at 12:57
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    \$\begingroup\$ @LvW, it actually isn't necessary since, typically, we assume the input is driven by a source. If a voltage source, then \$v = v_S\$. If a current source, then \$i = i_S\$. The result that \$v = -iR\$ or that \$v_o = 2v\$ is independent of these details. \$\endgroup\$ Commented May 30, 2014 at 14:34

6 Answers 6

  1. Op-amp always behaves as a differential amplifier and the behavior of circuit depends on the feedback network . If negative feedback dominates, the circuit works in linear region. Else if positive feedback dominates, then in saturation region.
  2. I think the condition \$V^+ = V^-\$, the virtual short principle, is valid only when the negative feedback dominates. So if you are not sure that negative feedback dominates, consider op-amp as a differential amplifier. To analyze the circuit, find \$V^+\$ and \$V^-\$ in terms of \$V_{in}\$ and \$V_{out}\$. Then substitute in the following formula, $$V_{out} = A_v(V^+-V^-)$$ calculate \$V_{out}/V_{in}\$ and then apply the limit \$A_v\rightarrow\infty\$
  3. Now, net feedback is negative if \$V_{out}/V_{in}\$ is finite. Else if \$V_{out}/V_{in} \rightarrow \infty\$, then the net feedback is positive.

From the circuit given in the question, $$V^+ = V_{in}\ \text{and}\ V^- = V_{out}/2$$ $$V_{out} = A_v(V_{in} - V_{out}/2)$$ $$\lim_{A_v\rightarrow\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\rightarrow\infty}\frac{A_v}{1+A_v/2} = 2$$ $$V_{out} = 2V_{in}$$ \$V_{out}/V_{in}\$ is finite and net feedback is negative.

\$\mathrm{\underline{Non-ideal\ source:}}\$
In the above analysis, \$V_{in}\$ is assumed to be an ideal voltage source. Considering the case when \$V_{in}\$ is not ideal and has an internal resistance \$R_s\$. $$V^+ = V_{out}+(V_{in}-V_{out})f_1\ \text{ and }\ V^- = V_{out}/2$$ where, \$f_1 = \dfrac{R}{R+R_s}\$ $$V_{out} = A_v(V_{out}/2+(V_{in}-V_{out})f_1)$$ $$V_{out}(1-A_v/2+A_vf_1) = A_vf_1V_{in}$$ $$\lim_{A_v\rightarrow\infty}\frac{V_{out}}{V_{in}} = \lim_{A_v\rightarrow\infty}\frac{f_1}{\frac{1}{A_v}-\frac{1}{2}+f_1}$$ $$\frac{V_{out}}{V_{in}} = \frac{f_1}{f_1-\frac{1}{2}}$$

case1: \$R_s\rightarrow 0,\ f_1\rightarrow 1,\ V_{out}/V_{in}\rightarrow 2\$

case2: \$R_s\rightarrow R,\ f_1\rightarrow 0.5,\ V_{out}/V_{in}\rightarrow \infty\$

\$%case3: R_s \rightarrow \infty,\ f_1 \rightarrow 0,\ V_{out}/V_{in} \rightarrow 0\$

The output is finite in case1 and so net feedback is negative in these conditions (\$R_s < R\$). But at \$R_s = R\$, negative feedback fails to dominate.

Case1 is the normal working of this circuit but it is not used as an amplifier with gain 2. If we connect this circuit as a load to any circuit, this circuit can act as a negative load (releases power instead of absorbing).

Continuing with the analysis, the current through \$R\$ (from in to out) is, $$I_{in}=\frac{V_{in}-V_{out}}{R}=\frac{-V_{in}}{R}$$ calculating the equivalent resistance \$ R_{eq}\$ $$R_{eq} = \frac{V_{in}}{I_{in}} = -R$$

This circuit can act as negative impedance load or it act as a negative impedance converter.

