Ch2: Common Op Amps Configurations

Op Amps Table of Contents

Unity Gain Buffer

To build a unity gain buffer, you simply connect the input to the non-inverting pin and feedback the output to the inverting pin. It’s a very simple circuit and gives you an amplifier circuit with a gain of 1. This might not seem like a very useful feature, but check out this video to learn more about this circuit:

This video describes the connections and purposes of a unity gain buffer and shows why an infinite open loop voltage gain leads to unity gain when you have negative feedback.

Unity Gain Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000, and the inverting input connected directly to its output terminal. In other words, write an equation describing the output voltage of this op-amp ( $V_{out}$ ) for any given input voltage at the noninverting input ( $V_{in(+)}$ ):

Then, once you have an equation written, solve for the over-all voltage gain ( $A_V = {V_{out} \over V_{in(+)}}$ ) of this amplifier circuit, and calculate the output voltage for a noninverting input voltage of +6 volts.

File Num: 00927

Answer

V_{out} = 100,000(V_{in(+)} - V_{out})

(I’ve left it up to you to perform the algebraic simplification here!)

A_V = {100,000 \over 100,001} = 0.99999

For an input voltage of +6 volts, the output voltage will be +5.99994 volts.

Notes

The significant point of this question is that students see the over-all voltage gain of the opamp radically attenuated from 100,000 to approximately 1. What is not so evident is just how stable this new voltage gain is, which is one of the purposes for employing negative feedback.

Question 2. (Click on arrow for answer)

How much effect will a change in the op-amp’s open-loop voltage gain have on the overall voltage gain of a negative-feedback circuit such as this?

If the open-loop gain of this operational amplifier were to change from 100,000 to 200,000, for example, how big of an effect would it have on the voltage gain as measured from the noninverting input to the output?

File Num: 02288

Answer

The different in overall voltage gain will be trivial.

Follow-up question: what advantage is there in building voltage amplifier circuits in this manner, applying negative feedback to a “core” amplifier with very high intrinsic gain?

Notes

Work with your students to calculate a few example scenarios, with the old open-loop gain versus the new open-loop gain. Have the students validate their conclusions with numbers!

Negative feedback is an extremely useful engineering principle, and one that allows us to build very precise amplifiers using imprecise components. Credit for this idea goes to Harold Black, an electrical engineer, in 1920’s. Mr. Black was looking for a way to improve the linearity and stability of amplifiers in telephone systems, and (as legend has it) the idea came to him in a flash of insight as he was commuting on a ferry boat.

An interesting historical side-note is that Black’s 1928 patent application was initially rejected on the grounds that he was trying to submit a perpetual motion device! The concept of negative feedback in an amplifier circuit was so contrary to established engineering thought at the time, that Black experienced significant resistance to the idea within the engineering community. The United States patent office, on the other hand, was inundated with fraudulent “perpetual motion” claims, and so dismissed Black’s invention at first sight.

Question 3. (Click on arrow for answer)

Complete the table of voltages for this opamp “voltage follower” circuit:

$$\vbox{\offinterlineskip \halign{\strut \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil \vrule \cr \noalign{\hrule}

V_{in}

V_{out}

\cr\noalign{\hrule}

0 volts & 0 volts \cr

\noalign{\hrule}+5 volts & \cr\noalign{\hrule}+10 volts & \cr\noalign{\hrule}+15 volts & \cr\noalign{\hrule}+20 volts & \cr\noalign{\hrule}-5 volts & \cr\noalign{\hrule}-10 volts & \cr\noalign{\hrule}-15 volts & \cr\noalign{\hrule}-20 volts & \cr\noalign{\hrule} } % End of \halign }$$ % End of \vbox

File Num: 02289

Answer

$$\vbox{\offinterlineskip \halign{\strut \vrule \quad\hfil # \ \hfil & \vrule \quad\hfil # \ \hfil \vrule \cr \noalign{\hrule}

V_{in}

V_{out}

\cr\noalign{\hrule}

0 volts & 0 volts \cr

\noalign{\hrule}+5 volts & +5 volts \cr\noalign{\hrule}+10 volts & +10 volts \cr\noalign{\hrule}+15 volts & +15 volts \cr\noalign{\hrule}+20 volts & +15 volts \cr\noalign{\hrule}-5 volts & -5 volts \cr\noalign{\hrule}-10 volts & -10 volts \cr\noalign{\hrule}-15 volts & -15 volts \cr\noalign{\hrule}-20 volts & -15 volts \cr\noalign{\hrule} } % End of \halign }$$ % End of \vbox

Follow-up question: the output voltage values given in this table are ideal. A real opamp would probably not be able to achieve even what is shown here, due to idiosyncrasies of these amplifier circuits. Explain what would probably be different in a real opamp circuit from what is shown here.

Notes

A common mistake I see students new to opamps make is assuming that the output voltage will magically attain whatever value the gain equation predicts, with no regard for power supply rail voltage limits.

Another good follow-up question to ask your students is this: “How much voltage is there between the two input terminals in each of the situations described in the table?” They will find that the “golden rule” of closed-loop opamp circuits can be violated!

If students have difficulty answering the follow-up question, drop these two hints: (1) Rail-to-rail output swing and (2) Latch-up.

Question 4. (Click on arrow for answer)

This operational amplifier circuit is often referred to as a voltage buffer, because it has unity gain (0 dB) and therefore simply reproduces, or “buffers,” the input voltage:

What possible use is a circuit such as this, which offers no voltage gain or any other form of signal modification? Wouldn’t a straight piece of wire do the same thing? Explain your answers.

File Num: 03801

Answer

While this circuit offers no voltage gain, it does offer current gain and impedance transformation. Much like the common-collector (or common-drain) single transistor amplifier circuits which also had voltage gains of (near) unity, opamp buffer circuits are useful whenever one must drive a relatively “heavy” (low impedance) load with a signal coming from a “weak” (high impedance) source.

Notes

I have found that some students have difficulty with the terms “heavy” and “light” in reference to load characteristics. That a “heavy” load would have very few ohms of impedance, and a “light” load would have many ohms of impedance seems counter-intuitive to some. It all makes sense, though, once students realize the terms “heavy” and “light” refer to the amount of current drawn by the respective loads.

Ask your students to explain why the straight piece of wire fails to “buffer” the voltage signal in the same way the the opamp follower circuit does.

Question 5. (Click on arrow for answer)

For all practical purposes, how much voltage exists between the inverting and noninverting input terminals of an op-amp in a functioning negative-feedback circuit?

File Num: 00930

Answer

Zero volts

Notes

Ask your students to explain why there will be (practically) no voltage between the input terminals of an operational amplifier when it is used in a negative feedback circuit.

Question 6. (Click on arrow for answer)

Just as certain assumptions are often made for bipolar transistors in order to simplify their analysis in circuits (an ideal BJT has negligible base current, $I_C = I_E$ , constant $\beta$ , etc.), we often make assumptions about operational amplifiers so we may more easily analyze their behavior in closed-loop circuits. Identify some of these ideal opamp assumptions as they relate to the following parameters:

Magnitude of input terminal currents:
Input impedance:
Output impedance:
Input voltage range:
Output voltage range:
Differential voltage (between input terminals) with negative feedback:

File Num: 02749

Answer

Magnitude of input terminal currents: infinitesimal
Input impedance: infinite
Output impedance: infinitesimal
Input voltage range: never exceeding +V/-V
Output voltage range: never exceeding +V/-V
Differential voltage (between input terminals) with negative feedback: infinitesimal

Notes

Just in case your students are unfamiliar with the words infinite and infinitesimal, tell them they simply mean “bigger than big” and “smaller than small”, respectively.

Question 7. (Click on arrow for answer)

The purpose of this circuit is to provide a pushbutton-adjustable voltage. Pressing one button causes the output voltage to increase, while pressing the other button causes the output voltage to decrease. When neither button is pressed, the voltage remains stable:

After working just fine for quite a long while, the circuit suddenly fails: now it only outputs zero volts DC all the time.

An experienced technician first checks the power supply voltage to see if it is within normal limits, and it is. Then, the technician checks the voltage across the capacitor. Explain why this is a good test point to check, and what the results of that check would tell the technician about the nature of the fault.

File Num: 03773

Answer

Checking for voltage across the capacitor will tell the technician what voltage the op-amp follower is being “told” to reproduce at the output.

Challenge question: why do you suppose I specify a CA3130 operational amplifier for this particular circuit? What is special about this opamp that qualifies it for the task?

Notes

Knowing where to check for critical signals in a circuit is an important skill, because it usually means the difference between efficiently locating a fault and wasting time. Ask your students to explain in detail the rationale behind checking for voltage across the capacitor, and (again, in detail) what certain voltage measurements at that point would prove about the nature of the fault.

