Intro: Common Emitter Amplifier
A Common Emitter Amplifier is a remarkably simple, but also very useful, type of transistor circuit. The core purpose of this type of circuit is the ability to amplify a voltage applied to the base of a capacitor. This amplification has many useful applications, particularly in acoustics, wherein a relatively weak AC electrical signal can be amplified before being converted to audio, so that the resulting sound is easier to hear.
This instructable will show you the basics of setting up this circuit, so let’s get started.
1. DC Voltage source. I used one that output a constant 15 Volts.
2. An AC function generator capable of producing a sine wave of 0.5 V amplitude. I used a BK Precision 4011A 5mHz model.
3. A capacitor to filter out high-frequency outputs (including the DC part of the function generator’s output). I used one with a Farad value of 100 nF.
4. 4 Resistors. I used ones of 5.6 kΩ, 56 kΩ, 6.8 kΩ, and 680 Ω.
5. A transistor. I used a Q1 2N3904 model.
6. Wires to attach everything.
7. Breadboards to hold the wires and other items in place without the need to solder anything.
8. An oscilloscope capable of reading and displaying your circuits output voltage sine wave.
Step 1: Voltage Divider
The first step in building this circuit is simply creating a voltage divider to supply a set voltage to the base of your transistor. The steps for this are simple enough. Simply wire your 15 V DC voltage source in series with two resistors. Wiring the negative end of your voltage source and the end of your second resistor to ground completes the circuit. The above picture also shows a small wire coming off from between the two resistors that will attach to the base of your transistor in the next step.
This voltage divider applies a voltage of
[ 15V * (5.6 kΩ / 5.6 kΩ + 56 kΩ) = 1.36 V]
to the base of the transistor.
Step 2: Adding and Powering the Transistor
The above picture is pretty self-explanatory, but at this point, we want to attach the wire coming off from between our two resistors from earlier to the base of our transistor. We then attach a grounded 680 Ω resistor to the collector of the transistor. We note here that the voltage at the base undergoes a drop of 0.6 V before becoming the voltage at the collector, so our 1.36 V becomes 0.76 V. This 0.76 V then drops across our 680 Ω resistor, creating a current of about 1.1 mA.
It is important to note here that because we’re using a transistor, it can be assumed that ALL the current coming off the emitter (1.1 mA) is produced by the collector. Thus, if we wire a 6.8 kΩ resistor between the 15 V power supply and the collector, we will see ~1.1 mA of current running through that resistor. This makes from a voltage drop of 7.6 V across that resistor, which subtracted form our original 15 V is 7.4 V. This value of DC voltage portion of output voltage allows for the maximum room for the AC portion to fit into.
Please note also that our choice of resistors here is what creates the value of “Gain” that our AC voltage will undergo. Simply put, the Gain will be equal to the value of the top resistor over that of the bottom resistor. In this case, Gain = 6.8 kΩ / 680 Ω = 10.
Step 3: Adding Our AC Voltage and Reading the Output:
At this point we will finally attach the AC voltage source which we want to amplify. To do this, simply wire your (grounded) AC source (I used a function generator giving a sine wave of amplitude 0.5 V) and your capacitor to a point in between the two resistors which make up the voltage divider portion of this circuit. Then, wire a probe from your oscilloscope to the collector of your transistor. This probe will measure the output voltage and will plot it on your oscilloscope. If you know your oscilloscopes well enough, you should also be able to wire a second channel just after your capacitor, thus allowing you to plot the input AC voltage alongside your output AC voltage. Doing this, and with a voltage gain of 10, you should see a reading on your oscilloscope that looks somewhat like in this slide's second picture.
As can be seen, the amplitude of the output voltage is 10X higher than the input, although the output also seems to have negative amplitudes wherever the input has positive, and vice-versa. Fortunately, this happens by design with this circuit, so it is not a sign of something wrong! For things that might go wrong, however, we turn to our final step: troubleshooting!
Step 4: Troubleshooting!
A few problems might occur in the making of this circuit. Most of them have to do with the wiring itself, so make sure that all of your connections have been made properly and all your resistors are in the right place! If the circuit you’ve built does not match the diagrams given, it will not work right.
Another problem might occur if the spikes of your AC output voltage exceed 15 V or drop below 0 V, as this particular circuit is not designed to handle spikes of this magnitude. Generally, you should be keeping the amplitude of your AC output below 7.5 V. With a AC input of 0.5 V amplitude and gain of 10, however, this shouldn’t be a problem.
Another problem that might occur is clipping, which occurs when the voltage bing applied to the base is higher than the output voltage. Due to the nature of an amplifier making the output voltage much higher than the base voltage however, this shouldn’t occur except at very low values of gain, and certainly not at a gain of 10. So unless you’re attempting to achieve a gain significantly less than 10, you shouldn’t see any clipping.
And that should do it! Enjoy playing with your new common emitter amplifier, and thanks for reading!