  • \$\begingroup\$ Thanks for your answer. That's an interesting method which has the advantage of working every time as it's the exact formula of what the opamp is doing as far as I know. Could you analyse the aforementioned circuit with that method so that we can compare the results obtained with the other methods? \$\endgroup\$ Commented May 31, 2014 at 11:28
  • \$\begingroup\$ @MisterMystère There is no need to analyze the circuit in the question. Input-output relation is already given. But let me try... \$\endgroup\$
    – nidhin
    Commented May 31, 2014 at 15:13
  • \$\begingroup\$ Honestly I took a random circuit from Google images to illustrate the question and serve as an example. I don't have a particular problem, this is for personal improvement. But seeing that others have developed their methods, I would like to compare them. \$\endgroup\$ Commented May 31, 2014 at 17:22
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    \$\begingroup\$ @MisterMystère Thank you and LvW for pointing out the errors. Case3 should be \$V_{out}/V_{in}\rightarrow 0\$. It does not go into saturation. Try simulating this. \$\endgroup\$
    – nidhin
    Commented Jun 2, 2014 at 17:23
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    \$\begingroup\$ @MisterMystère and nidhin, the circuit nidhin has simulated and linked to for verification of case 3 has the op-amp 'upside down'; the op-amp input terminals are opposite that of the circuit above. The circuit simulated is stable for \$R_S > R\$ and unstable for \$R_S < R\$ which is precisely the opposite of the NIC circuit analyzed. The analysis above for case 3 is incorrect and the simulated circuit is not the circuit analyzed. i.sstatic.net/gcuEi.png \$\endgroup\$ Commented Jun 3, 2014 at 13:24

How does the opamp change its behaviour depending on the feedback?

The ideal opamp behaviour itself is unchanged; it is the circuit's behaviour that is different.

Isn't it something in the lines of the voltage added increases the error instead of reducing it in the case of + feedback?]

That's correct as far as it goes. If we perturb (or disturb) the input voltage, negative feedback will act to attenuate the disturbance while positive feedback will act to amplify the disturbance.

How can we analyse circuits where both are present?

As usual, assume there is net negative feedback which implies that the non-inverting and inverting input voltages are equal. Then, check you result to see if, in fact, negative feedback exists.

I'll demonstrate by solving your example circuit.

Write, by inspection

$$v_+ = v_o + iR$$

$$v_- = v_o \frac{R_1}{R_1 + R_1} = \frac{v_o}{2}$$

Set these two voltage equal and solve

$$v_o + iR = \frac{v_o}{2} \rightarrow v_o = -2Ri$$

which implies

$$v_o = 2v_+ = 2v $$

This is a good thing because we expect that this is a non-inverting amplifier and indeed, we get a positive voltage gain. Interestingly, the input resistance is negative: \$\frac{v}{i} = -R\$.

However, if we add an additional resistor \$R_S\$ in series with the input, we can run into trouble.

In that case, the equation for the non-inverting input voltage becomes

$$v_+ = v_S \frac{R}{R_S + R} + v_o \frac{R_S}{R_S + R} $$

which implies

$$v_o = \frac{2R}{R - R_S}v_S $$

Note that when \$R_S < R\$, the voltage gain is positive as expected from a non-inverting amplifier.

However, when \$R_S > R\$, the voltage gain is negative for a non-inverting amplifier which is a red flag that something is wrong with our assumptions.

The wrong assumption is that there is negative feedback present and it was that assumption which licensed us to set the non-inverting and inverting input voltages equal in the analysis.

Note that the voltage gain goes to infinity as \$R_S\$ approaches \$R\$ from below. Indeed, there is no net feedback when \$R_S = R\$; the negative and positive feedbacks cancel. This is the 'boundary' between net negative feedback and net positive feedback.

Is this method of picking up on red flags always valid to determine the limit between net positive and negative feedback?

What I did, in this case, was to make an assumption, solve the circuit under that assumption, and check the solution for consistency with the assumption. This is a generally valid technique.

The assumption was, in this case, that net negative feedback is present which implies that the op-amp input terminal voltages are equal.

When we solved the circuit in the 2nd case, we found that the net negative feedback assumption is valid only when \$R_S \lt R\$. If \$R_S \ge R\$, there is no or positive feedback and, thus, no reason to constrain the input terminal voltages to be equal.

