I was reading through some notes a while ago and am struggling conceptually with the ideal op-amp philosophy. This started when I looked at the top figure below, I originally solved it using basic voltage divider/thevenin equivalent techniques. But when I looked at how my professor did the problem, he used the fact that the virtual ground at the inverting terminal means R3 is || with R2. Again this made sense to me, but I wanted to look at some other problems to understand this result which leads to the middle and bottom problems. I wanted to know why the load resistor on the op-amp in the middle problem isn't the same as shorting it to the virtual ground, sort of like we did in the top problem? Also, why does the professors technique in the first one work whereas it doesn't work in the middle/bottom?


simulate this circuit – Schematic created using CircuitLab


simulate this circuit

  • \$\begingroup\$ Why are you assuming that the output is 0V ? \$\endgroup\$ – Nedd Apr 20 '16 at 4:47
  • \$\begingroup\$ I am never assuming that the output is 0V, I am simply assuming we have an ideal opamp, and that the inverting terminal is 0V. \$\endgroup\$ – Mathephysicist Apr 20 '16 at 4:50
  • \$\begingroup\$ OK, seems you're labeling convention is just a bit confusing. "Vout" would be a much better convention \$\endgroup\$ – Nedd Apr 20 '16 at 4:53
  • \$\begingroup\$ @Mathephysicist: You've used a zero, V0, rather than an 'o', VO. Details matter. \$\endgroup\$ – Transistor Apr 20 '16 at 8:08

For calculating the voltage and current of R3 (of the second circuit) you could say that it is virtually in parellel with R2. But that is where the similarity ends.

A major part of analyzing an op-amp circuit is to use the feedback current flowing to (or from) the -input pin position to determine the circuit operation. In this negative amplifier configuration the feedback current is equal and opposite of the input current, this keeps the -input pin at a virtual ground (equal to the +input pin).

In the second circuit, the load resistor is "shorted" to the actual ground, so the current through that resistor does not affect the feedback section at all.

In the first circuit R2 being shorted to ground does affect the feedback current.

So your progression from the first circuit, to the second, then the third does not follow the correct idea in terms of the feedback current, so each circuit will operate differently.


One of the properties of an ideal op amp is that it can drive any load. In other words, the ideal op amp will supply whatever output current is necessary to achieve the output voltage set by the input and the feedback network.

Consider the middle circuit with and without a load resistor. Without a load resistor (i.e. \$R_{\text{L}} \to \infty\$) there is clearly only one path for the output current to flow -- into the feedback resistor, and this current also flows entirely into the input resistor since no current flows into an ideal op amp's inputs. This sets the relationship between the input and output voltages for the inverting amplifier. Adding a finite load resistance doesn't affect the feedback network nor the relationship between input and output -- it just means that the op amp needs to supply more output current (the usual current into the feedback network, as well as the current into the load resistor to satisfy Ohm's Law). Since an ideal op amp can output any current necessary, the load resistor does not affect the transfer function.

In the first and third circuits there are resistors added to the feedback network. The feedback is changed so the transfer function is changed. The first circuit adds \$R_2\$, which shunts current away from the normal path from input to output. The third circuit adds a second path for output current to flow to the op amp's input -- but this second current also flows through the input resistor \$R_4\$ so the transfer function is changed. The additional output current flowing into the load resistor in the second circuit doesn't actually flow through the input resistor.

Adding a load resistor does not affect the current flowing through the node at the op amp's inverting input. Adding resistors to the feedback network does, and that's why the transfer function is affected. Treating \$R_2\$ and \$R_3\$ in parallel in the first circuit is a convenient way to calculate the current flowing through the resistors, but they're not actually in parallel.


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