Linear Computational Circuitry
Analog Integrated Circuits
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:
 Carefully measure and record all component values prior to circuit construction.
 Draw the schematic diagram for the circuit to be analyzed.
 Carefully build this circuit on a breadboard or other convenient medium.
 Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements onebyone on the diagram.
 Mathematically analyze the circuit, solving for all voltage and current values.
 Carefully measure all voltages and currents, to verify the accuracy of your analysis.
 If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully recalculate the values and remeasure.
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.
Suppose I set a potentiometer to the 30% position, so that the wiper is 30% of the way up from the bottom:

Describe the mathematical relationship between V_{in} and V_{out} with the potentiometer in this position. What does this suggest about the usefulness of a potentiometer as an analog computational element?
Ideally, an inverting amplifier circuit may be comprised of just one opamp and two resistors, as such:

However, if high accuracy is desired, a third resistor must be added to the circuit, in series with the other opamp input:

Explain what this “compensation” resistor is compensating for, and also what its value should be.
We know that an opamp connected to a voltage divider with a voltage division ratio of [1/2] will have an overall voltage gain of 2, and that the same circuit with a voltage division ratio of [2/3] will have an overall voltage gain of 1.5, or [3/2]:

There is definitely a mathematical pattern at work in these noninverting opamp circuits: the overall voltage gain of the circuit is the mathematical inverse of the feedback network’s voltage gain.
Building on this concept, what do you think would be the overall function of the following opamp circuits?

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!
Rewrite the equation for this differential amplifier circuit, assuming that the marked resistor fails open:

The purpose of this circuit is to provide a pushbuttonadjustable 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. Essentially, it is an active integrator circuit, with resistors R_{1} and R_{2} and capacitor C_{1} setting the rates of integration:

After working just fine for quite a long while, the circuit suddenly fails: now it only outputs zero volts DC all the time. Identify at least one component or wiring failure that would account for the circuit’s faulty behavior, and explain why that fault would cause the circuit to act like this.

Integrator circuits may be understood in terms of their response to DC input signals: if an integrator receives a steady, unchanging DC input voltage signal, it will output a voltage that changes with a steady rate over time. The rate of the changing output voltage is directly proportional to the magnitude of the input voltage:

A symbolic way of expressing this input/output relationship is by using the concept of the derivative in calculus (a rate of change of one variable compared to another). For an integrator circuit, the rate of output voltage change over time is proportional to the input voltage:

A more sophisticated way of saying this is, “The timederivative of output voltage is proportional to the input voltage in an integrator circuit.” However, in calculus there is a special symbol used to express this same relationship in reverse terms: expressing the output voltage as a function of the input. For an integrator circuit, this special symbol is called the integration symbol, and it looks like an elongated letter “S”:

Here, we would say that output voltage is proportional to the timeintegral of the input voltage, accumulated over a period of time from time=0 to some point in time we call T.
“This is all very interesting,” you say, “but what does this have to do with anything in real life?” Well, there are actually a great deal of applications where physical quantities are related to each other by timederivatives and timeintegrals. Take this water tank, for example:

One of these variables (either height H or flow F, I’m not saying yet!) is the timeintegral of the other, just as V_{out} is the timeintegral of V_{in} in an integrator circuit. What this means is that we could electrically measure one of these two variables in the water tank system (either height or flow) so that it becomes represented as a voltage, then send that voltage signal to an integrator and have the output of the integrator derive the other variable in the system without having to measure it!
Your task is to determine which variable in the water tank scenario would have to be measured so we could electronically predict the other variable using an integrator circuit.