Now, it may not be clear why there is positive feedback when \$R_S \gt R\$. Recall the setup for deriving the negative feedback equation:

enter image description here

Here, we subtract a scaled version of the output voltage from the input voltage and feed this difference \$V_{in} - \beta V_{out}\$ to the input of the amplifier.

Clearly, this assumes \$\beta\$ is positive in order that there be a difference between the input and scaled output voltages.

The well known result is

$$V_{out} = \frac{A_{OL}}{1 + \beta A_{OL}} V_{in}$$

and, in the limit of infinite gain \$A \rightarrow \infty\$

$$V_{out} = \frac{1}{\beta}V_{in}$$

Comparing this equation with the result for the 2nd case above, see that

$$\beta = \frac{R - R_S}{2R}$$

from which it immediately follows that we have net negative feedback only when \$R_S \lt R\$.

There is some discussion in the comments about the conclusion for case 3, \$R_S > R\$, in the accepted answer. Indeed, the analysis for case 3 is not correct.

As shown above, if we assume the op-amp input terminal voltages are equal, we find a solution where

$$v_o = \frac{2R}{R - R_S} v_S$$

Now assume, for example, that \$R_S = 2R\$ then

$$v_o = -2v_S$$

And, in fact, one can verify that this is a solution where the op-amp input terminal voltages are equal

$$v_+ - v_- = 0$$

However, if we perturb the output slightly

$$v_o = -2v_S + \epsilon$$

The voltage across the op-amp input is perturbed to

$$v_+ - v_- = \frac{\epsilon}{6}$$

which is in the same 'direction' as the disturbance. Thus, this is not a stable solution since the system will 'run away' from the solution if disturbed.

Contrast this with the case that \$R_S < R\$. For example, let \$R_S = \frac{R}{2}\$. Then

$$v_o = 4v_S$$

Perturb the output

$$v_o = 4V_S + \epsilon$$

and find that the op-amp input voltage is perturbed to

$$v_+ - v_- = -\frac{\epsilon}{6}$$

This is in the opposite direction as the disturbance. Thus, this is a stable solution since the system will 'run back' to the solution if disturbed.

  • \$\begingroup\$ Thanks for the clear answer. Is this method of picking up on red flags always valid to determine the limit between net positive and negative feedback? Is the limit that brutal or is there a blurry limit? \$\endgroup\$ Commented May 31, 2014 at 11:24
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    \$\begingroup\$ @MisterMystère, I will work on an addendum to my answer to address your comment later. \$\endgroup\$ Commented May 31, 2014 at 12:04
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    \$\begingroup\$ @MisterMystère, see the addendum to my answer. \$\endgroup\$ Commented May 31, 2014 at 18:25
  • \$\begingroup\$ Thanks again, that's really an excellent answer. It was really tough to decide which answer to accept, but I went for nidhin's mainly because he could use the reputation (that's a water drop in a lake for you). See you around on SE. \$\endgroup\$ Commented Jun 2, 2014 at 8:55
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    \$\begingroup\$ @MisterMystère: Are you aware that nidhin´s answer is NOT correct in all cases? He wrote:"The output is finite in cas1 and case3 so net feedback is negative in these conditions". Apparently, this is false for case 3. In this case, the circuit is unstable and the result "-2" is wrong. Instead, the opamp goes into saturation. \$\endgroup\$
    – LvW
    Commented Jun 2, 2014 at 10:41

It's still useful to analyse this as a linear situation where you can assume that -Vin always equals +Vin. I'm going to redraw to show the input voltage going through a resistor because as the OP has shown it in his diagram "v" could be assumed to be a voltage source and therefore the effect of "R" is of no consequence: -


simulate this circuit – Schematic created using CircuitLab

\$V_X = (V_{IN} - V_{OUT})(\dfrac{R2}{R1+R2})+ V_{OUT}\$

And also: -

\$V_X = V_{OUT}(\dfrac{R4}{R3+R4})\$ (because the two op-amp inputs are the same i.e. still a linear analysis)

