Transistor as an amplifier (common emitter)
In a CE amplifier, a small base-current change gives a β times larger collector-current change. Across RL this gives a large output, 180° out of phase with the input.
What an amplifier does
A microphone turns your voice into a tiny voltage, too weak to drive a loudspeaker. An amplifier makes such a signal bigger without changing its shape. The ratio of output to input voltage is the voltage gain AV = Vo/Vi. In the same way, current gain AI = io/ii and power gain AP = Po/Pi. The extra power does not come from nowhere. It is taken from the DC supply, so energy is conserved.
How the CE amplifier works
- 1An npn transistor is used in CE mode. VCC reverse biases the collector through the load RL, and VBB forward biases the base.
- 2Biasing: VBB is set so the base current IB sits in the middle of the linear range, between cut-off and saturation. This is the operating point.
- 3The signal vs is added in series with VBB. It makes the base current swing by Δib = vs/ri about IB.
- 4The collector current swings by Δic = β Δib, which is much larger.
- 5This current flows through RL, so the output voltage swings by Δvo = −Δic × RL: a large voltage.
- 6★When the input rises, IC rises, the drop across RL rises and VCE falls. So output and input are 180° out of phase. Too large a signal reaches cut-off or saturation and distorts the output.
★ key step
Deriving the voltage gain
- 1Input: Δib = vs/ri, so vs = Δib × ri.
- 2Output current: Δic = β Δib, so vs = Δic × ri/β.
- 3Kirchhoff's law for the output loop: VCC = VCE + IC RL.
- 4VCC is constant, so 0 = ΔVCE + ΔIC RL. This gives Δvo = ΔVCE = −Δic RL.
- 5AV = Δvo/vs = −Δic RL ÷ (Δic ri/β) = −β RL/ri.
- 6★So, AV = −β RL/ri = −gm RL, with gm = β/ri. The minus sign means a 180° phase change, and AP = β × AV.
★ key step
How to draw the CE amplifier
- Draw an npn transistor. Join the emitter to the common (earth) line.
- Base side: the signal source vs in series with the bias battery VBB, joined to the base. Mark it as the input.
- Collector side: the load RL from the collector to the + of VCC, and the − of VCC to the common line.
- Take the output vo between the collector and the common line.
- Beside it, sketch the input as a small sine wave and the output as a larger, upside-down sine wave.
Worked example: gains of a CE amplifier
A transistor with β = 60 and input resistance ri = 1.5 kΩ is used in CE mode with RL = 3 kΩ. Find (a) gm, (b) the voltage gain, (c) the power gain, and (d) the output for a 10 mV input.
- (a) gm = β/ri = 60 ÷ 1500 Ω = 0.04 S, that is 40 mA per volt.
- (b) AV = −β RL/ri = −60 × 3000 Ω ÷ 1500 Ω = −120. Its size is 120, with a 180° phase change.
- (c) AP = β × AV = 60 × 120 = 7200.
- (d) vo = AV × vi = 120 × 10 mV = 1200 mV = 1.2 V, upside down compared with the input.
(a) 0.04 S; (b) −120; (c) 7200; (d) 1.2 V, out of phase with the input
The loudspeaker system at a wedding or a temple festival uses transistor amplifiers. They turn a microphone's tiny voltage into one strong enough to drive big speakers, using power from the mains.
5 marks: “Draw the circuit of an npn transistor amplifier in CE mode. Explain how it amplifies, and show that its voltage gain is −β RL/ri.” Circuit with VBB and vs at the base and RL to VCC at the collector. Bias at the middle of the linear range; Δib = vs/ri; Δic = βΔib; VCC = VCE + IC RL gives Δvo = −Δic RL; so AV = −β RL/ri, with a 180° phase change. Power comes from the DC supply.
2 marks: “In a CE amplifier, the input is 20 mV and the output is 2 V. Find the voltage gain. If the current gain is 50, find the power gain.” AV = 2 V ÷ 0.02 V = 100. AP = β × AV = 50 × 100 = 5000.
2 marks: “Why are the input and output voltages of a CE amplifier 180° out of phase?” A rise in input raises IB and IC. The drop IC RL rises, so VCE = VCC − IC RL falls. The output falls as the input rises.
Base wiggles, collector wiggles β times, output flips: AV = −β RL/ri.