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Transistor switch

Curricula: Sek I/II, A-level (Electronics), electronics courses

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Aim

See how a small base current controls a large collector current, and measure the current gain of the transistor.

The bench

A 6 V source. Base circuit: a switch, a 4.7 kΩ resistor and an ammeter — to the base of an NPN transistor (\(\beta\) = 100). Collector circuit: a 6 V / 0.5 W lamp and an ammeter — to the collector. The emitter is connected to the negative terminal of the source. The switch is closed.

Procedure

  1. Record the readings of both ammeters and find the ratio of the collector current to the base current.
  2. Open the switch, then close it again. Watch the lamp.
  3. In the settings of the base resistor, replace 4.7 kΩ with 47 kΩ. Record both currents again and find their ratio.
  4. In the transistor's settings, change the current gain \(\beta\) from 100 to 200.
Expected result

The base current is about 1.1 mA, the collector current about 82 mA — the lamp burns almost at full brightness. The current ratio is about 73, less than \(\beta\): the transistor is in saturation, and the collector current is now limited by the lamp, not by the base. With no base current the transistor is off, and the lamp goes out. With the 47 kΩ resistor the base current is about 0.11 mA, the collector current about 11 mA, and the lamp does not glow although current flows: the filament is heated only to 960 K; the current ratio equals \(\beta\) = 100. With \(\beta\) = 200 the collector current doubles, to 23 mA, while the base current hardly changes. In a real transistor \(\beta\) is not something you choose: it differs between transistors of the same type, from 100 to 300, and the current ratio with the 47 kΩ resistor will show the \(\beta\) of your particular transistor.

Questions

  1. How many times smaller is the current that switches the lamp on than the lamp's own current?
  2. Why is the current ratio less than \(\beta\) with the 4.7 kΩ resistor, but equal to it with 47 kΩ?
  3. What is the largest base resistor that still puts the transistor into saturation with this lamp?
Answers
  1. About 73 times: a base current of about 1.1 mA switches on a lamp current of about 82 mA.
  2. With 4.7 kΩ the base current of 1.1 mA would allow \(\beta \cdot I_B\) ≈ 110 mA, more than the lamp passes at 6 V, about 82 mA. The transistor is saturated: the voltage between collector and emitter is small, the lamp and the source set the current, and extra base current does not raise it. With 47 kΩ, \(\beta \cdot I_B\) ≈ 11 mA is far below what the lamp would pass, the transistor works in the active mode, and the collector current equals \(\beta\) times the base current.
  3. Saturation needs \(\beta \cdot I_B\) of at least the lamp current, so \(I_B\) ≥ 82 mA / 100 = 0.82 mA. The base resistor has about 1.1 mA · 4.7 kΩ ≈ 5.2 V across it, so \(R_B\) ≤ 5.2 V / 0.82 mA ≈ 6.3 kΩ, about 6 kΩ. In practice a resistor several times smaller is used, to keep a margin for the spread of \(\beta\).