Op amps are high gain differential amplifiers.  Like the discrete differential amplifiers that you built last week, op amps have two inputs $V_+$ and $V_-$, and one output, $V_\mathrm{out}$. Unlike discrete differential amplifiers, op amps have essentially infinite differential gain (gains over $10^7$ are readily available), negligible common mode gain, extremely high input impedance, very low output impedance, small temperature drifts and inconsequential piece-to-piece variation. Furthermore, they are insensitive to power supply fluctuations, are difficult to burn out, cheap, and available in countless varieties; one vendor, Digikey, sells over 30,000 varieties.

Note that the op amp must also be connected to power; in our case to both $+12\,\mathrm{V}$ and $-12\,\mathrm{V}$.  The power leads are typically not drawn.

The simplest op amp circuit is the follower.  The first golden rule implies that the input impedance is infinite. The second rule is more interesting; since $V_-$ is wired to $V_\mathrm{out}$ , these two voltages must be equal. Likewise, $V_+=V_\mathrm{in}$. The only way that the two inputs can be at the same voltage, $V_+=V_-$, is for $V_\mathrm{out}=V_\mathrm{in}$. Consequently the op amp behaves like a perfect voltage follower.

Understanding how the second rule is actually satisfied is particularly easy for the follower. If the output $V_\mathrm{out}$were to deviate low, $V_+-V_- = V_\mathrm{in}-V_\mathrm{out}>0$, the amplifier would drive its output $V_\mathrm{out}$ higher, reestablishing $V_\mathrm{out}=V_\mathrm{in}$.  If the output $V_\mathrm{out}$ were to deviate high, $V_+-V_-<0$, and the amplifier would drive its output lower, once again reestablishing $V_\mathrm{out}=V_\mathrm{in}$.  If the input voltage $V_\mathrm{in}$ were to change, $V_\mathrm{out}=V_\mathrm{in}$ would be similarly reestablished.

Looked at in a bit more detail, one might be bothered by a seeming inconsistency.  Since $V_+-V_- = V_\mathrm{in}-V_\mathrm{out}=0$ in equilibrium, and the output $V_\mathrm{out}$ is proportional to $V_+-V_- $, one might think that the output $V_\mathrm{out}$ must also be zero, not the desired $V_\mathrm{in}$.  But remember that an ideal op amp’s gain is infinite. Since zero times infinity is undefined, it is not contradictory to set $V_+-V_- =0$ and $V_\mathrm{out}\ne0$. More realistically, the op amp’s finite gain means that the error voltage $V_+-V_- $ must be infinitesimally positive for $V_\mathrm{in}>0$,  and $V_\mathrm{out}$ will settle to some value infinitesimally less than $V_\mathrm{in}$. Likewise, $V_\mathrm{out}$ will be infinitesimally greater than $V_\mathrm{in}$ for $V_\mathrm{in}<0$.  Most precisely,  $V_+-V_-=V_\mathrm{out}/G$, and $V_\mathrm{out}=V_\mathrm{in}/(1+1/G)$ for an op amp with gain $G$. Since $G$ is enormous for most op amps, the follower is essentially perfect.

Another simple op amp circuit is the inverting amplifier.  Here the inverting input $V_+$ is tied to ground.   Since the second golden rule insists that $V_+-V_- $, the negative input $V_-$must also be at ground. As this ground is enforced by the op amp itself rather than by any physical connection to ground, it is often called a virtual ground. The current flowing towards the $V_-$ junction is then easy to calculate: $I=V_\mathrm{in}/R_1$. Since the first golden rule asserts that no current flows into the input itself, all of this current $I$ must flow out through the resistor $R_2$.  Remembering that the $V_-$ end of $R_2$ is at virtual ground, the output of the op amp must be at $V_\mathrm{out}=-R_2I=-(R_2/R_1)V_\mathrm{in}$. Thus, the gain of the circuit is $-R_2/R_1$.

