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Operational Amplifiers

An operational amplifier, commonly called an op amp, is a cascaded, direct coupled, transistor device that is used to perform mathematical operations through electrical analog signals. It is a high input impedance, high gain voltage amplifier with a flat frequency response up to several kHz. There are an odd number of directly coupled stages to prevent positive feedback from causing oscillation and instability. An emitter follower is used as the last stage to provide low output impedance and, therefore, eliminate any loading effect. An operational amplifier can, and sometimes is, actually built using individual electronic components, but this section will use the term op amp from this point on and assume that it is of the integrated circuit type. Op amps usually consist of three main sections, as shown in the block diagram of Figure 1. Notice that the differential amplifier has two input terminals. Recall that a differential amplifier functions to amplify any difference of potential between the two inputs. The symbol used in electronic schematics to represent an op amp is shown in Figure 2.

Figure 1: Operational Amplifier Block Diagram

Figure 2: Operational Amplifier Symbol

The input terminals have been labeled "Inverting" and "non-inverting" to describe what the op amp output polarity is with respect to the input polarity. The term "non-inverting" means that the polarity is not reversed, or non-inverted. The use of both a positive supply and a negative supply allows the op amp to produce either a positive or a negative output. A general understanding of the three main sections is beneficial in understanding the basic characteristics and limitations of an op amp. Figure 3 shows a simplified internal circuit of a typical op amp.

Figure 3: Simplified Op Amp Internal Circuit

The input signals to the inverting and non-inverting terminals are directly coupled to high impedance, differential amplifier composed of transistors Q` and Q. Transistor Q} senses the negative and positive supply voltages and functions as a constant current source to regulate against voltage supply variations. Transistors Q? and Qr form another differential amplifier to provide additional voltage gain. Transistor Qq controls the biasing of transistors Qo and Qp, which are connected as emitter-followers to provide a high current gain and low output impedance. Diodes CR` and CR provide temperature stability. The actual operating characteristics of an op amp come very close to the characteristics of an "ideal amplifier." A typical op amp compares with these ideal characteristics, as indicated in Table 1

Table 1: Comparison of Op Amp Characteristics

As shown, one of the most limiting characteristics of a real op amp is the output current capability. The low value is predominately due to the size and, therefore, the heat-dissipating capability. Op amps are normally considered voltage devices. If significant amounts of current are needed, a power or current amplifier is usually added as the next stage after an op amp.


Open Loop Configuration

As shown in Figure 4, Open Loop is a term that means there is no circuit path to form a closed loop from the output back to the input.

Figure 4: Open Loop Op Amp Circuit

When a voltage potential is applied to the input terminals, a voltage difference exists across the internal differential amplifier and the op amp output increases. Because the positive voltage is applied to the inverting input terminal with respect to ground, the output is negative with respect to ground. As described earlier, the gain of the op amp is approximately 20,000. This means that if the input is equal to 1 millivolt, the op amp attempts to increase the output to negative 20 volts. This can be shown mathematically using the standard gain equation shown below:

The negative polarity is due to the inverting action of the op amp. In reality, the output does not increase to negative 20 volts. The voltage cannot be larger than the supply voltage, which as shown is positive and negative 15 volts. What does happen is the op amp "saturates" at an output voltage very near the negative supply voltage. Saturation is a term that means the amplifier output stops increasing, even if the input continues to increase. Figure 5 shows the resultant output voltage for both a positive and negative voltage input to the open loop op amp circuit. The output at saturation does not quite reach the supply voltage amplitude due to small internal voltage drops in the op amp output circuit. The circuit functions to produce an output that is either saturated at a positive or negative voltage value when it is sent a small negative or positive input voltage.

Figure 5: Open Loop Amp Circuit Waveforms

Closed Loop Configuration

Closed Loop is a term that means there is a circuit path to form a closed loop from the output back to the input. The circuit that provides this path is called a feedback circuit, or feedback loop. Feedback can exist as either positive feedback or negative feedback, as shown in Figure 6.

