 1. Introduction
 2. Differential Amplifiers
 2.1. Basic Differential Amplifier Circuit
 2.2. The TwoInput, SingleOutput, Difference Amplifier
 2.3. Typical Differential Amplifier Circuit
 2.4. SingleInput, SingleOutput, Differential Amplifier
 2.5. SingleInput, DifferentialOutput, Differential Amplifier
 2.6. DifferentialInput, DifferentialOutput, Differential Amplifier
 3. Operational Amplifiers
1. Introduction
If you were to make a quick review of the subjects discussed in this book up to this point, you would see that you have been given a considerable amount of information about amplifiers. You have been shown what amplification is and how the different classes of amplifiers affect amplification. You also have been shown that many factors must be considered when working with amplifiers, such as impedance, feedback, frequency response, and coupling. With all this information behind you, you might ask yourself "what more can there be to know about amplifiers?"
There is a great deal more to learn about amplifiers. Even after you finish this chapter you will have only "scratched the surface" of the study of amplifiers. But, you will have prepared yourself for the remainder of the NEETS. This, in turn, should prepare you for further study and, perhaps, a career in electronics.
As in chapter 2, the circuits shown in this chapter are intended to present particular concepts to you. Therefore, the circuits may be incomplete or not practical for use in an actual piece of electronic equipment. You should keep in mind the fact that this text is intended to teach certain facts about amplifiers, and in order to simplify the illustrations used, complete operational circuits are not always shown.
In this chapter two types of special amplifiers are discussed. These are: DIFFERENTIAL AMPLIFIERS and OPERATIONAL AMPLIFIERS. These are called special amplifiers because they are used only in certain types of equipment.
The names of each of these special amplifiers describe the operation of the amplifier, NOT what is amplified.
A differential amplifier is an amplifier that can have two input signals and/or two output signals. This amplifier can amplify the difference between two input signals. A differential amplifier will also "cancel out" common signals at the two inputs.
One of the more interesting aspects of an operational amplifier is that it can be used to perform mathematical operations electronically. Properly connected, an operational amplifier can add, subtract, multiply, divide, and even perform the calculus operations of integration and differentiation. These amplifiers were originally used in a type of computer known as the "analog computer" but are now used in many electronic applications.
These brief descriptions of the two special amplifiers are intended to provide you with a general idea of what these amplifiers are and how they can be used. The remaining sections of this chapter will provide you with more detailed information on these special amplifiers.
2. Differential Amplifiers
A differential amplifier has two possible inputs and two possible outputs. This arrangement means that the differential amplifier can be used in a variety of ways. Before examining the three basic configurations that are possible with a differential amplifier, you need to be familiar with the basic circuitry of a differential amplifier.
2.1. Basic Differential Amplifier Circuit
Before you are shown the operation of a differential amplifier, you will be shown how a simpler circuit works. This simpler circuit, known as the DIFFERENCE AMPLIFIER, has one thing in common with the differential amplifier: It operates on the difference between two inputs. However, the difference amplifier has only one output while the differential amplifier can have two outputs.
By now, you should be familiar with some amplifier circuits, which should give you an idea of what a difference amplifier is like. In NEETS, Module 7, you were shown the basic configurations for transistor amplifiers. Figure 31 shows two of these configurations: the common emitter and the common base.
In Figure 1 a commonemitter amplifier is shown. The output signal is an amplified version of the input signal and is 180 degrees out of phase with the input signal. Figure 2 is a commonbase amplifier. In this circuit the output signal is an amplified version of the input signal and is in phase with the input signal. In both of these circuits, the output signal is controlled by the basetoemitter bias. As this bias changes (because of the input signal) the current through the transistor changes. This causes the output signal developed across the collector load (R2) to change. None of this information is new, it is just a review of what you have already been shown regarding transistor amplifiers.
Note

Bias arrangements for the following explanations will be termed basetoemitter. In other publications you will see the term emittertobase used to describe the same bias arrangement. 
2.2. The TwoInput, SingleOutput, Difference Amplifier
If you combine the commonbase and commonemitter configurations into a single transistor amplifier, you will have a circuit like the one shown in Figure 3. This circuit is the twoinput, singleoutput, difference amplifier.
In Figure 3, the transistor has two inputs (the emitter and the base) and one output (the collector). Remember, the current through the transistor (and therefore the output signal) is controlled by the basetoemitter bias. In the circuit shown in Figure 3, the combination of the two input signals controls the output signal. In fact, the DIFFERENCE BETWEEN THE INPUT SIGNALS determines the basetoemitter bias.
For the purpose of examining the operation of the circuit shown in Figure 3, assume that the circuit has a gain of 10. This means that for each 1volt change in the basetoemitter bias, there would be a 10volt change in the output signal. Assume, also, that the input signals will peak at 1volt levels (+1 volt for the positive peak and 1 volt for the negative peak). The secret to understanding this circuit (or any transistor amplifier circuit) is to realize that the collector current is controlled by the basetoemitter bias. In other words, in this circuit the output signal (the voltage developed across R3) is determined by the difference between the voltage on the base and the voltage on the emitter.
