Overview
The inverting amplifier inverts and amplifies the input signal. Its transfer equation is Vout = -Vin * Rf / Rin. Using this equation, design is straightforward, but the inverting amplifier has an inherent drawback: its input impedance is not high. In circuit design it is generally preferable for an amplifier to present a high input impedance so it does not draw current from the signal source and thereby alter the signal or the amplification result.
Basic Example
Consider the inverting amplifier shown below. According to Vout = -Vin * Rf / Rin, the gain is 10k / 1k = 10. Keep this result in mind.
Virtual Short and Input Impedance
Using the op amp assumptions of virtual short and virtual open, the noninverting input is grounded at 0 V. By the virtual short, the inverting input is also at 0 V. The right side of Rin is therefore effectively at 0 V. The op amp's Thevenin equivalent is shown below.
Viewing the amplifier as a black box from the Vin source, Rin determines the input impedance. Therefore the input impedance equals Rin, which in this example is 1k.
Effect of Source Resistance on Gain
Because Rin gives the inverting amplifier a low input impedance, the amplifier can load the signal source and affect the amplification. Real signal sources are not ideal voltage sources; they have source resistance. Assume the source resistance Rz is 1k. The modified schematic is shown below.
In this configuration Rz and Rin are in series. The effective gain becomes Vout = -Vin * Rf / (Rz + Rin). With Rf = 10k, Rin = 1k and Rz = 1k, the gain is 10k / (1k + 1k) = 5, not 10 as previously calculated.
Thus the intended gain of 10 is not achieved; the error can be large and the gain will vary with the source resistance Rz. This explains the origin of the inverting amplifier's sensitivity to source impedance.
Polarity of the Output
Because the inverting input is at 0 V, the current direction shown in the figure indicates that Vout is negative. This is the reason for the negative sign in the inverting amplifier formula: the input voltage is inverted. The transfer equation can be derived from the current directions in the schematic.
