Electronic – How does the op-amp stabilize

Peter Bennett's got a point about the minimum and maximum voltages that can be present at the inputs, but I'm not certain it's the problem here. First off, we can look at the behaviour of your op-amp. With no input, it's being driven to the rails, as it's called. Basically, an op-amp's maximum range is it's positive and negative supply voltage, less a small amount, referred to as it's headroom. For modern op-amps, the headroom is usually less than a volt. For older op-amps, like the 741, the headroom required for the internal circuitry is much more. This should be specified on the specification sheet, if you're curious. I recommend http://www.datasheetcatalog.com/ as a source for that, although you can always get datasheets at the manufacturer's website. That describes the output voltage you're seeing from the op-amp. Rail voltage less some amount of headroom, likely a volt or so. It doesn't describe why you're seeing it, though. For this, you'll need to learn about how an op-amp actually works, and where a real op-amp differs from it's ideal model.

While I'm not certain of what your electronics background is, it seems like you're expecting the output of your op-amp to behave something like \$v^+ - v^- = v_{out}\$. That's an okay mental model of the op-amp, but it's missing one element, the gain. A better mathematical model is \$K(v^+ - v^-) = v_{out}\$, where K is a number. A really big number. Like, 100,000 or so (it's in the spec-sheet. Look for the Gain-Bandwidth Product). A real op-amp can't output at a voltage outside the range of it's supply voltage, so it's being driven to the rails hard. What you can do (and I'll add some schematics later) is add negative feedback to the circuit, which decreases the gain to a more manageable level. For a straight difference, you want the gain to be 1.

Finally, we want to know why the op-amp is outputting a voltage at all. For that, you need to know even more about the non-idealities of op-amps. Basically, everything inside that puppy is transistors. If I remember the 741 schematic, you're looking at 14 or so BJTs. I think wikipedia has an article that shows the schematic, if you're so inclined. At the input of the 741 is what's called a differential amplifier. This takes the difference between the two inputs, and outputs the result, where it's later buffered, most likely. The problem is that this takes at least four BJT transistors, and every transistor's going to be different. This creates small little dc biases in the differential amplifier, so even if you were to connect the \$v^+\$ and \$v^-\$ pins together, you would still see the op-amp driven to the rails. With a gain of 100,000, even 1 microvolt of difference is a big error.

As an aside, the 741 is an eight-pin package. Check out the spec-sheet, and you'll see that two of them, if I recall, are labelled null offset. These pins are actually used to slightly offset the differential amplifier, removing that error in the output. Newer op-amps are a lot more precise, and don't need to offer that functionality anymore. I still wouldn't recommend nulling the op-amp without reducing that gain first, though.

I'm going to hit you with something big.

^{simulate this circuit – Schematic created using CircuitLab}

WHAM! I stole that diagram from http://www.electronics-tutorials.ws/opamp/opamp_5.html . Give that whole string of articles a read, it's really great. This is an op-amp circuit that can be used to take the difference of two inputs. It's a bit of a handful to analyze if you don't have much of an electronics background. From a quick glance in my old textbooks, the output of that circuit is something on the order of $$v_{out} = \frac{v_2*R_1(R_3+R_4) - v_1*R_4(R_1+R_2)}{R_3(R_1+R_2)}$$ This is a fairly elementary circuit, and you should be able to find it easily in any university or college-level textbook, or online. Basically, if you want your subtraction operation with a gain of 1, just make all the resistors the same (I recommend 10K resistors). The equation reduces to \$v_{out} = v_2 - v_1\$.

At first, the principle of "virtual ground" can be applied during DESIGN of opamp-based amplifiers. This simplifies calculations - and the error is in most cases acceptable. Error? Yes - because there is always a differential voltage between both opamp inputs, which is exactly Vdiff=Vout/Aol. (Aol=open-loop gain of the opamp). Because of the large values for Aol (1E4...1E6 for lower frequencies) this diff. voltage Vdiff is in the µV range.

However, because this is not true for larger frequencies, the closed-loop gain will deviate from the calculated value for rising frequencies.

Regarding your last sentence: Yes - introducing additional delay in the feedback path will cause additional phase shift - and this can lead to instability/oscillations.

EDIT: "...until the virtual ground is re-established and the cycle repeats."

I suppose, with the above cited sentence you are asking for something like a "sequence" which leads to the steady-state conditions after applying an input signal, correct? This is, indeed, a question which deserves some explanations.

Example: Inverting opamp-based amplifier with a gain of "-2". Input: +1V step (t=0).

At the very beginning (t>0), the feedback is not yet active and the output will jump to the maximum negative voltage (supply rail). Now the feedback network causes the inverting terminal to become negative - and the output starts to go to positive voltages. However, this will not continue again and again because the opamp has internal delay elements (causing bandwidth limitations and phase shift). That means: The output does not "jump" to other values but it takes some time to reach the upper rail. But, in reality, the output will NOT reach the upper rail because on the way to the maximum positive output the output voltage crosses some finite negative values - and for an output value of app. Vout=-1.999V there will be an equilibrium between input and output. Explanation:

Vout=-1.999V and Vin=+1V cause a very small voltage between both resistors (at the inv. input terminal) which - when multiplied with Aol - is exactly the assumed output voltage (in the example: Vout=-1.999V.) This equilibrium state is stable.