===== Related Reading =====
Hayes and Horowitz: Pages 169 - 177
[[https://www.physics.dcu.ie/~bl/anacont.html|Lawless]]: Chapters 30 - 35
[[https://docs.google.com/document/d/1DVyIvG5UbYleYklLkTbdDl6y_ZJKycffqKg-2ZhyGro/copy|Lab Template]]
====== Transistors II: Amplification ======
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In the last lab, you used transistors as a switch, and to solve impedance issues. In this lab, we'll use transistors to //amplify// voltage signals instead. As a reminder, your transistor's emitter, base, and collector (E, B, and C) are oriented as follows:
| {{:phylabs:lab_courses:phys-226-wiki-home:lab_4_transistors_i:bjtpinout.png?200|}} | {{:phylabs:lab_courses:phys-226-wiki-home:lab_4_transistors_i:bjtpackageorientation.png?200|}} |
| A mapping of package pins to the circuit symbol | Mapping package pins to the physical device. \\ Note that TO-92 refers to the physical package the transistor is in, \\ and is one of [[https://en.wikipedia.org/wiki/List_of_integrated_circuit_packaging_types#Transistor,_diode,_small-pin-count_IC_packages|many, many]] commonly used forms. |
====== ======
====== Followers and amplifiers ======
Last time we focused on how to use a transistor mostly as a binary device, turning a light on or off. This lab will focus on a transistor's use as an amplifier, something impossible for passive components alone. We can use transistors for amplifying signals, which is something that's often relevant when gathering data from sensors. Conversely, we can also use transistors to amplify the power provided to a circuit element, which is important when driving things like speakers, motors, heaters, or electromagnets. While most circuits serving either of these purposes will be more complex than the ones we build here, transistors are absolutely required for many electronic devices to perform their functions.
As a reminder, here's the shorthand for the transistor's behaviors:
{{pdfjs 600px >:phylabs:lab_courses:phys-226-wiki-home:spring2020-lab-4-transistors-i:transistor_behavior.pdf|BJT transistor behavior summary (click to download pdf)}}
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In this section, we examine three different output voltages from a transistor circuit that is based on the common-emitter amplifier.
===== DC Conditions =====
To begin, consider the following circuit.
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:dc_bias_cea.png}} |
> Predict the values of $V_{out,1}$, $V_{out,2}$, and $V_{out,3}$ would be if $V_{in}$ were 0V.
If the input is 0V, then the base won't be higher voltage than the ground connection made from the emitter resistors.
The transistor will be in cutoff, and thus there will be no current through any part of this circuit.
This means that in the lower branch, $V_{out,1}$ and $V_{out,2}$ will be 0V (since there's no current through either 1k, their voltage can't change from ground)
In the upper branch, $V_{out,3}$ will be 15V (again, there's no current through the 10k resistor and thus the voltage across it doesn't change.)
Use your DC power supply for both $V_{in}$ (of 0V) and the 15V connection, then build and test the circuit.
> Record your circuit's outputs, and resolve any discrepancies between your measurements and predictions.
Now, suppose that $V_{in}$ were instead 2V.
> Predict the values of $V_{out,1}$ , $V_{out,2}$, and $V_{out,3}$ in this instance.
Check your predictions and record the measured values for all three voltages
In this case, the transistor will be forward-active.
As a result, the transistor's emitter $V_{out,1}$ will be 0.6V lower than the base, resulting in $V_{out,1}$ being 1.4 V.
Next, there's effectively a voltage divider formed by the bottom two resistors, thus $V_{out,2}$ will be half of $V_{out,1}$. $V_{out,2}$ = 0.7 V
Finally, we know that the emitter and collector current are the same.
Since that current produces an 0.7V change over a 1k resistor, it will produce a 7V change over the 10k resistor.
This means that $V_{out,3}$ will be 15-7 = 8V.
Some students will instead explicitly calculate the current, this is fine and will get the same result.
===== AC Conditions =====
Now suppose $V_{in}$ is a sine wave with the following properties:
* 1kHz frequency
* 1V peak-to-peak amplitude
* 0V dc offset
> //Predict// the shape and characteristics (frequency, amplitude, and offset) of $V_{out,1}$ and briefly explain.
Check your prediction from and note how, if at all, $V_{out,1}$ differs from your prediction.
The first thing to ask here is: when will the input voltage be high enough to forward bias the transistor's base emitter (BE) junction?
The answer is just whenever it is above 0.6V.
Thus, when the input is above 0.6V, the emitter voltage will be $V_{in} - 0.6 V$
Otherwise, the transistor will be in cutoff and the emitter voltage will stay at zero
In effect, you'd expect to see some little humps corresponding to only the tops of the sine waves.
Now suppose you change your input such that it has a 2V dc offset.