Figure 4 shows this twoinput, singleoutput amplifier with input signals that are equal in amplitude and 180 degrees out of phase. Input number one has a positive alternation when input number two has a negative alternation and vice versa.
The circuit and the input and output signals are shown at the top of the figure. The lower portion of the figure is a comparison of the input signals and the output signal. Notice the vertical lines marked "T0" through "T8." These represent "time zero" through "time eight." In other words, these lines provide a way to examine the two input signals and the output signal at various instants of time.
In Figure 4 at time zero (T0) both input signals are at 0 volts. The output signal is also at 0 volts. Between time zero (T0) and time one (T1), input signal number one goes positive and input signal number two goes negative. Each of these voltage changes causes an increase in the basetoemitter bias which causes current through Q1 to increase. Increased current through Q1 results in a greater voltage drop across the collector load (R3) which causes the output signal to go negative.
By time one (T1), input signal number one has reached +1 volt and input signal number two has reached −1 volt. This is an overall increase in basetoemitter bias of 2 volts. Since the gain of the circuit is −10, the output signal has decreased by 20 volts. As you can see, the output signal has been determined by the difference between the two input signals. In fact, the basetoemitter bias can be found by subtracting the value of input signal number two from the value of input signal number one.
Between time one (T1) and time two (T2), input signal number one goes from +1 volt to 0 volts and input signal number two goes from −1 volt to 0 volts. At time two (T2) both input signals are at 0 volts and the basetoemitter bias has returned to 0 volts. The output signal is also 0 volts.
Between time two (T2) and time three (T3), input signal number one goes negative and input signal number two goes positive. At time three (T3), the value of the basetoemitter bias is −2 volts.
This causes the output signal to be +20 volts at time three (T3).
Between time three (T3) and time four (T4), input signal #1 goes from −1 volt to 0 volts and input signal #2 goes from +1 volt to 0 volts. At time four (T4) both input signals are 0 volts, the bias is 0 volts, and the output is 0 volts.
During time four (T4) through time eight (T8), the circuit repeats the sequence of events that took place from time zero (T0) through time four (T4).
You can see that when the input signals are equal in amplitude and 180 degrees out of phase, the output signal is twice as large (40 volts peak to peak) as it would be from either input signal alone (if the other input signal were held at 0 volts).
Figure 5 shows the twoinput, singleoutput, difference amplifier with two input signals that are equal in amplitude and in phase.
Notice, that the output signal remains at 0 volts for the entire time (T0  T8). Since the two input signals are equal in amplitude and in phase, the difference between them (the basetoemitter bias) is always 0 volts. This causes a 0volt output signal.
If you compute the bias at any time period (T0  T8), you will see that the output of the circuit remains at a constant zero.
For example:
From the above example, you can see that when the input signals are equal in amplitude and in phase, there is no output from the difference amplifier because there is no difference between the two inputs. You also know that when the input signals are equal in amplitude but 180 degrees out of phase, the output looks just like the input except for amplitude and a 180degree phase reversal with respect to input signal number one. What happens if the input signals are equal in amplitude but different in phase by something other than 180 degrees? This would mean that sometimes one signal would be going negative while the other would be going positive; sometimes both signals would be going positive; and sometimes both signals would be going negative. Would the output signal still look like the input signals? The answer is "no," because Figure 6 shows a difference amplifier with two input signals that are equal in amplitude but 90 degrees out of phase. From the figure you can see that at time zero (T0) input number one is at 0 volts and input number two is at 1 volt. The basetoemitter bias is found to be +1 volt.
This +1volt bias signal causes the output signal to be −10 volts at time zero (T0). Between time zero (T0) and time one (T1), both input signals go positive. The difference between the input signals stays constant. The effect of this is to keep the bias at +1 volt for the entire time between T0 and T1. This, in turn, keeps the output signal at 10 volts.
Between time one (T1) and time two (T2), input signal number one goes in a negative direction but input signal number two continues to go positive. Now the difference between the input signals decreases rapidly from +1 volt. Halfway between T1 and T2 (the dotted vertical line), input signal number one and input signal number two are equal in amplitude. The difference between the input signals is 0 volts and this causes the output signal to be 0 volts. From this point to T2 the difference between the input signals is a negative value. At T2:
From time two (T2) to time three (T3), input signal number one goes negative and input signal number two goes to zero. The difference between them stays constant at 1 volt. Therefore, the output signal stays at a +10volt level for the entire time period from T2 to T3. At T3 the bias condition will be:
Between T3 and T4 input signal number one goes to zero while input signal number two goes negative. This, again, causes a rapid change in the difference between the input signals. Halfway between T3 and T4 (the dotted vertical line) the two input signals are equal in amplitude; therefore, the difference between the input signals is 0 volts, and the output signal becomes 0 volts. From that point to T4, the difference between the input signals becomes a positive voltage. At T4:
(The sequence of events from T4 to T8 are the same as those of T0 to T4.)
As you have seen, this amplifier amplifies the difference between two input signals. But this is NOT a differential amplifier. A differential amplifier has two inputs and two outputs. The circuit you have just been shown has only one output. Well then, how does a differential amplifier schematic look?