Equating the two formulas for \$V_X\$ we get: -

\$V_{OUT}(\dfrac{R4}{R3+R4}) = (V_{IN} - V_{OUT})(\dfrac{R2}{R1+R2})+ V_{OUT}\$

Rearranging we get: -

\$V_{OUT}(-1 +\dfrac{R2}{R1+R2} +\dfrac{R4}{R3+R4})= V_{IN}(\dfrac{R2}{R1+R2})\$

Sanity check - in the normal case when R2 is infinite the equation boils down to: -

\$V_{OUT}(-1 +1 +\dfrac{R4}{R3+R4})= V_{IN}(1)\$ and we see that: -

\$\dfrac{V_{OUT}}{V_{IN}} = 1+\dfrac{R3}{R4}\$ so that's OK and going back to the equation: -

\$V_{OUT}(-1 +\dfrac{R2}{R1+R2} +\dfrac{R4}{R3+R4})= V_{IN}(\dfrac{R2}{R1+R2})\$ we see that: -

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{-\dfrac{R2}{R1+R2}}{1-\dfrac{R2}{R1+R2}-\dfrac{R4}{R3+R4}}\$

Clearly we approach a "problem" (i.e. infinite gain) when the denominator heads towards zero and this happens when: -

\$\dfrac{R2}{R1+R2} + \dfrac{R4}{R3+R4} = 1\$

So hopefully this makes sense. Normally, for linear operations the circuit gain is dependant on all four resistors but, if the ratios of the resistors are as above, the gain is infinite.

  • \$\begingroup\$ Yes - I agree to the above result. However, I would suggest to use another form of the result: Vout/Vin=+[R2/(R1+R2)]/[R4/(R3+R4)-R1/(R1+R2)]. This form allows a quick analysis of the circuit´s properties. The gain must be positive (the + input is energized) and the circuit is stable as long as the negative feedback dominates. Otherwise, the result would be negative which is inconsistent. The stability limit is for pos. feedback equal to neg.feedback . \$\endgroup\$
    – LvW
    Commented May 30, 2014 at 14:19
  • \$\begingroup\$ @LvW I'm struggling with seeing your formula = the Vout/Vin I got dude \$\endgroup\$
    – Andy aka
    Commented May 30, 2014 at 14:27
  • \$\begingroup\$ I must admit,I don`t understand the contents of your comment ("dude" ?) \$\endgroup\$
    – LvW
    Commented May 30, 2014 at 14:41
  • \$\begingroup\$ @LvW dude is just a friendly name! I don't see how my formula can equal your formula! \$\endgroup\$
    – Andy aka
    Commented May 30, 2014 at 14:58
  • \$\begingroup\$ Simply set: 1-[R2/(R1+R2)]=[R1/(R1+R2)]. \$\endgroup\$
    – LvW
    Commented May 30, 2014 at 15:09

Because the question was: How to analyze? Here comes a way to analyze such a circuit which is relatively quick and easy:

From the classical feedback formula (H. Black) we know that for an idealized opamp with infinite open-loop gain the closed-loop gain is simply (see the circuit diagram with four resistors in one of the answers):

$$A_{cl} = -\frac{H_f}{H_r}$$

(\$H_f\$: Forward damping factor; \$H_r\$: feedback factor.)

Both functions can be easily derived from the circuit:

$$H_f = \frac{R_2}{R_1+R_2}$$


$$H_r = \frac{R_1}{R_1+R_2} - \frac{R_4}{R_3+R_4}$$

Hence, the result is

$$A_{cl} = \dfrac{\dfrac{R_2}{R_1+R_2}}{\dfrac{R_4}{R_3+R_4}-\dfrac{R_1}{R_1+R_2}}$$

It is worth mentioning that the advantage of the circuit is the following: We can select a desired stability margin and/or use non-compensated opamps for lower gain values (data sheet: stable for gain>Acl, min only).

Justification: From the expressions above one can derive that it is possible to match the feedback factor to the corresponding open-loop gain (for a certain stability margin) - without restrictions to the closed-loop gain value. One can regard this method as a special kind of "external frequency compensation".

With other words: I can choose less feedback (good for stability) and - at the same time - a small value for closed-loop gain Acl.

