I'm a Computer Science major, and have a class called Analog and Digital Circuits.
I'm learning about the op-amp now. The inverting op-amp has something that I just can't comprehend. I understand the algebra that leads to Vout = (-R2/R1)×Vin, but the feedback Vout is going into V-. V- and V+ are about 0. How can Vout be 0 V, and another, bigger value at the same time?
Such fundamental questions have always made me think about the philosophy of negative feedback op-amp circuits looking for where it all starts. This is the kind of fancy story I wrote these days, in which I show the evolution of these circuits. For the purposes of intuitive understanding, I have used descriptive and figurative names of the devices under consideration but have shown their relationship to conventional names.
I have built the circuit solutions step by step and examined them using CircuitLab. To keep the schematics simple and clear, I have not used named nodes and ordinary DC simulation but a live DC simulation. So hover your mouse over elements and their terminals to see the quantity values.
As a result of much observations and reflection, I have come to the conclusion that the initial negative feedback circuit is a voltage follower with zero input voltage. To make this "circuit", we just need to connect the output of an inverting amplifier to its input.
simulate this circuit – Schematic created using CircuitLab
There is hardly a simpler "circuit" than this, because it consists of only an amplifier, which may not even be an op-amp with a differential input but a humble amplifier with a single-ended input (if there was one in the CircuitLab library, I would use it). And what is most interesting about this most elementary implementation is that no high gain amplifier is needed. If you open the op-amp properties in CircuitLab, and start setting the gain lower and lower, you will see that the circuit works (produces zero voltage) even at a gain of 1. This is because there is no disturbance and the op-amp does its job effortlessly. It watches the input intently, and every time something tries to change the zero voltage, it changes its output voltage in the opposite direction (actually the two voltages are the same).
It is a bit hard to imagine what we can use such an output-only circuit (zero voltage source) for... maybe as a virtual ground. So let's try to find an input where to apply an input voltage. And here, as above, with much thought I eventually realized that this actually means finding an entry point where we can perturb the op-amp... disturb its "peace":-) Since it has set itself the goal of maintaining zero voltage, any attempt to change this voltage represents a disturbance to it, which it tries to compensate for. This forces it to change its output voltage in proportion to the disturbance. For example, we can insert the "disturbing" input voltage between the input and the output.
At the first moment, the op-amp fails to respond and its output voltage remains zero. Because we have connected the positive terminal of the input source to the output, the op-amp input voltage (at the inverting input) becomes negative. So the op amp begins to vary its output voltage in a positive direction until the ratio of its output to input voltage equals its open-loop gain. Open the op-amp properties in CircuitLab and set various gains from one to several hundred thousand (1 ÷ 300,00 or more); then hover your mouse over the op-amp input and output.
As you can see, as the gain increases the op-amp keeps the ratio between its output and (differential) input voltage equal to its open-loop gain. The op-amp input voltage remains negligibly small and the output voltage is almost equal to the circuit input voltage Vin.
The voltage follower we "invented" is great, but there is a problem - the input voltage source is "floating" (not connected to ground). A simple rule of thumb is that out of three devices connected in series, at least one must be floating. Now it is the turn of the op-amp input - ie. to ground the input source we need a differential input. Ah... that is why it was invented back in the day!
As above, at the first moment, the op-amp fails to respond and its output voltage remains zero. Because the op-amp input voltage (at the non-inverting input) becomes positive, the op amp begins to vary its output voltage in a positive direction until the ratio of its output to (differential) input voltage equals its open-loop gain.
Now set various gains and hover your mouse over the op-amp input and output. As you can expect, the op-amp input voltage remains negligibly small and the output voltage is almost equal to the circuit input voltage Vin.
Then run a DC sweep simulation varying the op-amp open gain. For example, in the graph below, it only changes from one to 100 and yet the op-amp input voltage drops to 10 mV. You can only imagine what it will be like when amplified 300 thousand times!
There is a solution that combines the advantages of the above two - to disturb the voltage follower through a current source; then both the source is grounded and the op-amp input is single-ended. For this purpose, we can try to change the op-amp input voltage (disturb the op-amp) by the input voltage source Vin through a resistor R1. Figuratively speaking, the input source is trying to "pull up" the op-amp inverting input through a resistor R1 but the op-amp output does not let it by "pulling down" to ground the inverting input through a conductor with zero resistance (a piece of wire). As a result, nothing happens.
