Don’t just sit there! Build something!! 
Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuitbuilding exercises, follow these steps:
Avoid using the model 741 opamp, unless you want to challenge your circuit design skills. There are more versatile opamp models commonly available for the beginner. I recommend the LM324 for DC and lowfrequency AC circuits, and the TL082 for AC projects involving audio or higher frequencies.
As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistor values between 1 kΩ and 100 kΩ.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another timesaving technique is to reuse the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
Let the electrons themselves give you the answers to your own “practice problems”!
It has been my experience that students require much practice with circuit analysis to become proficient. To this end, instructors usually provide their students with lots of practice problems to work through, and provide answers for students to check their work against. While this approach makes students proficient in circuit theory, it fails to fully educate them.
Students don’t just need mathematical practice. They also need real, handson practice building circuits and using test equipment. So, I suggest the following alternative approach: students should build their own “practice problems” with real components, and try to mathematically predict the various voltage and current values. This way, the mathematical theory “comes alive,” and students gain practical proficiency they wouldn’t gain merely by solving equations.
Another reason for following this method of practice is to teach students scientific method: the process of testing a hypothesis (in this case, mathematical predictions) by performing a real experiment. Students will also develop real troubleshooting skills as they occasionally make circuit construction errors.
Spend a few moments of time with your class to review some of the “rules” for building circuits before they begin. Discuss these issues with your students in the same Socratic manner you would normally discuss the worksheet questions, rather than simply telling them what they should and should not do. I never cease to be amazed at how poorly students grasp instructions when presented in a typical lecture (instructor monolog) format!
A note to those instructors who may complain about the “wasted” time required to have students build real circuits instead of just mathematically analyzing theoretical circuits:
What is the purpose of students taking your course?
If your students will be working with real circuits, then they should learn on real circuits whenever possible. If your goal is to educate theoretical physicists, then stick with abstract analysis, by all means! But most of us plan for our students to do something in the real world with the education we give them. The “wasted” time spent building real circuits will pay huge dividends when it comes time for them to apply their knowledge to practical problems.
Furthermore, having students build their own practice problems teaches them how to perform primary research, thus empowering them to continue their electrical/electronics education autonomously.
In most sciences, realistic experiments are much more difficult and expensive to set up than electrical circuits. Nuclear physics, biology, geology, and chemistry professors would just love to be able to have their students apply advanced mathematics to real experiments posing no safety hazard and costing less than a textbook. They can’t, but you can. Exploit the convenience inherent to your science, and get those students of yours practicing their math on lots of real circuits!
In a commonemitter transistor amplifier circuit, the presence of capacitance between the collector and base terminals  whether intrinsic to the transistor or externally connected  has the effect of turning the amplifier circuit into a lowpass filter, with voltage gain being inversely proportional to frequency:

Explain why this is. Why, exactly, does a capacitance placed in this location affect voltage gain? Hint: it has something to do with negative feedback!
The capacitance provides a path for an AC feedback signal to go from the collector to the base. Given the inverting phase relationship between collector voltage and base voltage, this feedback is degenerative.
Students should realize that this is no hypothetical question. Intrinsic capacitance does indeed exist between the collector and base of a bipolar junction transistor (called the Miller capacitance), and this has a degenerating effect on voltage gain with increasing frequency. If time permits, you may wish to discuss how the commoncollector and commonbase amplifier configurations naturally avoid this problem.
Which of the following amplifier circuits will be most affected by the basecollector capacitance (shown here as an externallyconnected 10 pF capacitor) as frequency increases? Explain why.

