I'm aware of the theory of a "virtual ground" at the summing point of an op amp feedback network and the inverting input terminal for the amp itself.

According to the theory, the virtual ground at the summing point does not have a path through the amp's input resistance to the grounded non-inverting terminal of the amp.

Never-the-less, the theory says that current does flow through the negative feedback resistor (and the input resistor as well) to the virtual ground.

So if the virtual ground is not really connected to ground, what is the actual path for the current through the input and feedback resistors?

 

The actual path is through the input and feedback resistors. Current through the input resistor is nulled by an equal and opposite current through the feedback resistor...or did I misunderstand your question?

"Virtual ground" is just a way to remember that the inverting input is always the same voltage as the non-inverting input in a closed loop system.

 

So in the ideal op-amp case, no current actually flows into the input terminal. The op-amp output automatically adjusts - by virtue of the feedback - to ensure the input terminal voltages are (ideally) the same.

 

As #12 said, the feedback resistor keeps the DC current path open. When the output is high, the current flows From output TO the non-inverting input and when output is low, current is drawn away from the inverting input.

 

As #12 said, the feedback resistor keeps the DC current path open. When the output is high, the current flows From output TO the non-inverting input and when output is low, current is drawn away from the inverting input.

Speaking in terms of, "conventional flow", right?

That always makes me stop and think twice because I think equally well in both modes and I have to remember the rules to figure out which one you're using when you phrase things like that.

 

According to the theory of virtual ground, the inverting and non-inverting terminals of the amp are at the same voltage. Furthermore, if the input resistance between these terminals is infinite, no current can flow into (or between them).

The gain calculation relies on a voltage drop across the input and feedback resistors and in order to get a drop across the resistors, currents must flow through them to the virtual ground.

However, in addition to a voltage source, a current requires a closed circuit. It does not make sense to say that if two points (the inverting and non-inverting terminals) are at the same voltage, that also implies another supposedly isolated point at a different voltage can cause a current to flow.

 

Just ignore the words, "virtual ground" for the moment.
If one volt is applied to a 1k resistor and the other end of that resistor is connected to point A, then negative two volts is applied to a 2 k resistor and the other end of that 2k resistor is connected to point A, what is the voltage at point A? It's zero. It's virtually ground.

Where did the negative 2 volts come from? From the output of the op-amp. How does the op-amp know to do that? Because a real op-amp is not an ideal op-amp. There really is an internal connection between the input pins and the output pin.

 

Just ignore the words, "virtual ground" for the moment.
If one volt is applied to a 1k resistor and the other end of that resistor is connected to point A, then negative two volts is applied to a 2 k resistor and the other end of that 2k resistor is connected to point A, what is the voltage at point A? It's zero. It's virtually ground.

Where did the negative 2 volts come from? From the output of the op-amp. How does the op-amp know to do that? Because a real op-amp is not an ideal op-amp. There really is an internal connection between the input pins and the output pin.

This is similar to the explanation that I developed.

If you assume there is some connection between the inv. and non-inv. input terminals (and some current can flow), I was able to successfully derive the formula for gain control with feedback.

 

Please try not to drive yourself crazy trying to reconcile "ideal" models with real parts.
Human models are almost always incomplete, but it is physically impossible for electrons to lie.

 

I'm a bit surprised one has to allow for some non-ideal behavior to do an analysis of this type.
There are many cases in the teaching of circuit theory where ideal controlled sources are routinely included to test the student's analytical skills.
Granted, one may have to deal with limiting cases as parameters such as gain & input impedance become infinite. But that's generally manageable with some algebraic manipulation and testing what happens as the limits are approached.

With respect to the idea of virtual ground - this is really only a truly valid notion when the op-amp is considered to be ideal - specifically infinite open loop gain. Otherwise, one must assume a finite difference in potential exists between the amplifier input terminals.

 

Referring to the previous post, a formal mathematical approach does assume the op amp has finite input resistance and finite gain.

Then the formula is refined by applying the "limit" as the input resistance and the gain approach infinity. However my approach is to derive the formulas using practical reasoning rather than using rather abstract mathematical analysis.

In other words, "Don't make things more complicated than they have to be".

 

This is similar to the explanation that I developed.

If you assume there is some connection between the inv. and non-inv. input terminals (and some current can flow), I was able to successfully derive the formula for gain control with feedback.

You don't need to have any current flow and, even if you do, it doesn't address your question because the amount of current that flows is minuscule compared to the current flowing in the input and feedback paths.

You asked where the current that flows in the feedback resistor goes to (or comes from). As has been said already, it is sourced or sunk by the output of the opamp.

Using the phrase "virtual ground" can get you in trouble because the only reason that the inverting input is at 0V in this case is because the non-inverting input is grounded. In general, you want to think in terms of a "virtual short" between the two input terminals -- it's a short in the sense that both are at the same voltage (whatever it turns out to be) and it is virtual in the sense that no current can actually flow between the two (via the opamp input terminals).

 

Referring to the previous post, a formal mathematical approach does assume the op amp has finite input resistance and finite gain.

Then the formula is refined by applying the "limit" as the input resistance and the gain approach infinity. However my approach is to derive the formulas using practical reasoning rather than using rather abstract mathematical analysis.

In other words, "Don't make things more complicated than they have to be".

There's no reason to assume finite input resistance at all or, for that matter, to assume finite gain and take a limit.

What formulas are you trying to derive? For the gain of an opamp configured in the classic inverting amplifier configuration in which the non-inverting input is grounded and with an input resistor Ro between the signal input, Vin, and the inverting input and a feedback resistor, Rf, between the inverting input and the output?

Reasoning according to the ideal opamp model - the gain is infinite which requires that in order to have a finite output there must be no voltage difference between the two input terminals. The input resistance of both terminals is infinite, hence zero current flows into or out of either input. Thus the current that flows from the signal input through the input resistor is

Iin = (Vin-0V)/Ro

The current that flows from the inverting input through the feedback resistor is

If = (0V-Vout)/Rf.

Since these currents are the same, namely Iin=If, we have

Vout = -(Rf/Rin)Vin

It doesn't get much simpler than that.

 

Speaking in terms of, "conventional flow", right?

That always makes me stop and think twice because I think equally well in both modes and I have to remember the rules to figure out which one you're using when you phrase things like that.

Yes, I meant conventional flow.

Back to @Glenn Holland 's issues.

If you want to think of how this whole thing works, just look at a differential amplifier - the so-called "long tail pair" with PNP transistors. The emitters are connected to each other and those emitters are fed with a constant current of, for example, 100 uA. The bases of each PNP are your inverting and non-inverting op amp. Only the first PNP is connected to the op amp output!



Now, nothing says the voltage at each input has to be the same (and it is not when an op amp is connected as a comparitor with open loop gain). The way an op amp is designed, however, is to self regulate and TRY to keep both inputs at the same voltage (no matter what that voltage may be (within the operating limits of the op amp). Some people call that "virtual ground" but that is a stupid term from my point of view. Now, an op amp cannot self-regulate without some external circuitry or "feedback".

In theory, our two PNP transistors have infinite gain and the output of the first PNP would only slam back and forth between the limits of the power supply every time the input of that first PNP goes above or below the input voltage on the second PNP (inverting input) like a comparitor.

To combat this infinite gain, feedback is used so it can amplify an input signal. The output is connected to the input of the second PNP. Now, the output can control the current flowing through the second PNP. Since the two PNPs SHARE a constant current source, the output forces some current through that second PNP and limits the current flowing through the first PNP and, therefore, limits the gain of the first PNP.

So, if you need to think about what is happening in an op amp, think about it with the proper components - not as an overly simple resistor.