Inverting Amplifier Circuit

An inverting amplifier is one where the output is inverted, or phase shifted by 180^o compared to the input. This is another simple circuit, but a little more complicated than the unity gain buffer. See this video to learn more:

This video shows why the amplifier is called an “inverting” one and shows how to figure out the voltage gain based on the input and feedback resistor. It also includes an LTSpice simulation to demonstrate how the amplifier works.

Introduction to Inverting Op Amp Circuits

Non-Inverting Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Trace the directions for all currents in this circuit, and calculate the values for voltage at the output ( $V_{out}$ ) and at test point 1 ( $V_{TP1}$ ) for several values of input voltage ( $V_{in}$ ):

Vin	Vtp1	Vout
0.0V
0.4V
1.2V
3.4V
7.1V
10.8V

Then, from the table of calculated values, determine the voltage gain ( $A_V$ ) for this amplifier circuit.

File Num: 02467

Answer

Vin	Vtp1	Vout
0.0V	0.0V	0.0V
0.4V	0.0V	-0.4V
1.2V	0.0V	-1.2V
3.4V	0.0V	-3.4V
7.1V	0.0V	-7.1V
10.8V	0.0V	-10.8V

A_V = 1 (ratio) = 0 dB

Follow-up question: the point marked “TP1” in this circuit is often referred to as a virtual ground. Explain why this is, based on the voltage figures shown in the above table.

Notes

Some texts describe the voltage gain of an inverting voltage amplifier as being a negative quantity. I tend not to look at things that way, treating all gains as positive quantities and relying on my knowledge of circuit behavior to tell whether the signal is inverted or not. In my teaching experience, I have found that students have a tendency to blindly follow equations rather than think about what it is they are calculating, and that strict adherence to the mathematical signs of gain values only encourages this undesirable behavior (“If the sign of the gain tells me whether the circuit is inverting or not, I can just multiply input voltage by gain and the answer will always be right!”).

This strategy is analogous to problem-solving in electromagnetics, where a common approach is to use math to solve for the absolute values of quantities (potential, induced voltage, magnetic flux), and then to use knowledge of physical principles (Lenz’ Law, right-hand rule) to solve for polarities and directions. The alternative — to try to maintain proper sign convention throughout all calculations — not only complicates the math but it also encourages students to over-focus on calculations and neglect fundamental principles.

Question 2. (Click on arrow for answer)

Calculate the overall voltage gain of this amplifier circuit ( $A_V$ ), both as a ratio and as a figure in units of decibels (dB). Also, write a general equation for calculating the voltage gain of such an amplifier, given the resistor values of $R_1$ and $R_2$ :

File Num: 02458

Answer

A_V = 1 = 0

A_V = {R_1 \over R_2} \hbox{ (expressed as a ratio, not dB)}

Follow-up question \#1: sometimes the voltage gain equation for an amplifier of this type is given in the following form:

A_V = -{R_1 \over R_2}

What is the significance of the negative sign in this equation? Is it really necessary, or does it communicate an important concept?

Follow-up question \#2: manipulate the gain equation for this amplifier circuit to solve for the value of resistor $R_1$ .

Notes

Whether inverting amplifier gains are expressed as negative or positive quantities seems to be a matter of taste, from surveying introductory textbooks on the subject. I prefer to stick with absolute (positive) gain values and consider signal inversion separately.

Question 3. (Click on arrow for answer)

Calculate all voltage drops and currents in this circuit, complete with arrows for current direction and polarity markings for voltage polarity. Then, calculate the overall voltage gain of this amplifier circuit ( $A_V$ ), both as a ratio and as a figure in units of decibels (dB):

File Num: 02468

Answer

A_V

= 0.468 = -6.594 dB

Notes

Operational amplifier circuits provide a great opportunity to review basic concepts of DC circuits: voltage drops, polarity, current directions, Ohm’s Law, Kirchhoff’s Voltage Law, Kirchhoff’s Current Law, etc. This circuit is no exception. Emphasize the fact that a great many opamp circuits may be comprehensively analyzed merely with knowledge of these fundamental principles and the characteristics of an ideal opamp (zero input current, infinite open-loop gain, unlimited output voltage swing, zero voltage between input terminals when negative feedback is in effect).

Some students may arrive at the wrong gain figure because they blindly followed a formula with $R_1$ and $R_2$ shown as variables, plugging in this circuit’s values for $R_1$ and $R_2$ without considering which resistor is which (is $R_1$ the feedback resistor or is $R_2$ ?). This is by design, as I want students to learn to think about what they are doing rather than thoughtlessly follow instructions.

Question 4. (Click on arrow for answer)

Determine both the input and output voltage in this circuit:

File Num: 02732

Answer

V_{in}

= -10 V.

V_{out}

= 24 V

Follow-up question: how do we know that the input voltage in this circuit is negative and the output voltage is positive?

Notes

Ask your students how they solved this problem, sharing techniques and strategies to help other students know where to begin and where to proceed from there.

Question 5. (Click on arrow for answer)

The equation for voltage gain ( $A_V$ ) in a typical inverting, single-ended opamp circuit is as follows:

A_V = {R_{1} \over R_{2}}

Where,

R_1

is the feedback resistor (connecting the output to the inverting input)

R_2

is the other resistor (connecting the inverting input to voltage signal input terminal)

Suppose we wished to change the voltage gain in the following circuit from 3.5 to 4.9, but only had the freedom to alter the resistance of $R_{2}$ :

Algebraically manipulate the gain equation to solve for $R_2$ , then determine the necessary value of $R_2$ in this circuit to give it a voltage gain of 4.9.

File Num: 02708

Answer

R_2 = {R_1 \over A_V}

For the circuit shown, $R_2$ would have to be set equal to 1.571 k $\Omega$ .

Notes

Nothing more than a little algebra to obtain the answers for this question!

Question 6. (Click on arrow for answer)

Calculate the necessary resistor value ( $R_1$ ) in this circuit to give it a voltage gain of 15:

File Num: 02729

Answer

R_1

= 1.467 k

\Omega

Notes

Ask your students how they solved this problem, especially since it is fairly safe to say that they didn’t find the equation directly solving for $R_1$ in any book. Algebraic manipulation is necessary to take the standard voltage gain equation and put it into a form suitable for use answering this question.

Question 7. (Click on arrow for answer)

Calculate the necessary resistor value ( $R_1$ ) in this circuit to give it a voltage gain of 7.5:

File Num: 02730

Answer

R_1

= 62.25 k

\Omega

Notes

Question 8. (Click on arrow for answer)

Calculate the output voltage of this op-amp circuit (using negative feedback):

Also, calculate the DC voltage gain of this circuit.

File Num: 00932

Answer

V_{out} =

-8.1 volts

A_V =

5.4

Follow-up question: the midpoint of the voltage divider (connecting to the inverting input of the op-amp) is often called a virtual ground in a circuit like this. Explain why.

Notes

It is important that students learn to analyze the op-amp circuit in terms of voltage drops and currents for each resistor, rather than just calculate the output using a gain formula. Detailed, Ohm’s Law analysis of op-amp circuits is essential for analyzing more complex circuitry.

The “virtual ground” question is an important one for the sake of rapid analysis. Once students understand how and why there is such a thing as a “virtual ground” in an op-amp circuit like this, their analysis of op-amp circuits will be much more efficient.

Question 9. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

File Num: 02470

Answer

Stage 1: $A_V$ = 1.5 = 3.522 dB

Stage 2: $A_V$ = 1.545 = 3.781 dB

Total: $A_V$ = 2.318 = 7.303 dB

Notes

Not only does this question review calculation of voltage gain for inverting amplifier circuits, but it also reviews decibel calculations (for both single and multi-stage amplifiers). Discuss how the decibel figures for each stage add to equal the total decibel gain, whereas the ratios multiply.

Question 10. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

File Num: 02471

Answer

Stage 1: $A_V$ = 1.766 = 4.940 dB

Stage 2: $A_V$ = 0.455 = -6.848 dB

Total: $A_V$ = 0.803 = -1.909 dB

Notes

Question 11. (Click on arrow for answer)

Operational amplifier circuits employing negative feedback are sometimes referred to as “electronic levers,” because their voltage gains may be understood through the mechanical analogy of a lever. Explain this analogy in your own words, identifying how the lengths and fulcrum location of a lever relate to the component values of an op-amp circuit:

File Num: 00933

Answer

The analogy of a lever works well to explain how the output voltage of an op-amp circuit relates to the input voltage, in terms of both magnitude and polarity. Resistor values correspond to moment arm lengths, while direction of lever motion (up versus down) corresponds to polarity. The position of the fulcrum represents the location of ground potential in the feedback network.

