Once thought to have outlived their usefulness, operational amplifiers are as popular today in control applications as they ever were.

While many predicted the demise of operational amplifiers (op-amps) because of the development of digital integrated circuits, these devices are as popular today as they ever were. In fact, most solid-state analog control systems have long relied on op-amps as essential building blocks. These devices perform useful tasks, including addition, multiplication, differentiation, integration, and feedback control.

Op-amps are still a primary building block for analog systems, performing tasks like amplification, active filtering, and signal transformation. In digital systems, op-amps are used in buffers, analog-to-digital converters, digital-to-analog converters, and regulated power supplies, to name a few applications.

Op-amp characteristics

The name "operational amplifier" stems from the op-amp's ability to perform mathematical operations. Thus, it's somewhat of a misnomer, considering the wide variety of op-amp applications today.

Functionally, the op-amp is actually a differential voltage amplifier. As shown in Fig. 1, an op-amp takes two input voltages and produces an output voltage that's proportional to the difference between the two input voltages. The inputs are called the inverting and non-inverting inputs. An increase in voltage at the inverting input will cause a decrease in output voltage, while an increase in voltage at the non-inverting input will, as you might expect, cause an increase in output voltage.

A standard op-amp is powered by a dual-voltage [+ or -]15VDC power supply; however, many op-amps are now available with single voltage supplies.

Op-amps are usually found under linear integrated circuits in catalogs or data books.

The op-amp's popularity stems from its simple characteristics, which make predictable circuit design fairly easy. These characteristics are as follows.

Very high voltage gain. The difference in voltage between the inputs is multiplied by the voltage gain to determine the output. Op-amps are capable of voltage gains of more than 100,000. The output cannot exceed the positive power supply voltage or go below the negative supply voltage, so gain is limited by the magnitude of the incoming signal. For gains of more than 15,000, for example, the input signal (the difference between the inputs) must be below 1 millivolt to avoid saturation.

High input impedance. The input impedance of an op-amp is defined as the effective resistance between the input terminals, and this is usually above 1 megohm. Virtually no current flows between the two input terminals. This high input impedance makes the op-amp particularly well suited for the amplification of the sensitive low-level input signals typically associated with instrument field devices like strain gauges and thermocouples.

Low output impedance. The output impedance is defined as the effective internal resistance in series with the output, and it's usually less than 100 ohms. When external components are added, the op-amp's output resistance is effectively reduced and is sometimes considered to be near zero in practical circuits. There is still a limitation on the output current that an op-amp can provide, however, and in most circuits the output current is less than 50 mA.

High common-mode rejection. Because op-amps amplify the difference between the two inputs, any signal that is superimposed equally on the inputs will be ignored. This property is important for instrumentation design because any noise that's present on both conductors of a long field cable (60 Hz induced noise, for example) will be ignored by the op-amp, which will cleanly amplify the true signal.

Basic amplifier circuits

Many practical amplifiers can be constructed using op-amps. We'll discuss the two most common ones. As you'll see, these circuits act as basic building blocks from which more complicated circuits can be created.

Inverting amplifier. The inverting amplifier, the easiest circuit to understand, is shown in Fig. 2 (on page 26). To create one, the non-inverting input of an op-amp is grounded and an input resistor ([R.sub.1]) is placed in series with the inverting input. Another resistor ([R.sub.F]), called the feedback resistor, is connected between the outut and the inverting input. Some form of feedback (connection between the output and input terminals) is used in most practical op-amp circuits. The output voltage ([V.sub.o]) of the inverting amplifier, as a function of the input voltage, can be found by using the following equation:

[V.sub.o] = (- [R.sub.F] / [R.sub.1]) [V.sub.1]

The gain of the inverting amplifier, which is the ratio shown within the parenthesis, is the negative of the ratio of the feedback resistance to the input resistance.

Non-inverting amplifier. The non-inverting amplifier, as shown in Fig. 3, is an inverting amplifier flipped over with the inverting input grounded. The input signal is connected directly to the non-inverting input of the op-amp. The output voltage of the non-inverting amplifier can be found by using the following equation:

[V.sub.o] = ([R.sub.F] + [R.sub.1] / [R.sub.1]) [V.sub.1]

The gain of the non-inverting amplifier is the ratio shown in parentheses in the above equation and is dependent on the values of the input and feedback resistors.

Mathematical operations

Op-amps are capable of performing mathematical operations, a fact that was taken advantage of in early analog computers. Although computations are done primarily with digital hardware in today's world, there are still times when analog computation is useful in signal processing and control systems.

Weighted-sum adder. One common computational op-amp circuit is the weighted-sum adder, which is shown in Fig. 4. It's built by connecting several signals to the input of an inverting amplifier, with each signal passing through its own input resistor. The inverting input of the op-amp functions as a summing junction in this circuit, and the output voltage of the weighted-sum adder can be found by using the following equation:

[V.sub.o] = (- [R.sub.F] / [R.sub.1]) [V.sub.1] + (- [R.sub.F] / [R.sub.2]) [V.sub.2] + (- [R.sub.F] / [R.sub.3]) [V.sub.3]

Note that the output is the weighted sum of the input voltages, and that the weights are simply the negatives of the ratios between the feedback resistor and each of the input resistors. This circuit easily can be tailored to provide a weighted sum of several signals with whatever relative weights are desired. If a non-inverted output is desired, the weighted-sum adder can simply be followed with another inverting amplifier with unity gain.

