Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and outp

III. Long Answer Questions · Q9

Q. Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.

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Concept understanding — Transistor as Amplifier

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:

A_v = \frac{\Delta V_{out}}{\Delta V_{in}} = -g_m R_C

Tip

The gain is simply transconductance times load resistance. A bigger $R_C$ gives more gain, but too large a resistor starves the transistor of operating current — there's always a trade-off.

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:

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.
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:

A_v = -\frac{I_C}{V_T} \cdot R_C

Watch out

The gain formula assumes the transistor stays in the active region throughout the signal swing. If the output tries to swing below about $0.2\ \text{V}$ (saturation) or above the supply voltage (cutoff), the transistor stops amplifying and the signal distorts.

Why It Works (Step by Step)

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).
This larger $I_C$ flows through $R_C$ , increasing the voltage drop across it.
Since the collector is connected to the supply through $R_C$ , the collector voltage drops by the same amount.
The result: a small positive input produces a large negative output — amplification with inversion.

Important

The transistor does not create energy. The amplified output signal gets its power from the DC supply ( $V_{CC}$ ). The transistor merely modulates this DC power in proportion to the small input signal. That's why it's called an amplifier, not an energy source.

A Concrete Example …

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Step 1 (Circuit). A single-stage common-emitter amplifier uses one NPN transistor biased via resistors off the supply $+V_{CC}$ , with the Q-point set near the middle of the DC load line for maximum undistorted swing; coupling capacitor $C_1$ passes the AC input signal $V_S$ onto the base while blocking DC, emitter bypass capacitor $C_E$ gives the amplified AC signal a low-reactance path around the emitter resistor, and output coupling capacitor $C_C$ passes the amplified signal onward while again blocking DC; collector resistor $R_C$ converts changes in $I_C$ into an output voltage.

Step 2 (Governing relations). $I_C=\beta I_B$ , and by Kirchhoff's voltage law in the output loop, $V_{CE}=V_{CC}-I_CR_C$ .

Step 3 (Positive half cycle). $V_S$ increases the forward voltage across the emitter-base junction, so $I_B$ increases, and $I_C$ increases $\beta$ times; the larger $I_C$ increases the drop $I_CR_C$ across $R_C$ , which DECREASES $V_{CE}$ -- the output swings negative while the input swings positive.

Step 4 (Negative half cycle). $V_S$ decreases the forward voltage, so $I_B$ and $I_C$ decrease, the drop across $R_C$ decreases, and $V_{CE}$ INCREASES -- the output swings positive while the input swings negative. …

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Q.Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.

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