Notes

I found this analogy in one of the best books I’ve ever read on op-amp circuits: John I. Smith’s Modern Operational Circuit Design. Unfortunately, this book is out of print, but if you can possibly obtain a copy for your library, I highly recommend it!

Question 12. (Click on arrow for answer)

Compare and contrast inverting versus noninverting amplifier circuits constructed using operational amplifiers:

How do these two general forms of opamp circuit compare, especially in regard to input impedance and the range of voltage gain adjustment?

File Num: 02469

Answer

The noninverting configuration exhibits a far greater input impedance than the inverting amplifier, but has a more limited range of voltage gain: always greater than or equal to unity.

Notes

Just a simple comparison between amplifier configurations, nothing more. Ask your students to elaborate on the inverting amplifier’s range of gain adjustment: how does it differ from the noninverting configuration?

Question 13. (Click on arrow for answer)

What possible benefit is there to adding a voltage buffer to the front end of an inverting amplifier, as shown in the following schematic?

File Num: 02472

Answer

The voltage buffer raises the amplifier’s input impedance without altering voltage gain.

Notes

Discuss with your students how this is very common: using a voltage buffer as an impedance transformation (or isolation) device so that a weak (high-impedance) source is able to drive an amplifier.

Question 14. (Click on arrow for answer)

The junction between the two resistors and the inverting input of the operational amplifier is often referred to as a virtual ground, the voltage between it and ground being (almost) zero over a wide range of circuit conditions:

If the operational amplifier is driven into saturation, though, the “virtual ground” will no longer be at ground potential. Explain why this is, and what condition(s) may cause this to happen.

Hint: analyze all currents and voltage drops in the following circuit, assuming an opamp with the ability to swing its output voltage rail-to-rail.

File Num: 02473

Answer

Any input signal causing the operational amplifier to try to output a voltage beyond either of its supply rails will cause the “virtual ground” node to deviate substantially from ground potential.

Notes

Before students can answer this question, they must understand what saturation means with regard to an operational amplifier. This is where the “hint” scenario comes into play. Students failing to grasp this concept will calculate the voltage drops and currents in the “hint” circuit according to standard procedures and assumptions, and arrive at an output voltage well in excess of +15 volts. Resolving this paradox will lead to insight, and hopefully to a more realistic set of calculations.

Question 15. (Click on arrow for answer)

There is something wrong with this amplifier circuit. Despite an audio signal of normal amplitude detected at test point 1 (TP1), there is no output measured at the “Audio signal out” jack:

Next, you decide to check for the presence of a good signal at test point 3 (TP3). There, you find 0 volts AC and DC no matter where the volume control is set.

From this information, formulate a plan for troubleshooting this circuit, answering the following questions:

What type of signal would you expect to measure at TP3?
What would be your next step in troubleshooting this circuit?
Are there any elements of this circuit you know to be working properly?
What do you suppose would be the most likely failure, assuming this circuit once worked just fine and suddenly stopped working all on it’s own?

File Num: 02474

Answer

The correct voltage signal at TP3 should be an audio waveform with significant crossover distortion (specifically, a vertical “jump” at each point where the waveform crosses zero volts, about 1.4 volts peak to peak). I’ll let you figure out answers to the other questions on your own, or with classmates.

Notes

I have found that troubleshooting scenarios are always good for stimulating class discussions, with students posing strategies for isolating the fault(s) and correcting one another on logical errors. There is not enough information given in this question to ensure a single, correct answer. Discuss this with your students, helping them to use their knowledge of circuit theory and opamps to formulate good diagnostic strategies.

Non-Inverting Amplifier Circuit

A non-inverting amplifier does not invert that output (as you probably suspected). It does use negative feedback like the inverting amplifier to give a gain that is controllable by the external resistors. Check out this video for the details.

This video shows the configuration of a non-inverting op amp circuit and shows how to derive the equation for the gain of the amplifier. It even includes some simulations to show the op amp in action

Non-Inverting Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000, and the inverting input connected to a voltage divider on its output terminal (so the inverting input receives exactly one-half the output voltage). In other words, write an equation describing the output voltage of this op-amp ( $V_{out}$ ) for any given input voltage at the noninverting input ( $V_{in(+)}$ ):

Then, once you have an equation written, solve for the output voltage if the noninverting input voltage is -2.4 volts.

File Num: 00928

Answer

V_{out} = 100,000(V_{in(+)} - {1 \over 2}V_{out})

(I’ve left it up to you to perform the algebraic simplification here!)\vskip 20pt

For an input voltage of -2.4 volts, the output voltage will be -4.7999 volts.

Follow-up question: what do you notice about the output voltage in this circuit? What value is it very close to being, in relation to the input voltage? Does this pattern hold true for other input voltages as well?

Notes

Your students should see a definite pattern here as they calculate the output voltage for several different input voltage levels. Discuss this phenomenon with your students, asking them to explain it as best they can.

Question 2. (Click on arrow for answer)

File Num: 02457

Answer

A_V = 2 = 6.02

A_V = {R_1 \over R_2} + 1 \hbox{ (expressed as a ratio, not dB)}

Follow-up question: explain how you could modify this particular circuit to have a voltage gain (ratio) of 3 instead of 2.

Notes

Nothing special here — just some practice with voltage gain calculations.

Question 3. (Click on arrow for answer)

What would have to be altered in this circuit to increase its overall voltage gain?

File Num: 00931

Answer

The voltage divider would have to altered so as to send a smaller proportion of the output voltage to the inverting input.

Notes

Ask your students to explain how they would modify the voltage divider in this circuit to achieve the goal of a smaller voltage division ratio. This should be trivial, but it is always good to review basic principles of electricity even when “deep” into a more advanced topic.

Question 4. (Click on arrow for answer)

File Num: 02459

Answer

A_V

= 1.468 = 3.335 dB

Notes

Question 5. (Click on arrow for answer)

File Num: 02460

Answer

A_V

= 4.704 = 13.449 dB

Follow-up question: how much input impedance does the -2.35 volt source “see” as it drives this amplifier circuit?

Notes

The follow-up question is important because it showcases one of the great advantages of using noninverting opamp amplifier circuits as voltage signal amplifiers: extremely high input impedance. This would be a good opportunity to review typical input impedance values for operational amplifiers, by showing datasheets for some typical opamps and for some non-typical (i.e. MOSFET input) opamps.

Question 6. (Click on arrow for answer)

$\int f(x) dx$ Calculus alert!

You are part of a team building a rocket to carry research instruments into the high atmosphere. One of the variables needed by the on-board flight-control computer is velocity, so it can throttle engine power and achieve maximum fuel efficiency. The problem is, none of the electronic sensors on board the rocket has the ability to directly measure velocity. What is available is an altimeter, which infers the rocket’s altitude (it position away from ground) by measuring ambient air pressure; and also an accelerometer, which infers acceleration (rate-of-change of velocity) by measuring the inertial force exerted by a small mass.

The lack of a “speedometer” for the rocket may have been an engineering design oversight, but it is still your responsibility as a development technician to figure out a workable solution to the dilemma. How do you propose we obtain the electronic velocity measurement the rocket’s flight-control computer needs?

File Num: 02702

Answer

One possible solution is to use an electronic integrator circuit to derive a velocity measurement from the accelerometer’s signal. However, this is not the only possible solution!

Notes

This question simply puts students’ comprehension of basic calculus concepts (and their implementation in electronic circuitry) to a practical test.

Question 7. (Click on arrow for answer)

Calculate the necessary resistor value ( $R_1$ ) in this circuit to give it a voltage gain of 30:

File Num: 02725

Answer

R_1

= 1.345 k

\Omega

Notes

Question 8. (Click on arrow for answer)

Calculate the necessary resistor value ( $R_1$ ) in this circuit to give it a voltage gain of 10.5:

File Num: 02724

Answer

R_1

= 76.95 k

\Omega

Notes

Question 9. (Click on arrow for answer)

Determine both the input and output voltage in this circuit:

File Num: 02726

Answer

V_{in}

= 10 V.

V_{out}

= 46 V

Notes

Ask your students how they solved this problem, sharing techniques and strategies to help other students know where to begin and where to proceed from there.

Question 10. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

File Num: 02727

Answer

$A_V$ = 4.3 = 12.669 dB

$A_V$ = 6.745 = 16.579 dB

$A_V$ = 29.002 = 29.249 dB

Notes

Question 11. (Click on arrow for answer)

How much effect will a change in the op-amp’s open-loop voltage gain have on the overall voltage gain of a negative-feedback circuit such as this?

File Num: 00929

Answer

The different in overall voltage gain will be trivial.