Figure 6: Positive and Negative Feedback

Positive feedback exists when a portion of the output is sent back to a summing junction such that it is of the same polarity, and aids the input signal. As shown in the left side of Figure 6, a positive input to the non-inverting terminal causes a positive output, which is also sent back to the summing junction to add to the original positive input. Positive feedback is used in ordinary amplifiers to increase their gain. Op amps already have a sufficiently high gain and, therefore, positive feedback is not needed. Negative feedback exists when a portion of the output is sent back to a summing junction such that it is of opposite polarity and opposes the input signal. As shown in the right side of Figure 6, a positive input to the inverting terminal causes a negative output, which is also sent back to the summing junction to oppose the original positive input. This opposition reduces the input signal. Negative feedback is used with op amps to reduce the gain to a more usable level, or a controllable level. As the negative feedback is increased, the gain is decreased, and vice versa.


Without negative feedback, an op amp goes into saturation with only a small input applied. Figure 7 shows an op amp comparator circuit that uses this characteristic. Assume initially that the signal applied to the inverting input is a positive 2 volts to exactly match the reference voltage. The op amp sees zero differentials across the input terminals and, therefore, the output is zero volts. At time T1, the input increases to a positive 2.1 volts. The op amp sees a positive 0.1 volt differential and, due to the high gain value, the op amp goes into negative saturation. At time T2, the input is positive 1.9 volts. The op amp sees a negative 0.1 volt differential, and the op amp goes into positive saturation.


Figure 7: Comparator


Figure 8 shows an op amp buffer circuit that makes use of negative feedback. A buffer provides signal isolation between two other circuits. Op amps with their high input impedance are well suited to provide this isolation. A good buffer also should not attenuate the signal. An op amp buffer circuit has a gain of 1 or unity, and so results in no signal loss or gain.

Figure 8: Buffer

Initially, the positive 3 volts applied to the non-inverting input causes the op amp to see a 3-volt potential difference across the inputs. The output then rapidly increases in the positive direction. A negative feedback loop has been provided which sends back a voltage equal to the output voltage. When the output voltage reaches positive 3 volts, the potential difference across the inputs is zero volts, and the output stops increasing. At this point, the op amp is "satisfied"; it has increased its output sufficiently to reduce the input differential to zero. As can be seen, the output voltage exactly follows the input voltage. Another name then for a buffer is a voltage follower. Buffers are used to isolate input signals from the following circuits.

Inverting Amplifier

Figure 9 shows a basic inverting amplifier. Input resistor R` has been provided to isolate the summing junction from the input source. Feedback resistor R has been provided to isolate the summing junction from the output. These isolations allow the summing junction to be at a voltage potential that is different from either the input or the output. These two properties are important, since the circuit gain essentially becomes a factor of the input and output resistances.

Figure 9: Inverting Amplifier

At this point, it is important to keep in mind the difference between the operational circuit and the operational amplifier. The operational amplifier is represented by the triangle-like symbol, while the operational circuit includes the resistors and any other components as well as the operational amplifier. Ein is the input to the operational circuit. However, the signal at the inverting input of the operation amplifier is determined by the feedback signal as well as the circuit input signal.

The term inverting amplifier refers to the output polarity change. A signal applied to the inverting input produces a signal at the output terminal that is 180 degrees inverted. The feedback signal is a portion of the output signal, and is therefore also 180 degrees out of phase with the input signal. Whenever the input signal goes positive, the output signal and the feedback signal go negative. The result is that the inverting input to the operational amplifier is always very close to 0 volts when the non-inverting input is connecting to ground. This is because in a closed loop operation the inverting and non-inverting inputs constantly attempt to cancel out any difference in potential.

Because the op amp essentially draws no current, all the current flows through feedback resistor Rf and input resistor Rin to the input voltage. The output voltage rises until the voltage felt between the input pins of the op amp approaches zero. The amount of voltage required to drive the voltage between the input pins to zero is determined by the ratio of Rf/Rin.

Kirchhoffs Voltage Law for closed loops can be used to determine the voltage potential at the summing junction. Voltages in a loop must algebraically equal zero when summed together.

