# Why can't the following configuration be considered negative feedback?

In the following circuit I have kept the resistor ratio such that beta factor of negative terminal (part of output will come on inverting terminal by voltage division), is higher than that of positive terminal. So net feedback should be negative. But my opamp is still hitting the rail. What is the reason?

• Looks like V2 is set to -2V and it should be set to 2V ( so that the -ve terminal is set 2V below GND ) Commented Jun 23 at 18:01
• @Antonio51 how do you calculate that at a glance? Commented Jun 23 at 18:09
• @YellowYeti yes you are right, thank you! Commented Jun 23 at 18:11
• @needbrainscratched So you are good now that the supply rail is set right? By the way, just because, I find $V_{+} =\frac13 V_{ _\text{OUT} }$ and $V_{-} = \frac23 V_{ _\text{OUT} }+\frac13 V_3$ and these must equal each other, so $-\frac13 V_{_\text{OUT}}=\frac13 V_3$ or $V_{_\text{OUT}}=-V_3$. Commented Jun 23 at 23:42
• @needbrainscratched, thanks for the response! I expected you to show interest and started preparing my answer. So far I've made schematics and I'm currently writing the explanations. I'm putting it in this unfinished form so you can think about the circuit idea. The name of the circuit is a little scary - "voltage inversion negative impedance converter" (VINIC), but I will try to dispel the mystery around it :-) Commented Jun 24 at 14:17

# A chance to understand the circuit

It is a well-known fact that problems drive the progress of humanity. Here is how a wrongly connected power source gave us a chance to ponder over this weird circuit.

# Making a perfect ammeter

The best way to understand such a sophisticated circuit solution is to solve some simple practical problem with it. An interesting problem is how to turn an imperfect ammeter (with some internal resistance) into an "ideal" ammeter with zero resistance with the help of the negative resistance.

## Imperfect voltage source

Imagine you have an imperfect (1 V) voltage source Vin with a high (1 kΩ) internal resistance Rin (in the shematic below it is represented by a 1 kΩ resistor in series). If you connect a (perfect) voltmeter to its terminals, it will show exactly 1 V Vout because no current flows and there is no voltage drop across Rin.

simulate this circuit – Schematic created using CircuitLab

## Perfect ammeter

If we short-circuit such a "bad" voltage source, the so-called short-circuit current will flow. It is an important parameter of the real voltage source, and since we do not always know the exact value of the internal resistance, it is of interest to measure the current. For this purpose, we insert a perfect ammeter IA, and it shows IA = Vin/Rin = 1 mA (Ohm's law).

simulate this circuit

## Imperfect ammeter...

But if our ammeter is imperfect and for example has a high internal resistance of 1 kΩ, it is added to the source internal resistance, and the current we read is less - IA = Vin/(Rin + RA) = 0.5 mA. Also, the voltage drop across the imperfect ammeter is not zero as above but 0.5 V. To set the ammeter resistance, open the CircuitLab parameters window and enter 1 kΩ.

simulate this circuit

## ... improved by negative resistance

Obviously, the ammeter resistance is something harmful (loss) and we need to compensate it with something useful (gain). We can imagine it as a "negative resistance" -RA which has the same value as the positive ammeter resistance RA, but instead of losing voltage I.R across it, it adds the same voltage I.R to the input voltage Vin. For now, this is just an idea that we have yet to implement, but nothing prevents us from testing it in CircuitLab. We know that the resistances of resistors in series add up, but in this case they will subtract and cancel each other out. So we connect another 1 kΩ resistor in series and set its resistance negative. Also, we connect another voltmeter in parallel to -RA.

simulate this circuit

The result is amazing: 1 V voltage is lost on the ammeter but the negative transistor adds 1 V voltage, and the net voltage is zero!

## Virtual short

The combination of an ammeter with positive resistance and the same negative resistance behaves as a "piece of wire"! We can call it "virtual short" because it is not true wire but contains also a voltage source. Thus we have made a perfect ammeter by helping an imperfect one.

simulate this circuit

## Negative impedance converter with...

The above was conceptual. But now the question arises how to make a negative resistance. It has already become clear that, in fact, the negative resistor is not a resistor, but a variable source that changes its voltage in proportion to the current. So we need to make a voltage source copy the behavior of the resistor but reverse the voltage. To do this, we can double the voltage drop across a positive resistor and add it as a voltage. Since the same current flows through both resistors, I.R - 2I.R = -I.R.

... behavioral voltage source. We can do it in CircuitLab simulations by the so-called "behavioral voltage source" 2VR whose voltage is two times higher than VR. Thus it acts as a -2 kΩ "resistor". 1 kΩ is neutralized by the "sample" positive resistor R=RA so the net resistance is negative (-1 kΩ).

simulate this circuit

... fixed-gain amplifier. In electronic circuits, voltage sources are usually amplifiers. So, we can represent the "doubling voltage source" by an amplifier with a gain of 2. Note that its input is differential and the output is single-ended.

simulate this circuit

... op-amp. The best implementation is through an op-amp circuit with feedback. However, negative feedback alone is not enough here; positive is also necessary. This should be considered into further...

simulate this circuit

# Why positive feedback?

This is an extremely interesting question to which I will first answer verbally.

• If we introduce negative feedback into a moderately high gain (inverting) amplifier, we can virtually decrease its input resistance.

• If we start increasing its gain all the way to "infinity", its input resistance will drop to almost zero (the so-called virtual ground), but we can't do more than that with negative feedback.

• For the resistance to change sign we need to start introducing positive feedback. It forces the voltage at the inverting input to drop below zero.

The following phenomenon occurs:

• If we increase the current through a positive resistor, the voltage across it increases in the positive direction.

• In the case of negative resistor, however, the voltage changes its polarity and "moves" in the negative direction. This means it, becoming negative, is added to the input voltage (travelling the loop, the signs are - +, - +).

I have also created an AI-enhanced version of this answer in a dialog with artificial intelligence.

Calculated gain with Maple sheet

• What in the world? I didn't know Maple could do that... Very interesting. So it automatically scans the picture from the URL and can provide an equation? Commented Jun 23 at 20:33
• No. You have to write the equations and then it solves these ... But there is Syrup under Maple that can do that. Just draw figure ... Commented Jun 24 at 7:17

V2 should be changed to +2V because you are connecting negative terminal of the battery to Negative power supply of the opamp.

If you are connecting positive terminal of the battery to Negative power supply of the opamp then you need to set the power supply as -2V.

Gain of this circuit can be calculated as follows.

V(-) = ((Vout20)/30) + ((Vin10)/30) ; Voltage at the inverting input

V(+) = ((Vout*10)/30) ;Voltage at the non inverting terminal

In a negative feedback system V(-) = V(+) .

If you equate the above two equations you will get

Vout = -Vin