Understanding the Virtual Earth Approximation in Op-Amps

[gabloammar]

Homework Statement

I'm studying op-amps at the moment, and I came across a statement and I don't understand it.

'To understand how the inverting amplifier works, you need to understand the concept of the virtual Earth approximation. In this approximation the potential at the inverting input (-) is very close to 0 V. Why is this true? There are two steps in the argument.

1. The op-amp multiplies the difference in potential between the inverting and non-inverting inputs, V^- and V⁺, to produce the output voltage V_out. Because the open-loop voltage gain is very high, the difference between V- and V+ must be almost zero.

2. The non-inverting input (+) is connected to the zero volt line so V⁺ = 0. Thus V^- must be close to zero and the inverting input (-) is almost at Earth potential.'

I get the second point, and I even understand most of what the first point is trying to tell me. But, see, it says that for there to be a large gain, there has to be a small potential difference between the inverting and non-inverting inputs. But isn't it correct that G=V^out/Vⁱⁿ?

Doesn't that equation mean that the smaller the difference in the two values the smaller the gain? [e.g. 100/10 is larger than 10/10.]

What am I not getting here? I'm sure the book's correct and I'm wrong but I'm not seeing what exactly I'm getting wrong. So can someone please help me a little?

 

[gneill]

Suppose that the intrinsic (open-loop) gain of a real op-amp happened to be 10⁷, and that the amplifier circuit it is built into happens to yield a gain of G = -5 (don't ask how the gain is set right now, you'll be finding that out soon enough).

At the input to this circuit a 1V source is connected so that we expect -5V at the output. What would the voltage difference between V^- and V⁺ have to be in order to produce -5V at the output? Is there a relevant (for analysis or design purposes) difference between this voltage and zero when compared to the voltages between other points in the circuit?