Electronic – Current direction in a multivibrator

I guess by "complete the feedback loop" you mean "hold the inverting and noninverting inputs at the same voltage". This is basically the op-amp's only goal in life, and given suitable negative feedback, it will accomplish it. If it can't, then it will drive the output into one supply rail or the other attempting to do so.

So, why can IC1 accomplish this, while the astable multivibrator can not? Let's consider the essential components of each:

^{simulate this circuit – Schematic created using CircuitLab}

Now consider the definition of capacitance:

$$ I(t) = C\frac{\mathrm dV(t)}{\mathrm dt} $$

It might make a little more sense algebraically re-arranged:

$$ \frac{\mathrm dV(t)}{\mathrm dt} = \frac{I(t)}{C} $$

That says, "the rate of change of current with respect to time is equal to current divided by the capacitance". So, if you put 1A through a 1F capacitor, voltage changes at a rate of 1V/s. If you increase the current or decrease the capacitance, voltage will change faster. To get voltage to change instantly, you need infinite current or zero capacitance.

For IC1, it's easy for the op-amp to respond to any change in the input. The voltage across a capacitor wants to remain constant -- it takes time and current to change it. If in some instant the input voltage increases by 1V, the output can increase by 1V, and instantly the inverting input also increases by 1V, and the two inputs have the same voltage. Mission accomplished.

But what about IC3? Say the input increases by 1V instantly. What can the opamp do? It can increase the output voltage, but the voltage across C2 (and thus, at the inverting input) can not change instantly. To change it instantly would require infinite current. But that's impossible, because the current the op-amp can drive through the capacitor is limited by R1.

So instead, the op-amp will do the best it can and saturate the output at the positive supply rail. Eventually, it will manage to charge C2 to match the voltage at the input, and the output voltage will go to 0V.

To make an astable multivibrator, you add positive feedback so as the output starts to settle to 0V the input voltage also changes. Thus, IC3 (with positive feedback added) can never accomplish its goal. It's always trying to catch up, and every time it succeeds, it starts another cycle.

It is not dependent upon how the current flows at the output of the opamp.

We can treat the output pin as having a very low output impedance.

The calculations determine by how much the inverting input changes in voltage due to the current flowing into (or out of) the inverting input itself. This is determined by the effective resistance of the feedback network.

By putting the same value resistor on the non-inverting input we can compensate for that error.

For example if R1 and R2 were both 2K, the effective resistance at the input would be 1K. (the two are effectively in parallel and the output pin is assumed to have zero resistance).

If the amplifier had an input bias current of 1uA this would cause a 1mV change in the voltage at the input that would cause an error as the output would have to change by 2mV to make the non-inverting input match the inverting input.

If however we put a 1K resistor in series with the non-inverting input as well, it would also change its voltage by 1mV in the same direction and cancel out the error.