The ideal inverting operational amplifier with voltage shunt feedback from output terminal to the inverting input terminal feedback impedance Z_{f} and with non-inverting terminal grounded configuration is shown is figure 1. 25+ verified electronics projects using op-amp 741.

This forms the basic inverting op-amp. It is “Inverting” because our signal input comes to the “-” input, and there has the opposite sign to the output. The feedback arrangement used here forms the voltage shunt feedback. Hence as per theory of feedback amplifiers, voltage gain with feedback A_{Vf} is given by,

…..(1)

Equation 1 may be established in an alternative way as below:

- We make use of Rule I (The output attempts to make the voltage difference between the inputs zero i.e. make V
_{d}zero). Thus, since point B is at ground potential, then as per rule I, point A is also at ground potential. Thus, there exist a virtual ground or virtual short circuit at the input of the Op-Amp indicated by double lines with arrow head in figure 2.

- Using Rule II (The inputs draw no current), no current flows through the input of the Op-amp. This implies that the full current I flows through Z
_{f}and nothing is diverted to the input of the Op-Amp.

Hence voltage -IZ_{f} across Z_{f} is the output voltage V_{0} while voltage IZ_{1} across Z_{1} is the input voltage V_{i}. Hence voltage gain A_{Vf} is given by,

This agree with equation 1.

What is input impedance?

Now point A is always at zero volt because of the virtual ground. Hence Z_{in} = Z_{1}.

The circuit is known as an inverting amplifier. Its one undesirable feature is the low input impedance, particularly for amplifiers with large closed loop voltage gain where Z_{1} tends to be rather small. This undesirable feature is remedied in non-inverting amplifier.

Practical Inverting Op-Amp Equation 1 giving A_{Vf} is valid only for infinity voltage gain A_{v}. next we consider a practical inverting Op-Amp shown in figure 3. Here the Op-Amp of figure 1 is replaced by its practical small signal model with , and . Here A_{v} represents the open circuit or unloaded voltage gain.

We now apply Miller’s theorem and thereby replace Z_{f} by two impedances, an impedance across the input and an impedance across the output resulting in the modified model of Figure 4.

Then from Figure 4 we get the following expression for the closed loop gain,

…..(2)

Where is the voltage gain talking into account the loading effect of Z_{f}.

Then A_{V} is given by,

…..(3)

From Equation 3 we note that if R_{0} = 0 or Y_{f} = 0 (i.e. ), the loading effect of Z_{f} gets removed and A_{V} equal A_{v}. Also as , |A_V| also tends to infinity and . Thus, this value of A_{Vf} agrees with that given by Equation 1.

The output impedance is very small .

Problem regarding inverting amplifier.

Determine the output voltage for the inverting amplifier shown in figure 1, if (a) v_{in} = 20 mV dc (b) V_{in} = -50 uV peak sine wave.

Assume that the op-amp is a 741.

- V
_{O}= -AV_{in}= -(2)x(10^{5})x(20)x(10^{-3}) = -4000V

This is the theoretical value; the actual value will be a negative saturation voltage of -14V.

- V
_{O}= -AV_{in}= -(2)x(10^{5})x(-50)x(10^{-6}) = 10V peak sine wave

This means that the output is a sine wave, since it is less than the output voltage swing of +/- 14V or 28V peak to peak.