We know that the output of an integrator circuit is proportional to the timeintegral of the input voltage:


But how do we turn this proportionality into an exact equality, so that it accounts for the values of R and C? Although the answer to this question is easy enough to simply look up in an electronics reference book, it would be great to actually derive the exact equation from your knowledge of electronic component behaviors! Here are a couple of hints:


If an object moves in a straight line, such as an automobile traveling down a straight road, there are three common measurements we may apply to it: position (x), velocity (v), and acceleration (a). Position, of course, is nothing more than a measure of how far the object has traveled from its starting point. Velocity is a measure of how fast its position is changing over time. Acceleration is a measure of how fast the velocity is changing over time.
These three measurements are excellent illustrations of calculus in action. Whenever we speak of “rates of change,” we are really referring to what mathematicians call derivatives. Thus, when we say that velocity (v) is a measure of how fast the object’s position (x) is changing over time, what we are really saying is that velocity is the “timederivative” of position. Symbolically, we would express this using the following notation:

Likewise, if acceleration (a) is a measure of how fast the object’s velocity (v) is changing over time, we could use the same notation and say that acceleration is the timederivative of velocity:

Since it took two differentiations to get from position to acceleration, we could also say that acceleration is the second timederivative of position:

“What has this got to do with electronics,” you ask? Quite a bit! Suppose we were to measure the velocity of an automobile using a tachogenerator sensor connected to one of the wheels: the faster the wheel turns, the more DC voltage is output by the generator, so that voltage becomes a direct representation of velocity. Now we send this voltage signal to the input of a differentiator circuit, which performs the timedifferentiation function on that signal. What would the output of this differentiator circuit then represent with respect to the automobile, position or acceleration? What practical use do you see for such a circuit?
Now suppose we send the same tachogenerator voltage signal (representing the automobile’s velocity) to the input of an integrator circuit, which performs the timeintegration function on that signal (which is the mathematical inverse of differentiation, just as multiplication is the mathematical inverse of division). What would the output of this integrator then represent with respect to the automobile, position or acceleration? What practical use do you see for such a circuit?

A familiar context in which to apply and understand basic principles of calculus is the motion of an object, in terms of position (x), velocity (v), and acceleration (a). We know that velocity is the timederivative of position (v = [dx/dt]) and that acceleration is the timederivative of velocity (a = [dv/dt]). Another way of saying this is that velocity is the rate of position change over time, and that acceleration is the rate of velocity change over time.
It is easy to construct circuits which input a voltage signal and output either the timederivative or the timeintegral (the opposite of the derivative) of that input signal. We call these circuits “differentiators” and ïntegrators,” respectively.

Integrator and differentiator circuits are highly useful for motion signal processing, because they allow us to take voltage signals from motion sensors and convert them into signals representing other motion variables. For each of the following cases, determine whether we would need to use an integrator circuit or a differentiator circuit to convert the first type of motion signal into the second:
 Converting velocity signal to position signal: (integrator or differentiator?)
 Converting acceleration signal to velocity signal: (integrator or differentiator?)
 Converting position signal to velocity signal: (integrator or differentiator?)
 Converting velocity signal to acceleration signal: (integrator or differentiator?)
 Converting acceleration signal to position signal: (integrator or differentiator?)
Also, draw the schematic diagrams for these two different circuits.
This active integrator circuit processes the voltage signal from an accelerometer, a device that outputs a DC voltage proportional to its physical acceleration. The accelerometer is being used to measure the acceleration of an athlete’s foot as he kicks a ball, and the job of the integrator is to convert that acceleration signal into a velocity signal so the researchers can record the velocity of the athlete’s foot:

During the setup for this test, a radar gun is used to check the velocity of the athlete’s foot as he does come practice kicks, and compare against the output of the integrator circuit. What the researchers find is that the integrator’s output is reading a bit low. In other words, the integrator circuit is not integrating fast enough to provide an accurate representation of foot velocity.
Determine which component(s) in the integrator circuit may have been improperly sized to cause this calibration problem. Be as specific as you can in your answer(s).

How much current (I) would have to be forced through the resistor in order to generate an output voltage of 5 volts?

At what rate would V_{in} have to increase in order to cause this amount of current to go “through” the capacitor, and thereby cause 5 volts to appear at the V_{out} terminal? What does this tell us about the behavior of this circuit?