> Predict the peak-to-peak amplitude of $V_{out,1}$ as well as its dc offset. Will $V_{out,1}$ be in phase or 180° out of phase with $V_{in}$?
> Predict the peak-to-peak amplitude of $V_{out,2}$ as well as its dc offset. Will $V_{out,2}$ be in phase or 180° out of phase with $V_{in}$?
> Predict the peak-to-peak amplitude of $V_{out,3}$ as well as its dc offset. Will $V_{out,3}$ be in phase or 180° out of phase with $V_{in}$?
Check your predictions and resolve any inconsistencies.
The DC offsets will be what we found earlier: 1.4, 0.7, and 8 V respectively.
Quick answer for amplitudes: the AC gain of $V_{out,1}$ is 1, the gain of $V_{out,2}$ is 0.5, and the gain of $V_{out,3}$ is -5 (out of phase)
The pk-pk amplitudes are then 1V, 0.5V, and 5V
== Reasoning: ==
$V_{out,1}$ is the output from a follower circuit, and whatever change we make to the base voltage will be reflected in the emitter voltage.
$V_{out,2}$ is always half of $V_{out,1}$, so it the amplitude will likewise be half that of the input: 0.5V
For both of these points, the voltage will be in phase with $V_{in}$ because they are directly proportional to it
$V_{out,3}$ will change by a factor of 5 compared to the change at $V_{out,1}$, since the same current through the pair of 1k resistors passes through the 10k resistor.
This results in the predicted peak-to-peak amplitude being 5V in this instance.
However, increasing the base voltage results in a larger drop in voltage across the 10k, so $V_{out,3}$ is 180° out of phase in this instance.
Keep your circuit built, you'll be using it again soon.
====== The role of blocking capacitors and biasing networks ======
As you may have noticed, getting our amplifier to function for AC signals requires that we have some bias on the signals. When the input comes from the function generator this is something that we can add ourselves, but for other instances we'll need to make circuits that can create that offset for us. A capacitor (often referred to as a **blocking capacitor**) and a biasing network of resistors (similar to a voltage divider) are often used together to accomplish this task. In this part of the lab, we explore the role of a blocking capacitor and biasing network.
Using Thevenin's theorem
Earlier on in lecture, you may have encountered Thevenin's theorem, which states that a circuit made up of only resistors and sources can be simplified to a single source in series with a single resistor. Let's go through the process here by starting with just the voltage divider part of our circuit:
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:thevenin_1.png}} |
The most straightforward way to find the Thevenin equivalent is by determining two things: The voltage at the output when no load is attached, and the current through the output when it is shorted. Let's do the voltage first because that's simple:
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:thevenin_2.png}} |
The open-circuit voltage $V_{OC}$ is equal to $15 V *\dfrac{4.7 k\Omega}{4.7k\Omega+47k\Omega} = 15 V * \frac{1}{11} \approx 1.4 V$
Next, we find out how the circuit behaves when the output is shorted:
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:thevenin_3.png}} |
The short-circuit current $I_{SC}$ is just what we'd get when we run 15V through the $47 k\Omega$ resistor.
With these two things in hand, we use Ohm's law to find the Thevenin equivalent resistance to be $V_{OC} = I_{SC} R_{TH}$ ; substituting in numbers we get $R_{TH} = 4.3 k\Omega$
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:thevenin_4.png}} |
| The Thevenin equivalent of our voltage divider |
That's it! That's all it takes to convert our voltage divider into one source and one resistor whose output behaves exactly the same as the original. But the circuit we have doesn't look like anything familiar. Maybe if we turned it upside-down:
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:thevenin_5.png}} |
| The same circuit, flipped. |
Now let's add our capacitor back in and see what that does.
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:equivalent_filter.png}} |
| A circuit equivalent to our original |
If this looks familiar, you're right! This is basically a high-pass filter, except the resistor goes to 1.4 V instead of ground. The rest of this exercise is about seeing what difference that makes.
Consider the circuit below.
> For $V_{in}$ = + 5V DC, //predict// the value of $V_{out}$. (//Hint:// What is the impedance of a capacitor when $f = 0$?)
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:biased_filter.png}} |
Assemble the circuit and check your predictions. Resolve any discrepancies.
When the input is DC, the capacitor will effectively act as an open circuit.
As a result, this will just act as a voltage divider between the two resistors.
$V_{out}$ will then be 15V * 1/11 ≈ 1.4V
Now suppose $V_{in}$ is a sine wave with the following properties:
* 1kHz frequency
* 2V peak-to-peak amplitude
* 0V dc offset
> //Predict and sketch// what you will see if you observe $V_{out}$ on the scope. Explain.
(//Hint:// Look at the section on Thévenin equivalence for how to analyze this circuit as a high-pass filter.)