2.3. Typical Differential Amplifier Circuit
Figure 7 is the schematic diagram of a typical differential amplifier. Notice that there are two inputs and two outputs. This circuit requires two transistors to provide the two inputs and two outputs. If you look at one input and the transistor with which it is associated, you will see that each transistor is a commonemitter amplifier for that input (input one and Q1; input two and Q2). R1 develops the signal at input one for Q1, and R5 develops the signal at input two for Q2. R3 is the emitter resistor for both Q1 and Q2. Notice that R3 is NOT bypassed. This means that when a signal at input one affects the current through Q1, that signal is developed by R3. (The current through Q1 must flow through R3; as this current changes, the voltage developed across R3 changes.) When a signal is developed by R3, it is applied to the emitter of Q2. In the same way, signals at input two affect the current of Q2, are developed by R3, and are felt on the emitter of Q1. R2 develops the signal for output one, and R4 develops the signal for output two.
Even though this circuit is designed to have two inputs and two outputs, it is not necessary to use both inputs and both outputs. (Remember, a differential amplifier was defined as having two possible inputs and two possible outputs.) A differential amplifier can be connected as a singleinput, singleoutput device; a singleinput, differentialoutput device; or a differentialinput, differentialoutput device.
Q1. How many inputs and outputs are possible with a differential amplifier? Q2. What two transistor amplifier configurations are combined in the singletransistor, twoinput, singleoutput difference amplifier? Q3. If the two input signals of a difference amplifier are in phase and equal in amplitude, what will the output signal be? Q4. If the two input signals to a difference amplifier are equal in amplitude and 180 degrees out of phase, what will the output signal be? Q5. If only one input signal is used with a difference amplifier, what will the output signal be? Q6. If the two input signals to a difference amplifier are equal in amplitude but neither in phase nor 180 degrees out of phase, what will the output signal be?
2.4. SingleInput, SingleOutput, Differential Amplifier
Figure 8 shows a differential amplifier with one input (the base of Q1) and one output (the collector of Q2). The second input (the base of Q2) is grounded and the second output (the collector of Q1) is not used.
When the input signal developed by R1 goes positive, the current through Q1 increases. This increased current causes a positivegoing signal at the top of R3. This signal is felt on the emitter of Q2. Since the base of Q2 is grounded, the current through Q2 decreases with a positivegoing signal on the emitter. This decreased current causes less voltage drop across R4. Therefore, the voltage at the bottom of R4 increases and a positivegoing signal is felt at the output.
When the input signal developed by R1 goes negative, the current through Q1 decreases. This decreased current causes a negativegoing signal at the top of R3. This signal is felt on the emitter of Q2. When the emitter of Q2 goes negative, the current through Q2 increases. This increased current causes more of a voltage drop across R4. Therefore, the voltage at the bottom of R4 decreases and a negative going signal is felt at the output.
This singleinput, singleoutput, differential amplifier is very similar to a singletransistor amplifier as far as input and output signals are concerned. This use of a differential amplifier does provide amplification of a.c. or d.c. signals but does not take full advantage of the characteristics of a differential amplifier.
2.5. SingleInput, DifferentialOutput, Differential Amplifier
In chapter one of this module you were shown several phase splitters. You should remember that a phase splitter provides two outputs from a single input. These two outputs are 180 degrees out of phase with each other. The singleinput, differentialoutput, differential amplifier will do the same thing.
Figure 9 shows a differential amplifier with one input (the base of Q1) and two outputs (the collectors of Q1 and Q2). One output is in phase with the input signal, and the other output is 180 degrees out of phase with the input signal. The outputs are differential outputs.
This circuit’s operation is the same as for the singleinput, singleoutput differential amplifier just described. However, another output is obtained from the bottom of R2. As the input signal goes positive, thus causing increased current through Q1, R2 has a greater voltage drop. The output signal at the bottom of R2 therefore is negative going. A negativegoing input signal will decrease current and reverse the polarities of both output signals.
Now you see how a differential amplifier can produce two amplified, differential output signals from a singleinput signal. One further point of interest about this configuration is that if a combined output signal is taken between outputs number one and two, this single output will be twice the amplitude of the individual outputs. In other words, you can double the gain of the differential amplifier (single output) by taking the output signal between the two output terminals. This singleoutput signal will be in phase with the input signal. This is shown by the phantom signal above R5 (the phantom resistor connected between outputs number one and two would be used to develop this signal).
2.6. DifferentialInput, DifferentialOutput, Differential Amplifier
When a differential amplifier is connected with a differential input and a differential output, the full potential of the circuit is used. Figure 10 shows a differential amplifier with this type of configuration (differentialinput, differentialoutput).
Normally, this configuration uses two input signals that are 180 degrees out of phase. This causes the difference (differential) signal to be twice as large as either input alone. (This is just like the twoinput, singleoutput difference amplifier with input signals that are 180 degrees out of phase.)
Output number one is a signal that is in phase with input number two, and output number two is a signal that is in phase with input number one. The amplitude of each output signal is the input signal multiplied by the gain of the amplifier. With 180degreeoutofphase input signals, each output signal is greater in amplitude than either input signal by a factor of the gain of the amplifier.
