☰ Contents · Physics

Semiconductor devices

Lessons 44–45 · 2 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 10, 1st edition. Niso Poligraf Publishing House, Tashkent, 2017
44

Semiconductor devices (diode, transistor) and their use in technology

Textbook: pp. 161–165
GoalExplain the one-way conductivity of a p–n junction, the operation of a semiconductor diode and transistor, and their uses in technology.
New words
p–n junction · p–n o‘tishbarrier (depletion) layer · berkituvchi qatlamtransistor · tranzistorrectifier · to‘g‘rilagich
Explanation

If part of one crystal is n-type and the other p-type, at the boundary electrons fill holes and a thin barrier (depletion) layer almost free of carriers forms. With forward bias (the p-region to the positive terminal, the n-region to the negative) the layer narrows, carriers cross it and the current is large; with reverse bias (p negative, n positive) the layer widens and the current is very small. So a device with one p–n junction – a semiconductor diode – conducts mainly in one direction (in a silicon diode the current rises sharply above a forward voltage of about 0.6–0.7 V). A diode is used to rectify alternating current: a half-wave rectifier conducts during the positive half-cycle. A device with two p–n junctions is a transistor (p–n–p or n–p–n): its terminals are the emitter, base and collector. The emitter–base junction is forward-biased and the base–collector junction reverse-biased; because the base is very thin most carriers from the emitter reach the collector. A small change of base current causes a large change of collector current – the current and signal are amplified (β = Ic/Ib). Transistors and diodes are packed in integrated circuits (microchips) in thousands to billions (computers, phones).

Worked examples
In a transistor Ib = 0.05 mA and Ic = 2 mA: the gain is β = Ic/Ib = 2/0.05 = 40. The emitter current is Ie = Ib + Ic = 2.05 mA. If Ib rises to 0.1 mA, Ic ≈ 4 mA.
A silicon diode is connected forward to a 6 V source with a series resistor R; 0.6 V drops across the diode and the current should be 10 mA: R = (6 – 0.6)/0.01 = 540 Ω. Connected in reverse, the current is almost zero.
Class activity

Only by calculation and drawing: sketch the forward and reverse connection of a p–n junction, marking the barrier width and the current; label the three terminals of a transistor on a diagram. Connect real circuits only with the teacher and at low voltage.

Practice
1
Why does a diode conduct in one direction?
2
A transistor has Ib = 0.1 mA, Ic = 3 mA. What is β?
3
Silicon diode, source 5 V, 0.6 V on the diode, current 20 mA. What series R (Ω)?
4
Which junction of a transistor is forward-biased and which reverse-biased?