Op amps come in several different standard packages. Our op amps come in a package called an 8 pin DIP (dual inline package), shown at right. DIP pins are numbered sequentially in the counterclockwise direction. Pin 1 is at the end with the half-moon cutout; looking at the IC from the top (leads pointing away), pin 1 is the first pin counterclockwise from the cutout. This pin is sometimes indicated by a small dot. Note that the breadboard channel is designed so that the op amp neatly straddles the channel as shown to the right:

By convention, a DIP is always oriented so that the cut out is to the left (or up, if the dip is rotated by 90°.)   Many of you will be tempted to ignore this convention, and you will get away with it for circuits that have only one or two op amps.  But by the time you build circuits with ten op amps, as you will in later labs, randomly orienting the op amps will lead you to much confusion; better to get in the habit of orienting them properly from the beginning.  Furthermore, orienting the circuits properly will make it much easier and faster for the GSI's to help you with debugging.

The pins on most op amps are connected as shown to the right.  N/C stands for no connection, and should be left floating. The BAL pins are used to eliminate output voltage offsets caused by improper matching of the input transistors.  These pins will not be used in this lab, and should also be left floating.

The op amp power leads $+V$ and $-V$ must be connected for the op amp to work.  Remember that the power leads are not normally drawn on circuit schematics. They must be connected to the +12 and -12 volt power supplies! 

Op amp circuits love to undergo parasitic oscillations. Always use power supply decoupling capacitors placed close to the op amps.  $0.1\,\mu\mathrm{F}$ caps are a good choice.

Semiconductor devices like photocells, photodiodes and phototransistors can react to illumination. Photocells sometimes called solar cells, convert light to electrical power, and are becoming increasingly cheap and practical. Photodiodes do not produce power, but are good light detectors. They are very similar to normal diodes, and are always operated reversed-biased. Remember that reverse-biased diodes do not conduct because of the absence of free carriers at their pn junctions; when a reversed biased photodiode is illuminated, photons create electron hole pairs in the junction region. The holes and electrons are swept in the appropriate direction by the junction’s electric field, causing a temporary, leakage-like current to flow through the junction. As the number of electron hole pairs created is proportional to the number of photons hitting the junction, and hence to the intensity of the light illuminating the junction, the total leakage current will be proportional to the light intensity.

Any pn junction can show photo activity. Bizarre effects sometimes occur in our 1N4448 diodes because their glass packages allow light to strike the diode junctions.  In this lab however, we will use a device optomized for light detection: the OP802SL bipolar phototransistor.

Bipolar transistors come in two classes, NPN and PNP transistors, and are three leaded devices; the leads are called the collector, the base, and the emitter. The schematic for an NPN phototransistor is shown to the right.  Our phototransistors come in a standard TO‑18 package, modified to include a plastic lens which focuses light onto the base junction. The emitter is the lead nearest the case tab, and the collector is the lead 180° from the emitter. The base is the lead in between. 

For a bipolar transitor, the collector is analogous to a JFET transistor’s drain, the base is analogous to a JFET’s gate, and the emitter is analogous to a JFET’s source.  Bipolar transitor operation is roughly similar to JFET operation, but there are some very significant differences.  The general use and operation of these transistors will be briefly discussed in the next lab.

Despite our phototransistor being a three leaded device, we will leave the base unattached and connect only two of its leads, as if it were a photodiode.  Even though it is not wired, the base increases the photransitor's light sensitivity compared to that of a simple photodiode.  Shown to the right are typical curves showing the phototransistor output current between the collector and emitter, as a function of the collector to emitter voltage, for different illumination levels.  Note that for collector to emitter voltages between 1 and 10V, the current is approximately independent of the collector emitter voltage; thus the phototransistor looks almost like a current source. 

 Build the following comparator circuit, with $R_1$and $R_2 $ temporarily set to zero.  For simplicity the V₊ input is connected to the $+5\,\mathrm{V}$ supply, setting the comparator threshold at $+5\,\mathrm{V}$.More generally, the V₊ input could be connected to any DC or AC signal. Show that the output switches between approximately $-12\,\mathrm{V}$ and $+12\,\mathrm{V}$ when the potentiometer voltage goes through $+5\,\mathrm{V}$. Look at the output on the scope. Can you adjust the potentiometer to obtain output voltages intermediate to $\pm12\,\mathrm{V}$? Why is this hard to do? Can you set a lower bound on the op amp gain?

