Inverting Op Amp: Why does current flowing into output seem to magically reappear at ground?

Question

I'm trying to understand the flow of current through the inverting op-amp circuit. I've created a simulation:

The thing that is baffling me is: How come the values of PR1 (op-amp output) and PR2 (ground) are the same? How can the current flow INTO the op-amp output, and then OUT of ground??? Are they somehow connected internally?

Edit 1: I added the resistor between output and ground.

Edit 2: Added the current probes for power supply. Now I'm even more confused...

Answer 1

Trace both current paths from output, and you find they both end up at ground. Via R3 directly, and via R2, R1 and V1.

So if the output sinks current I, that current must come from ground. Simple as that...

The more interesting question, is what happens to it inside the opamp. From the opamp output there is no direct connection, internally, to ground. Instead, there are connections via both the V+ and V- terminals and your external voltage sources V2,V3 to ground. So inside the opamp, the output current ought to appear as an imbalance between the currents supplied by V2 and V3, and that imbalance should return the missing current to ground, closing the loop.

One clue is the supply currents themselves which are identical, enormous (for a 741) and look like worst-case ratings. A more realistic value for these would be 3-5mA (+ the output current on V3 in this case), up to a maximum of 24mA, at which point the opamp would fail to deliver all the required current (say, 20mA) at the output.

So it looks as if your opamp model is too simplistic to reflect this behaviour, which you would undoubtedly observe in a real circuit. Is that Robert Pease I hear cackling in the background?

Answer 2

I'm pretty sure the Boyle macromodel for 741 is causing this. In Op Amp Applications Handbook (ed. Walt Jung) p. 742 the MPZ model (well AD's flavor thereof) has this praise

output load current is correctly reflected in the supply currents. This feature is a significant improvement over the Boyle macromodel [...]

And here's the [741] Boyle schematic from Some Practical Aspects of SPICE Modeling for Analog Circuit by E. Kennedy (published in Analog Circuit Design, Art, Science, and Personalities Edited by Jim Williams).

The article has a longer description, but it's pretty obvious just from the schematic that the output current is generated relative to the ground not to the rails. The paper says "most of the open-loop gain is obtained with the VCIS dependent-generator G_b", which you can see is connecting the output to ground (via R01).

There's another obvious DC path from output to ground via (R01, R02); in fact that's how the opamp output impedance is simulated.

Finally the two voltage sources near the output (Vp and Vn) are not simulating a push-pull stage (as claimed in another answer here). They just limit the output voltage so that it saturates to some value[s] chosen to be less than the rails. This is explicitly said in the paper on page 305 and even repeated on page 309. Also, Vp an Vn are not controlled, but fixed sources.

Answer 3

The current flowing from the ground node only has two paths- along R1 and R2, or through R3. Once these meet at PR1, the only other way they can go is the op-amp output. So the current at ground has to be equal to the current at the output. Mathematically, you could demonstrate this with Kirchoff's current law, which states that the currents into a node must sum to zero.

Answer 4

An op amp output is a pair of transistors, BJTs or FETS, in a push-pull configuration that source or sink current from the power pins. Something like:

^{simulate this circuit – Schematic created using CircuitLab}

So in your post, the output circuit is:

^{simulate this circuit}

That current has to return to the other side of the power source - through the GND.