effect of inverting input on circuit load?

[stellar]

when analyzing a circuit, what is the effect of an opamp utilitzing the inverting input? I am confused by this....there is no voltage at the input... say you have a source with two 10k resistors in paralell, one follows to ground and the other to an inverting input of an opamp, what load does the source see? 5k? or is it 10k? or something else? do you treat the opamp as ground or as a loose end?

[Led]

If they're in parallel it should be 5k but I've always been unsure about this paticular situation and would like to hear what the more experienced among us say.

[JohnRoberts]

Simple or first order analysis is that the - input is a virtual earth so pretty much same as connected to ground.

As you approach the gain bandwidth limits of the opamp this virtual earth gets softer as it takes a larger error voltage at that input to swing the feedback network in opposition.

With most audio applications at audio frequencies you can ignore gain bandwidth effects on this characteristic.

JR

[David Kulka]

The above answers are good, let me just add that the inverting input is not zero impedance unless a feedback resistor is connected. Otherwise it's super high impedance, just like the noninverting input.

The virtual ground thing is little mind boggling but one way to make sense of it is this: Imagine a unity gain opamp. +5 volts is connected to the inverting input via a 10k resistor. There will be -5 volts on the output, which also has a 10k resistor connected to the inverting input.

The junction of the two equal resistors -- one connected to +5 volts and one connected to -5 volts -- must be zero volts, thus there is a virtual ground at that point. No matter how many resistors you add or what value they are, the same principle will hold true.

[Matt Syson]

Hi
I am not sure if it will help as the answers above are good and pleasantly simplistic, but another way to think is in terms of the CURRENT flowing into / out of the junction of input resistor, feedback resistor and the inverting input to the op amp. Accepting that when taken to frequency extremes and the fact that an op amp will draw some current (usually many times smaller than the resistor currents so gets ignored) it follows Kirchoff's law that the CURRENTS into any 'node' sum to zero. So the example of 5 volts through a 10K resistor causes a current of 0.05mA to flow. Ignoring any current into the op amp, there MUST be 0.05mA coming from the feedback resistor. If this is also 10K then by implication, the 'other end' of the feedback resistor (the output of the op amp) must be at MINUS 5 Volts. If the feedback resistor were changed to say 20K ohms, then to cause 0.05mA to flow then the output of the op amp must be 10 Volts.
The relevent thing is that the Voltage at the inverting input must be the same as the NON inverting input, and is usually held at ground, both AC and DC, thus it gets the name 'virtual earth'.
Kirchoff's law IS followed at all times but since we usually get bored counting the zeros in front of the current flowing into the op amp and the effects of capacitance when talking about actually implementing the thing get rather 'involved' we tend to think of the situation at a low AC or DC condition for the basic calculations.
This should now be as clear as mud!
Matt S

[JohnRoberts]

I'm afraid some of the explanations could be confusing (while perhaps correct).

The basic characteristic of an opamp configured with negative feedback, is that it's output will swing however it can to hold the - input at the same voltage as the + input. In a virtual earth inverter or summing topology, the + input is tied to ground so the output will drive the feedback network to hold the - input at ground also. More complicated analysis may be useful for calculating gains and such, but the answer to your question really is this simple.

JR