# OP-AMP Positive/Negative Feedback together

I've read several articles about the application of OP-AMP negative feedback. As far as an application for positive feedback, I came across hysteresis and it makes sense.

Are other uses for positive feedback other than hysteresis?

I am also not sure why in some applications they use both the positive and negative feedback path?

I've leanred that if the net gain on the negative feedback path is more than the net gain on the positive feedback path, then it acts as an OP-AMP with negative feedback, and vice versa.

If we tend to get an overall negative feedback or an overall positive feedback, then what is the point of using both paths at the same time, at the first place? Why do we even bother ourselves to use both paths, if at the end all we care about is to have either a net positive or negative feedback?

See my answer here for an example of why you might want to use negative and positive feedback at the same time. The 598.3K resistor in the positive feedback path maintains a constant current through the variable resistor and the negative feedback path determines the gain (output volts per ohm of resistance of the variable resistor).

To see how this yields a constant current, consider Scott's generalized NIC answer. This is also a NIC- creating a negative resistance to cancel out the incremental effect of the 100K resistor R2, so the current remains constant.

Looking into the non-inverting input of the op-amp (without R2 and the variable resistor connected, but with R4 in place), the resistance looks like:

Referring to the NIC schematic, we have:-

R1 = 100K || 25.68K = 20.43K R2 = 122.2K R3 = 598.3K

$R_{IN} =$ - 598.3K $\cdot$ $20.43K \over 122.2K$ = -100K, which exactly cancels out the effect of the 100K resistor R2 (R2 on the original schematic).

• could yo please explain how does the positive path maintain a constant current? – Rudy01 Feb 12 '14 at 2:19
• Please see my edits above. – Spehro Pefhany Feb 12 '14 at 2:33
• Thanks, it started making more sense. But, I am still not sure where 39.78uA current came from, and why at the output of the op-amp, the voltage varies from 0-5? – Rudy01 Feb 12 '14 at 18:54
• If you assume the op-amp is at 5V when the variable resistor is at 40K, you can easily calculate the current. The voltage varies because of the gain G=1+122.2/20.43 = 6.981. 39.78uA * (40K -22K) * 6.981 = 5.00V. R5 and R6 (2 degrees of freedom) are chosen so that 0V and 5V are the 'zero' and 'full scale' outputs. R1 is arbitrary, chosen to be reasonable. – Spehro Pefhany Feb 12 '14 at 18:59
• Okay, it makes a lot of sense. And, I was able to compute all the numbers; thanks a lot. However, one last thing that I have a question is that it seems like for a NIC circuit we have a case where the power supply will be sinking current, and that is why we get the negative sign for the current (Is), as explained here en.wikipedia.org/wiki/Negative_impedance_converter. But in this example, the power supply will always be sourcing current ! So does this mean that even though NIC have a negative sign for Is, it may not necessarily sink current all the time !?! – Rudy01 Feb 12 '14 at 22:34

@Rudy01, you have asked an excellent question that deserves an exact answer...

The positive feedback can be used for amplifying purposes as well... and it was used in the past to increase the gain of the then amplifiers having fixed, stable but low gain. Nowadays the situation is just the opposite - we have (op)amps with enormous but not exactly specified gain. For this reason, we use a negative feedback to reduce their excessive gain... and make it stable and exactly specified. So, the negative feedback is used, as a rule, in any amplifier.

Now let's see when we add an additional positive feedback... or, as you asked, "why in some applications they use both the positive and negative feedback path?" Of course, the current inversion NIC - INIC (in the Wikipedia picture above) is the most typical example of using this mixing technique. As the best way to understand circuits is to build them, I suggest to do it with this odd circuit.

The idea of creating this kind of negative resistance (we can name it voltage controlled) is extremely simple - we connect an opposing voltage source (in the simplest case, 2V) via a positive resistor (R) to the input voltage source (V). As a result, the current through the positive resistor is reversed (thus the name INIC)... and the resistor as though produces instead to consume a current V/R. To implement this opposing voltage source, we need a fixed-gain non-inverting amplifier... and we make it by applying a negative feedback to an op-amp that reduces its excessive gain to the finite value 1 + R2/R1.

Then, to obtain the negative resistance, we should apply this opposing voltage back to the input voltage source. So, we connect the op-amp output via a resistor (R3 in the picture) with its input. Now, if the input voltage source is perfect (Rin = 0), nothing new (regarding the feedback) will happen - a negative resistor -R will be connected to the perfect input voltage source... and it will pass back current V/R through it... but there will be no positive feedback. Usually, voltage sources are not perfect (they have some internal resistance), or we intentionally connect a positive resistance in series to the input source... so, in this cases, a positive feedback appears... and we have the both kinds of feedback... We have only to make the negative feedback dominate over the positive one to keep the circuit stable (to avoid the "hysteresis" as you said in your question)...

Now about the most interesting part of this discussion where you asked, "How does the positive path maintain a constant current?"... and commented, "but it is only changing the direction in the positive feeback path, correct? The current througth the source is always the same direction, correct?"... and simply you wanted to know the meaning of all this...

First at all, Spehro Pefhany should note that this circuit is known as Howland current source (pump). It was invented in the early 60's simultaneously by Howland and Deboo. Simply speaking, it is a mixture of two opposite but equivalent resistors - positive (R2 = 100 kom) and negative (RN = -100 kom, the whole INIC). They completely neutralize each other and the result is an infinite (differential!) resistance connected in series with the input voltage source (+5 V). So, the whole combination (excluding only the rheostat) acts as a perfect current source suplying the rheostat...

Like negative feedback, in most cases, the negative resistance is something "positive" (useful, "helping"). Think of the combination Vin + R2 as an imperfect current source that is affected (disturbed) by the rheostat (this current source "wants" to pass a current Vin/R2 = 5/100 = 50 mA through the rheostat but the current is less - Vin/(R2 + Rrh)):

I = (Vin - Vrh)/R2 = Vin/R2 - Vrh/R2 = Idesired - Ierror

As you can see, the part (a kind of current) Vrh/R2 is the error, and the negative resistor adds exactly this current. Thus the negative resistor "helps" the input source in its "striving" to pass the "desired" current through the rheostat by adding an additional current to it. So, in this circuit, both the input source and INIC (the op-amp output through R4) pass currents in the same direction through a common load - the rheostat.

Now about the discrepancy with the Wikipedia article. There the simplest case is shown - a voltage source (with positive voltage) "loaded" by an INIC. In this simple case, really the INIC "pushes" back (sources) the current into the input source since there is no load to divert the current. I have illustrated this situation in the picture below by means of voltage bars and current loops; so you can see where currents flow:

If the input source had a negative voltage, the INIC would sink the current from the source:

I have explained in detail this topic in my stories below; maybe they would be useful for you:

Revealing the Mystery of Negative Impedance

How to Compensate Resistive Losses by Parallel Connected Negative Resistor

Linear Mode of Current Inversion NIC

Yes. You can use positive feedback techniques to make a negative impedance converter. http://en.m.wikipedia.org/wiki/Negative_impedance_converter

Such techniques can be used to do things like null out the large capacitance of a microelectrode so you can recover the high frequency response.