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Common-emitter amplifier

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

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Aim

Build a transistor amplifier stage, find its operating point and its voltage gain.

The bench

A 9 V source powers an NPN transistor (\(\beta\) = 100): the collector through a 2.2 kΩ resistor, the base through a 470 kΩ bias resistor, the emitter through a 470 Ω resistor to the common wire. An AC voltage source (sine wave, 0.2 V amplitude, 1 kHz) feeds the signal to the base through a 10 µF capacitor. Oscilloscope: CH1 — the output of the AC source (100 mV/div), CH2 — the collector (500 mV/div, AC coupling), 0.5 ms/div.

Procedure

  1. After the bench opens, wait a couple of seconds: the capacitor charges and the operating point settles. Compare the two traces: the peak-to-peak value and the phase of the signal at the input and at the collector.
  2. Find the voltage gain: the ratio of the peak-to-peak value at the collector to the peak-to-peak value at the input.
  3. On CH2 select DC coupling and 2 V/div, and estimate the DC voltage at the collector.
  4. Increase the amplitude of the AC source to 1 V and switch CH1 to 0.5 V/div.
  5. In the emitter resistor's settings, change 470 Ω to 220 Ω, and set the amplitude of the AC source back to 0.2 V.
Expected result

The signal at the collector is larger than the input and in antiphase with it: about 1.8 V peak to peak against 0.4 V, a gain of about 4.5 — close to the ratio of the collector and emitter resistances, 2.2 kΩ / 470 Ω. The DC voltage at the collector is about 5.4 V. By calculation the collector current is about 1.6 mA and the emitter is at about 0.77 V — the bench does not show these values. At an amplitude of 1 V the output sine wave is clipped: at the top the transistor cuts off, at the bottom, near the emitter voltage, the transistor goes into saturation. With a 220 Ω emitter resistor the gain roughly doubles, and the operating point shifts. A single bias resistor sets the base current, and the collector current is \(\beta\) times larger, so the operating point depends on \(\beta\). Real transistors of the same type have different \(\beta\), from 100 to 300. Set Current gain β = 300 in the transistor's settings: the bias now asks for a collector current of about 4 mA, more than the 2.2 kΩ and 470 Ω resistors can pass, and the transistor saturates at about 3.3 mA. Without a signal the collector sits at about 1.7 V, only 0.2 V above the emitter; with the 0.2 V signal the lower half-waves are already clipped, and the collector voltage averages about 2.7 V. That is why real amplifiers bias the base with a divider made of two resistors.

Questions

  1. Why is the output signal in antiphase with the input?
  2. Why is a capacitor needed between the AC source and the base?
  3. Why can't the gain be doubled simply by using a collector resistor twice as large, 4.4 kΩ? Where would the operating point end up?
Answers
  1. When the input voltage rises, the base current and the collector current grow, and so does the voltage drop across the 2.2 kΩ collector resistor. The collector voltage is 9 V minus this drop, so it falls when the input rises.
  2. The capacitor passes the 1 kHz signal (its reactance is only about 16 Ω) and blocks direct current. The other terminal of the AC source is on the common wire, so without the capacitor the source would hold the base near 0 V through its low resistance: the bias from the 470 kΩ resistor would be lost, and the transistor would stay off, because a 0.2 V signal is too small to open it.
  3. The collector current is set by the base current and stays about 1.6 mA, so a 4.4 kΩ resistor would drop about 1.6 mA · 4.4 kΩ ≈ 7.0 V, and the collector would sit at about 2 V, only about 1.2 V above the emitter. The operating point would move close to saturation. The output, now about 3.6 V peak to peak, would not fit into the room left below it, and its lower half-waves would be clipped already at the 0.2 V input.