When an output signal is taken between the two output terminals of the amplifier (as shown by the phantom connections, resistor, and signal), the combined output signal is twice as great in amplitude as either signal at output number one or output number two. (This is because output number one and output number two are 180 degrees out of phase with each other.) When the input signals are 180 degrees out of phase, the amplitude of the combined output signal is equal to the amplitude of one input signal multiplied by two times the gain of the amplifier.
When the input signals are not 180 degrees out of phase, the combined output signal taken across output one and output two is similar to the output that you were shown for the twoinput, singleoutput, difference amplifier. The differential amplifier can have two outputs (180 degrees out of phase with each other), or the outputs can be combined as shown in Figure 10.
In answering Q7 through Q9 use the following information: All input signals are sine waves with a peaktopeak amplitude of 10 millivolts. The gain of the differential amplifier is 10.
Q7. If the differential amplifier is configured with a single input and a single output, what will the peaktopeak amplitude of the output signal be? Q8. If the differential amplifier is configured with a single input and differential outputs, what will the output signals be? Q9. If the singleinput, differentialoutput, differential amplifier has an output signal taken between the two output terminals, what will the peaktopeak amplitude of this combined output be? In answering Q10 through Q14 use the following information: A differential amplifier is configured with a differential input and a differential output. All input signals are sine waves with a peaktopeak amplitude of 10 millivolts. The gain of the differential amplifier is 10. Q10. If the input signals are in phase, what will be the peaktopeak amplitude of the output signals? Q11. If the input signals are 180 degrees out of phase with each other, what will be the peaktopeak amplitude of the output signals? Q12. If the input signals are 180 degrees out of phase with each other, what will the phase relationship be between (a) the output signals and (b) the input and output signals? Q13. If the input signals are 180 degrees out of phase with each other and a combined output is taken between the two output terminals, what will the amplitude of the combined output signal be? Q14. If the input signals are 90 degrees out of phase with each other and a combined output is taken between the two output terminals, (a) what will the peaktopeak amplitude of the combined output signal be, and (b) will the combined output signal be a sine wave?
3. Operational Amplifiers
An OPERATIONAL AMPLIFIER (OP AMP) is an amplifier which is designed to be used with other circuit components to perform either computing functions (addition, subtraction) or some type of transfer operation, such as filtering. Operational amplifiers are usually highgain amplifiers with the amount of gain determined by feedback.
Operational amplifiers have been in use for some time. They were originally developed for analog (nondigital) computers and used to perform mathematical functions. Operational amplifiers were not used in other devices very much because they were expensive and more complicated than other circuits.
Today many devices use operational amplifiers. Operational amplifiers are used as d.c. amplifiers, a.c. amplifiers, comparators, oscillators (which are covered in NEETS, Module 9), filter circuits, and many other applications. The reason for this widespread use of the operational amplifier is that it is a very versatile and efficient device. As an integrated circuit (chip) the operational amplifier has become an inexpensive and readily available "building block" for many devices. In fact, an operational amplifier in integrated circuit form is no more expensive than a good transistor.
3.1. Characteristics of an Operational Amplifier
The schematic symbols for an operational amplifier are shown in Figure 11 and in Figure 12. Figure 11 shows the power supply requirements while Figure 12 shows only the input and output terminals. An operational amplifier is a special type of highgain, d.c. amplifier. To be classified as an operational amplifier, the circuit must have certain characteristics. The three most important characteristics of an operational amplifier are:

Very high gain

Very high input impedance

Very low output impedance
Since no single amplifier stage can provide all these characteristics well enough to be considered an operational amplifier, various amplifier stages are connected together. The total circuit made up of these individual stages is called an operational amplifier. This circuit (the operational amplifier) can be made up of individual components (transistors, resistors, capacitors, etc.), but the most common form of the operational amplifier is an integrated circuit. The integrated circuit (chip) will contain the various stages of the operational amplifier and can be treated and used as if it were a single stage.
3.2. Block Diagram of an Operational Amplifier
Figure 13 is a block diagram of an operational amplifier. Notice that there are three stages within the operational amplifier.
The input stage is a differential amplifier. The differential amplifier used as an input stage provides differential inputs and a frequency response down to d.c. Special techniques are used to provide the high input impedance necessary for the operational amplifier.
The second stage is a highgain voltage amplifier. This stage may be made from several transistors to provide high gain. A typical operational amplifier could have a voltage gain of 200,000. Most of this gain comes from the voltage amplifier stage.
The final stage of the OP AMP is an output amplifier. The output amplifier provides low output impedance. The actual circuit used could be an emitter follower. The output stage should allow the operational amplifier to deliver several milliamperes to a load.
Notice that the operational amplifier has a positive power supply (+V_{CC}) and a negative power supply (−V_{EE}). This arrangement enables the operational amplifier to produce either a positive or a negative output.