The amplifier with the larger collector resistance will be affected more by the feedback capacitance, because its naturally greater voltage gain produces a larger voltage signal to be fed back to the base, for any given level of input signal.
The purpose of this question is to get students to see, on a discrete component level, that for commonemitter amplifier there is a tradeoff between maximum gain and maximum operating frequency. This question foreshadows the concept of GainBandwidth Product (GBW) in operational amplifier circuits.
A common problem encountered in the development of transistor amplifier circuits is unwanted oscillation resulting from parasitic capacitance and inductance forming a positive feedback loop from output to input. Often, these parasitic parameters are quite small (nanohenrys and picofarads), resulting in very high oscillation frequencies.
Another parasitic effect in transistor amplifier circuits is Millereffect capacitance between the transistor terminals. For commonemitter circuits, the basecollector capacitance (C_{BC}) is especially troublesome because it introduces a feedback path for AC signals to travel directly from the output (collector terminal) to the input (base terminal).
Does this parasitic basetocollector capacitance encourage or discourage highfrequency oscillations in a commonemitter amplifier circuit? Explain your answer.
The presence of C_{BC} in a commonemitter circuit mitigates highfrequency oscillations.
Note that I chose to use a the word “mitigate” instead of give the answer in more plain English. Part of my reasoning here is to veil the given answer from immediate comprehension so that students must think a bit more. Another part of my reasoning is to force students’ vocabularies to expand.
A student connects a model CA3130 operational amplifier as a voltage follower (or voltage buffer), which is the simplest type of negative feedback opamp circuit possible:

With the noninverting input connected to ground (the midpoint in the split 6/6 volt power supply), the student expects to measure 0 volts DC at the output of the opamp. This is what the DC voltmeter registers, but when set to AC, it registers substantial AC voltage!
Now this is strange. How can a simple voltage buffer output alternating current when its input is grounded and the power supply is pure DC? Perplexed, the student asks the instructor for help. Öh,” the instructor says, “you need a compensation capacitor between pins 1 and 8.” What does the instructor mean by this cryptic suggestion?
Some opamps are inherently unstable when operated in negativefeedback mode, and will oscillate on their own unless “phasecompensated” by an external capacitor.
Followup question: Are there any applications of an opamp such as the CA3130 where a compensation capacitor is not needed, or worse yet would be an impediment to successful circuit operation? Hint: some models of opamp (such as the model 741) have builtin compensation capacitors!
Your students should have researched datasheets for the CA3130 opamp in search of an answer to this question. Ask them what they found! Which terminals on the CA3130 opamp do you connect the capacitor between? What size of capacitor is appropriate for this purpose?
Given the fact that some opamp models come equipped with their own builtin compensation capacitor, what does this tell us about the CA3130’s need for an external capacitor? Why didn’t the manufacturer simply integrate a compensation capacitor into the CA3130’s circuitry as they did with the 741? Or, to phrase the question more directly, ask your students to explain what disadvantage there is in connecting a compensation capacitor to an opamp.
Some operational amplifiers come equipped with compensation capacitors built inside. The classic 741 design is one such opamp:

