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Operational amplifier circuits
As you may have noticed previously, working with transistors can be difficult: you have to consider the gain you want, set an appropriate bias point, and design your input stage such that it doesn't filter your signal. Fortunately for us people have already packaged entire combinations of transistor circuits that will do most of the hard work for us in the form of operational amplifiers.
There are many, many different operational amplifiers, each tailored to slightly different situations. To start with we'll use the LF411, which is a good all-purpose model.
There's some more text after the numbers
The text after the chip name is part of the device marking code. The suffix typically contains information about allowed temperature ranges, the “grade” of the chip, and the physical packaging. The codes are going to be idiosyncratic, varying from company to company. The most common special grades you might see are things rated for automotive or military use; reliability demands are a bit more stringent for a toaster versus your car's cruise control.
In this instance, CP is used to denote that the chip is rated for 0-70${}^\circ$C temperatures and that it is in a plastic, dual-inline package. TI also sells smaller variants, ones rated for -40-85${}^\circ$C, and parts shipped on a reel for large-scale manufacturing; each option with a different letter code.
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| Pin connections for the LF411 op-amp chip. Pins are typically numbered starting with 1 in the top-left of the diagram and continuing counter-clockwise around a chip. Note that pins 1, 5, and 8 aren't used for this part; this is to keep a common footprint and make mass production simpler. | How to place an Integrated Circuit (IC) chip on a breadboard. Note that the notch on the top of the diagram corresponds to a notch or circle on the physical chip, and that the chip should be placed across the channel in your breadboard. |
While there are many, many subtleties to op-amp behavior, we'll focus on a model that will get us the furthest without getting bogged down in details.
Op-amp golden rules:
- There is no current into either the inverting input ($V_-$) or the non-inverting input ($V_+$ )
- There can (and typically is) current at the output and the power connections. Kirchhoff would be sad otherwise.
When there is negative feedback (i.e. there is a path for current from the output to the inverting input), then:
- The output $V_{out}$ will do what is needed for there to be no potential difference between the inverting input $(V_-)$ and the non-inverting input $(V_+)$
You can get a long ways with these two rules of thumb, but there are some important common caveats:
- $V_{out}$ can never be higher than the positive power rail $V_{CC+}$ nor lower than the negative power rail $V_{CC-}$
- Op-amps can manipulate voltages, but they don't create them.
- Often it is limited to being a volt or two lower, but this depends heavily on the specific device
- If there is not negative feedback, then
- $V_{out} \rightarrow V_{CC+}$ if $V_+ > V_-$
- $V_{out} \rightarrow V_{CC-}$ if $V_+ < V_-$
There are many more limitations on the behavior of real op-amps, but this simple model is enough to understand and design a lot of useful circuits.
Setting up an Op-Amp
When working with transistors, we usually just used the power supply for providing a positive voltage and ground. The op-amp circuits you'll be using are designed to use a bi-polar supply, meaning both a positive and negative voltage, as power. This means setting the low side of one variable supply to ground and the high side of the other variable supply to ground. To set facilitate this, we can use a metal jumper to make the connections for us. The method is shown below.
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| Unscrew the ground terminal's cover | Place the bar over the ground terminal, then loosen the adjacent + and - terminal covers |
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| Rotate the bar under the adjacent terminal covers and tighten them | Replace the ground terminal cover |
Now the leftmost red terminal will be positive with respect to ground and the right black terminal will be negative with respect to ground.
Non-Inverting Amplifiers
To start, consider the op-amp circuit shown below:
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| A diagram of an op-amp non-inverting amplifier circuit. The numbers next to the five different connections refer to the pin connection on the physical chip; this isn't standard but we've included it here to make the initial construction simpler. |
You should use your power supply for the the $\pm 15 V$ rail connections here. Refer to the previous labs for instructions on how to set up a negative voltage.
For the input, use a sine wave with the following parameters:
- 1 kHz frequency
- 1V pk-pk amplitude
- 0V offset
Build the circuit on your breadboard, and then observe both $V_{in}$ and $V_{out}$ on the scope.
Measure the amplitudes of both signals, and include a sketch or screenshot of the two signals.
Calculate the measured gain $g = \dfrac{V_{out}}{V_{in}}$ for this circuit. Is it close to what you'd expect?