Integrator. Another common computational op-amp circuit is the integrator. Integration is useful in control systems as well as in signal processing applications, where an op-amp provides a very simple method for integrating a signal in real time. The integrator is constructed by putting a capacitor instead of a resistor in the feedback loop, as shown in Fig. 5.

The charging effect of the capacitor in the feedback loop makes the output of the amplifier time-dependent and causes the integration effect.

Differentiator. A computational op-amp circuit that performs the opposite function to that of the integrator is the differentiator. This circuit takes the derivative of a signal in real time. To make a differentiator, the capacitor is moved from the feedback loop to the input as shown in Fig. 6. In this case, the charging effect of the capacitor makes the amplifier sensitive to the rate of change, or derivative, of the input signal with time.

Instrumentation applications

Many of the general-purpose op-amp circuits described above have been widely used in instrumentation and control systems. Analog controllers use inverting and non-inverting amplifiers, integrators, and differentiators to perform proportional-integral-derivative (PID) control. The weighted-sum adder has been used extensively in DC drive control systems.

Let's talk about some additional op-amp circuits that are designed specifically for instrument applications, particularly for interfacing to the low-level signals typically generated by field sensors.

Voltage follower. The voltage follower, as shown in Fig. 7, is the simplest of op-amp circuits. It produces the identical voltage on the output as is present at the input. The feedback loop guarantees that the non-inverting input is equal to the output. If the output is too low, for example, the op-amp will amplify the difference between the input and output voltage tremendously, driving the output up. The op-amp will not stabilize until the output voltage exactly matches the input voltage.

The voltage follower is used as a buffer, and it has the advantage of providing a very high input impedance because no external resistors are connected to the non-inverting input. This makes the voltage follower especially useful for buffering sensitive low-level signals.

Differential amplifier. Up to this point, all of the op-amp circuits we've discussed are single-ended; i.e., the input signal is connected between a terminal and ground. However, many instrument signals are differential in nature, where the important information is contained in the difference in voltage between two points. These signals require the op-amp to be used in differential mode. This allows you to take advantage of the common-mode rejection capability that is so important for noise rejection in instrument circuits.

A differential amplifier circuit, as shown in Fig. 8, is usually balanced, so that each input presents the same impedance to the system. That's why the input and feedback resistors are shown with the same values for each input, and these values must be nearly identical for accurate operation.

As its name suggests, the differential amplifier amplifies the difference between the input voltages, and the output voltage can be found by using this equation:

[V.sub.o] = [R.sub.F] / [R.sub.1] ([V.sub.2] - [V.sub.1])

One drawback of the differential amplifier is the relatively low input impedance. Although op-amps themselves have very high input impedances, the external resistors required in the differential amplifier significantly reduce this impedance. This makes the differential amplifier unsuitable for signals from devices like strain gauges, which require very high input impedances.

Instrumentation amplifier. The instrumentation amplifier, as shown in Fig. 9, solves the previously discussed low input impedance problem. By incorporating two voltage followers [ILLUSTRATION FOR FIGURE 7 OMITTED] as input buffers, the instrumentation amplifier behaves exactly like the differential amplifier, on which it is based, but has very high input impedance.

As its name suggests, the instrumentation amplifier was designed specifically for interfacing to field sensors used in instrumentation.

Bridge amplifier. Another circuit designed for instrument applications is the bridge amplifier. Many sensitive measurement circuits utilize a balanced resistance bridge as a measurement device. The bridge is constructed of three identical resistors and another resistance that is nearly equal but varies with the sensor measurement. An op-amp can be inserted into the balanced bridge, as shown in Fig. 10, to form the bridge amplifier circuit. As with the balanced differential amplifier, the bridge amplifier requires resistors of nearly identical value to perform accurately.

The output of the bridge amplifier can be found by using the following equation:

[V.sub.o] = ([Delta]R / 2R) [V.sub.s]

Note that the output is directly proportional to the change in resistance ([Delta]R) in the variable-resistance leg of the bridge.

Voltage-to-current converter. Because op-amps are voltage amplifiers, most instruments operate internally on voltage signals. When the need comes to communicate to the outside world, however, current loop signals are more common (4-20 mA, for example). The voltage-to-current converter, as shown in Fig. 11, is designed to generate a current signal from a voltage input. Although the voltage input is single-ended, balanced input and feedback resis-tors are used for accuracy.

The output (load) current ([I.sub.L]) generated by the voltage-to-current converter can be found by using the following equation:

[I.sub.L] = (- 1 / [R.sub.1]) [V.sub.1]

Note that the load current is directly proportional to the input voltage.


We should mention at this point that the simplest possible circuits are presented here for clarity, and additional circuitry may be necessary for some applications. In particular, circuitry for biasing, balancing, and temperature compensation may be required to improve the accuracy of some op-amp circuits.

Ryan G. Rosandich, Ph.D., is Assistant Professor, Dept. of Industrial Engineering, University of Minnesota-Duluth, Duluth, Minn.