By carefully tweaking the potentiometer, you should be able to get the output voltage to fluctuate between two values at one pot setting.  These fluctuations are related to tiny fluctuations in the $+5\,\mathrm{V}$ power supply.  If the output doesn't fluctuate, try adjusting the resistor values: $R_1=560\,\mathrm{k}$ $R_2=390\,\mathrm{k}$ may work, but these values may require tweaking if fully turning the pot doesn't cause the output voltage to switch.  It may help to put a $0.1\,\mu\mathrm{F}$ across the LF356 power inputs (pins 4 and 7).

 Prove that the fluctuations are at $60\,\mathrm{Hz}$ by line-syncing the scope.  Line sync is a trigger option, similar to aux and found on the same menu, that sync's the scope trigger to the $60\,\mathrm{Hz}$ oscillations of the $120\,\mathrm{V}$ power line.  Noise and flucutations are often at this frequency, and, when examined on a scope set to line sync, will appear to partially freeze on the scope's display.

Note that this circuit does not use feedback, and the golden rules do not apply. 

The fluctuations that you observed in the previous problem can be eliminated by using positive feedback to provide hysteresis. Build the circuit at right, first with the $+5\,\mathrm{V}$ power supply.  Show that the output cannot be made to fluctuate. Then disconnect the $+5\,\mathrm{V}$ supply from the V_in input, and connect instead a $500\,\mathrm{Hz}$, $20\,\mathrm{Vpp}$ triangle wave.  Look at both the input and the output on the scope. You shold get traces that look like the image at right, showing the triangular input on channel 1 and the square wave output on channel 2.

Look carefully at the input voltage at which the transitions in the output occur. You should observe that the output voltage switches high at an input voltage that is higher than the input voltage where the output switches low; this is the desired hysteretic effect.  Determine the hysteresis (i.e. the difference between the turn-on and turn-off voltages.)  It may be helpful to use the scope cursors.

Note that this circuit uses positive feedback, and the golden rules do not apply;  this is one of the very few times in this course that we will use positive feedback.

Construct the follower like that show at right.  Set $R=0$.  By looking at $V_\mathrm{out}$, investigate the follower’s performance for a variety of input signals of different shapes, frequencies, and amplitudes.

With a $1\,\mathrm{Vpp}$, $1\,\mathrm{kHz}$ sine wave input , vary $R$. Can you set a lower bound on the input impedance? Hint: Imagine that the op amp input looks like a voltage divider.  Varying $R$ is like varying the top resistor in the divider.   You may find that with sufficiently large resistors, say near $10\,\mathrm{M}\Omega$, you observe anomalies. Do these anomalies go away if you lower the frequency of your input signal to, say, $50\,\mathrm{Hz}$?

Build the non-inverting amplifier shown at right, and measure its gain.  How does its gain compare with the design gain calculated in Pre-lab question 3?  Now carefully measure the resistor values and calculate the expected gain.  How does this gain compare with the observed gain?

Build the inverting amplifier shown at right.  This amplifier has an unusual, and under some circumstances deleterious, $10\,\mathrm{k}$ output resistor.  This resistor degrades the normally very low op amp output impedance to  $10\,\mathrm{k}$.  This is a fairly high output impedance. Nonetheless, feedback will cure the deleterious effects of the high impedance in some circumstances.

With the $1\,\mathrm{k}$ load resistor $R_\mathrm{L}$ temporarily removed, what is the gain of the circuit for a  $0.1\,\mathrm{Vpp}$, $1\,\mathrm{kHz}$ sine wave? For a $1\,\mathrm{Vpp}$  sine wave? (Make sure that the signal generator output is set to High Z, and make sure that your smaller signal really is $0.1\,\mathrm{V}_\mathrm{pp}$.)  Is the gain what you would expect for both cases? Now insert $R_\mathrm{L}$ and repeat.  Does the amplifier still work for both input voltages? Is the gain the same with and without the load resistor at both input voltages?  You might wish to also examine the response of the circuit with a triangular wave.

What causes the circuit failure at $1\,\mathrm{Vpp}$ with the load resistor? Hint: Think about what the output voltage of the LF356 would have to be for this circuit to work with a $1\,\mathrm{Vpp}$ input.

Feeback normally works beautifully, but it cannot fix fundamental component limitations.

