"... so the whole combination of the attenuator and the op-amp begins acting as an amplifier."

Can we really say that "the whole combination ...begins acting as an amplifier" ?

I rather think, we did nothing else than reducing the gain of an amplifier (opamp) by using an attenuator in the negative feedback path.

Up to now, I have no answer to the "greater challenge"-question.

But Lutz, Armstrong managed to answer this "challenge" still in 1912:)

Armstrong managed to answer this "challenge" still in 1912:)

So - surprisingly, he was even more ingenial than me. What was his solution?

It was based on the positive feedback...

Dear Lutz,

To be clear, let me explain (in a fully intuitive manner:) what I mean when asking this question.

I will illustrate my thoughts with an extremely simple example of an amplifier with 100% positive feedback that is implemented by a non-inverting amplifier with fixed-gain of 0.9 (see the attached picture and also the question, "What is the equilibrium point of a feedback system?"). For simplicity, the input voltage source is floating... so there is no need of a differential input. I have fabricated this conceptual circuit for the purposes of this question but it will do a good job also in the question, "What is the equilibrium point of a feedback system?".

In the beginning, assume that the input voltage of the whole circuit VIN, the input voltage of the fixed-gain amplifier VINA and the output voltage VOUT are zero... and the system is at rest.

Now imagine the circuit input voltage VIN sharply increases to 1 V. The amp input voltage VINA also  increases to 1 V... the amp output voltage VOUT - to 0.9 V... adds to VIN... so VINA increases to 1.9... and so on... During this process, the fixed-gain amplifier is actually not an amplifier... it is sooner an integrator... so VOUT/VINA ≠ 0.9.

The question is, "When will this process stop?"

My answer is - when the ratio between the amp output and input voltage becomes equal to its fixed gain, i.e. VOUT/VINA = 0.9. Then the voltages will be: VIN = 1 V, VINA = 10 V and VOUT = 9 V... the whole circuit will be at standstill... and the "equilibrium" will be reached.

Cyril - thank you. Of course, the result is not a surprise because the loop gain

Dear Lutz,

Thank you for the support of this fundamental question! I have asked it with the idea to complement all the questions about the positive feedback phenomenon by considering this less discussed its side.

Initially, the question was something like, "Can we make a low-gain amplifier act as a high-gain one?" Subsequently, I decided to make it more interesting, engaging, and even a little "provocative"... expecting that, in this way, I will provoke the interest of participants...

Although my intuition is very well developed over the years, I always experience difficulty when trying to imagine how this "magic" is possible - to obtain a high-gain amplifier by using a low-gain one.

Perhaps these difficulties are due to the "vicious circle" in which we find ourselves when trying to trace the path of the signal that circulates along the loop. Or maybe the problem is that my notion about amplification is of a multiplicative operation while here we use an additive operation...

As you have noted, to focus on the "pure idea" here, I have simplified extremely the feedback network... so the amp gain and loop gain are the same... Thus there is only one low-gain (imagine < 1!) amplifier ("attenuator":) and a "piece of wire" in this arrangement... but still it can amplify thousand times (>>> 1!)... What a magic! The expense of what is this? Maybe there is some "regenerative accumulation" here? Or something else?

The paradox here is that this amplifier actually is not a voltage amplifier - it decreases (attenuates) the voltage VINA at its input... but, regarding the circuit input voltage VIN, the output voltage VOUT is increased (amplified)!

I am no expert in active filters, but remember that there was a whole class of circuits realized exactly with such fixed-low-gain amplifiers instead with the perfect extremely high-gain op-amps. Maybe they are based on the same amplifying idea? Perhaps Josef will take a position on this issue?

What a magic! The expense of what is this? Maybe there is some "regenerative accumulation" here? Or something else?

Cyril - here is my attempt to explain the "magic effect": You are assuming an ACTIVE attenuator - that means: Amplifier with ideal properties (infinite input resistance, zero output resistance, gain of 0.9). 

Hence, we need a supply voltage that allows an output of 9 volts (referenced to ground!) - for example +/- 12 volts. Because of the supply voltages, the circuit now has a ground reference.

But we must not overlook the fact that the "input voltage" of 1V has NO ground reference. Hence, our classical gain approach output/input does not apply anymore. And we are not allowed to view the 9V output as a result of a ground referenced input.

I think, this clarifies the paradox.

Regarding active filters: No, I don`t think that the behaviour as discussed here has any relation to active filter topologies. The well-known unity-gain Sallen-Key filters have a basic gain value of unity - and the filter action (eventually with gain peaking) is caused by positive feedback only.