We know that the output of a differentiator circuit is proportional to the timederivative of the input voltage:


But how do we turn this proportionality into an exact equality, so that it accounts for the values of R and C? Although the answer to this question is easy enough to simply look up in an electronics reference book, it would be great to actually derive the exact equation from your knowledge of electronic component behaviors! Here are a couple of hints:


You are part of a team building a rocket to carry research instruments into the high atmosphere. One of the variables needed by the onboard flightcontrol 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 (rateofchange 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 flightcontrol computer needs?

A Rogowski Coil is essentially an aircore current transformer that may be used to measure DC currents as well as AC currents. Like all current transformers, it measures the current going through whatever conductor(s) it encircles.
Normally transformers are considered AConly devices, because electromagnetic induction requires a changing magnetic field ([(d φ)/dt]) to induce voltage in a conductor. The same is true for a Rogowski coil: it produces a voltage only when there is a change in the measured current. However, we may measure any current (DC or AC) using a Rogowski coil if its output signal feeds into an integrator circuit as shown:

Connected as such, the output of the integrator circuit will be a direct representation of the amount of current going through the wire.
Explain why an integrator circuit is necessary to condition the Rogowski coil’s output so that output voltage truly represents conductor current.

A Rogowski coil has a mutual inductance rating of 5 μH. Calculate the size of the resistor necessary in the integrator circuit to give the integrator output a 1:1 scaling with the measured current, given a capacitor size of 4.7 nF:

That is, size the resistor such that a current through the conductor changing at a rate of 1 amp per second will generate an integrator output voltage changing at a rate of 1 volt per second.
Practical integrator circuits must have a compensating resistor connected in parallel with the capacitor, in the feedback loop. Typically, this resistor value is very large: about 100 times as large as R_{in}.

Describe why this is a necessity for accurate integration. Hint: an ideal opamp would not need this resistor!
Practical opamp integrator and differentiator circuits often cannot be built as simply as their “textbook” forms usually appear:

The voltage gains of these circuits become extremely high at certain signal frequencies, and this may cause problems in real circuitry. A simple way to “tame” these high gains to moderate levels is to install an additional resistor in each of the circuits, as such:

The purpose of each resistor is to “dominate” the impedance of the RC network as the input signal frequency approaches the point at which problems would occur in the ideal versions of the circuits. In each of these “compensated” circuits, determine whether the value of the compensation resistor needs to be large or small compared to the other resistor, and explain why.
Of course, this solution is not without problems of its own. By adding this new resistor to each circuit, a halfpower (3 dB) cutoff frequency point is created by the interaction of the compensation resistor and the capacitor, as predicted by the equation f_{c} = [1/(2 πR_{comp}C)]. The value predicted by this equation establishes a practical limit for the differentiation and integration functions, respectively. Operating on the wrong side of the frequency limit will result in an output waveform that is not the true timederivative or timeintegral of the input waveform. Determine whether the f_{c} value constitutes a low frequency limit or a high frequency limit for each circuit, and explain why.

The chain rule of calculus states that:

Similarly, the following mathematical principle is also true:

It is very easy to build an opamp circuit that differentiates a voltage signal with respect to time, such that an input of x produces an output of [dx/dt], but there is no simple circuit that will output the differential of one input signal with respect to a second input signal.
However, this does not mean that the task is impossible. Draw a block diagram for a circuit that calculates [dy/dx], given the input voltages x and y. Hint: this circuit will make use of differentiators.
Challenge question: draw a full opamp circuit to perform this function!
Draw the appropriate wires in this diagram to complete the circuit for a differentiator circuit (the opposite function of an integrator circuit):

The particular opamp used in this circuit is an LM324 (quad) operational amplifier. Of course, you need only show how one of the four opamps within the IC may be connected to form a differentiator.
Analog computers have been all but replaced by digital computers in modern electronic systems. Yet, analog computational circuits still enjoy certain advantages over digital circuits. Describe what some of the limitations of analog computers are, and why these limitations have led to their obsolescence. Also, discuss some of the advantages of analog computers, and why a designer might still choose to use an analog computational circuit in a modern system.
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