BJT and MOSFET Amplifiers

BJT operating regions and biasing, stability, small-signal hybrid-pi model, CE, CB and CC amplifiers, MOSFET characteristics, CS amplifier gain, frequency response and multistage amplifiers.

Transistors amplify small signals. Design has two steps: set a stable DC operating point (biasing), then find the small-signal gain and impedances.

BJT

  • Regions: active (amplifier: BE forward, BC reverse), saturation and cut-off (switch).
  • I_C = β I_B, I_E = I_B + I_C, α = β/(1 + β).
  • Voltage-divider bias gives the best stability against β changes; stability factor S = ∂I_C/∂I_CO.
  • Thermal runaway is prevented by emitter resistance and heat sinks.

Small-signal (hybrid-π) model: g_m = I_C/V_T, r_π = β/g_m, r_o = V_A/I_C.

ConfigurationVoltage gainInput resistanceOutput resistanceUse
Common emitter−g_m R_C (high, inverting)r_π (medium)R_CVoltage amplifier
Common base+g_m R_C≈ 1/g_m (low)HighHigh frequency, current buffer
Common collector (emitter follower)≈ 1HighLowBuffer

An unbypassed emitter resistor R_E gives negative feedback: gain ≈ −R_C/R_E, more stable but lower.

MOSFET

  • Saturation (amplifier region): I_D = (k/2)(V_GS − V_T)² (k = μC_ox W/L), for V_DS ≥ V_GS − V_T.
  • Linear (triode) region: I_D = k[(V_GS − V_T)V_DS − V_DS²/2].
  • g_m = k(V_GS − V_T) = √(2k I_D) = 2I_D/(V_GS − V_T).
  • Common source gain = −g_m (R_D ∥ r_o). Very high input resistance (insulated gate).
  • Enhancement vs depletion; n-channel vs p-channel; CMOS uses both.

Frequency response

  • Low-frequency cut-off from coupling and bypass capacitors; high-frequency cut-off from junction capacitances.
  • Miller effect: a capacitance C between input and output of an inverting amplifier with gain −A appears at the input as C(1 + A), which limits bandwidth.
  • Gain–bandwidth product is roughly constant.
  • Cascading n identical stages: overall gain multiplies, bandwidth shrinks.

Worked example

A BJT has I_C = 2 mA (V_T = 25 mV) and R_C = 2 kΩ in a CE stage.

  • g_m = 2/25 = 80 mA/V; gain = −80 × 10⁻³ × 2000 = −160.

Practice

GATE-styleNAT · numerical answer

A BJT has β = 100 and I_C = 1 mA (V_T = 25 mV). What is r_π in kΩ?

Show answer and explanationAnswer: 2.5

g_m = 1/25 = 40 mA/V; r_π = β/g_m = 100/0.04 = 2500 Ω.

GATE-styleNAT · numerical answer

A MOSFET with k = 4 mA/V² operates at V_GS − V_T = 0.5 V. What is its drain current in mA?

Show answer and explanationAnswer: 0.5

I_D = (k/2)(V_GS − V_T)² = 2 × 0.25 = 0.5 mA.

GATE-styleNAT · numerical answer

An inverting amplifier with gain −99 has a 2 pF capacitance between input and output. What is the Miller input capacitance in pF?

Show answer and explanationAnswer: 200

C_in = C(1 + A) = 2 × 100 = 200 pF.

GATE-styleMCQ · one correct answer

Which BJT configuration has a voltage gain close to 1 and a low output resistance?

Show answer and explanationAnswer: C

The emitter follower (CC) is a buffer: gain ≈ 1, high input and low output resistance.