In today’s digital world, everything is about ones and zeros, on and off, true and false. Digital electronics dominate the field, and as such, most electronics projects involve switches, LEDs, microcontrollers, and the like. To spice things up, I decided to mess around with some analog electronics, and the simplest project I could think of was to make a basic amplifier with a bipolar junction transistor (BJT).

 

The 2N3904 is a common general-purpose NPN-type BJT.

The plan was to use a function generator to output a small signal and then measure a (hopefully) amplified version of that wave on the oscilloscope. Simple enough, right? Here’s the bipolar junction transistor amplifier circuit that I put together.

 

 

Attempt #1 - Bipolar Junction Transistor Amplifier Schematic:

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>>> My thinking here was that my function generator inputs a small signal with low current to the base, and the transistor would amplify it. I vaguely remembered that the gain of the amplifier would be equal to the value of the collector resistor divided by the emitter resistor, so the gain should be about 45. Let’s plug the input and output into the oscilloscope channels and see what we get.

Observation & learning from 1st attempt of BJT amplifier output:

Yikes! Not exactly what I was expecting. Now, this is slightly embarrassing considering I took an entire university course on transistors, op-amps, diodes, and the like, but hey, it’s been a few years, and every student knows that realization that you forgot everything after the exam. The last time I used a BJT in a project, I was essentially using it as a switch, so that’s sort of where my head was at with this silly configuration. I was quickly reminded that using a transistor as an amplifier is a little more involved. So, I looked at some articles online for a quick refresher and sure enough, all the knowledge started flowing back.

 

There are two main problems with my BJT amplifier circuit:

1) I am trying to amplify the signal to an amplitude that is larger than my supply voltage, which is not possible. My theoretical gain of 45 would amplify my 1 V input to a 45 V output, so my wave is being clipped at the 5 V supply voltage. This is why my output wave looks so strange.

2) I haven’t ensured that the BJT is in the active region, which is essential when making an amplifier. (For the BJT to operate in the active region, the condition is that the base-emitter junction should be forward-biased while the base-collector junction is reverse-biased.)

 

To review, BJTs have three main operating regions: active, saturation, and cutoff. Bipolar Junction Transistor (BJT) Active Mode: The active region is characterized by two conditions being satisfied: the collector voltage being greater than the base voltage, and the base voltage being about 0.7 V greater than the emitter voltage. We will need to add some extra resistors with values that will produce the desired operating point. This process is called biasing. There are many ways to do this, but I find H-biasing to be straightforward. This will make our circuit look like this:

 

BJT Biasing - Transistor Biasing Calculations:

The two resistors we added create a voltage divider, which allows us to set the base voltage by varying the ratio between the resistor values.

 

Let’s aim for a more modest gain of 10. We’ll choose a 10 kΩ resistor for the collector and 1 kΩ for the emitter. Let’s also use a 20 V power source so that we can avoid the clipping issue. The first thing we need to do is figure out the collector and emitter currents, but we can approximate these as equal because the emitter current is the sum of the collector and base currents, and the base current is very small.

 

Now, a little bit of math. Optimal amplification happens when the voltage between the collector and emitter is half of our supply voltage, so we want it to be around 10 V. Since our supply is 20 V, this means that the remaining 10 V is consumed by the collector and emitter resistors. Using Ohm’s Law, 10 V divided by the combined resistance of 11 kΩ gives us a collector (and emitter) current of 0.9 mA. We can use this to determine that the collector voltage is 20 V – (0.9 mA) (10 kΩ) = 11 V. Similarly, the emitter voltage is 0 + (0.9 mA) (1 kΩ) = 0.9 V. And since our base voltage needs to be 0.7 V higher than that, 0.9 V + 0.7 V = 1.6 V.

 

Now we can set the resistor values in our voltage divider to produce a base voltage of 1.6 V, which is about one-tenth of our supply voltage, so we need our top resistor to be around 10 times larger than the bottom resistor. We’ll choose 100 kΩ and 10 kΩ.

 

One more thing that I forgot in the first go around: the AC input and output should be isolated from the DC sources used for biasing! To do this, we need to add capacitors at the input and output. The idea here is that small capacitors charge up quickly and then act like an open circuit for DC current but act like a short circuit for AC current, so only the AC can cross the capacitors.

 

 

Attempt #2 - Bipolar Junction Transistor Amplifier on Breadboard: Setting it up on the board looks like this.

 

 

Now, when I measure both the input and output on the two channels of the oscilloscope:

We amplified our wave! A few more steps than I thought it would take, but not too difficult once I knew what I was missing. This type of amplifier is called a common-emitter amplifier (CE amplifier), and one of its characteristics is that in addition to being amplified, our wave was also inverted. You can see that the high points of the input wave are the low points of the output wave.