(Ein) + (Erf ) + (E summing junction to ground) = 0

In this case, the output continues to increase until the feedback current is large enough to cause the voltage drop across Rf to be 5 volts. At that point, the input voltage will be reduced to 0 volts at the summing junction because the opposing 5 volts across Rin totally cancels out the 5 volt input. This can be verified by using Kirchhoffs Voltage Law. Proceeding clockwise form the negative terminal of Ein, voltages are assigned to the equation as:

(-5V) + (+5V) + (Es.j. - gnd.) = 0

(Es.j. - gnd.) = 0

When the summing junction is at 0 volts, the op amp is satisfied. The output then stops increasing and remains at an output level to maintain the summing junction at 0 volts. Kirchhoff can again be used to determine what the amplitude of voltage output is at this point. By knowing the voltage across R` must be 5 volts to cause the summing junction to be 0 volts, the current flow through R` is determined to be:

ERU/RU = I thru RU

{1k}/{1k} = IRU

IRU = 5 mA

This 5 mA also flows through R1 to give a voltage drop of:

The voltage loop can be analyzed at this point using the equation of:

(Ein) + (ER`) + (ER1) + (Eout) = 0

Proceeding clockwise from the negative terminal of Ein, the voltages are assigned to the equation as:

(-5V) +(+5V) + (+5V) + (Eout) = 0

Eout = -5V

Figure 10 shows the inverting op amp with these voltage values assigned.

Figure 10: Inverting Amplifier at Final Output

The gain of the inverting amplifier is represented by: The negative sign represents the inverting action of the op amp. Figure 11 shows another inverting amplifier with a different value resistor in the feedback loop.

Figure 11: Inverting Amplifier

The voltage across R` must be equal to the input voltage for the summing junction to be 0 volts, and the op amp to be satisfied. Initially, the positive 2 volts is sensed at the inverting terminal and the op amp output increases negative. As the output increases, the feedback current through R` increases. When the voltage across R` equals 2V to cancel the input voltage, the current through R` is:

The 2 mA flows through R to cause a voltage of:

I x R` = ER1

(2mA) x (5k) = ER1

ER1 = 10V

Again, using Kirchhoff:

(Ein) + (ER`) + (ER1) + (Eout = 0

(-2V) + (+2V) +(+10V) + (Eout) = 0

Eout = -10V

The gain of this inverting amplifier is:

Notice that by only changing the value of the resistor in the feedback loop, the gain of the inverting amplifier was changed, and a larger output was provided. The output voltage is dependent on the ratio of the resistance in the feedback loop versus the resistance in the input circuit. For inverting amplifiers, this is expressed as:

Eout = -Ein R1/R`

Transposing the output voltage equation to:

Non-Inverting Amplifier

Figure 12 shows a non-inverting amplifier. The term Non-Inverting refers to the output polarity matching the input polarity.

Figure 12: Non-Inverting Amplifier

For the op amp to be satisfied in this circuit, the voltage at the summing junction must be a positive 2 volts. The feedback current through Rf provided by the output when the op amp is satisfied is:

This 2 mA flows through Rin to provide a voltage of:

Using Kirchhoff, and proceeding from the grounded side of R`:

The output voltage to ground is also the voltage seen across R1 plus R` with respect to ground, and the input voltage equals the voltage across R`. Therefore, the gain of the non-inverting amplifier is:

Summing Amplifier

Figure 13 shows a summing amplifier. For the op amp to be satisfied in this circuit, the summing junction must be at 0 volts. To obtain this, each input resistor must have a voltage drop equal to its respective input voltage, and all the current flowing in the input resistors must also flow through R?. The resultant voltages and currents are as shown.

Figure 13: Summing Amplifier

Any one of the input circuits can be used with the voltage across R? to determine the output voltage:

(E1) + (ER1) + (ER?) + (Eout)= 0

(-2V) + (+2V) + (+6V) + (Eout)= 0

Eout = -6V

This 3-input summer is really just an inverting amplifier that has two additional input circuits. Recall that the output voltage for an inverting amplifier is:

The output voltage determined by the three inputs is:

Each of the inputs contributes to the output according to the gain it sees:

If it is desired to use only a portion of one or more input signals to determine a desired output voltage, the gain for the one or more inputs could be changed by changing the value of their respective input resistors. This is called "input scaling." For example, if it is desired that input E` has a larger effect on the output as compared to the other inputs, the resistance value of R` could be changed as shown in Figure 14.

Figure 14: Scaling Amplifier

Difference Amplifier

Figure 15 shows a difference amplifier.