  • \$\begingroup\$ Thanks for answering. I assume with this method you separate linear from saturated mode by Acl going very high, but how high? Could you explain more about how to get the Hf and Hr factors generally speaking (transfer function from Vo to Vin on both pads?)? \$\endgroup\$ Commented May 31, 2014 at 11:36
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    \$\begingroup\$ In my opinion, using the Hf and Hr factors is the most efficient way to analyse (complicated or involved) opamp circuits. Definitions are as follows: Hf is the portion of the input voltage that appears across the opamp input in case we set Vout=0. Accordingly, Hr is the portion of the output voltage that appears across the opamp input (V+ - V-) in case the input voltage is set to zero. This is simply an application of the superposition theorem. \$\endgroup\$
    – LvW
    Commented Jun 1, 2014 at 8:27
  • \$\begingroup\$ Thanks for your very good answer; but I went for nidhin's answer which is more detailed and intuitive. You're right about the voltage source though, but as I said it was only an illustration example, I didn't know at that time anyone would actually try to solve it. Up to next time \$\endgroup\$ Commented Jun 2, 2014 at 8:58
  • \$\begingroup\$ I' d like to add something to your justification part. By matching the feedback factor and open loop gain we may actually create a self-oscillating circuit, as is the case with the known circuit that has a an op amp connected to a Wien bridge. \$\endgroup\$
    – Shemafied
    Commented Jan 11, 2016 at 22:48

I joined this forum yesterday, after I came across your interesting discussion in Google.

Your thoughts are wonderful and I fully support them. My point is just that they are based more on a detailed and sometimes formal analysis of the INIC circuit (what it does) than on the disclosure of its philosophy (why it does this). So I will try to roughly fill that gap with my comment.

We can consider this circuit from two perspectives: first - as a circuit with only input and no output (a load with negative resistance); second - as a circuit with input and output (an amplifier with mixed feedback).

Negative load. Beginning from the early 90's, I spent a lot of effort to reveal and explain in an easy and intuitive way the first perspective. If you are interested and patient enough, you can familiarize yourself with the resources I created in Web; I described them in detail in two questions asked by me in ResearchGate - What is negative impedance? and What is the basic idea behind the negative impedance converter? For those who do not have patience to read all of this, here is a very brief explanation.

The circuit behaves as an active load (dynamic voltage source with internal resistance R) that reverses the current through the resistor R (in the original Wikipedia picture) and "pushes" it back to the input source. In this way, it converts the resistor R (originally consuming a current) into a negative "resistor" -R (producing a current). It does this by opposing (through the resistor) a reverse and higher (2V) voltage to the input voltage (V). This is the output voltage of the operational amplifier and it is not used here... but still the circuit has an output... and, although it sounds strange, it is its input! Simply the circuit behaves like a source that attacks back the input source...

Amplifier with mixed feedback. According to me, this is the subject of the question asked here. As described in the comments above, this circuit is an amplifier with negative feedback, which is partially neutralized by a weaker positive feedback. But what is the point of that?

In general, the positive feedback increases the gain of the imperfect amplifiers and it is used in the past (remember Armstrong's regenerative idea). But in our case, the op-amp has a huge gain and this is not necessary. Then what is the point of using positive feedback here?

My speculation is that we can use it to decrease the ratio R3/R4 (in the second figure) in the case of INIC or R2/R1, in the case of VNIC (when the input voltage is applied to the inverting input). As a result, the resistors R2 and R3 can be low resistive.

In this amp application, the op-amp output is the circuit output. But as above, this amplifier has another output... and this is its input... so the circuit can act as an exotic 1-port amplifier...

Reinventing INIC by CircuitLab experiments

I decided to expand on my answer from nine years ago by revealing the secret of the "Wikipedia circuit" because it is not just a circuit, but rather a concept. It represents one of the possible ways to obtain a negative resistance from a positive one by reversing the current direction; hence the name current-inversion negative impedance converter (INIC). Legendary circuits and techniques such as Howland current source, load canceller, bootstrapping, etc. are built on it. So let's see what this powerful idea is with the help of which we can modify resistance.

I suggest that we do it in the form of step-by-step CircuitLab experiments with which we can follow the evolution of the idea.