The two input terminals are labeled "inverting input" (−) and "noninverting input" (+). The operational amplifier can be used with three different input conditions (modes). With differential inputs (first mode), both input terminals are used and two input signals which are 180 degrees out of phase with each other are used. This produces an output signal that is in phase with the signal on the noninverting input. If the noninverting input is grounded and a signal is applied to the inverting input (second mode), the output signal will be 180 degrees out of phase with the input signal (and onehalf the amplitude of the first mode output). If the inverting input is grounded and a signal is applied to the noninverting input (third mode), the output signal will be in phase with the input signal (and onehalf the amplitude of the first mode output).
Q15. What are the three requirements for an operational amplifier? Q16. What is the most commonly used form of the operational amplifier? Q17. Draw the schematic symbol for an operational amplifier. Q18. Label the parts of the operational amplifier shown in Figure 14.
3.3. ClosedLoop Operation of an Operational Amplifier
Operational amplifiers can have either a closedloop operation or an openloop operation. The operation (closedloop or openloop) is determined by whether or not feedback is used. Without feedback the operational amplifier has an openloop operation. This openloop operation is practical only when the operational amplifier is used as a comparator (a circuit which compares two input signals or compares an input signal to some fixed level of voltage). As an amplifier, the openloop operation is not practical because the very high gain of the operational amplifier creates poor stability. (Noise and other unwanted signals are amplified so much in openloop operation that the operational amplifier is usually not used in this way.) Therefore, most operational amplifiers are used with feedback (closedloop operation).
Operational amplifiers are used with degenerative (or negative) feedback which reduces the gain of the operational amplifier but greatly increases the stability of the circuit. In the closedloop configuration, the output signal is applied back to one of the input terminals. This feedback is always degenerative (negative). In other words, the feedback signal always opposes the effects of the original input signal. One result of degenerative feedback is that the inverting and noninverting inputs to the operational amplifier will be kept at the same potential.
Closedloop circuits can be of the inverting configuration or noninverting configuration. Since the inverting configuration is used more often than the noninverting configuration, the inverting configuration will be shown first.
3.3.1. Inverting Configuration
Figure 15 shows an operational amplifier in a closedloop, inverting configuration. Resistor R2 is used to feed part of the output signal back to the input of the operational amplifier.
At this point it is important to keep in mind the difference between the entire circuit (or operational circuit) and the operational amplifier. The operational amplifier is represented by the trianglelike symbol while the operational circuit includes the resistors and any other components as well as the operational amplifier. In other words, the input to the circuit is shown in Figure 15, but the signal at the inverting input of the operational amplifier is determined by the feedback signal as well as by the circuit input signal.
As you can see in Figure 15, the output signal is 180 degrees out of phase with the input signal. The feedback signal is a portion of the output signal and, 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 of this is that the inverting input to the operational amplifier is always very close to 0 volts with this configuration. In fact, with the noninverting input grounded, the voltage at the inverting input to the operational amplifier is so small compared to other voltages in the circuit that it is considered to be VIRTUAL GROUND. (Remember, in a closedloop operation the inverting and noninverting inputs are at the same potential.)
Virtual ground is a point in a circuit which is at ground potential (0 volts) but is NOT connected to ground. Figure 16, Figure 17, and Figure 18 show an example of several circuits with points at virtual ground.
In Figure 16, V1 (the lefthand battery) supplies +10 volts to the circuit while V2 (the righthand battery) supplies 10 volts to the circuit. The total difference in potential in the circuit is 20 volts.
The total resistance of the circuit can be calculated:
Now that the total resistance is known, the circuit current can be calculated:
The voltage drop across R1 can be computed:
The voltage at point A would be equal to the voltage of V1 minus the voltage drop of R1.
To check this result, compute the voltage drop across R2 and subtract this from the voltage at point A. The result should be the voltage of V2.
It is not necessary that the voltage supplies be equal to create a point of virtual ground. In Figure 17 V1 supplies +1 volt to the circuit while V2 supplies 10 volts. The total difference in potential is 11 volts. The total resistance of this circuit (R1 + R2) is 11 ohms. The total current (I_{T}) is 1 ampere. The voltage drop across R1 (E_{R1} = R_{1} × I_{T}) is 1 volt. The voltage drop across R2 (E_{R2} = R2 × I_{T}) is 10 volts. The voltage at point A can be computed:
So point A is at virtual ground in this circuit also. To check the results, compute the voltage at V2.
You can compute the values for Figure 18 and prove that point A in that circuit is also at virtual ground.
The whole point is that the inverting input to the operational amplifier shown in Figure 15 is at virtual ground since it is at 0 volts (for all practical purposes). Because the inverting input is at 0 volts, there will be no current (for all practical purposes) flowing into the operational amplifier from the connection point of R1 and R2.
Given these conditions, the characteristics of this circuit are determined almost entirely by the values of R1 and R2. Figure 19 should help show how the values of R1 and R2 determine the circuit characteristics.
Note

It should be stressed at this point that for purpose of explanation the operational amplifier is a theoretically perfect amplifier. In actual practice we are dealing with less than perfect. In the practical operational amplifier there will be a slight input current with a resultant power loss. This small signal can be measured at the theoretical point of virtual ground. This does not indicate faulty operation. 