Why is this circuit called a non-inverting amplifier?
Suppose that you swapped the position of the 22k and 10k resistors.
Predict what the gain would be for this instance, and briefly explain
Test your prediction, briefly describe what you observe, and account for any discrepancies.
Keep the op-amp on your board, you'll be using it for upcoming circuits.
Detour: Potentiometers
Before building the next circuit, we're going to introduce a new element: the potentiometer. Potentiometers are made of some resistive material that has fixed connections at either end and a movable contact point. This is shown in the schematic diagram as a resistor with an arrow pointed partway along its length:
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| A photograph of your potentiometer, with connections annotated. The 1, 2, and 3 refer to the three pin connections. This one has been soldered to a tiny bit of PCB to make it easier to use. | The schematic symbol for a potentiometer |
Find a 1k potentiometer (or 10k or 100k, the specifics aren't critical) and place it in your breadboard. Connect three wires to the three terminals, and then use your multimeter to test its behavior.
Does the resistance between terminal 1 and 2 increase, decrease, or stay the same when the screw is turned?
Does the resistance between terminal 1 and 3 increase, decrease, or stay the same when the screw is turned?
Okay, we can adjust a resistor. Let's use it in our amplifier and see what happens!
Consider the circuit shown below:
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| An adjustable amplifier circuit. Note the blue numbers indicate the potentiometer pins |
Predict what gains you can achieve with this circuit, and briefly explain
Build the circuit and test the limits of its gain using the potentiometer. Resolve any discrepancies with your predictions.
Followers
As you observed with the potentiometer circuit, we can make a non-inverting amplifier such that it has a gain of 1. Like its transistor counterpart, this is called a follower circuit. Let's make one intentionally by connecting our op-amp as follows:
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After building this, test the follower with various inputs. You may also (carefully) test reducing the $\pm$15 V power rails.
Under what conditions does the op-amp circuit follow?
How are the requirements different from transistor followers?
Inverting Amplifiers
Next, let's construct an inverting amplifier circuit, shown below:
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| An Inverting Amplifier circuit. |
NOTICE: The depiction of the non-inverting and inverting inputs has been flipped; this is not uncommon in op-amp circuits so always check the symbol.
For the input, use a sine wave with the following parameters:
- 1 kHz frequency
- 1V pk-pk amplitude
- 0V offset
Construct the circuit and observe both $V_{in}$ and $V_{out}$ on the scope, measuring the amplitudes of both signals.
Don't forget to include a sketch or screenshot of the two signals in your report
Find the gain $g = \dfrac{V_{out}}{V_{in}}$ for this circuit. Is it close to what you'd expect?
Why is this circuit called an inverting amplifier?
Suppose that you swapped the position of the 22k and 10k resistors.
Predict what the gain would be for this instance, and briefly explain
Test your prediction, briefly describe what you observe, and account for any discrepancies.
Current to Voltage Conversion
By replacing one of our resistors with a circuit whose current depends on physical parameters, we can use the op-amp to convert this current to a voltage. Since measuring voltages is typically easier than measuring current (you don't have to break a circuit and put a meter in), this sort of circuit can be useful to us when we want to do something like measure light levels with a phototransistor.
Phototransistors
As you might guess from the name, phototransistors act like bjt transistors that react to light. Specifically, photons are able to pass through the clear packaging and strike the base layer of the transistor, knocking electrons free via the photoelectric effect. These electrons create a small base current, which is then amplified to a much larger current through the collector and emitter. More photons yield more current so this is a proportional measurement, unlike the photoresistor.
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| Our phototransistor and its schematic symbol. Note the flattened edge belongs to the emitter side. |
With this in mind, let's see what happens when we replace the 10k resistor in our last circuit with a phototransistor, as shown below:
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| A light to current converter, constructed with an op-amp. Note that the phototransistor's base isn't connected to anything. |
Construct and test the circuit
In what way does $V_{out}$ vary as the light level increases? (You may want to use your phone's flashlight here). Is it what you'd expect?
Predict how this circuit's sensitivity to light would change if the resistance was increased.
Test your prediction, and resolve any discrepancies.
Portions of this page are adapted from “Flexible Resources for Analog Electronics” by Stetzer and Van De Bogart.