Follow-up question: what advantage is there in building voltage amplifier circuits in this manner, applying negative feedback to a “core” amplifier with very high intrinsic gain?

Notes

Work with your students to calculate a few example scenarios, with the old open-loop gain versus the new open-loop gain. Have the students validate their conclusions with numbers!

Question 12. (Click on arrow for answer)

A simple “follower” circuit that boosts the current-output ability of this noninverting amplifier circuit is a set of bipolar junction transistors, connected together in a “push-pull” fashion like this:

However, if connected exactly as shown, there will be a significant voltage error introduced to the opamp’s output. No longer will the final output voltage (measured across the load) be an exact 3:1 multiple of the input voltage, due to the 0.7 volts dropped by the transistor in active mode:

There is a very simple way to completely eliminate this error, without adding any additional components. Modify the circuit accordingly.

File Num: 00935

Answer

If you understand why this circuit works, pat yourself on the back: you truly understand the self-correcting nature of negative feedback. If not, you have a bit more studying to do!

Notes

The answer is not meant to be discouraging for those students of yours who do not understand how the solution works. It is simply a “litmus test” of whether or not your students really comprehend the concept of negative feedback. Although the change made in the circuit is simple, the principle is a bit of a conceptual leap for some people.

It might help your students understand if you label the new wire with the word sense, to indicate its purpose of providing feedback from the very output of the circuit, back to the opamp so it can sense how much voltage the load is receiving.

Question 13. (Click on arrow for answer)

Suppose a technician is checking the operation of the following electronic circuit:

She decides to measure the voltage on either side of resistor R1 with reference to ground, and obtains these readings:

On the top side of R1, the voltage with reference to ground is -5.04 volts. On the bottom side of R1, the voltage with reference to ground is -1.87 volts. The color code of resistor R1 is Yellow, Violet, Orange, Gold. From this information, determine the following:

Voltage across R1 (between top to bottom):
Polarity (+ and -) of voltage across R1:
Current (magnitude) through R1:
Direction of current through R1:

Additionally, explain how this technician would make each one of these determinations. What rules or laws of electric circuits would she apply?

File Num: 02733

Answer

Voltage across R1 (between top to bottom): 3.17 volts
Polarity (+ and -) of voltage across R1: (-) on top, (+) on bottom
Current (magnitude) through R1: 67.45 $\mu$ A
Direction of current through R1: upward, following conventional flow

Follow-up question: calculate the range of possible currents, given the specified tolerance of resistor R1 (67.45 $\mu$ A assumes 0\% error).

Challenge question: if you recognize the type of circuit this is (by the part number of the IC “chip”: TL082), identify the voltage between pin 3 and ground.

Notes

This is a good example of how Kirchhoff’s Voltage Law is more than just an abstract tool for mathematical analysis — it is also a powerful technique for practical circuit diagnosis. Students must apply KVL to determine the voltage drop across R1, and then use Ohm’s Law to calculate its current.

If students experience difficulty visualizing how KVL plays a part in the solution of this problem, show them this illustration:

By the way, the answer to the challenge question may only be realized if students recognize this circuit as a noninverting opamp voltage amplifier. The voltage at pin 3 (noninverting input) will be the same as the voltage at pin 2 (inverting input): -1.87 volts.

Question 14. (Click on arrow for answer)

There is something wrong with this amplifier circuit. Note the relative amplitudes of the input and output signals as measured by an oscilloscope:

This circuit used to function perfectly, but then began to malfunction in this manner: producing a “clipped” output waveform of excessive amplitude. Determine the approximate amplitude that the output voltage waveform should be for the component values given in this circuit, and then identify possible causes of the problem and also elements of the circuit that you know cannot be at fault.

File Num: 02465

Answer

V_{out}

(ideal) = 1.01 volts RMS

I’ll let you determine possible faults in the circuit! From what we see here, we know the power supply is functioning (both +V and -V rails) and that there is good signal getting to the noninverting input of the opamp.

Notes

There is definitely more than one possible cause for the observed problem. Discuss alternatives with your students, involving them in the diagnosis process. Ask them why we know that certain elements of the circuit are functioning as they should? Of the possible causes, which are more likely, and why?

Question 15. (Click on arrow for answer)

Calculate the voltage gain for each stage of this amplifier circuit (both as a ratio and in units of decibels), then calculate the overall voltage gain:

File Num: 02728

Answer

$A_V$ = 1.702 = 4.62 dB

$A_V$ = 5.136 = 14.213 dB

$A_V$ = 8.743 = 18.833 dB

Notes

Summing Amplifier Circuit

A summing amplifier has multiple inputs and sums them (adds them) all together to create the output. Sometimes, the inputs have equal weightings and sometimes they don’t. Sometimes the output is inverted and sometimes it is non-inverted. See this video to find out about the different configurations

This video describes summing amplifiers and shows the two main configurations. Simulations show that the input voltages do not need to be in phase or even the same frequency, the summer will add them together even if they aren’t sinusoidal.

Summing Amplifier Practice Problems

Question 1. (Click on arrow for answer)

The simple resistor network shown here is known as a passive averager. Describe what the word “passive” means in this context, and write an equation describing the output voltage ( $V_d$ ) in terms of the input voltages ( $V_a$ , $V_b$ , and $V_c$ ):

Hint: there is a network theorem that directly applies to this form of circuit, and it is known as Millman’s Theorem. Research this theorem and use it to generate your equation!

File Num: 01001

Answer

“Passive” means that the circuit contains no amplifying components.

V_d = {{V_a + V_b + V_c} \over 3}

Notes

Students need to realize that even passive circuits are able to model (some) mathematical functions! Ask your students if they can think of any network analysis methods to easily calculate the output voltage ( $V_d$ ) of this circuit, given the input voltages. There is one theorem in particular that works very well for this particular circuit.

Question 2. (Click on arrow for answer)

Add an op-amp circuit to the output of this passive averager network to produce a summer circuit: an operational circuit generating an output voltage equal to the sum of the four input voltages. Then, write an equation describing the whole circuit’s function.

\vskip 30pt Should be image 01002x01

File Num: 01002

Answer

V_{sum} = V_a + V_b + V_c + V_d

Notes

The equation for this circuit is simple enough as to require no explanation. How your students derived this equation, from the base equation of a passive averager network, on the other hand, is worth discussion. Discuss with them the necessary gain of the op-amp circuit, and how this gain figure converts an averaging function into a summing function.

Question 3. (Click on arrow for answer)

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

File Num: 02515

Answer

Follow-up question: what would be required to get this circuit to output the exact sum of the four input voltages?

Notes

This question not only provides practice analyzing the behavior of a summer circuit, but also analyzing the behavior of a passive averager circuit. If your students need some refreshing on how to analyze the passive averager, you might want to review Millman’s Theorem with them.

Question 4. (Click on arrow for answer)

Determine all current magnitudes and directions, as well as voltage drops, in this circuit:

File Num: 02523

Answer

Follow-up question: what would be required to get this circuit to output the exact sum of the four input voltages?

Notes

Question 5. (Click on arrow for answer)

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor $R_1$ fails open:
Solder bridge (short) across resistor $R_3$ :
Resistor $R_4$ fails open:
Resistor $R_5$ fails open:
Solder bridge (short) across resistor $R_5$ :
Resistor $R_6$ fails open:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03780

Answer

Resistor $R_1$ fails open: $V_{out$ becomes equal to ${4 \over 3}$ the sum of voltages $V_2$ , $V_3$ , and $V_4$ .}

Solder bridge (short) across resistor $R_3$ : $V_{out$ becomes equal to 4 times $V_3$ .}

Resistor $R_4$ fails open: $V_{out$ becomes equal to ${4 \over 3}$ the sum of voltages $V_1$ , $V_2$ , and $V_3$ .}

Resistor $R_5$ fails open:>Circuit operates as an averager, not a summer.

Solder bridge (short) across resistor $R_5$ : $V_{out$ saturates in a positive direction.}

Resistor $R_6$ fails open: $V_{out$ saturates in a positive direction.}

Notes

The purpose of this question is to approach the domain of circuit troubleshooting from a perspective of knowing what the fault is, rather than only knowing what the symptoms are. Although this is not necessarily a realistic perspective, it helps students build the foundational knowledge necessary to diagnose a faulted circuit from empirical data. Questions such as this should be followed (eventually) by other questions asking students to identify likely faults based on measurements.

Question 6. (Click on arrow for answer)

Determine the amount of current from point A to point B in this circuit:

File Num: 02516

Answer

I

= 6.5 mA

Notes

This question, while being an application of Kirchhoff’s Current Law, is also a prelude to an inverting summer circuit, where an opamp takes that 6.5 mA (total) current and converts it into an output voltage.