The input signal causes current to flow through R1. (Only the positive half cycle of the input signal is shown and will be discussed.) Since the voltage at the inverting input of the operational amplifier is at 0 volts, the input current (I_{in}) is computed by:
The output signal (which is opposite in phase to the input signal) causes a feedback current (Ifdbk) to flow through R2. The lefthand side of R2 is at 0 volts (point A) and the righthand side is at E_{out}. Therefore, the feedback current is computed by:
(The minus sign indicates that E_{out} is 180 degrees out of phase with E_{in} and should not be confused with output polarity.)
Since no current flows into or out of the inverting input of the operational amplifier, any current reaching point A from R1 must flow out of point A through R2. Therefore, the input current (I_{in}) and the feedback current (I_{fdbk}) must be equal. Now we can develop a mathematical relationship between the input and output signals and R1 and R2.
Mathematically:
If you multiply both sides of the equation by R1:
If you divide both sides of the equation by E_{out}:
By inverting both sides of the equation:
You should recall that the voltage gain of a stage is defined as the output voltage divided by the input voltage:
Therefore, the voltage gain of the inverting configuration of the operational amplifier is expressed by the equation:
(As stated earlier, the minus sign indicates that the output signal is 180 degrees out of phase with the input signal.)
3.3.2. Noninverting Configuration
Figure 20 shows a noninverting configuration using an operational amplifier. The input signal (E_{in}) is applied directly to the noninverting (+) input of the operational amplifier. Feedback is provided by coupling part of the output signal (E_{out}) back to the inverting () input of the operational amplifier. R1 and R2 act as voltage divider that allows only a part of the output signal to be applied as feedback (E_{fdbk}).
Notice that the input signal, output signal, and feedback signal are all in phase. (Only the positive alternation of the signal is shown.) It may appear as if the feedback is regenerative (positive) because the feedback and input signals are in phase. The feedback is, in reality, degenerative (negative) because the input signals is applied to the noninverting input and the feedback signal is applied to the inverting input, (Remember, that the operational amplifier will react to the difference between the two inputs.)
Just as in the inverting configuration, the feedback signal is equal to the input signal (for all practical purposes). This time, however, the feedback signal is in phase with the input signal.
Therefore:
Given this condition, you can calculate the gain of the stage in terms of the resistors (R1 and R2).
The gain of the stage is defined as:
The feedback signal (E_{fdbk}) can be shown in terms of the output signal (E_{out})
and the voltage divider (R1 and R2). The voltage divider has the output signal
on one end and ground (0 volts) on the other end. The feedback signal is that
part of the output signal developed by R1 (at point A). Another way to look at
it is that the feedback signal is the amount of output signal left (at point A)
after part of the output signal has been dropped by R2. In either case, the
feedback signal (E_{fdbk}) is the ratio of R1 to the entire voltage divider (R1
R2) multiplied by the output signal (E_{out}).
Mathematically, the relationship of the output signal, feedback signal, and voltage divider is:
If you divide both sides of the equation by E_{out}:
By inverting both sides of the equation:
Separating the righthand side:
Remember:
Therefore, by substitution:
You can now see that the gain of the noninverting configuration is determined by the resistors. The formula is different from the one used for the inverting configuration, but the gain is still determined by the values of R1 and R2.
3.4. Bandwidth Limitations
As with most amplifiers, the gain of an operational amplifier varies with frequency. The specification sheets for operational amplifiers will usually state the openloop (no feedback) gain for d.c. (or 0 hertz). At higher frequencies, the gain is much lower. In fact, for an operational amplifier, the gain decreases quite rapidly as frequency increases.
Figure 21 shows the openloop (no feedback) frequencyresponse curve for a typical operational amplifier. As you should remember, bandwidth is measured to the halfpower points of a frequency response curve. The frequencyresponse curve shows that the bandwidth is only 10 hertz with this configuration. The UNITY GAIN POINT, where the signal out will have the same amplitude as the signal in (the point at which the gain of the amplifier is 1), is 1 megahertz for the amplifier. As you can see, the frequency response of this amplifier drops off quite rapidly.
Figure 21 is the openloop frequencyresponse curve. You have been told that most operational amplifiers are used in a closedloop configuration. When you look at the frequencyresponse curve for a closedloop configuration, one of the most interesting and important aspects of the operational amplifier becomes apparent: The use of degenerative feedback increases the bandwidth of an operational amplifier circuit.
This phenomenon is another example of the difference between the operational amplifier itself and the operationalamplifier circuit (which includes the components in addition to the operational amplifier). You should also be able to see that the external resistors not only affect the gain of the circuit, but the bandwidth as well.
You might wonder exactly how the gain and bandwidth of a closedloop, operationalamplifier circuit are related. Figure 22 should help to show you the relationship. The frequencyresponse curve shown in Figure 22 is for a circuit in which degenerative feedback has been used to decrease the circuit gain to 100 (from 100,000 for the operational amplifier). Notice that the halfpower point of this curve is just slightly above 10 kilohertz.
Now look at Figure 23. In this case, more feedback has been used to decrease the gain of the circuit to 10. Now the bandwidth of the circuit is extended to about 100 kilohertz.