Positive resistance

In the beginning, a 1 kΩ "positive" resistor consumes 1 mA current from 1 V voltage source.


simulate this circuit – Schematic created using CircuitLab

Increased resistance

The straightforward way to increase (eg, double) its resistance is to replace it with a 2 kΩ resistor. But we can apply a small trick to artificially increase the resistance connecting in series and in the opposite direction a "behavioral" voltage source V1 with 0.5*V voltage. It is subtracted from the main voltage V and the current decreases twice. So the main source V has the illusion that the resistance has doubled (R' = V/I = 2 kΩ).


simulate this circuit

A more practical implementation is through an "amplifier" with a gain of 0.5 (I have set such a gain to an op-amp from the CircuitLab library).


simulate this circuit

Infinite resistance

If we continue to increase the additional voltage V1, the current will decrease more and more and the resistance R' will increase. When the two voltages equalize, the current stops flowing and the resistance R' becomes infinite (we can simulate it by a following behavioral voltage source V1 = V).


simulate this circuit

This famous circuit trick figuratively named "bootstrapping", can be implemented by an amplifier with a fixed gain of 1...


simulate this circuit

... that is usually made by an op-amp with a 100% negative feedback (op-amp follower).


simulate this circuit

Negative resistance

And now comes the most interesting thing, for the sake of which we did everything up to here. If we continue to increase the additional voltage, e.g. we double it (V1 = 2*V), the current reverses its direction and enters the source V as if the resistance does not consume but produces current... it has become negative (R' = V/-I = -1 kΩ).


simulate this circuit

We can implement it by an amplifier with a fixed gain of 2...


simulate this circuit

... that is made by an op-amp non-inverting amplifier.


simulate this circuit

Is there feedback?

Neither positive nor negative feedback

The amplifier output is connected through the resistor R to its non-inverting input, but this does not automatically mean that there is positive feedback. If the input voltage source is perfect (with zero internal resistance), the amplifier output will not be able to change the voltage at the non-inverting input, and there will be no positive feedback. Since the amplifier is with fixed gain, there is no need for negative feedback.

Only positive feedback

If the input voltage source is imperfect (with some internal resistance) or there is an input resistor in series, then the amp output can change the non-inverting input voltage; so there is positive feedback. For stability, the feedback loop gain must be less than 1. As above, there is no need for negative feedback. See for example Schematic 4.2.

Both positive and negative feedback

The op-amp non-inverting amplifier implementation needs negative feedback. For stability, it must dominate over the positive feedback. See for example Schematic 4.3 and the application below.


Howland current source

Now that we have uncovered the secret of INIC, let's consider one of its most famous applications - the Howland current source. Let's follow the evolution of this great idea.

"Perfect" current source: The simplest current source is just a resistor (Rin) in series to a voltage source (Vin). If we measure its current by a perfect ammeter (with zero resistance), the current is exactly as Ohm's law says - IL = Vin/Rin = 1 mA.


simulate this circuit

Imperfect current source: But if we measure the current by an imperfect ammeter (e. g. with 1 kΩ resistance RL), the current will be lower - IL = Vin/(Rin+RL) = 0.5 mA.


simulate this circuit

Let's sweep RL from zero to 10 kΩ to see its impact on the current.



Improved current source: The Howland idea is brilliant: to neutralize the 1 kΩ positive resistance (Rin) with -1 kΩ negative resistance (R1, R2, R and OA). The result is infinite resistance, i.e. constant load current IL. Let's investigate it by sweeping the load (ammeter) resistance RL from 0 to 10 kΩ. Note that in the schematic below, RL = 1 kΩ (to see or change it, open the IL parameters window).


simulate this circuit

As you can see in the graph below, when RL increases from 0 to 10 kΩ, the voltage across the load increases from 0 to 10 V but the current does not change. Do not pay attention to the yellow curve; it is only to "cheat" the simulator autoscaling.

STEP 5.1.3a

STEP 5.1.3b

Helped current source: The trick behind the Howland current source is really clever - making a perfect current source (the whole circuit) by helping an imperfect current source (Vin and Rin) with another auxiliary current source (R1, R2, R and OA).

