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Designing Circuits with Operational Amplifiers

Module by: Don Johnson

Summary: (Blank Abstract)

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.

Figure 1
opamp
opamp (opamp12a.png)
Figure 2
opamp
opamp (opamp12b.png)

When we take advantage of the op-amp's characteristics—large input impedance, large gain, and small output impedance

point of interest:

Although the op-amp circuit model contains only resistors, we still use the word impedance to describe their values. A resistor is a special case of an impedance!
mdash;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=10-5 v v 10 5 v making the current i in =10-11 a 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 2. 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. Do check to make sure the results you obtain are consistent with the assumptions.

Example 1

Figure 3: Two-source, single-output op-amp circuit example.
Two Source Circuit
Two Source Circuit (opamp13.png)

Let's try this analysis technique on a simple extension of the inverting amplifier configuration shown in Figure 3. If either of the source-resistor combinations were not present, the inverting amplifier remains, and we know that transfer function. By superposition, we know that the input-output relation is

v out =- R F R 1 v in ( 1 ) - R F R 2 v in ( 2 ) v out R F R 1 v in ( 1 ) R F R 2 v in ( 2 ) (1)
When we start from scratch, the node joining the three resistors is at the same potential as the reference ( e≈0 e 0 ) and the sum of currents flowing into that node is zero. Thus, the current i i flowing in the resistor R F R F equals v in ( 1 ) R 1 + v in ( 2 ) R 2 v in ( 1 ) R 1 v in ( 2 ) R 2 . Because the feedback resistor is essentially in parallel with the load resistor, the voltages must satisfy v=- v out v v out . In this way, we obtain the input-output relation given above.

What utility does this circuit have? Can the basic notion of the circuit be extended without bound?

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