The transistor and its characteristics

A small turn of a big tap

A small turn of a tap can change a large flow of water. A transistor does this for an electric current. A tiny current into one terminal controls a current that is tens or hundreds of times larger in the other two. This one idea lets a transistor amplify a signal and also switch a current on and off.

A note about the syllabus. The latest NCERT text for Class 12 leaves the junction transistor out of its chapter, though the older edition and many state boards and university courses keep it. A-Level and IB courses meet it as a current amplifier and as a switch. The gates of lesson 3 are made of it, so we study it here.

Structure

A junction transistor is a crystal with three regions and two junctions. There are two types: npn and pnp. The three regions have different doping and size.

• The emitter is heavily doped. It supplies the charge carriers.
• The base is very thin (about a micrometre) and lightly doped. It controls the flow.
• The collector is moderately doped and larger in area. It collects the carriers and loses the heat.

The symbol shows the emitter with an arrow. In the npn symbol the arrow points away from the base. In the pnp symbol it points towards the base. The arrow gives the direction of the conventional current in the emitter.

How it works (npn)

In normal use, which we call the active region, the emitter-base junction is forward biased ($V_{BE} \approx 0.7$ V for silicon) and the collector-base junction is reverse biased.

1. The forward bias sends electrons from the emitter into the base.
2. The base is thin and lightly doped, so only a few electrons, about $1\ \%$ to $5\ \%$, meet a hole and recombine. These make the small base current $I_B$.
3. The rest reach the collector junction. The field of the reverse-biased junction sweeps them into the collector. They make the collector current $I_C$.

So $I_C$ is nearly as large as $I_E$, and $I_B$ is small. The three currents obey

The share of the emitter current that reaches the collector is $\alpha$, and the ratio of the collector current to the base current is $\beta$:

For example, $\alpha = 0.98$ gives $\beta = 0.98/0.02 = 49$. Values of $\alpha$ lie between $0.95$ and $0.995$, and $\beta$ between about $20$ and a few hundred. For a small change we use the ac values:

They are close to the dc values, so we write $\beta$ for both unless the question says otherwise.

The common-emitter circuit

In the most used circuit the emitter is common to the input and the output. The base loop has a supply $V_{BB}$ and a resistor $R_B$. The collector loop has a supply $V_{CC}$ and a resistor $R_C$. Kirchhoff's rule gives two equations:

Two graphs describe a transistor in this circuit.

• The input characteristic is $I_B$ against $V_{BE}$ at a fixed $V_{CE}$. It looks like the curve of a forward diode. The input resistance is $r_i = \left(\dfrac{\Delta V_{BE}}{\Delta I_B}\right)_{V_{CE}}$, between a few hundred ohm and a few kilo-ohm.
• The output characteristic is $I_C$ against $V_{CE}$ at a fixed $I_B$, one curve for each $I_B$. The output resistance is $r_o = \left(\dfrac{\Delta V_{CE}}{\Delta I_C}\right)_{I_B}$, which is large, tens of kilo-ohm, because the curves are nearly flat.

The output characteristic has three regions.

• Cut-off: the emitter junction is not forward biased and the collector junction is reverse biased. $I_B \approx 0$ and $I_C \approx 0$.
• Active: the emitter junction is forward biased and the collector junction is reverse biased. $I_C \approx \beta I_B$, and the curve is nearly flat.
• Saturation: both junctions are forward biased. $V_{CE} \approx 0.2$ V and $I_C < \beta I_B$.

A worked example. A silicon transistor has $\beta = 80$. The supply is $10$ V and it is joined to the base through $390\ \text{k}\Omega$ and to the collector through $1.5\ \text{k}\Omega$. Find $I_B$, $I_C$, $I_E$, $\alpha$ and $V_{CE}$.

1. The base current. $I_B = \dfrac{10\ \text{V} - 0.7\ \text{V}}{390\ \text{k}\Omega} = 23.8\ \mu\text{A}$.
2. The collector current. $I_C = \beta I_B = 80 \times 23.8\ \mu\text{A} = 1.91$ mA.
3. The emitter current. $I_E = I_B + I_C = 0.0238\ \text{mA} + 1.908\ \text{mA} = 1.93$ mA.
4. The ratio $\alpha$. $\alpha = \dfrac{I_C}{I_E} = \dfrac{1.908}{1.932} = 0.988$.
5. The collector voltage. $V_{CE} = 10\ \text{V} - 1.908\ \text{mA} \times 1.5\ \text{k}\Omega = 7.14$ V. This is above $0.2$ V, so the transistor is in the active region, as we assumed.

Set the base current to $0$. Where is the dot? Then raise it in steps. At what base current does the dot leave the flat part of the curve?

Practise with new numbers

Try it in Python

The program changes the base resistor of a circuit with $\beta = 100$, $V_{CC} = 12$ V and $R_C = 2.2\ \text{k}\Omega$. For each, it finds the currents, $V_{CE}$ and the region.

Common mistakes

• Saying $I_C = I_E$ exactly. The base current is the difference, $I_E = I_B + I_C$.
• Thinking the base current is large. It is the smallest of the three, a percent or two of $I_E$.
• Forgetting the sign of the junction bias. In the active region the emitter junction is forward and the collector junction is reverse.
• Using $I_C = \beta I_B$ in saturation. There the collector current is smaller than $\beta I_B$.
• Mixing up the arrow of the symbol. The arrow is on the emitter. It points out for npn and in for pnp.
• Using $V_{CE} = V_{CC}$ in the active region. $R_C$ drops $I_C R_C$ of the supply.

Remember these

• Emitter: heavily doped. Base: thin and lightly doped. Collector: larger. $I_E = I_B + I_C$.
• $\alpha = \dfrac{I_C}{I_E}$, $\beta = \dfrac{I_C}{I_B}$, $\beta = \dfrac{\alpha}{1 - \alpha}$, $\alpha = \dfrac{\beta}{1 + \beta}$.
• Active: emitter forward, collector reverse, $I_C \approx \beta I_B$. Saturation: both forward, $V_{CE} \approx 0.2$ V. Cut-off: no base current.
• $r_i = \left(\dfrac{\Delta V_{BE}}{\Delta I_B}\right)_{V_{CE}}$, $r_o = \left(\dfrac{\Delta V_{CE}}{\Delta I_C}\right)_{I_B}$, $\beta_{ac} = \left(\dfrac{\Delta I_C}{\Delta I_B}\right)_{V_{CE}}$.

References

• NCERT Physics Class 12, Semiconductor Electronics, the older edition on the junction transistor (transistor action and characteristics).
• Horowitz and Hill, The Art of Electronics, third edition, Chapter 2, Transistors.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A transistor has $I_B = 50\ \mu$A and $I_C = 4.0$ mA. What are $I_E$ and $\beta$?

2. A transistor has $\alpha = 0.99$. What is $\beta$?

3. Why is the base made thin and lightly doped?

4. In a transistor circuit, $V_{CE}$ is $0.2$ V and $I_B$ is large. In which region is it?