Transistor as switch and amplifier — common emitter configuration

Question

Explain how a transistor works as (a) a switch and (b) an amplifier in the common emitter (CE) configuration. Draw the circuit diagram and define current gain $\beta$ .

(NCERT Class 12 — Semiconductor Electronics)

Solution — Step by Step

In common emitter configuration, the emitter is common to both input and output circuits. The base-emitter junction is forward biased, and the base-collector junction is reverse biased.

Three currents: $I_E = I_B + I_C$ (emitter = base + collector).

Current gain: $\beta = \frac{I_C}{I_B}$ (typically 50-300 for common transistors).

A small base current $I_B$ controls a much larger collector current $I_C$ .

In the active region, a small change in base current produces a large change in collector current:

\Delta I_C = \beta \times \Delta I_B

Voltage gain: $A_V = \frac{\Delta V_{out}}{\Delta V_{in}} = -\beta \times \frac{R_C}{R_B}$

The negative sign means the output is 180° out of phase with the input (CE configuration inverts the signal). The magnitude of voltage gain can be much greater than 1 — that is amplification.

For amplification, the transistor must operate in the active region (base-emitter forward biased, base-collector reverse biased).

A transistor switches between two states:

OFF state (cutoff): $I_B = 0$ (or very small). The transistor is like an open switch — no current flows from collector to emitter. $V_{CE} \approx V_{CC}$ (full supply voltage across transistor).

ON state (saturation): $I_B$ is large enough that $I_C$ is at its maximum (limited by $R_C$ and $V_{CC}$ ). The transistor is like a closed switch — $V_{CE} \approx 0$ . Current flows freely.

No intermediate state is used. The transistor is either fully ON or fully OFF — digital behaviour.

Why This Works

A transistor amplifies because of the asymmetry in doping and geometry. The base is very thin and lightly doped. Most charge carriers injected from the emitter pass straight through to the collector instead of exiting through the base. So a tiny base current “gates” a much larger emitter-to-collector current.

As a switch, we exploit the two extreme operating points: cutoff (no conduction) and saturation (maximum conduction). This binary behaviour is the foundation of all digital electronics — every logic gate, every processor uses transistors as switches.

Alternative Method

You can also understand the switching action through the transfer characteristic ( $V_{out}$ vs $V_{in}$ graph). For low $V_{in}$ : transistor is off, $V_{out} = V_{CC}$ (high). For high $V_{in}$ : transistor saturates, $V_{out} \approx 0$ (low). This is an inverter — the simplest logic gate (NOT gate).

For CBSE boards, draw a clean CE circuit diagram with both batteries ( $V_{BB}$ for input, $V_{CC}$ for output), both resistors ( $R_B$ and $R_C$ ), and label all three terminals. The diagram carries 1-2 marks by itself. Also label the current directions — $I_B$ into base, $I_C$ into collector, $I_E$ out of emitter.

Common Mistake

Students confuse current gain $\beta$ (CE configuration, $I_C/I_B$ ) with $\alpha$ (CB configuration, $I_C/I_E$ ). The relation is $\beta = \alpha/(1 - \alpha)$ . Since $\alpha$ is close to 1 (say 0.98), $\beta$ is large (49 in this case). In board exams, if they ask for “current gain in CE configuration,” they mean $\beta$ , not $\alpha$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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