The relationship between circuit gain and bandwidth in an operationalamplifier circuit can be expressed by the GAINBANDWIDTH PRODUCT (GAIN × BANDWIDTH = UNITY GAIN POINT). In other words, for operationalamplifier circuits, the gain times the bandwidth for one configuration of an operational amplifier will equal the gain times the bandwidth for any other configuration of the same operational amplifier. In other words, when the gain of an operationalamplifier circuit is changed (by changing the value of feedback or input resistors), the bandwidth also changes. But the gain times the bandwidth of the first configuration will equal the gain times the bandwidth of the second configuration. The following example should help you to understand this concept.
The frequencyresponse curves shown in Figure 21, Figure 22, and Figure 23 have a gainbandwidth product of 1,000,000. In Figure 21, the gain is 100,000 and the bandwidth is 10 hertz. The gainbandwidth product is 100,000 times 10 (Hz), or 1,000,000. In Figure 22, the gain has been reduced to 100 and the bandwidth increases to 10 kilohertz. The gainbandwidth product is 100 times 10,000 (Hz) which is also equal to 1,000,000. In Figure 23 the gain has been reduced to 10 and the bandwidth is 100 kilohertz. The gainbandwidth product is 10 times 100,000 (Hz), which is 1,000,000. If the gain were reduced to 1, the bandwidth would be 1 megahertz (which is shown on the frequencyresponse curve as the unitygain point) and the gainbandwidth product would still be 1,000,000.
Q19. What does the term "closedloop" mean in the closedloop configuration of an operational amplifier? In answering Q20, Q21, and Q23, select the correct response from the choices given in the parentheses. Q20. In a closedloop configuration the output signal is determined by (the input signal, the feedback signal, both). Q21. In the inverting configuration, the input signal is applied to the (a) (inverting, noninverting) input and the feedback signal is applied to the (b) (inverting, noninverting) input. Q22. In the inverting configuration, what is the voltage (for all practical purposes) at the inverting input to the operational amplifier if the input signal is a 1volt, peaktopeak sine wave? Q23. In the inverting configuration when the noninverting input is grounded, the inverting input is at (signal, virtual) ground. Q24. In a circuit such as that shown in Figure 19, if R1 has a value of 100 ohms and R2 has a value of 1 kilohm and the input signal is at a value of + 5 millivolts, what is the value of the output signal? Q25. If the unitygain point of the operational amplifier used in question 24 is 500 kilohertz, what is the bandwidth of the circuit? Q26. In a circuit such as that shown in Figure 20, if R1 has a value of 50 ohms and R2 has a value of 250 ohms and the input signal has a value of +10 millivolts, what is the value of the output signal? Q27. If the openloop gain of the operational amplifier used in question 26 is 200,000 and the open loop bandwidth is 30 hertz, what is the closed loop bandwidth of the circuit?
3.5. Applications of Operational Amplifiers
Operational amplifiers are used in so many different ways that it is not possible to describe all of the applications. Entire books have been written on the subject of operational amplifiers. Some books are devoted entirely to the applications of operational amplifiers and are not concerned with the theory of operation or other circuits at all. This module, as introductory material on operational amplifiers, will show you only two common applications of the operational amplifier: the summing amplifier and the difference amplifier. For ease of explanation the circuits shown for these applications will be explained with d.c. inputs and outputs, but the circuit will work as well with a.c. signals.
3.5.1. Summing Amplifier (Adder)
Figure 24 is the schematic of a twoinput adder which uses an operational amplifier. The output level is determined by adding the input signals together (although the output signal will be of opposite polarity compared to the sum of the input signals).
If the signal on input number one (E1) is +3 volts and the signal on input number two (E2) is +4 volts, the output signal (E_{out}) should be −7 volts [(+3 V) + (+4 V) = +7 V and change the polarity to get −7 V].
With +3 volts at E1 and 0 volts at point A (which is at virtual ground), the current through R1 must be 3 milliamperes.
Mathematically:
(The + sign indicates a current flow from right to left.)
By the same sort of calculation, with +4 volts at E2 and 0 volts at point A the current through R2 must be 4 milliamps.
This means that a total of 7 milliamps is flowing from point A through R1 and R2. If 7 milliamps is flowing from point A, then 7 milliamps must be flowing into point A. The 7 milliamps flowing into point A flows through R3 causing 7 volts to be developed across R3. With point A at 0 volts and 7 volts developed across R3, the voltage potential at E_{out} must be a −7 volts. Figure 25 shows these voltages and currents.
An adder circuit is not restricted to two inputs. By adding resistors in parallel to the input terminals, any number of inputs can be used. The adder circuit will always produce an output that is equal to the sum of the input signals but opposite in polarity. Figure 26 shows a fiveinput adder circuit with voltages and currents indicated.
The previous circuits have been adders, but there are other types of summing amplifiers. A summing amplifier can be designed to amplify the results of adding the input signals. This type of circuit actually multiplies the sum of the inputs by the gain of the circuit.
Mathematically (for a threeinput circuit):
If the circuit gain is 10:
The gain of the circuit is determined by the ratio between the feedback resistor and the input resistors. To change Figure 24 to a summing amplifier with a gain of 10, you would replace the feedback resistor (R3) with a 10kilohm resistor. This new circuit is shown in Figure 27.
