Define bandwidth of an amplifier.

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Q. Define bandwidth of an amplifier.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 3mImportance★★★★★

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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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The bandwidth of an amplifier is the frequency range between its lower and upper half-power (cut-off) frequencies, i.e. the band over which the gain stays within $1/\sqrt2$ of its maximum value.

Frequency response of an amplifier

The voltage gain of a practical amplifier is not constant at all frequencies; it is nearly constant over a mid-frequency range but falls off at very low and very high frequencies (due to coupling/bypass capacitors at low frequencies and stray/junction capacitances at high frequencies). A plot of gain versus frequency is called the frequency-response curve.

Definition of bandwidth

The two frequencies at which the gain falls to $\dfrac1{\sqrt2}$ (about 70.7%) of the maximum (mid-band) gain are called the lower cut-off frequency $f_1$ and the upper cut-off frequency $f_2$ . These are also called the half-power frequencies, because the power delivered is proportional to the square of the voltage gain, and $\left(\dfrac1{\sqrt2}\right)^2=\dfrac12$ , i.e. the power falls to half its maximum value at these points.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set A1 markMCQ
$Q. Instrument used to increase input voltage/current is called (A) oscillator (B) amplifier (C) diode (D) rectifier$
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The device used to increase (magnify) an input voltage or current is an amplifier → option (B).

An amplifier raises the amplitude of a weak input signal to a larger output level, drawing the extra energy from a DC power supply. A rectifier only converts AC to DC, a diode is a t …

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CBSE 2019Set ANNUAL1 mark
$Q. Draw a neat labelled circuit diagram showing the use of an n-p-n transistor as a common emitter aplifier.$
›Reveal solutionSolution
Figure — Labelled common-emitter (CE) npn-transistor amplifier circuit

CE amplifier: input at base-emitter (via $C_1$ ), biasing resistors $R_1,R_2$ set the base voltage, $R_E$ + $C_E$ stabilise/bypass the emitter, $R_C$ is the collector load, output is taken across $R_C$ via $C_2$ — it inverts and amplifies the signal.

Circuit description (n-p-n transistor, common-emitter configuration):

The emitter terminal is common to both the input and output circuits (grounded through $R_E$ ).

DC biasing: Two resistors $R_1$ (from $V_{CC}$ to base) and $R_2$ (from base to ground) form a potential divider that fixes the base voltage and hence the quiescent (no-signal) base current $I_B$ , collector current $I_C = \beta I_B$ , and $V_{CE}$ .

Collector circuit: A load resistor $R_C$ connects the collector to the supply $V_{CC}$ ; the output voltage is developed across $R_C$ as $I_C$ varies.

Emitter resistor $R_E$ provides thermal stability (stabilises the operating point against temperature drift) and is bypassed by a capacitor $C_E$ (so it does not reduce the AC gain).

Input coupling capacitor $C_1$ connects the AC input signal $v_i$ to the base while blocking DC from the source; output coupling capacitor $C_2$ connects $R_C$ to the load while blocking the DC collector voltage. …
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Q.Define bandwidth of an amplifier.

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