How a transistor works as amplifier — common emitter configuration

Question

Explain how an NPN transistor amplifies a signal in the common emitter (CE) configuration. What are the current gain ( $\beta$ ), voltage gain, and phase relationship between input and output?

(CBSE 12 boards ask the CE amplifier circuit and gain formula; JEE Main tests the phase relationship)

Solution — Step by Step

In the common emitter configuration, the emitter is common to both input and output circuits. The input signal is applied between base and emitter, and the amplified output is taken between collector and emitter.

The base-emitter junction is forward biased (low resistance), and the collector-base junction is reverse biased (high resistance).

A small change in base current ( $\Delta I_B$ ) causes a large change in collector current ( $\Delta I_C$ ). This is because the thin, lightly-doped base allows most of the emitter current to pass through to the collector.

The current gain: $\beta = \frac{\Delta I_C}{\Delta I_B}$ (typically 50 to 300)

Since $I_E = I_B + I_C$ and $I_C \gg I_B$ , almost all emitter current reaches the collector.

The small input signal at the base ( $\Delta V_{BE}$ ) produces a large current change at the collector, which flows through a high load resistance $R_L$ , creating a large voltage change.

A_V = \frac{\Delta V_{out}}{\Delta V_{in}} = -\beta \times \frac{R_L}{R_{in}}

The negative sign indicates a 180° phase inversion — when input voltage increases, output voltage decreases.

The power gain combines both current and voltage amplification:

A_P = \beta \times A_V = \beta^2 \times \frac{R_L}{R_{in}}

This is why the CE configuration is the most widely used — it gives the best combination of current gain, voltage gain, and power gain.

flowchart LR
    A["Small AC input<br/>at Base-Emitter"] --> B["Small ΔI_B<br/>(microamps)"]
    B --> C["Transistor action<br/>β = I_C/I_B"]
    C --> D["Large ΔI_C<br/>(milliamps)"]
    D --> E["ΔI_C × R_L =<br/>Large ΔV_out"]
    E --> F["Amplified output<br/>180° phase inverted"]

Why This Works

The transistor is not “creating” energy — it is using the small base current as a control signal to modulate a much larger current from the collector power supply. The energy for amplification comes from the DC power supply ( $V_{CC}$ ), not from the input signal.

Think of it like a water tap — a small force on the handle (base signal) controls a large flow of water (collector current) from the tank (power supply).

Alternative Method

For numerical problems, remember the key relations: $I_E = I_B + I_C$ , $\beta = I_C/I_B$ , and $\alpha = I_C/I_E$ . The relationship between them: $\beta = \frac{\alpha}{1-\alpha}$ . If $\alpha = 0.98$ , then $\beta = 49$ . These conversions appear frequently in CBSE boards and JEE.

Common Mistake

Students often say the CE amplifier “amplifies without any change” to the signal. The 180° phase inversion is a critical feature — the output waveform is an inverted copy of the input. If a question asks about the phase relationship and you write “same phase,” you lose marks. In a multi-stage amplifier, two CE stages give 360° = 0° net phase shift (back to the original phase).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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