In the graph below, when the load resistance increases, the input current decreases but the "helping" current increases. As a result, their sum - the load current, stays constant.

STEP 5.1.3c

Load canceller

Here is an even more extravagant INIC application where a positive resistance is neutralized by an equivalent negative resistance.

Unloaded voltage divider: The humble potentiometer works precisely if it is not loaded (the voltmeter is perfect with extremely high resistance).


simulate this circuit

If the wiper is in the middle (K = 0.5), Vout = 0.5Vin.


Loaded voltage divider: But if we connect a low-resistance (only 100 Ω) load RL, the current increases and the output voltage drastically decreases...


simulate this circuit

... when Vin changes from zero to 10 V.


Helped voltage divider: The remedy is the same as in the Howland current source - to help the weak voltage divider by supplying the load with another current source. In terms of resistance, the 100 Ω positive resistance RL is completely neutralized by the 100 Ω negative resistance of the INIC (R1, R2, R and OA).


simulate this circuit

As a result, as though there is no load connected to the voltage divider, and the graph is the same as in the Schematic 5.2.1.


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    \$\begingroup\$ The negative-impedance load reminds me of a motor with excessive IR compensation. Normally, if a motor is trying to remain still, externally applying some clockwise torque will make it turn clockwise, though more slowly than if it weren't trying to remain still. If the motor is overcompensated, however, applying clockwise torque will make it turn counterclockwise. Very weird. \$\endgroup\$
    – supercat
    Commented Dec 16, 2014 at 19:50
  • \$\begingroup\$ Exactly! This is a very good electromechanical analogy of the op-amp circuit above (INIC) where the op-amp reverses the current and "blows" it back into the input source. Conversely, if the motor was overcompensated so that it accelerates in the same direction (clockwise), it would behave like the dual VNIC. \$\endgroup\$ Commented Dec 16, 2014 at 21:08
  • \$\begingroup\$ The overhelping (damaged) brake servo is another electromechanical (pneumatic, fluid) example of the VNIC - you just touch the brake pedal and the servo finishes the operation up to a full stop. I remember that years ago a friend of mine told me how he did a car crash just this way. \$\endgroup\$ Commented Dec 16, 2014 at 21:28
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    \$\begingroup\$ We use negative impedance amplifiers to zero out large capacitances associated w/ glass microelectrodes in physiological setups. We know what the output should look like, so we tweak the value to get it there. Things will oscillate if you get it too high, of course. \$\endgroup\$ Commented Dec 16, 2014 at 22:53
  • \$\begingroup\$ Although the initial question was more about knowing which behaviour was dominant if both positive and negative were present in any circuit (this one is only an example, actually it's the first circuit i've found on google images...), this is interesting thanks. \$\endgroup\$ Commented Dec 16, 2014 at 23:14

@supercat, your comment awakened my desire (deliberately suppressed by me) to think about these diabolical circuits:) Maybe you will not believe me, but I have been thinking on them from the early 90s... and I still continue thinking... Now I want to explain what is the meaning of the fact that this circuit (INIC) reverts the current direction and passes the current back through the resistor. We can observe three situations:

Ideal voltage source (Ri = 0) connected to INIC. There is no benefit from this arrangement, it simply passes a reverse current through the input source (really, if it is a rechargable battery, it will be charged).

Real voltage source (having some Ri) connected to INIC. The circuit passes a reverse current through the input source, creates a voltage drop across its Ri in addition to its internal voltage, and thus raises its external voltage.

Real voltage source and INIC connected to a common load Rl. This is the typical INIC application where it is connected with the input source in parallel to a common load. The INIC adds an additional current to the input current thus helping the input source. The Howland current source is a typical application of this idea.

A negative resistor (INIC) and an input source connected in parallel to a common load

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    \$\begingroup\$ Well made drawing. Off topic: it amazes me that people still use paper for anything else other than drafts and scribbles, especially round corners ;) However you may want to add to your previous post instead and delete this one, this forum is not designed to allow several posts from the same person. Just a gentle heads up. \$\endgroup\$ Commented Dec 16, 2014 at 23:21

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