Check your prediction and resolve any discrepancies.
Referring to the equivalent circuit found in the notes section, we'll consider the following:
| {{ :phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:equivalent_filter.png?400 |}} |
This is a high-pass filter (albeit biased around 0.9V instead of ground), and the cutoff frequency is just $F_{3dB} = 1/RC = 112 Hz$. This is far enough from our test frequency of 1kHz that we'll have minimal distortion from the filter network.
Note: This line of reasoning isn't obvious to students at all, so they're likely to need some help here.
Another way to think of this is by using the idea of a **gauge transformation**: as long as all the voltages in a circuit are change by the same amount, the behavior will be the same.
In that instance, we'd instead have our usual filter circuit, but we'd replace the inputs/outputs with $V_{in} - 1.4V$ and $V_{out} - 1.4V$.
We can then do the analysis for the filter circuit, and then add the 1.4V back to everything in the end.
Now suppose $V_{in}$ is a sine wave with the following properties:
* 1kHz frequency
* 2V peak-to-peak amplitude
* **-5V dc offset**
> Predict how, if at all, will $V_{out}$ change.
Check your prediction.
In this case, the behavior should be the same as before.
We've already concluded that a DC component of the signal won't show up on the output, so the same is true in this case.
I don't know if they'll have seen this in class or not, but there is a superposition theorem for circuits that states that you can analyze the effects of voltage sources on a circuit one at a time and then sum the results. Caveats apply, but in general this makes life with mixed signals much easier.
Note: Please retain this circuit; you will use it shortly.
====== Followers and Amplifiers ======
By combining the two circuits you've already built, you can now make an amplifier that works regardless of the bias (DC component) of you incoming signal.
Connect the output of your biasing network to the input of your amplifier circuit, as shown below:
| {{phylabs:lab_courses:phys-226-wiki-home:lab_5_transistors_ii:full_cea.png}} |
Suppose $V_{in}$ is (once again) 1 kHz sinusoidal signal with a peak-to-peak amplitude of 1 V and a dc offset of 0 V.
> //Predict// the peak-to-peak amplitude of $V_{out,3}$ as well as its dc offset. Will $V_{out,3}$ be in phase or 180° out of phase with $V_{in}$? Explain.
Check your predictions and resolve any inconsistencies.
As before, $V_{out,3}$ will change by a factor of 5 compared to the change at $V_{out,1}$.
Here, the amplitude of $V_{out,1}$ will be approximately the same of $V_{in}$, though there may be some decrease since we're a bit close to the filter's cutoff frequency.
The end result is that we'd still expect a 5V pk-pk amplitude signal at $V_{out,3}$, still out of phase with the input.
Now suppose $V_{in}$? is a 1 kHz sinusoidal signal with a peak-to-peak amplitude of 1 V and **//a dc offset of –5 V.//**
How, if at all, will $V_{out,3}$ change? Check your prediction.
Thanks to the biasing network, there won't be any change as any DC part of the input is removed.
You'll see this sort of network frequently on devices that need to accept signals from outside sources. Instead of trusting that whoever generated the signal also made the exact biasing voltages you need, you just strip off whatever is there and replace it.
For instance, audio signals are usually biased around 0V, which is good for speakers, but not so good for a transistor amplifier. However, the large DC offset for a transistor amplifier isn't great to have for speakers. Thus, you can end up with a circuit that creates a new bias point, does something to a signal, and then resets the bias to 0V at the output.
[[https://canvas.uchicago.edu/courses/64033/assignments/765095| Use this link to submit your report]]
====== Lab Recap ======
In this lab, we walked you though:
* The conditions required for a common emitter amplifier circuit to function
* The base has to always be 0.6V higher than the emitter
* The base has to always be slightly lower voltage than the collector
* The behavior of the common emitter amplifier circuit
* As measured from the collector, the gain of the circuit is $g = \dfrac{R_C}{R_E}$
* The output has a constant, DC bias on it
* The output has a 180 degree phase shift from the input
* How to bias a common emitter amplifier circuit
* A capacitor and resistor network can be used to add a dc offset to a signal
* The network is equivalent to a biased high pass filter
This sort of signal processing is fairly common in experimental physics. For instance, the detector of a Geiger counter produces a signal that's on top of several hundred volts DC. To actually interface this with other electronics, you usually need to strip this off with a high-pass filter to avoid irreparable damage to the semiconductors they're made of.
In a similar vein, it is incredibly common to amplify signals to be able to better measure them. For instance, a hall effect sensor might have an output of a few mV per mT. If you want to measure small fields, you'll want to amplify the signal enough to be able to measure it with a scope or other measurement device.
Portions of this page are adapted from "Flexible Resources for Analog Electronics" by Stetzer and Van De Bogart