Figure 15: Difference Amplifier

Input E1 is divided by R1 and R/. To determine the voltage at the non-inverting terminal, the voltage across R/ must be determined. The current through R/ is:

The 1 mA through R/ provides a positive 1-volt at the non-inverting terminal. For the op amp to be satisfied, a positive 1-volt at the inverting terminal summing junction must be provided. To obtain this, a 2-volt drop across R` must be provided by the feedback current:

The difference amplifier is essentially an inverting amplifier and a non-inverting amplifier providing an output that is the difference between the two input signals.


Operational amplifiers can be arranged so as to perform the mathematical function of integration. Integration is the process of summing the input signal over time. The op amp integrator is shown in Figure 16. Notice that this circuit looks like the inverting amplifier except that the feedback resistor has been replaced by a capacitor.

Figure 16: Integrator

An op amp integrator converts a voltage input to a ramp output, whose slope is a function of the magnitude of input voltage and the RC time constant. As the amplitude of the input waveform is increased, the resultant output at any point in time is increased, or simply the rate of change of the output is increased. A square-wave can be viewed as a step change in voltage. The integrator takes this step change and converts it to a slower rate of change. The new rate of change is dependent on the amplitude of the original step change.

A current entering the summing junction through R due to Ein, must be cancelled with a current from the output of the amplifier in order to keep the summing point at 0 volts. The only way to get a current through a capacitor is by having a changing voltage across it. A constantly increasing voltage on the output of the amplifier will produce such a constant current through the capacitor. Therefore, as long as current from Ein exists, Eout must continue to change.

At T1 the input step changes to a positive 1 volt. A voltage potential exists across the op amp input terminals, and the output increases in the negative direction. The capacitor attempts to charge to the potential of the input voltage, and the RC time constant determines the rate of which the capacitor charges. Therefore, the voltage across the capacitor at T2 is expressed as:

The 1-volt across the resistor opposes the input voltage and would appear to reduce the summing junction to 0 volts, which would stop any further output increase. However, once any capacitor is charged, charging current stops. Without charging current, there would be no voltage developed across the resistor to oppose the input voltage. This would result in the positive input again being applied to the op amp to cause a further negative increase in the output. In actuality, a constant current is needed through the resistor to oppose the input voltage. To obtain this, the voltage across the capacitor must be increased as seen by the capacitor voltage equation in terms of I:

Time is not a constant term; it is continually increasing. To provide a constant value of current I, the voltage V must be increased because capacitance C is a constant value.

The op amp Integrator circuit characteristically operates different than a simple series RC integrator circuit, in that the output does not stop at a value equal to the voltage across the capacitor. The output of the op amp continues to increase in the negative direction. The increasing voltage across the capacitor maintains a constant charging current through the resistor to develop a voltage that opposes the input. The output continues to increase until it reaches negative saturation. However, the output at the saturation amplitude is not the portion of the output that is useful. The portion of the output that is useful is the ramp voltage. Different amplitude input voltage causes different rate of change ramp output voltages. This can be shown by converting the integrator output equation:


A differentiator is basically an inverting amplifier with the input resistor replaced by a capacitor, as shown in Figure 17. Mathematically, the differentiation function is the inverse of integration. A differentiator accepts a ramp change input voltage and converts it to an output pulse, which is proportional to the rate of change of the input. If the slope of the input voltage is zero, the differentiator output is 0 V. If the slope of the input is constant, the output is constant. If the slope of the input is infinite, as in the edges of a square wave, the output is an infinite voltage.

Figure 17: Differentiator

Any change in input voltage will cause a current to flow through C, and this change will be cancelled by a current through R. As long as Ein is changing, Eout will be non-zero. The value of Eout depends upon the rate of change of Ein.

The output voltage is directly dependent on the amplitude of input voltage and the RC time constant. This determines how rapidly the capacitor charges and, therefore, the amplitude of the charging current through the resistor. When a ramp voltage is applied to a differentiator, the capacitor sees this as a constantly increasing input at a fixed or constant rate of change. Current flows through the resistor at a constant rate because the capacitor is being charged by the input in a linear manner.

Initially, the capacitor is not charged. At time T1 the input ramps positive at a rate of change equal to 1 volt per second. This positive going input is coupled by the uncharged capacitor to the op amp, and the output voltage is driven negative. The capacitor charges at a rate determined by the RC time constant, and the charging current flows through the resistor. At time T2, the capacitor has charged to 1 volt.