This lesson is beyond the Class 12 syllabus, which no longer includes the junction transistor. It is here because the logic gates of the next two lessons are built from transistor switches. Only two states matter, fully off and fully on; amplifiers are not treated. Read it for understanding; the other lessons of this chapter are the syllabus lessons.
Two junctions back to back
A junction transistor is a single crystal with three regions and two p-n junctions. In an n-p-n transistor a thin p-type layer lies between two n-type regions. The three regions have three names and three jobs. The emitter is heavily doped and supplies the charge carriers. The base in the middle is very thin, about a micrometre, and lightly doped. The collector is larger and gathers the carriers.
A p-n-p transistor is the mirror image, with an n-type base between two p-type regions; its carriers are holes. In the symbol, the arrow on the emitter shows the direction of conventional current. It points out of an n-p-n transistor and into a p-n-p one. This lesson uses only the n-p-n.
A junction transistor is a three-layer semiconductor device, n-p-n or p-n-p, with an emitter, a thin base and a collector. A small current into the base controls a much larger current between the collector and the emitter.
Biased for action
For the transistor to work, its two junctions are biased in opposite ways. The emitter-base junction is forward-biased, so electrons pour from the emitter into the base. The collector-base junction is reverse-biased, with the collector a few volts positive. The question is what happens to the electrons once they are inside the base.
Electrons from the emitter enter the p-type base, which is full of holes. What happens to most of them?
In the figure, nineteen electrons in twenty reach the collector and one leaves by the base lead. In a real transistor the share is even more uneven: about ninety-nine in a hundred reach the collector. Two features of the base make this happen. It is very thin, so the electrons cross it quickly. It is lightly doped, so there are few holes to meet on the way.
Three currents
The emitter current \(I_E\) is the whole stream of electrons. The few that recombine in the base make the small base current \(I_B\), and the rest make the collector current \(I_C\). By the current law at a junction, \(I_E = I_B + I_C\). The ratio of the collector current to the base current is the current gain, \(\beta\).
A transistor has \(\beta = 100\). A base current of \(20\,\mu\text{A}\) flows. Find the collector current and the emitter current.
\(I_C = \beta I_B = 100 \times 20\,\mu\text{A} = 2000\,\mu\text{A} = 2\,\text{mA}\).
\(I_E = I_B + I_C = 0.02 + 2 = 2.02\,\text{mA}\).
The base current is less than \(1\,\%\) of the emitter current, yet it sets the other two.
Values used: \(\beta = 100\).
In a transistor the emitter current is \(5\,\text{mA}\) and the collector current is \(4.95\,\text{mA}\). What is its current gain \(\beta\)?
A transistor with \(\beta = 150\) has a base current of \(40\,\mu\text{A}\). What is its emitter current, in mA?
Off: cut-off
Join a lamp between a supply and the collector, and the emitter to the return wire. Now drive the base from an input voltage through a resistor. If the input is below about \(0.7\,\text{V}\), the emitter junction is not forward-biased enough to conduct. No base current flows, and so no collector current either. The transistor is in cut-off: an open switch, and the lamp is dark.
On: saturation
Raise the input, and \(I_C = \beta I_B\) grows with it. But the collector current cannot grow for ever. The lamp and the supply allow at most \(V_{CC}/R_C\), where \(V_{CC}\) is the supply and \(R_C\) the resistance in the collector circuit. Once \(\beta I_B\) reaches this value, more base current adds nothing. The transistor is in saturation: a closed switch, with only about \(0.2\,\text{V}\) left across it.
The lamp and the supply allow at most \(6/60 = 100\,\text{mA}\). With \(\beta = 100\), that needs a base current of \(1\,\text{mA}\). The base resistor passes \(1\,\text{mA}\) when \(V_{in} - 0.7 = 2\,\text{V}\), at \(V_{in} = 2.7\,\text{V}\). So the lamp is dark below \(0.7\,\text{V}\) and fully on from \(2.7\,\text{V}\). Between these the lamp is only partly on.
A switching circuit is designed so that its input is always either well below \(0.7\,\text{V}\) or well above the saturation value. Then the transistor never stays between the two. It is either fully off or fully on, and wastes almost no power in either state. In cut-off no current flows through it. In saturation almost no voltage is left across it.
A transistor with \(\beta = 100\) has a \(10\,\text{k}\Omega\) base resistor and a \(1\,\text{k}\Omega\) collector resistor joined to a \(5\,\text{V}\) supply. The output is taken at the collector. Find the output for an input of \(5\,\text{V}\) and for an input of \(0\,\text{V}\).
Input \(5\,\text{V}\): \(I_B = (5 - 0.7)/10\,000 = 0.43\,\text{mA}\), and \(\beta I_B = 43\,\text{mA}\). The collector circuit allows only \(5/1000 = 5\,\text{mA}\), so the transistor saturates. Nearly all of the \(5\,\text{V}\) is across the resistor, and the output is close to \(0\,\text{V}\).
Input \(0\,\text{V}\): no base current, so the transistor is cut off. No current flows in the collector resistor, so there is no voltage across it, and the output sits at \(5\,\text{V}\).
A high input gives a low output, and a low input a high one. This is the NOT gate of the next lesson.
A transistor switch has a \(9\,\text{V}\) supply and a \(450\,\Omega\) load in its collector, and \(\beta = 100\). What is the least base current, in mA, that saturates it?
A transistor switch is already saturated. The base current is doubled. What happens to the collector current?
\(I_C = \beta I_B\) is not true in every state. It fails in cut-off, where both currents are zero, and in saturation, where \(I_C = V_{CC}/R_C\) is less than \(\beta I_B\). Check which state the transistor is in before using it.
In December 1947, at Bell Laboratories in New Jersey, John Bardeen and Walter Brattain pressed two gold contacts onto a crystal of germanium. A small current at one contact controlled a larger current at the other. A few weeks later William Shockley worked out the sandwich design of this lesson, the junction transistor, which was easier to make and to understand. The three shared the Nobel Prize in Physics for 1956.
The processor of a modern phone holds more than ten billion transistors, each far smaller than a bacterium. Each one is used as a switch, flipping between off and on billions of times a second. Every photo filter, every message and every calculation is done by these switches, joined into the logic gates of the next two lessons.
- Thinking the electrons from the emitter mostly leave by the base.The base is thin and lightly doped; about 99 in 100 electrons cross into the collector.
- Using \(I_C = \beta I_B\) in saturation.In saturation \(I_C = V_{CC}/R_C\). Extra base current changes nothing.
- Forgetting the base current in \(I_E\).\(I_E = I_B + I_C\); the emitter carries both.