....to continue my response to the "magic" amplifier:

After thinking for a while, my analysis of Cyril`s circuit (gain of 0.9) is as follows:

For the complete circuit we cannot define any gain because there is no external input for connecting a signal to be amplified. It is rather a circuit which has reached a fixed and stable operating point. As mentioned earlier, the potential of the various nodes can be defined with respect to a "ficticious ground" only. This is the common ground which the required power supply also must be referenced to.

A similar situation does exist for the shown BJT circuit (with positive feedback). The gain of the common emitter block is A

Very interesting chain of thoughts, Lutz...

Exactly, the "attenuator" is actually a power amplifier since it can provide a significant current. Indeed, its gain (< 1) looks strange... and this makes the question intriguing...

I just do not understand why the floating input source is such a problem in this conceptual circuit. In practice, it can be a microphone or some sensor whose signal should be amplified... and it would be amplified. So there is an "external input for connecting a signal to be amplified"... and the input voltage source VIN is connected exactly to this input.

The BJT circuit is very interesting. The power supply of 6 V acts as a constant input voltage?

Regarding your remarks about grounding, I will share that initially I had an intention to consider the classic 2-resistor circuit feedback amplifier (with grounded input source). I even have prepared a geometrical interpretation based on the voltage diagram across the resistors in a form of a "scenario" (we can use it to illustrate our reasoning).

In this way of presentation, I consider the feedback network of two descrete resistors in series as one linear potentiometer. For simplicity, R1 = R2.

1. Non-inverting fixed-gain amplifier without positive feedback (although the resistor R2 is connected between the output and input).

2. Non-inverting fixed-gain amplifier with positive feedback (the resistor R1 is included).

3. For comparison, an inverting fixed-gain amplifier without negative feedback (although the resistor R2 is connected between the output and input).

4. For comparison, an inverting fixed-gain amplifier with negative feedback (the resistor R1 is included).

Since R1 = R2, the gain A of the amp is close to 2.

I just do not understand why the floating input source is such a problem in this conceptual circuit. In practice, it can be a microphone or some sensor whose signal should be amplified... and it would be amplified.

Yes - I agree. If the attenuator is an ideal active amplifying device (gain < 1) a floating voltage source in the pos. feedback loop will be amplified and appears as a ground refrenced voltage at the amplifiers output. 

An interesting case is when A = 0.5; then the circuit acts as a kind of voltage follower with positive feedback.

I am sure that if I show this circuit to my students, some of them will "naively" ask me, "Can not we replace the 'amplifier' with a gain of 0.5 with a voltage divider with the same attenuation of 0.5 (two equal resistors in series)?" And I would have to answer in a convincing manner...

In a similar way, applying the powerful "floating input source idea", we can make the simplest voltage follower with negative feedback.

Let's compare the two "followers":

• The negative-feedback single-ended follower is implemented by an amplifier with extremely high (> 105) but not precise gain. Its transfer ratio can be only one; it does not (slightly) depend(s) on the amp gain. It can be made > 1 by introducing an attenuating negative feedback network. By means of the negative feedback here, the extremely high gain of the amplifier is decreased thousands of times... up to one.
• The positive-feedback single-ended follower is implemented by an amplifier with extremely low (< 1) but precise gain. Its transfer ratio can be made bigger than one without introducing any feedback network; it is determined entirely by the amp gain. By means of the positive feedback here, the extremely low gain of the amplifier is increased up to one...

Both followers have common features:

• They do not need an amplifier with a differential input (only a single-ended one)
• They can act as "inverters" as well (if we reverse the input source)

It turns out the floating idea is really a powerful idea as it is proposed by the famous Dan Sheingold in the enthusiastic issue The Lightning Empiricist of Philbrick Researches in the distant 60's. The only difference is that there the power supply is floating (at the end of the paper)... while here the input source is floating...

http://www.philbrickarchive.org/1964-1_v12_no1_the_lightning_empiricist.htm

Aparna, you want to grasp the idea behind the Sheingold's circuits?

If yes, you can make us familiar with it:)

It would be interesting to make a "historical comparison" between the two approaches:

• A century ago, the problem was how to obtain a high gain. Then they solved this problem by increasing the low gain of the existing amplifiers with the help of the positive feedback. Unfortunately, the gain inaccuracy also increased...
• Nowadays, it is not a problem to obtain a high gain... even we have it in excess. The problem now is that the gain is not precisely defined. We solve this problem by decreasing many times the high gain of the present amplifiers with the help of the negative feedback. As a result, the gain indeterminacy also decreases...

So, you suggest to use a gain of 1 instead of 0.5 in the positive feedback follower to increase the gain up to infinity:)