To help the input source, we can make it difficult for the op-amp by including a resistor R2 between its output and the inverting input. Thus, the two sources are placed under the same conditions. As in the game "tag of war", they "pull" the common point with the same "force" but in different directions - the input voltage source Vin "pulls up" through R1, and the op-amp output "pulls down" via R2.
We can make the inverting circuit more attractive if we replace the two discrete resistors R1 and R2 above with a linear potentiometer P.
With this arrangament we can obtain a classic family of curves Vout = f(Vin) at four wiper's positions (K = 0.2 - 0.8).
But it is most interesting to investigate the voltage distribution along the pot's resistive film (aka voltage diagram). Thus we can see in another spatial way the "static error" (small voltage) at the op-amp input caused by the insufficient open-loop gain. I have shown below how we can do it with the help of CircuitLab and my recent invention - the "copy" potentiometer P2. If we connect it in parallel to the existing potentiometer P1, and move its wiper from one to the other end, we can measure the local voltages inside the potentiometer P1. We can then run a DC sweep simulation with parameter K.P2 and observe the voltage distribution.
In this conceptual circuit, I have used a generic op-amp, currently set to a very low open loop gain of 10x. With it, I have made an inverting amplifier with a gain of -1, "moving" the P1's wiper in the middle (its r2 = r1, K = 0.5). I have also moved the P2's wiper in the middle (its r2 = r1, K = 0.5).
As we can see, when "moving" the P2's wiper from left to right (changing the potentiometer's transfer ratio K from 0 to 1), the voltage drop linearly decreases from 1 V to -1 V. This is the voltage diagram of P2 that is a copy of the P1's voltage diagram. So what we see in P2 is the same as in P1. We can open the op-amp properties and start changing the gain from 1 to 300,000 or more while watching the op-amp input voltage. Then run the DC sweep simulation with parameter K.P2. As a result, you will see the voltage diagram below where the middle wiper's location (K = 0.5) is indicated by a thin vertical line in black and the zero voltage level - by a horizontal line in red.
As you can see, the local voltages along P1 (P2) linearly decrease from 1 V to -00 mV. Why do not they reach -1 V?
The reason is the very small op-amp gain factor (only 10). We can figuratively imagine the voltage diagram as a kind of Archimedean "lever" - the input voltage "raises" it on the left, and the op-amp output voltage "lowers" it from the right. But because the op-amp has very little gain, it does not have enough power to pull it up enough. There is an op-amp input voltage ("static error") of 98 mV at the midpoint of the potentiometer that is quite different from the zero voltage of a virtual ground, i.e. there is no virtual ground. In fact, there is a virtual ground, but it has "entered" inside to the right of the P1's wiper. Also, the circuit (inverting amplifier) gain is -0.8 instead -1.
How does an inverting op-amp feed back 0 V when Vout is not 0?
The answer is, you assume an 'ideal' op-amp has infinite gain. When there is negative feedback, the feedback path completely nullifies any voltage difference between the two inputs. Thus, in the theoretical ideal op-amp case the input difference is always zero. (We'll get into how that can be, below.)
With this assumption, the inverting gain is indeed just -Rf/Rin.
In the real world, op-amp open-loop gain, \$Av\$, is not infinite, but it’s still very high. Nonetheless, real-world op-amps needs some input voltage difference to operate.
That is:
Say your op-amp has an open-loop \$Av\$ of 1 million (10^6). To obtain 1V on \$Vout\$, it needs to see a voltage difference of one microvolt at its Vp/Vn inputs.
In the presence of negative feedback, the op-amp strongly amplifies its input, trying its very best to cancel the input difference. But, being non-ideal, the op-amp still needs to see a little bit of input voltage to do it.
How little? That residual, not-fully-cancelled voltage will be equal to Vout/(open-loop gain), regardless of the negative feedback in use. That is,
A few things to notice here:
To understand the non-ideal case, you can play around with this idea in a simulator, here.
If you need a deeper dive, I recommend this MIT Open CourseWare curriculum which includes several modules on op-amps including this one.
Related: Why is op-amp output not zero if inputs have the same voltage?
In each of the following circuits, gain is \$\frac{V_{OUT}}{V_{IN}} = -\frac{R_2}{R_1} = -2\$. Check that each output and input conforms to this relationship:
simulate this circuit – Schematic created using CircuitLab
Notice how node Q in every case is 0V, as shown on VM1.
To understand why this is, I will draw each circuit without the op-amp, and instead of relying on an op-amp to impose some potential at OUT, I will do that myself, with sources V2 producing the same potentials that the op-amps above did:
Now each circuit is a simply a resistor potential divider, between two explicit potentials. In every case, the potential at Q, the feedback voltage which you are asking about, is still 0V.
