Q.Discuss the working of n-p-n transistor as common emitter amplifier. OR Explain the Energy bands of solids in detail.

Transistor as an Amplifier

Think of a transistor amplifier like a seesaw where you push down gently on one end and the other end swings up hard. Except here, the "push" is a tiny voltage change at the input, and the "swing" is a large voltage change at the output. The transistor itself is just a valve — it controls a big current using a small signal.

The Core Intuition

A transistor in common-emitter configuration has three terminals: base, emitter, and collector. The base-emitter junction behaves like a diode. A small change in base-emitter voltage ($\Delta V_{BE}$) causes a much larger change in collector current ($\Delta I_C$). This is the transistor's transconductance — the ratio $\Delta I_C / \Delta V_{BE}$, denoted $g_m$.

Now, if you place a resistor $R_C$ in the collector path, that changing current produces a changing voltage across $R_C$: $\Delta V_{out} = -\Delta I_C \cdot R_C$. The minus sign appears because when collector current increases, the voltage drop across $R_C$ increases, pulling the collector voltage down — so the output is inverted relative to the input.

The voltage gain is therefore:

The Precise Statement

A common-emitter transistor amplifier takes a small input voltage signal applied between base and emitter, and produces a larger (but inverted) output voltage signal between collector and emitter. The amplification relies on two things:

1. Transconductance ($g_m$): The transistor converts a voltage change at the input into a current change at the output. For a bipolar junction transistor (BJT) in active region, $g_m = I_C / V_T$, where $V_T \approx 25\ \text{mV}$ at room temperature and $I_C$ is the quiescent collector current.

2. Load resistor ($R_C$): This resistor converts the current change back into a voltage change. The output voltage swing is $-\Delta I_C \cdot R_C$.

The overall voltage gain is:

Why It Works (Step by Step)

1. A small increase in $V_{BE}$ (say $+5\ \text{mV}$) causes a large increase in $I_C$ (because the base-emitter diode is exponential in nature).
2. This larger $I_C$ flows through $R_C$, increasing the voltage drop across it.
3. Since the collector is connected to the supply through $R_C$, the collector voltage drops by the same amount.
4. The result: a small positive input produces a large negative output — amplification with inversion.

A Concrete Example …