Question 7. (Click on arrow for answer)

Determine the amount of current from point A to point B in this circuit, and also the output voltage of the operational amplifier:

File Num: 02517

Answer

I

= 6.5 mA

V_{out}

= -6.5 V

Notes

This question is best preceded by \#02516, which asks for students to solve for the current between A and B with no opamp in the circuit (simply grounded at point B). When students realize that point B is now a virtual ground instead of a real ground, they see that the same conclusion derived by Kirchhoff’s Current Law in the passive circuit is still valid in this active circuit, and that the result is an output voltage corresponding to that current.

Question 8. (Click on arrow for answer)

Write a mathematical equation for this op-amp circuit, assuming all resistor values are equal:

What is this circuit typically called?

File Num: 01003

Answer

c = -(a + b)

This type of circuit is typically called an inverting summer.

Follow-up question: explain why the addition of another resistor in this circuit is recommended for optimum accuracy, as shown in the following schematic.

Challenge question: write an equation describing the proper value of this extra resistor.

Notes

Ask your students about the proper resistor values for an inverting summer circuit. The choices of resistor values are definitely not the same for inverting summer and noninverting summer circuits alike! Discuss why the values are what they are in an inverting summer circuit (using Ohm’s Law to analyze the circuit’s function), emphasizing comprehension over rote memorization.

Question 9. (Click on arrow for answer)

Predict how the operation of this summer circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor $R_1$ fails open:
Resistor $R_2$ fails open:
Solder bridge (short) across resistor $R_3$ :
Resistor $R_4$ fails open:
Solder bridge (short) across resistor $R_4$ :

For each of these conditions, explain why the resulting effects will occur.

File Num: 03781

Answer

Resistor $R_1$ fails open: $V_{out$ becomes (inverted) sum of $V_2$ and $V_3$ only.}
Resistor $R_2$ fails open: $V_{out$ becomes (inverted) sum of $V_1$ and $V_3$ only.}
Solder bridge (short) across resistor $R_3$ : $V_{out$ saturates in a negative direction.}
Resistor $R_4$ fails open: $V_{out$ saturates in a negative direction.}
Solder bridge (short) across resistor $R_4$ : $V_{out$ goes to 0 volts.}

Notes

Question 10. (Click on arrow for answer)

Identify some of the distinguishing characteristics of inverting and noninverting summer circuits. How may you identify which is which, and how may you determine the proper resistor values to make each one work as it should?

File Num: 02520

Answer

I won’t directly answer the questions here, but I will give some hints. A noninverting summer circuit is composed of a passive voltage averager circuit coupled to a noninverting voltage amplifier with a voltage gain equal to the number of inputs on the averager. An inverting summer circuit is composed of a passive current summer node coupled to a current-to-voltage converter.

Notes

This question is designed to spur discussion amongst your students, exchanging ideas about each circuit’s defining characteristics. Having students explore each circuit type on their own, reaching their own conclusions about how to differentiate the two, is a far more effective way of making them understand the differences than simply telling them outright.

Difference Amplifier Circuit

A difference amplifier outputs a voltage that is the difference between two input voltages. The inputs might be equally weighted or they might not be. It depends on the values of the resistors around the op amp. Watch this video to learn more:

This video derives the equation for a difference amplifier given arbitrary values for the resistor around it. As you will see if the resistors are all the same or if R1/R3 are equal and R2/R4 are equal, you get a much easier to analyze circuit.

Difference Amplifier Practice Problems

Question 1. (Click on arrow for answer)

Complete the table of values for this opamp circuit, calculating the output voltage for each combination of input voltages shown:

V1	V2	Vout
0V	0V
+1V	0V
0V	+1V
+2V	+1.5V
+3.4V	+1.2V
-2V	+4V
+5V	+5V
-3V	-3V

What pattern do you notice in the data? What mathematical relationship is there between the two input voltages and the output voltage?

File Num: 02518

Answer

V1	V2	Vout
0V	0V	0V
+1V	0V	-1V
0V	+1V	+1V
+2V	+1.5V	-0.5V
+3.4V	+1.2V	-2.2V
-2V	+4V	+6V
+5V	+5V	0V
-3V	-3V	0V

Notes

Though it may be tedious to calculate the output voltage for each set of input voltages, working through all the voltage drops and currents in the opamp circuit one at a time, it shows students how they may be able to discern the function of an opamp circuit merely by applying basic laws of electricity (Ohm’s Law, KVL, and KCL) and the “golden assumptions” of negative feedback opamp circuits (no input currents, zero differential input voltage).

Question 2. (Click on arrow for answer)

This opamp circuit is known as a difference amplifier, sometimes called a subtractor. Assuming that all resistor values are equal in the circuit, write an equation expressing the output ( $y$ ) as a function of the two input voltages ( $a$ and $b$ ):

File Num: 01010

Answer

y = b - a

Notes

Work through some example conditions of input voltages and resistor values to calculate the output voltage using Ohm’s Law and the general principle of negative feedback in an opamp circuit (namely, an assumption of zero voltage differential at the opamp inputs). The goal here is to have students comprehend why this circuit subtracts one voltage from another, rather than just encourage rote memorization.

Question 3. (Click on arrow for answer)

How does the operation of this difference amplifier circuit compare with the resistor values given ( $2R$ = twice the resistance of $R$ ), versus its operation with all resistor values equal?

Describe what approach or technique you used to derive your answer, and also explain how your conclusion for this circuit might be generalized for all difference amplifier circuits.

File Num: 02525

Answer

It is very important that you develop the skill of “exploring” a circuit configuration to see what it will do, rather than having to be told what it does (either by your instructor or by a book). All you need to have is a solid knowledge of basic electrical principles (Ohm’s Law, Kirchhoff’s Voltage and Current Laws) and know how opamps behave when configured for negative feedback.

As for a generalized conclusion:

Notes

It is easy for you (the instructor) to show how and why this circuit acts as it does. The point of this question, however, is to get students to take the initiative to explore the circuit on their own. It is simple enough for any student to set up some hypothetical test conditions (a thought experiment) to analyze what this circuit will do, that the only thing holding them back from doing so is attitude, not aptitude.

This is something I have noticed over years of teaching: so many students who are more than capable of doing the math and applying well-understood electrical rules refuse to do so on their own, because years of educational tradition has indoctrinated them to wait for the instructor’s lead rather than explore a concept on their own.

Question 4. (Click on arrow for answer)

Predict how the operation of this difference amplifier circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor $R_1$ fails open:
Resistor $R_2$ fails open:
Solder bridge (short) across resistor $R_3$ :
Resistor $R_4$ fails open:
Solder bridge (short) across resistor $R_4$ :

For each of these conditions, explain why the resulting effects will occur.

File Num: 03782

Answer

Resistor $R_1$ fails open: $V_{out}$ becomes equal to ${1 \over 2}$ $V_2$ .}
Resistor $R_2$ fails open: $V_{out}$ saturates.}
Solder bridge (short) across resistor $R_3$ : $V_{out}$ becomes equal to $2 V_2 - V_1$ instead of $V_2 - V_1$ .}
Resistor $R_4$ fails open: $V_{out}$ becomes equal to $2 V_2 - V_1$ instead of $V_2 - V_1$ .}
Solder bridge (short) across resistor $R_4$ : $V_{out}$ becomes equal to $-V_1$ .}

Notes

Integrator Op Amp Circuit

The integrator circuit does exactly what you would think that it does…it takes the input voltage and integrates it to give the output voltage. Yes, circuits can do calculus. To find out how, see the video below

The integrator op amp circuit is an inverting op amp circuit with a resistor at the input and a capacitor for the feedback. This video goes through the analysis to see why it is called an integrator. The video also includes some simulations that show the integrator working and also not working

Differentiator Op Amp Circuit

Once again, the name of the circuit is the thing that it does. So this circuit differentiates, or takes the derivative of the input voltage to create the output voltage. Well, it doesn’t actually sit there and do the calculation, but the output voltage is proportional to the derivative of the input voltage and in this video, I show you why.

The differentiator op amp circuit does output the derivative of the input, but as you will see in this video, it is a little sensitive and unstable.

Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first.

David Williams (2022): First edits including videos

Practice Problem Copyright

All practice problems with a file num less than 4100 are Copyright 2003, Tony R. Kuphaldt, released under the Creative Commons Attribution License (v 1.0). All other files are Copyright 2022, David Williams, released under the Creative Commons Attribution License (V 4.0) This means you may do almost anything with this work, so long as you give proper credit.

To view a copy of the license, visit https://creativecommons.org/licenses/by/1.0/, or https://creativecommons.org/licenses/by/4.0/, or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public.