Drive the circuit at right with $0.1\,\mathrm{Vpp}$ and $1\,\mathrm{Vpp}$ sine waves. Record images of the output at $100\,\mathrm{Hz}$ and at $10\,\mathrm{kHz}$. You should observe that, under some circumstances, the diode-cap-resistor network greatly degrades the amplifier.  Note that the feedback loop does not include these circuit elements.

Explain the influence of the diode and the capacitor.

The circuit in 6.7 is frequency dependent and powerfully nonlinear. If the frequency dependent and nonlinear elements are included inside the feedback loop, the circuit will behave much better.

Change the feedback point to the position shown at right.  What is the output now with $0.1\,\mathrm{Vpp}$ and $1\,\mathrm{Vpp}$  inputs and at $100\,\mathrm{Hz}$ and $10\,\mathrm{kHz}$? Record typical images.  Is the circuit frequency dependent? Is it nonlinear?

You should observe that proper feedback can cleanup horribly nonlinear and frequency dependent behavior.   Feedback will take a poor quality op amp, like the "composite" op amp investigated here (the LF356 and the diode, resistor and capacitor) and make it function well.

Finish the design of this gain-equals-ten differential amplifier.  Construct the circuit.  Does it work? Measure the common mode gain, and the gain on each input with the other input grounded.  Do these gain agree with what you would predict?

Measuring the gain of one input with the other input grounded isn't quite the same as measuring the differential gain.  To make a true differential gain measurement, you need both your original input signal and its inverse.  Use an inverting op amp follower to generate this inverted signal, and feed both the orignal and the inverse into your differential amp. Now measure the true differential gain. 

Op amps can be used to sum signals. Construct the circuit at right, and show that its output equals$-(2V_A+V_B+V_C)$.

The Offset Adders which are part of your breadbboard box use a summing circuit similar to this with one input attached to a potentiometer.

Use the circuit at right to transform the current signal from the OP802SL  phototransistor into a voltage signal. 

How does the voltage signal change if you wave your hand over the phototransistor? Hook an LED up to the signal generator, and drive the LED with a 1kHz square wave. Place the LED next to the phototransistor.  Can you pick up its modulated light signal?  Prove that the circuit is detecting the light signal rather than some sort of electrical pickup signal by interposing a paper sheet between the LED and the phototransistor.  The paper, unlike your hand, is nonconductive and will not act like an antenna and influence the circuit.

Op amps can be used to make excellent voltage-controlled current sources. In the circuit at right, set the potentiometer output  to about $+3\,\mathrm{V}$, and measure the current through a load resistor $R_\mathrm{L}$  of $4.7\,\mathrm{k}\Omega$with a multimeter.  

Swap in different load resistors while measuring the current.  Find the range of $R_\mathrm{L}$ over which the current does not vary significantly.

Restore the load resistor to $4.7\,\mathrm{k}$, and vary the potentiometer voltage. How does the current change?

Assuming that the op amp has a finite gain $G_0$, but is otherwise ideal, what is the output impedance for the amplifier (the entire amplifier, including the $4.7\,\mathrm{k}\Omega$, $47\,\mathrm{k}\Omega$ and $10\,\mathrm{k}\Omega$ resistors) in 6.6? One way to determine the output impedance is illustrated at right.  One imagines hanging a voltage source $V$ on the output, and measuring the current $I$.   Then $Z_\mathrm{out}=V/I$.  To calculate the current, one needs the open loop gain $G_0$ (usually called $A_0$ on spec sheets).  This gain is typically very high, often in the millions near DC.  Since the positive input terminal is grounded, $V_\mathrm{out}=-GV_-$.  But $V_-$comes from input and feedback resistor voltage divider.  Once $V_\mathrm{out}$ is known, the current $I$ is just the sum of the current into the feedback $(47\,\mathrm{k}\Omega)$ resistor and the current into the $10\,\mathrm{k}\Omega$ resistor.  (Simplify your answer by taking advantage of the fact that $G_0$ is very large.) You should ignore the load resistance $R_\mathrm{L}$ throughout this calculation.

Your should find that the output resistance is very low.  So why does the op amp fail under certain conditions (large input drives) when the circuit is loaded?