Find the compensation capacitor in this schematic diagram, and identify how it provides frequencydependent negative feedback within the opamp to reduce gain at high frequencies.
Identifying the capacitor is easy: it is the only one in the whole circuit! It couples signal from the collector of Q_{17}, which is an activeloaded commonemitter amplifier, to the base of Q_{16}, which is an emitterfollower driving Q_{17}. Since Q_{17} inverts the signal applied to Q_{16}‘s base, the feedback is degenerative.
Answering this question will require a review of basic transistor amplifier theory, specifically different configurations of transistor amplifiers and their respective signal phase relationships.
Some operational amplifiers are internally compensated, while others are externally compensated. Explain the difference between the two. Hint: examples of each include the classic LM741 and LM101 operational amplifiers. Research their respective datasheets to see what you find on compensation!
The difference is the physical location of the compensating capacitor, whether it is a part of the integrated circuit or external to it.
Followup question: show how an external compensating capacitor may be connected to an opamp such as the LM101.
Ask your students to explain why we might wish to use either type of opamp when building a circuit. In what applications do they think an internallycompensated opamp would be better, and in what applications do they think an externallycompensated opamp would be preferable?
Define “GainBandwidth Product” (GBW) as the term applies to operational amplifiers.
GBW product is a constant value for most operational amplifiers, equal to the openloop gain of the opamp multiplied by the signal frequency at that gain.
There are other means of defining GainBandwidth Product, so do not be surprised if students present alternative definitions during the discussion time.
Define “UnityGain Bandwidth” (B_{1}) as the term applies to operational amplifiers.
UnityGain Bandwidth is the frequency at which an operational amplifier’s openloop voltage gain is equal to 1.
It does not require a great deal of insight to recognize that unitygain bandwidth (B_{1}) and gainbandwidth product (GBW) are pretty much the same thing. This would be a good point to bring up (in the form of a question!) for your students if you have already discussed GBW.
Explain the effect that compensation capacitance has on an operational amplifier’s gainbandwidth product (GBW). Does a larger compensation capacitance yield a greater GBW or a lesser GBW, and why?
The greater the amount of compensation capacitance in an opamp (either internal, or externally connected), the less the GBW product.
In this question, the really important aspect is not the answer given. What is important here is that students understand what GBW product is, and how it is affected by this thing we call “compensation capacitance” (another topic of research). The goal here is to get students to research these concepts and relate them together, so please do not be satisfied with any student answers that merely restate the answer given here! Ask students to explain what these terms and concepts mean, and to explain why the GBW product decreases with increased C_{comp}.
An important AC performance parameter for operational amplifiers is slew rate. Explain what “slew rate” is, and what causes it to be less than optimal for an opamp.
Slew rate is the maximum rate of change of output voltage over time \(\frac{dv}{dt}\)_{max} that an opamp can muster.
Followup question: what would the output waveform of an opamp look like if it were trying to amplify a square wave signal with a frequency and amplitude exceeding the amplifier’s slew rate?
The followup question is very important, as it asks students to apply the concept of a maximum [dv/dt] to actual waveshapes. This is often discussed by introductory textbooks, though, so it should not be difficult for students to find good information to help them formulate an answer.
It strikes some students as odd that opamps would have a constant slew rate. That is, when subjected to a stepchange input voltage, an opamp’s output voltage would quickly ramp linearly over time, rather than ramp in some other way (such as the inverse exponential curve seen in RC and RL pulse circuits):

Yet, this effect has a definite cause, and it is found in the design of the opamp’s internal circuitry: the voltage multiplication stages within operational amplifier circuits often use active loading for increased voltage gain. An example of active loading may be seen in the following schematic diagram for the classic 741 opamp, where transistor Q_{9} acts as an active load for transistor Q_{10}, and where transistor Q_{13} provides an active load for transistor Q_{17}:

Explain how active loading creates the constant slew rate exhibited by operational amplifier circuits such as the 741. What factors account for the linear ramping of voltage over time?
Active loads act as constantcurrent sources feeding a constant (maximum) current through any capacitances in their way. This leads to constant [dv/dt] rates according to the “Ohm’s Law” equation for capacitors:

Followup question: based on what you see here, determine what parameters could be changed within the internal circuitry of an operational amplifier to increase the slew rate.
This question provides good review of fundamental capacitor behavior, and also explains why opamps have slew rates as they do.
Calculate the impedance (in complex number form) “seen” by the AC signal source as it drives the passive integrator circuit on the left, and the active integrator circuit on the right. In both cases, assume that nothing is connected to the V_{out} terminal:


The most important thing to be learned here is that the opamp “isolates” the signal source from whatever impedance is in the feedback loop, so that the input impedance (in this case, the 10 kΩ resistor) is the only impedance “visible” to that source. This has profound effects on the phase relationship between the output signal and the input signal.
Calculate the phase angle of the current drawn from the AC signal source as it drives the passive integrator circuit on the left, and the active integrator circuit on the right. In both cases, assume that nothing is connected to the V_{out} terminal:

Θ = 46.7^{o} for the passive integrator current, while Θ = 0^{o} for the active integrator circuit.
The most important thing to be learned here is that the opamp “isolates” the signal source from whatever impedance is in the feedback loop, so that the input impedance (in this case, the 10 kΩ resistor) is the only impedance “visible” to that source. This has profound effects on the phase relationship between the output signal and the input signal.
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