You can prove for yourself that this is correct, by employing the equation for that potential at Q, \$V_Q\$:
$$ V_Q = V_{IN} + (V_{OUT}-V_{IN})\frac{R_1}{R_1+R_2} $$
Feedback is zero, whether it's the op-amp providing an output to achieve this state, or me or you or something else explicitly applying the appropriate \$V_{OUT}\$.
The reason is the resistor potential divider formed by R1 and R2. That's its job, to apply at node Q a potential somewhere in between \$V_{IN}\$ and \$V_{OUT}\$.
With the op-amp in place, and with Q being its inverting input, we call this negative feedback. If \$V_{OUT}\$ rises, so does \$V_Q\$, but only by some fraction of the change in \$V_{OUT}\$, a fraction determined by the ratio of R2 and R1. If \$V_{OUT}\$ falls, so does \$V_Q\$.
If \$V_Q\$ is only a few microvolts positive, the output would be many full volts negative (due to the op-amp's open loop gain \$A\$ being ridiculously large). If \$V_Q\$ were just a few microvolts negative, the output would be many volts positive.
The op-amp ultimately settles at some output for which the voltage applied at Q by the potential divider will be only microvolts. For all intents and purposes, zero.
In an inverting amplifier it is the current between the output and the inverting input that achieves negative feedback. The voltage at \$V_{OUT}\$ changes so that the voltage difference between the inverting and non-inverting inputs will become very small, ideally zero. When this happens, the current flowing through the feedback resistor will be just equal to the current flowing through the input resistor. If the non-inverting input happens to be connected to ground then the voltage at the inverting input will also be essentially zero.
EDIT: To clarify, I'm talking mostly about ideal op amp behavior. Check the specifications of any real op amp to ensure that the output swings all of the way to ground.
Look more closely at your statement
V- and V+ are aprox 0.
V+ is not "approx 0" it is normally wired directly to 0.
However the feedback voltage V- is not zero.
Check the "open loop gain" of your opamp : it isn't infinite; let's say it is about a million. And let's say Vout = -10V : to get that, V- must be 10 uV.
Unless you have a very expensive multimeter, that's close enough to 0V you simply can't measure it : but it's still not 0.
Consider the case where R1 = R2, so gain = -1. V- is simply the difference between Vin and Vout. It's not precisely 0, and therefore Vout isn't precisely -Vin, but it's close enough.
If you used an opamp with an open loop gain of 1000 or so, then V- would be about 10mV for the same -10V output, and you'd be able to measure that.
So your initial statement is wrong : it's a simplification that's near enough to make the analysis simple, but accurate enough to be within measurement limits.
simulate this circuit – Schematic created using CircuitLab
Consider the circuit above. An Op-amp has the property that if voltage at the non-inverting '+' input higher than that at the inverting '-' input (\$V_n\$ in this diagram) it will try to make the output (\$V_{out}\$ in this diagram) as high as possible. Similarly if voltage at the non-inverting '+' input lower than that at the inverting '-' input (\$V_n\$ in this diagram) it will try to make the output (\$V_{out}\$ in this diagram) as low as possible. Op-amps have a high but finite gain.
The second property to consider is both inputs take and supply negligible current and in most cases we assume this to be zero.
Lets assume the gain to be near infinite and set Vin to 1 volt. Since we have zero volts on the '+' input then if \$V_n\$ were >0V \$V_{out}\$ would go very low pulling \$V_n\$ negative which would cause \$V_{out}\$ to go very high. There is a balance point however where \$V_n = 0 \text{ volt}\$. This gives you the current in \$R_1\$, the same current must be following in \$R_2\$ so it should be clear that.
$$V_{out} = - V_{in} \cdot \frac{R_2}{R_1}$$
There are specs for maximum linear range for inputs (Vcm = V_common mode)and outputs (Vo max,min) before Op Amps can work as analog devices. This is especially true for bipolar junction transistor(BJT) type OP AMPs within 2V of the +/- supply voltages where something is usually saturated or cutoff.
(CMOS Type OP AMPs tend to be Rail to Rail (RR) more ideal and lower input but also lower output current)
Inverting "Negative Feedback"(NFB) always tries to force the input difference voltage =0 thru the NFB ratio. (e.g. Rfb/Rin)
When the output is in the analog or linear output voltage range, the input difference will always be 0 or defined by Vio, Iio spec (input-offset).
When any input changes, the Vout changes thru NFB ratio to keep the input error =0 according to your "analogic".