Ch1: Intro to Op Amps

Op Amps Table of Contents

Op Amp Basics

Operational amplifiers, also known as op amps, are semiconductor devices characterized with two high impedance inputs, a very high differential voltage gain and a low impedance output. By using external feedback components, a wide variety of different types of circuits can be built. Op amps may very well be the most useful device in analog circuitry because of the wide range of signal processing tasks it can accomplish. They are affordable, durable, and come in a wide variety of packages.

The diagram below shows the circuit symbol for a basic op amp

Together, the non-inverting (V_in+)pin and the inverting (V_in-)pin make up a differential input. V_S+ and V_S- are the positive and negative power supply respectively. And surprise, surprise V_out is the output. The general equation for the op amp relating inputs to outputs is:

$V_{out}=A_{vol}[V_{(in+)}-V_{(in-)}]$

A_vol is the open loop voltage gain and is typically a very big value and for an ideal op amp, it is infinite.

Video: Intro to Op Amps

To find out more, check out this video:

Op Amp Basics Practice Problems

Question 1. (Click on arrow for answer)

An operational amplifier is a particular type of differential amplifier. Most op-amps receive two input voltage signals and output one voltage signal:

Here is a single op-amp, shown under two different conditions (different input voltages). Determine the voltage gain of this op-amp, given the conditions shown:

Also, write a mathematical formula solving for differential voltage gain ( $A_V$ ) in terms of an op-amp’s input and output voltages.

File Num: 00848

Answer

A_V =

530,000

A_V = {\Delta V_{out} \over \Delta (V_{in2} - V_{in1})}

Follow-up question: convert this voltage gain figure (as a ratio) into a voltage gain figure in decibels.

Notes

The calculations for voltage gain here are not that different from the voltage gain calculations for any other amplifier, except that here we’re dealing with a differential amplifier instead of a single-ended amplifier.

A differential voltage gain of 530,000 is not unreasonable for a modern operational amplifier! A gain so extreme may come as a surprise to many students, but they will discover later the utility of such a high gain.

Question 2. (Click on arrow for answer)

Many op-amp circuits require a dual or split power supply, consisting of three power terminals: +V, -V, and Ground. Draw the necessary connections between the 6-volt batteries in this schematic diagram to provide +12 V, -12 V, and Ground to this op-amp:

File Num: 00880

Answer

Notes

I encourage your students to learn how to power op-amp circuits with interconnected batteries, because it really helps to build their understanding of what a “split” power supply is, as well as allow them to build functioning op-amp circuits in the absence of a quality benchtop power supply.

Question 3. (Click on arrow for answer)

The 8-pin Dual-Inline-Package (DIP) is a common format in which single and dual operational amplifiers are housed. Shown here are the case outlines for two 8-pin DIPs. Draw the internal op-amp connections for a single op-amp unit, and for a dual op-amp unit:

You will need to research some op-amp datasheets to find this information. Examples of single op-amp chips include the LM741, CA3130, and TL081. Examples of dual op-amp chips include the LM1458 and TL082.

File Num: 00874

Answer

Notes

Ask your students to reveal their information sources, and what specific models of op-amp they researched.

Question 4. (Click on arrow for answer)

In this circuit, an op-amp turns on an LED if the proper input voltage conditions are met:

Trace the complete path of current powering the LED. Where, exactly, does the LED get its power from?

File Num: 00801

Answer

The arrows shown in this diagram trace “conventional” current flow, not electron flow:

Notes

The important thing to note here is that the load current does not pass through either of the op-amp’s input terminals. All load current is sourced by the op-amp’s power supply! Discuss the importance of this fact with your students.

Question 5. (Click on arrow for answer)

What does it mean if an operational amplifier has the ability to “swing its output rail to rail”? Why is this an important feature to us?

File Num: 00844

Answer

Being able to “swing” the output voltage “rail to rail” means that the full range of an op-amp’s output voltage extends to within millivolts of either power supply “rail” (+V and -V).

Challenge question: identify at least one op-amp model that has this ability, and at least one that does not. Bring the datasheets for these op-amp models with you for reference during discussion time.

Notes

Discuss what this feature means to us as circuit builders in a practical sense. Ask those students who tackled the challenge question to look up the output voltage ranges of their op-amp models. Exactly how close to +V and -V can the output voltage of an op-amp lacking “rail-to-rail” output capability “swing”?

Question 6. (Click on arrow for answer)

Write the transfer function (input/output equation) for an operational amplifier with an open-loop voltage gain of 100,000. In other words, write an equation describing the output voltage of this op-amp ( $V_{out}$ ) for any combination of input voltages ( $V_{in(+)}$ and $V_{in(-)}$ ):

File Num: 00925

Answer

V_{out} = 100,000(V_{in(+)} - V_{in(-)})

Notes

The concept of a “transfer function” is very useful, and this may be your students’ first exposure to the idea. It is a phrase used quite often in engineering applications, and may denote an equation, a table of numbers, or a graph.

In this particular question, it is important that students know how to derive and use the basic transfer function for a differential amplifier. Challenge your students to express this function in a more general form, so that calculations may be made with different open-loop voltage gains.

Question 7. (Click on arrow for answer)

How much voltage would have to be “dialed up” at the potentiometer in order to stabilize the output at exactly 0 volts, assuming the opamp has no input offset voltage?

File Num: 00924

Answer

5 volts

Notes

This question is a basic review of an ideal differential amplifier’s function. Ask your students what voltage must be “dialed up” at the potentiometer to produce 0 volts at the output of the op-amp for several different voltages at the other input. If they don’t understand at first, they soon will after discussing these alternate scenarios.

Question 8. (Click on arrow for answer)

An op-amp has +3 volts applied to the inverting input and +3.002 volts applied to the noninverting input. Its open-loop voltage gain is 220,000. Calculate the output voltage as predicted by the following formula:

V_{out} = A_V \left( V_{in(+)} - V_{in(-)} \right)

How much differential voltage (input) is necessary to drive the output of the op-amp to a voltage of -4.5 volts?

File Num: 00926

Answer

V_{out} =

440 volts

Follow-up question: is this voltage figure realistic? Is it possible for an op-amp such as the model 741 to output 440 volts? Why or why not?

The differential input voltage necessary to drive the output of this op-amp to -4.5 volts is -20.455 $\mu$ V.

Follow-up question: what does it mean for the input voltage differential to be negative 20.455 microvolts? Provide an example of two input voltages ( $V_{in(+)}$ and $V_{in(-)}$ ) that would generate this much differential voltage.

Notes

Obviously, there are limitations to the op-amp formula for calculating output voltage, given input voltages and open-loop voltage gain. Students need to realize the practical limits of an op-amp’s output voltage range, and what sets those limits.

Ideal Op Amp

An ideal op amp has the following characteristics:

Infinite open loop voltage gain (A_vol)
Infinite input impedance. No current flows into inputs
Zero output impedance. No voltage drop at output due to loads
Infinite bandwidth. Works the same at all frequencies
No offset voltage. When the inputs are zero, the output is zero

Real op amps are designed to be as close to the ideal op amps as possible, and in practice, you generally (but not always) have to look pretty hard to see the non-ideal characteristics of a real op amp.

Comparator

A comparator is the simplest circuit that you can build with an op amp. It uses the enormous open loop voltage gain of an op amp to indicate which of the two inputs is at a higher voltage.

Video: Op Amp Comparators

This video describes how a comparator works and provides a few examples of the different comparator configurations:

Op Amp Comparators Practice Problems

Question 1. (Click on arrow for answer)

Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions:

File Num: 00803

Answer

In these illustrations, I have likened the op-amp’s action to that of a single-pole, double-throw switch, showing the “connection” made between power supply terminals and the output terminal.

Notes

Determining which “way” the output of an op-amp drives under different input voltage conditions is confusing to many students. Discuss this with them, and ask them to present any principles or analogies they use to remember “which way is which.”

Question 2. (Click on arrow for answer)

Determine the output voltage polarity of this op-amp (with reference to ground), given the following input conditions:

File Num: 03762

Answer

In these illustrations, I have likened the op-amp’s action to that of a single-pole, double-throw switch, showing the “connection” made between power supply terminals and the output terminal.

Notes

Question 3. (Click on arrow for answer)

In this circuit, a solar cell converts light into voltage for the opamp to “read” on its noninverting input. The opamp’s inverting input connects to the wiper of a potentiometer. Under what conditions does the LED energize?

File Num: 00872

Answer

The LED energizes under bright-light conditions, de-energizing when the light decreases below the threshold set by the potentiometer.

Follow-up question: determine what would have to be changed in this circuit to make the LED turn on when the solar cell becomes dark.

