# Introduction to Bipolar Junction Transistors

concept
The bipolar junction transistor (BJT) is one of the two basic semiconductor devices that will form the basis of this course (the other being the MOSFET, covered soon). The BJT (and MOSFET) can be used to amplify analog signals or to implement digital functions like AND, OR, NOT and so on. In this course we'll cover their use as analog amplifiers. BJTs are almost guaranteed to be found in any electronic device you have to perform some kind of amplification.
fact
The BJT is a 3-terminal device, compared to every other device we've seen up until this point which has only two terminals. The three terminals in the BJT are called the Collector, Base and Emitter as shown in the diagram. You'll notice a small black arrow drawn on the emitter terminal of the device, this is how you know which terminal is the emitter and which is the collector. The base terminal is always the one pointing into the perpendicular line.
fact
The arrow on the emitter can be either pointing out towards the terminal or in towards the "body" of the device. When pointed out the BJT is known as an NPN BJT, when pointed into the body it is a PNP BJT.
The terms NPN and PNP come from the types of semiconductor materials used in the BJT's construction. An NPN device has a base made from P type semiconductor and the collector and emitter made of N type semiconductor. A PNP BJT has an N type base and P type collector and emitter. The arrow points in the direction of conventional current flow. So an NPN BJT will have current flow out of the emitter while the PNP has current flow into the emitter. A BJT has 3 different "modes" of operation determined by the relative voltages of the collector, base and emitter terminals. We'll discuss the modes for NPN BJTs, for PNPs the only difference is the direction of current and the polarity of voltages (i.e. if we claim that $$V_{BE} \gt 0.7$$V just know that for a PNP transistor we need $$V_{BE} \lt 0.7$$V).
fact
The 3 modes of operation for a BJT are: Saturation Cutoff Forward Active

## Saturation

In saturation the transistor is operating like a switch that is closed. This is the mode used in logic circuits to represent "on". In saturation the current is constant and independent of the voltages on the transistor's nodes.
fact
For saturation to occur we need: $$V_B \gt V_C,\quad V_B \gt V_E$$ And $$V_{BE} \gt 600$$mV or so in order to "turn on" the transistor. We also have a "saturation voltage" between the collector and the emitter $$V_{CE(sat)} \approx 0.2$$V

## Cutoff

fact
In cutoff the BJT acts like an open circuit with zero currents coming into or out of any of the terminals. For cutoff: $$V_E \gt V_B,\quad V_C \gt V_B$$
Cutoff and saturation are the modes used when we want our BJT to act like a switch. This is used to build digital circuits or, when we switch it very quickly, to build circuits that can act like transformers for DC electronics. This is very important in power supplies.

## Forward Active Mode

fact
The main mode we care about is called the "Forward Active" mode, this is where the BJT works as an analog amplifier and is what we'll try to accomplish when we design circuits. For forward active mode: $$V_C \gt V_B \gt V_E$$
In practice we typically want $$V_{BE} = \approx 600$$mV $$- 800$$mV in order to "turn on" the transistor and get an adequate current flow through the base. In forward active the BJT can be thought of as a current amplifier. The current that runs from the base to the emitter is amplified and pushed through the collector and emitter. So if there is $$1\mu$$A entering the base in an NPN BJT and the current amplification of that transistor is $$100$$ we'll see $$100\mu$$A entering the collector and $$1\mu$$A entering the base and all $$101\mu$$A leaving the emitter.
fact
In equation form we say $$I_C = \beta I_B$$ where $$\beta$$ is the current amplification of the BJT and varies from device to device but is typically $$\approx 100 - 250$$
Kirchhoff's current law then shows us that $$I_E = I_C + I_B = (\beta + 1)I_B$$ In a PNP device the method is the same but $$1\mu$$A would be leaving the base making $$100\mu$$A leave the collector meaning $$101\mu$$A will have entered the emitter.
fact
We can also view the transistor as a voltage controlled current source with $$I_C = I_S \exp(\frac{V_{BE}}{V_T})$$. For the rest of this course we will assume that $$V_T = 26$$mV Where $$I_S$$ is a property of the individual transistor and is a number on the order of $$10^{-16}$$.
example

Determine the output voltage of the following circuit if $$I_S = 5\times 10^{-16}$$

From $$V_{BE} = 0.8$$ and the given $$I_S$$ we can calculate $$I_C = 11.5$$mA So $$V_o = 3$$V$$- I_C R_L = 3 - 1.15 = 1.85$$V
practice problems