The feedback configuration shown in Figure 2 is the most common op-amp circuit for obtaining what is known as an inverting amplifier. provides the exact input-output relationship. In choosing element values with respect to op-amp characteristics, we can simplify the expression dramatically.

• Make the load resistance, R L R L , much larger than R out R out . This situation drops the term 1 R L 1 R L from the second factor of Equation 1.
• Make the resistor, R R, smaller than R in R in , which means that the 1 R in 1 R in term in the third factor is negligible.

With these two design criteria, the expression (Equation 1) becomes Because the gain is large and the resistance R out R out is small, the first term becomes -1G 1 G , leaving us with

• If we select the values of R F R F and R R so that GR≫ R F ≫ G R R F , this factor will no longer depend on the op-amp's inherent gain, and it will equal -1 R F 1 R F .

Under these conditions, we obtain the classic input-output relationship for the op-amp-based inverting amplifier. Consequently, the gain provided by our circuit is entirely determined by our choice of the feedback resistor R F R F and the input resistor R R. It is always negative, and can be less than one or greater than one in magnitude. It cannot exceed the op-amp's inherent gain and should not produce such large outputs that distortion results (remember the power supply!). Interestingly, note that this relationship does not depend on the load resistance. This effect occurs because we use load resistances large compared to the op-amp's output resistance. Thus observation means that, if careful, we can place op-amp circuits in cascade, without incurring the effect of succeeding circuits changing the behavior (transfer function) of previous ones; see this problem.

As long as design requirements are met, the input-output relation for the inverting amplifier also applies when the feedback and input circuit elements are impedances (resistors, capacitors, and inductors).

Creating a specific transfer function with op-amps does not have a unique answer. As opposed to design with passive circuits, electronics is more flexible (a cascade of circuits can be built so that each has little effect on the others; see (Reference)) and gain (increase in power and amplitude) can result. To complete our example, let's assume we want a lowpass filter that emulates what the telephone companies do. Signals transmitted over the telephone have an upper frequency limit of about 3 kHz. For the second design choice, we require R F C=3.3×10-4 R F C 3.3-4 . Thus, many choices for resistance and capacitance values are possible. A 1 μF capacitor and a 330 Ω resistor, 10 nF and 33 kΩ, and 10 pF and 33 MΩ would all theoretically work. Let's also desire a voltage gain of ten: R F R=10 R F R 10 , which means R= R F 10 R R F 10 . Recall that we must have R< R in R R in . As the op-amp's input impedance is about 1 MΩ, we don't want R R too large, and this requirement means that the last choice for resistor/capacitor values won't work. We also need to ask for less gain than the op-amp can provide itself. Because the feedback "element" is an impedance (a parallel resistor capacitor combination), we need to examine the gain requirement more carefully. We must have | Z F |R<105 Z F R 10 5 for all frequencies of interest. Thus, R F |1+ⅈ2πf R F C|R<105 R F 1 2 f R F C R 10 5 . As this impedance decreases with frequency, the design specification of R F R=10 R F R 10 means that this criterion is easily met. Thus, the first two choices for the resistor and capacitor values (as well as many others in this range) will work well. Additional considerations like parts cost might enter into the picture. Unless you have a high-power application (this isn't one) or ask for high-precision components, costs don't depend heavily on component values as long as you stay close to standard values. For resistors, having values r10d r 10 d , easily obtained values of r r are 1, 1.4, 3.3, 4.7, and 6.8, and the decades span 0-8.

When we meet op-amp design specifications, we can simplify our circuit calculations greatly, so much so that we don't need the op-amp's circuit model to determine the transfer function. Here is our inverting amplifier.

When we take advantage of the op-amp's characteristics—large input impedance, large gain, and small output impedance—we note the two following important facts.

• The current i in i in must be very small. The voltage produced by the dependent source is 105 10 5 times the voltage v v. Thus, the voltage v v must be small, which means that i in =v R in i in v R in must be tiny. For example, if the output is about 1 v, the voltage v=10-5V v 10 5 V , making the current i in =10-11A i in 10 11 A . Consequently, we can ignore i in i in in our calculations and assume it to be zero.
• Because of this assumption—essentially no current flow through R in R in —the voltage v v must also be essentially zero. This means that in op-amp circuits, the voltage across the op-amp's input is basically zero.

Armed with these approximations, let's return to our original circuit as shown in Figure 5. The node voltage e e is essentially zero, meaning that it is essentially tied to the reference node. Thus, the current through the resistor R R equals v in R v in R . Furthermore, the feedback resistor appears in parallel with the load resistor. Because the current going into the op-amp is zero, all of the current flowing through R R flows through the feedback resistor ( i F =i i F i )! The voltage across the feedback resistor v v equals v in R F R v in R F R . Because the left end of the feedback resistor is essentially attached to the reference node, the voltage across it equals the negative of that across the output resistor: v out =-v=- v in R F R v out v v in R F R . Using this approach makes analyzing new op-amp circuits much easier. When using this technique, check to make sure the results you obtain are consistent with the assumptions of essentially zero current entering the op-amp and nearly zero voltage across the op-amp's inputs.