Notes

There is more than one way to accomplish the task posed by the follow-up question. Be sure to ask your students for their ideas on how to reverse the LED’s operation!

Question 4. (Click on arrow for answer)

A student is operating a simple comparator circuit and documenting the results in a table:

\settabs \+ \quad MMMM \quad & \quad MMMM \quad & \quad MMMM \quad & \quad MMMM \quad & \cr \+ \hfill &

V_{in(+)}

V_{in(-)}

V_{out}

\cr \+ \hfill & 3.00 V & 1.45 V & 10.5 V \cr \+ \hfill & 3.00 V & 2.85 V & 10.4 V \cr \+ \hfill & 3.00 V & 3.10 V & 1.19 V \cr \+ \hfill & 3.00 V & 6.75 V & 1.20 V \cr
\+ \hfill &

V_{in(+)}

V_{in(-)}

V_{out}

\cr \+ \hfill & 2.36 V & 6.50 V & 1.20 V \cr \+ \hfill & 4.97 V & 6.50 V & 1.21 V \cr \+ \hfill & 7.05 V & 6.50 V & 10.5 V \cr \+ \hfill & 9.28 V & 6.50 V & 10.4 V \cr
\+ \hfill &

V_{in(+)}

V_{in(-)}

V_{out}

\cr \+ \hfill & 10.4 V & 9.87 V & 10.6 V \cr \+ \hfill & 1.75 V & 1.03 V & 10.5 V \cr \+ \hfill & 0.31 V & 1.03 V & 10.5 V \cr \+ \hfill & 5.50 & 5.65 V & 1.19 V \cr

One of these output voltage readings is anomalous. In other words, it does not appear to be “correct”. This is very strange, because these figures are real measurements and not predictions! Perplexed, the student approaches the instructor and asks for help. The instructor sees the anomalous voltage reading and says two words: latch-up. With that, the student goes back to research what this phrase means, and what it has to do with the weird output voltage reading.

Identify which of these output voltage measurements is anomalous, and explain what “latch-up” has to do with it.

File Num: 00876

Answer

Latch-up occurs when one of the input voltage signals approaches too close to one of the power supply rail voltages. The result is the op-amp output saturating “high” even if it isn’t supposed to.

Challenge question: suppose we expected both input voltages to range between 0 and 10 volts during normal operation of this comparator circuit. What could we change in the circuit to allow this range of operation and avoid latch-up?

Notes

Ask your students what they found in their research on “latch-up,” and if this is an idiosyncrasy of all op-amp models, or just some.

Incidentally, the curved op-amp symbol has no special meaning. This symbol was quite popular for representing op-amps during their early years, but has since fallen out of favor. I show it here just to inform your students, in case they ever happen to encounter one of these symbols in an old electronic schematic.

Question 5. (Click on arrow for answer)

In this circuit, an op-amp turns on an LED if the proper input voltage conditions are met:

Trace the complete path of current powering the LED. Where, exactly, does the LED get its power from?

File Num: 00801

Answer

The arrows shown in this diagram trace “conventional” current flow, not electron flow:

Notes

Question 6. (Click on arrow for answer)

Trace the output waveform of this comparator circuit:

File Num: 00878

Answer

Follow-up question: explain what the phrase duty cycle means with reference to a “square” or “pulse” waveform.

Notes

During discussion, ask your students to explain how the output waveform of this comparator circuit comes to be, step by step. Ask them how they arrived at their solution, and if there is a way this AC/DC problem can be simplified to one that is DC only for easier analysis (determining what the output voltage will do for a certain set of input conditions).

Question 7. (Click on arrow for answer)

Calculate the amount of resistance that the thermistor much reach in order to turn the cooling fan on:

File Num: 04021

Answer

Thermistor resistance = 5.547 k $\Omega$

Notes

Ask your students how they arrived at their solution for this question. There is definitely more than one way to do it!

Question 8. (Click on arrow for answer)

Predict how the operation of this thermostat circuit (where the cooling fan motor is supposed to turn on when the temperature gets too high) will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Cable fails open:

Comparator $U_1$ fails with output saturated positive:

Resistor $R_1$ fails open:

Capacitor $C_1$ fails shorted:

Transistor $Q_1$ fails shorted (drain-to-source):

For each of these conditions, explain why the resulting effects will occur.

File Num: 03768

Answer

Cable fails open: Fan turns on and never turns off.

Comparator $U_1$ fails with output saturated positive: Fan turns on and never turns off.

Resistor $R_1$ fails open: Fan refuses to turn on.

Capacitor $C_1$ fails shorted: Fan refuses to turn on, transistor $Q_1$ likely fails due to overheating when it tries to energize fan.

Transistor $Q_1$ fails shorted (drain-to-source): Fan turns on and never turns off.

Notes

Question 9. (Click on arrow for answer)

Cable fails open:

Comparator $U_1$ fails with output saturated positive:

Resistor $R_1$ fails open:

Cable fails shorted:

Transistor $Q_1$ fails shorted (drain-to-source):

For each of these conditions, explain why the resulting effects will occur.

File Num: 03769

Answer

Cable fails open: Fan turns on and never turns off.

Comparator $U_1$ fails with output saturated positive: Fan refuses to turn on.

Resistor $R_1$ fails open: Fan refuses to turn on.

Cable fails shorted: Fan refuses to turn on.

Transistor $Q_1$ fails shorted (drain-to-source): Fan turns on and never turns off.

Notes

Question 10. (Click on arrow for answer)

Explain what a window comparator circuit is (sometimes called a window discriminator), and identify at least one practical application for one.

File Num: 03838

Answer

A “window comparator” circuit detects when a voltage falls between two different reference voltages. I’ll let you figure out some practical applications for such a circuit!

Notes

Ask your students where they found the answer for this question, and further explore some of the practical applications they offer.

Schmitt Trigger

Schmitt triggers are comparators with some built in hysteresis so that small changes around the input threshold are ignored.

Video: Schmitt Triggers

To find out more, check out this video:

Schmitt Trigger Practice Problems

Question 1. (Click on arrow for answer)

Determine the “trip” voltage of this comparator circuit: the value of input voltage at which the opamp’s output changes state from fully positive to fully negative or visa-versa:

Now, what do you suppose would happen if the output were fed back to the noninverting input through a resistor? You answer merely has to be qualitative, not quantitative:

For your information, this circuit configuration is often referred to as a Schmitt trigger.

File Num: 02294

Answer

With no feedback resistor, the “trip” voltage would be 9.21 volts. With the feedback resistor in place, the “trip” voltage would change depending on the state of the opamp’s output!

Follow-up question: describe what effect this changing “trip” voltage value will have on the operation of this comparator circuit.

Notes

Schmitt trigger circuits are very popular for their ability to “cleanly” change states given a noisy input signal. I have intentionally avoided numerical calculations in this question, so that students may concentrate on the concept of positive feedback and how it affects this circuit.

Question 2. (Click on arrow for answer)

A comparator is used as a high wind speed alarm in this circuit, triggering an audio tone to sound whenever the wind speed exceeds a pre-set alarm point:

The circuit works well to warn of high wind speed, but when the wind speed is just near the threshold level, every little gust causes the alarm to briefly sound, then turn off again. What would be better is for the alarm to sound at a set wind speed, then stay on until the wind speed falls below a substantially lower threshold value (example: alarm at 60 km/h, reset at 50 km/h).

An experienced electronics technician decides to add this functionality to the circuit by adding two resistors:

Explain why this circuit alteration works to solve the problem.

File Num: 01168

Answer

The added resistors provide positive feedback to the opamp circuit, causing it to exhibit hysteresis.

Challenge question: suppose you wished to increase the gap between the upper and lower alarm thresholds. What resistor value(s) would you have to alter to accomplish this adjustment?

Notes

A practical illustration for positive feedback in an opamp circuit. There is much to discuss here, even beyond the immediate context of positive feedback. Take for instance the oscillator circuit and on/off control transistor. For review, ask your students to explain how both these circuit sections function.

Question 3. (Click on arrow for answer)

Assume that the comparator in this circuit is capable of “swinging” its output fully from rail to rail. Calculate the upper and lower threshold voltages, given the resistor values shown:

V_{UT}

= \hskip 80pt

V_{LT}

File Num: 01169

Answer

V_{UT} =

+8 volts

V_{LT} =

-8 volts

Challenge question: how would you recommend we change the circuit to give threshold voltages of +6 volts and -6 volts, respectively?

Notes

Ask your students to explain what the terms “upper threshold” and “lower threshold” mean with regard to input voltage in a circuit such as this.

