Bipolar Junction Transistors (BJT)

The BJT amplifies current: a small base current controls a large collector current. Learn NPN/PNP structure, operating regions (cut-off, active, saturation), current gain β, voltage-divider biasing, DC operating point (Q-point), transconductance, and the BJT as a switch and amplifier.

Invented in 1947, the transistor started the electronics revolution. BJTs remain widely used in analog circuits, discrete drivers and RF. Understanding biasing — setting the DC operating point — is the key skill before building amplifiers.

An analogy

A BJT is like a water tap: a tiny force on the handle (base current) controls a large flow (collector current).

Regions

Cut-off: V_BE < 0.5 V, no current — switch OFF. Active: base–emitter forward, base–collector reverse — I_C = βI_B, used for amplification. Saturation: both junctions forward, V_CE ≈ 0.2 V — switch ON.

Biasing

Voltage-divider bias with an emitter resistor makes the Q-point nearly independent of β (which varies 100–400 between parts and with temperature). Rule: divider current ≈ 10 × I_B; V_E ≈ 1–2 V; V_CE ≈ V_CC/2 for maximum swing.

BJT as a switch

Drive the base with I_B ≥ I_C/β_forced (β_forced ≈ 10–20) to guarantee saturation. Add a flyback diode across inductive loads (relays).

Small-signal model

Hybrid-π: g_m = I_C/V_T, r_π = β/g_m, r_o = V_A/I_C. Common-emitter voltage gain ≈ −g_m R_C (with bypassed emitter). Early effect, Miller capacitance and thermal runaway are advanced design issues.

Formulas

• Currents: I_C = β I_B, I_E = I_B + I_C
• Thevenin base bias: V_TH = V_CC R₂/(R₁ + R₂), R_TH = R₁ ∥ R₂
• Base current: I_B = (V_TH − V_BE) / (R_TH + (β + 1) R_E)
• Collector–emitter voltage: V_CE = V_CC − I_C R_C − I_E R_E
• Transconductance: g_m = I_C / V_T

Worked example: Voltage-divider bias

Given: V_CC 12 V, R₁ 47 kΩ, R₂ 10 kΩ, R_C 4.7 kΩ, R_E 1 kΩ; β = 100

Find: Q-point and g_m.

• V_TH = 2.105 V, R_TH = 8.25 kΩ
• I_B = (2.105 − 0.7)/(8.25k + 101k) = 12.9 µA → I_C = 1.29 mA
• V_E = 1.30 V, V_C = 5.95 V → V_CE = 4.66 V
• g_m = 1.29 mA/25.85 mV = 49.8 mS

Answer: I_C ≈ 1.29 mA, V_CE ≈ 4.7 V (active); gain ≈ −234 with bypassed R_E

Worked example: Saturated switch

Given: Relay 12 V, 60 mA; Logic 3.3 V drive, forced β = 20

Find: Base resistor.

• I_B = 60/20 = 3 mA
• R_B = (3.3 − 0.7)/3 mA = 867 Ω → 820 Ω

Answer: 820 Ω + flyback diode across the relay

Common mistakes

• Wrong: Designing for exact β. — Right: Use emitter degeneration so the Q-point tolerates β spread.
• Wrong: No flyback diode on relay coils. — Right: Inductive kick can destroy the transistor.

Applications

• Relay and LED drivers: Saturated switches from MCU pins.
• Audio amplifiers: Discrete preamps and output stages.
• Current mirrors: Biasing inside analog ICs.

Key takeaways

• Cut-off, active, saturation.
• I_C = βI_B; bias with divider + R_E.
• g_m = I_C/V_T sets gain.
• Force saturation for switches.

Going further

• Hybrid-π model: Small-signal analysis with r_π, g_m, r_o.
• Thermal runaway: Emitter resistors prevent positive feedback heating.
• Darlington & Sziklai: High-gain compound transistors.