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:

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 many, many commonly used forms. |

In this section, we examine three different output voltages from a transistor circuit that is based on the common-emitter amplifier.

To begin, consider the following circuit.

Predict the values of $V_{out,1}$, $V_{out,2}$, and $V_{out,3}$ would be if $V_{in}$ were 0V.

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

Now suppose $V_{in}$ is a sine wave with the following properties:

- 1kHz frequency
- 1V peak-to-peak amplitude
- 0V dc offset

Predictthe 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.

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.

Keep your circuit built, you'll be using it again soon.

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.

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:

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:

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:

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 $R_{TH} = 4.3 k\Omega$

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:

Now let's add our capacitor back in and see what that does.

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,predictthe value of $V_{out}$. (Hint:What is the impedance of a capacitor when $f = 0$?)

Assemble the circuit and check your predictions. Resolve any discrepancies.

Now suppose $V_{in}$ is a sine wave with the following properties:

- 1kHz frequency
- 2V peak-to-peak amplitude
- 0V dc offset

Predict and sketchwhat 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.

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.

Note: Please retain this circuit; you will use it shortly.

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:

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.

Predictthe 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.

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.

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