Question 4. (Click on arrow for answer)

Assume that the comparator in this circuit is only capable of “swinging” its output to within 1 volt of its power supply rail voltages. Calculate the upper and lower threshold voltages, given the resistor values shown:

V_{UT}

= \hskip 80pt

V_{LT}

File Num: 02662

Answer

V_{UT} =

+2.093 volts

V_{LT} =

-2.093 volts

Notes

As many opamps and comparators are incapable of rail-to-rail output swings, this question is quite realistic.

Question 5. (Click on arrow for answer)

V_{UT}

= \hskip 80pt

V_{LT}

File Num: 02663

Answer

V_{UT} =

+4.087 volts

V_{LT} =

-4.087 volts

Notes

As many opamps and comparators are incapable of rail-to-rail output swings, this question is quite realistic.

Question 6. (Click on arrow for answer)

A student intends to connect a TL082 opamp as a voltage follower, to “follow” the voltage generated by a potentiometer, but makes a mistake in the breadboard wiring:

Draw a schematic diagram of this faulty circuit, and determine what the voltmeter’s indication will be, explaining why it is such.

File Num: 01148

Answer

Circuit schematic, as wired:

The output voltage will saturate at approximately +11 volts, or -11 volts, with the potentiometer having little or no effect.

Notes

Ask your students to characterize the type of feedback exhibited in this circuit. How does this type of feedback affect the opamp’s behavior? Is it possible for the opamp to function as a voltage follower, connected like this?

Question 7. (Click on arrow for answer)

Positive or regenerative feedback is an essential characteristic of all oscillator circuits. Why, then, do comparator circuits utilizing positive feedback not oscillate? Instead of oscillating, the output of a comparator circuit with positive feedback simply saturates to one of its two rail voltage values. Explain this.

File Num: 01172

Answer

The positive feedback used in oscillator circuits is always phase-shifted 360 $^{o}$ , while the positive feedback used in comparator circuits has no phase shift at all, being direct-coupled.

Notes

This is a challenging question, and may not be suitable for all students. Basically, what I’m trying to get students to do here is think carefully about the nature of positive feedback as used in comparator circuits, versus as it’s used in oscillator circuits. Students who have simply memorized the concept of “positive feedback causing oscillation” will fail to understand what is being asked in this question, much less understand the given answer.

Op Amp Oscillator

An square wave oscillator can easily be made by using a positive feedback circuit like the one used for the Schmitt trigger along with a negative feedback circuit using an RC circuit like this:

The voltage at the non-inverting pin is controlled by the output and the positive feedback network. The voltage at the inverting pin is controlled by a charging or discharging RC circuit. The capacitor will charge and discharge towards the voltage set at the non-inverting pin and when it reaches the voltage, the output will switch and the capacitor will then discharge or charge in the opposite direction. This automatic switching back and forth creates an oscillator

Video: Op Amp Oscillator

This video describes how the op amp oscillator works and derives an equation for how to calculate the frequency of the oscillator based on the R and C values

Operational Amplifiers - Relaxation Oscillators

Op Amp Oscillator Practice Problems

Question 1. (Click on arrow for answer)

This is a very common opamp oscillator circuit, technically of the relaxation type:

Explain how this circuit works, and what waveforms will be measured at points A and B. Be sure to make reference to RC time constants in your explanation.

File Num: 01171

Answer

You will measure a sawtooth-like waveform at point A, and a square wave at point B.

Challenge question: explain how you might go about calculating the frequency of such a circuit, based on what you know about RC time constant circuits. Assume that the opamp can swing its output rail-to-rail, for simplicity.

Notes

This circuit is best understood by building and testing. If you use large capacitor values and/or a large-value resistor in the capacitor’s current path, the oscillation will be slow enough to analyze with a voltmeter rather than an oscilloscope.

Question 2. (Click on arrow for answer)

A variation on the common opamp relaxation oscillator design is this, which gives it variable duty cycle capability:

Explain how this circuit works, and which direction the potentiometer wiper must be moved to increase the duty cycle (more time spent with the opamp output saturated at +V and less time spent saturated at -V).

File Num: 02673

Answer

Move the wiper up to increase the duty cycle.

Notes

Incidentally, the Schottky diodes are not essential to this circuit’s operation, unless the expected frequency is very high. Really, the purpose of the Schottky diodes, with their low forward voltage drops (0.4 volts typical) and minimal charge storage, is to make the opamp’s job easier at every reversal of output polarity. Remember that this circuit is not exploiting negative feedback! Essentially, it is a positive feedback circuit, and every voltage drop and nonlinearity in the capacitor’s current path will have an effect on capacitor charging/discharging.

Question 3. (Click on arrow for answer)

Most operational amplifiers do not have the ability to swing their output voltages rail-to-rail. Most of those do not swing their output voltages symmetrically. That is, a typical non-rail-to-rail opamp may be able to approach one power supply rail voltage closer than the other; e.g. when powered by a +15/-15 volt split supply, the output saturates positive at +14 volts and saturates negative at -13.5 volts.

What effect do you suppose this non-symmetrical output range will have on a typical relaxation oscillator circuit such as the following, and how might you suggest we fix the problem?

File Num: 02675

Answer

The duty cycle will not be 50\%. One way to fix the problem is to do something like this:

Follow-up question: explain how and why this solution works. Now you just knew I was going to ask this question the moment you saw the diagram, didn’t you?

Notes

Note that I added an additional resistor to the circuit, in series with the opamp output terminal. In some cases this is not necessary because the opamp is self-limiting in output current, but it is a good design practice nonetheless. In the event anyone ever swaps out the original opamp for a different model lacking overcurrent protection, the new opamp will not become damaged.

Question 4. (Click on arrow for answer)

Dual, or split, power supplies are very useful in opamp circuits because they allow the output voltage to rise above as well as sink below ground potential, for true AC operation. In some applications, though, it may not be practical or affordable to have a dual power supply to power your opamp circuit. In this event, you need to be able to figure out how to adapt your dual-supply circuit to single-supply operation.

A good example of such a challenge is the familiar opamp relaxation oscillator, shown here:

First, determine what would happen if we were to simply eliminate the negative portion of the dual power supply and try to run the circuit on a single supply (+V and Ground only):

Then, modify the schematic so that the circuit will run as well as it did before with the dual supply.

File Num: 02676

Answer

Here is one solution:

Here is another solution:

Follow-up question: now you just know what I’m going to ask next, don’t you? How do these modified circuits function?

Notes

Dual power supplies are a luxury in many real-life circumstances, and so your students will need to be able to figure out how to make opamps work in single-supply applications! Work with your students to analyze the function of the suggested solution circuit, to see how it is at once similar and different from its simpler, dual-supply forbear.

Question 5. (Click on arrow for answer)

Predict how the operation of this relaxation oscillator circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

Resistor $R_1$ fails open:

Solder bridge (short) across resistor $R_1$ :

Capacitor $C_1$ fails shorted:

Solder bridge (short) across resistor $R_2$ :

Resistor $R_3$ fails open:

For each of these conditions, explain why the resulting effects will occur.

File Num: 03798

Answer

Resistor $R_1$ fails open: Opamp output saturates either positive or negative.

Solder bridge (short) across resistor $R_1$ : Output voltage settles to 0 volts.

Capacitor $C_1$ fails shorted: Opamp output saturates either positive or negative.

Solder bridge (short) across resistor $R_2$ : Output voltage settles to 0 volts.

Resistor $R_3$ fails open: Output voltage settles to 0 volts.

Notes

Question 6. (Click on arrow for answer)

Identify at least two different component faults that would result in a change in duty cycle for this oscillator circuit:

File Num: 03799

Answer

A short in either of the two diodes would cause the duty cycle to change.

Follow-up question: what would happen if either of these two diodes failed open?

Notes

Ask your students to explain why the duty cycle would change as a result of either diode failing shorted. This is a good opportunity to further explore the operation of this oscillator circuit.

Contributors

Contributors to this chapter are listed in chronological order of their contributions, from most recent to first.

David Williams (2022): First edits including videos

Practice Problem Copyright

Operational Amplifiers

According to Texas Instruments, Op Amps are for Everyone. That is a tongue-in-cheek title for a popular and still relevant book on operational amplifiers, and while op amps are not for absolutely everyone, you do need to understand them if you are going to work with electronic circuits. They are popular and versatile components that have a high open loop gain, high input impedance and low output impedance. This section (work in progress) will have all the information, examples, and background that you need to know to begin to understand and use op amps in their many different configurations.

Ch1: Introduction to Op Amps

Ch2: Common Op Amp Configs

Feedback in Op Amps

“Nice!”

“Nice”

“Nice!”

Background picture (LM741 Schematic): Omegatron via wikipedia.org
Background picture (Op Amp Circuits): Inductiveload via wikipedia.org