If this circuit is designed correctly and the input voltages (E1 and E2) are +2 volts and +3 volts, respectively, the output voltage (E_{out}) should be:
To see if this output (−50 V) is what the circuit will produce with the inputs given above, start by calculating the currents through the input resistors, R1 and R2 (remember that point A is at virtual ground):
Next, calculate the current through the feedback resistor (R3):
(The minus sign indicates current flow from left to right.)
Finally, calculate the voltage dropped across R3 (which must equal the output voltage):
As you can see, this circuit performs the function of adding the inputs together and multiplying the result by the gain of the circuit.
One final type of summing amplifier is the SCALING AMPLIFIER. This circuit multiplies each input by a factor (the factor is determined by circuit design) and then adds these values together. The factor that is used to multiply each input is determined by the ratio of the feedback resistor to the input resistor. For example, you could design a circuit that would produce the following output from three inputs (E1, E2, E3):
Using input resistors R1 for input number one (E1), R2 for input number two (E2), R3 for input number three (E3), and R4 for the feedback resistor, you could calculate the values for the resistors:
Any resistors that will provide the ratios shown above could be used. If the feedback resistor (R4) is a 12kilohm resistor, the values of the other resistors would be:
Figure 28 is the schematic diagram of a scaling amplifier with the values calculated above.
To see if the circuit will produce the desired output, calculate the currents and voltages as done for the previous circuits.
With:
the output should be:
Calculate the current for each input:
Calculate the output voltage:
You have now seen how an operational amplifier can be used in a circuit as an adder, a summing amplifier, and a scaling amplifier.
3.5.2. Difference Amplifier (Subtractor)
A difference amplifier will produce an output based on the difference between the input signals. The subtractor circuit shown in Figure 29 will produce the following output:
Normally, difference amplifier circuits have the ratio of the inverting input resistor to the feedback resistor equal to the ratio of the noninverting input resistors. In other words, for Figure 29:
and, by inverting both sides:
For ease of explanation, in the circuit shown in Figure 29 all the resistors have a value of 1 kilohm, but any value could be used as long as the above ratio is true. For a subtractor circuit, the values of R1 and R3 must also be equal, and therefore, the values of R2 and R4 must be equal. It is NOT necessary that the value of R1 equal the value of R2.
Using Figure 29, assume that the input signals are:
The output signal should be:
E_{out} = E2  E1 E_{out} = (+12V)  (+3V) E_{out} = +9V
To check this output, first compute the value of R2 plus R4:
With this value, compute the current through R2 (I_{R2}):
(indicating current flow from left to right)
Next, compute the voltage drop across R2 (E_{R2}):
Then compute the voltage at point B:
Since point B and point A will be at the same potential in an operational amplifier:
Now compute the voltage developed by R1 (E_{R1}):
Compute the current through R1 (I_{R1}):
Compute the voltage developed by R3 (E_{R3}):
Add this to the voltage at point A to compute the output voltage (E_{out}):
As you can see, the circuit shown in Figure 29 functions as a subtractor. But just as an adder is only one kind of summing amplifier, a subtractor is only one kind of difference amplifier. A difference amplifier can amplify the difference between two signals. For example, with two inputs (E1 and E2) and a gain of five, a difference amplifier will produce an output signal which is:
The difference amplifier that will produce that output is shown in Figure 30. Notice that this circuit is the same as the subtractor shown in Figure 29 except for the values of R3 and R4. The gain of this difference amplifier is:
Then, for a difference amplifier:
With the same inputs that were used for the subtractor, (E1 = + 3 V; E2 = + 12 V) the output of the difference amplifier should be five times the output of the subtractor (E_{out} = + 45 V).
Following the same steps used for the subtractor:
First compute the value of R2 plus R4:
With this value, compute the current through R2 (I_{R2}):
Next, compute the voltage drop across R2 (E_{R2}):
Then, compute the voltage at point B:
Since point A and point B will be at the same potential in an operational amplifier:
Now compute the voltage developed by R1 (E_{R1}):
Compute the current through R1 (I_{R1}):
Compute the voltage developed by R3 (E_{R3}):
Add this voltage to the voltage at point A to compute the output voltage (E_{out}):
This was the output desired, so the circuit works as a difference amplifier.
Q28. What is the difference between a summing amplifier and an adder circuit? Q29. Can a summing amplifier have more than two inputs? Q30. What is a scaling amplifier? Refer to Figure 31 in answering Q31 through Q33.
Q31. What type of circuit is Figure 31? Q32. If: E1 = 2V, and: E2 = +6V, then E~out~ = ? Q33. What is the difference in potential between the inverting () and noninverting () inputs to the operational amplifier when: E1 = +6V, and E2 = +2V Q34. What is the difference between a subtractor and a difference amplifier? Q35. Can a difference amplifier have more than two inputs? Refer to Figure 32 in answering Q36 through Q38.
Q36. What type of circuit is Figure 32? Q37. If: E1 = 5V, and: E2 = +11V, then E~out~ = ? Q38. What is the difference in potential between the inverting (íDQGnoninverting () inputs to the operational amplifier when: E1 = +2V, and E2 = +4V