In one line
A CE amplifier is biased at a Q-point in the active region; a small AC input at the base varies the base current, which varies the collector current \( \beta \) times, producing a large voltage swing across the collector resistor that is 180° out of phase with the input.
Explained simply
A transistor amplifier is like a tap controlled by a tiny lever. The lever (input signal) moves a little; the water flow (collector current) changes a lot; and passing that flow through a resistor turns it into a big voltage change. To swing both up and down without hitting "fully off" or "fully on", the tap is first set half open: that is biasing.
Technically, biasing sets DC values \( I_{CQ} \) and \( V_{CEQ} \) (the Q-point), usually near the middle of the DC load line, using a voltage-divider bias with an emitter resistor for stability. Coupling capacitors pass AC and block DC; an emitter bypass capacitor increases AC gain.
How it works
- Input signal \( v_i \) is coupled to the base through \( C_1 \). It adds to the DC base-emitter voltage.
- Base current varies by a small \( i_b \); collector current varies by \( i_c = \beta i_b \).
- Collector voltage \( V_C = V_{CC} - I_CR_C \). When \( I_C \) rises, \( V_C \) falls, so the output is inverted (180° phase shift).
- \( C_2 \) passes only the AC part of \( V_C \) to the load.
- If the Q-point is too near cut-off or saturation, or the input is too large, the output is clipped (distortion).
- Frequency response: gain falls at low frequencies (due to coupling and bypass capacitors) and at high frequencies (due to internal capacitances). The mid-band range is used.
Key formula
\( A_v = \dfrac{v_o}{v_i} \approx -\dfrac{R_C}{r_e}, \quad r_e \approx \dfrac{25\text{ mV}}{I_E} \) (emitter resistor bypassed, no load)
\( A_v \) = voltage gain (no unit; minus sign = phase inversion), \( R_C \) = collector resistance (\( \Omega \)), \( r_e \) = AC emitter resistance (\( \Omega \)), \( I_E \) = DC emitter current (A), 25 mV ≈ thermal voltage at room temperature.
Quick example
With \( I_E = 1 \) mA, \( r_e \approx 25\,\Omega \); with \( R_C = 2.5 \) kΩ, \( A_v \approx -100 \). A 10 mV microphone signal becomes about 1 V, inverted. Real uses: preamplifier stages in audio systems, radios and sensors. (Illustration only.)
Common mistakes
- Biasing the transistor in saturation or cut-off and expecting linear amplification.
- Forgetting the 180° phase reversal in the CE amplifier.
- Thinking a transistor creates energy; the extra signal power comes from the DC supply \( V_{CC} \).
How MU asks it
Numericals are not expected in this module (MU syllabus), so questions are theory with diagrams. Typical: "Explain the application of a BJT as an amplifier" or "Explain the working of a CE amplifier with circuit diagram and waveforms" (5–8 marks). For full marks draw the voltage-divider-biased CE circuit with \( C_1 \), \( C_2 \), \( C_E \), show input and inverted output waveforms, explain the Q-point on the load line, and state the role of each component.