BJT Current Relationships

Every BJT has exactly three terminal currents, and they obey Kirchhoff's current law with no exceptions. Getting these relationships right lets you solve any BJT DC bias problem in under two minutes.

Core Concept

KCL applied at the BJT gives IE = IC + IB. This is not an approximation. It is exact. The emitter current equals the sum of collector and base currents. In a well-designed amplifier circuit, IC is around 100 times IB, so IE and IC differ by less than 1%.

The parameter β (or hFE) is the DC current gain. It equals IC divided by IB. It is a fixed number for a given operating point, but changes with temperature and IC magnitude. GATE problems usually give you a fixed β value and expect you to calculate the other currents.

The parameter α is the common-base current gain. It equals IC divided by IE. Since IC is slightly less than IE, α is always less than 1. The formulas connecting α and β are used frequently in GATE: α = β / (β + 1) and β = α / (1 - α). Memorize both.

Key Equations

KCL at emitter: IE = IC + IB. Valid for both NPN and PNP, in all regions of operation.

DC current gain: β = IC / IB. So IC = β × IB and IB = IC / β. For the BC547, β ranges from 100 to 600.

Common-base gain: α = IC / IE. Conversion: α = β / (β + 1) and β = α / (1 - α). Also: IE = IC / α = IC × (β + 1) / β.

Given:
  IB = 40 µA
  β = 80
  (BC547 with β on the lower side)

Why this formula:
  IC = β × IB (active region), IE = IC + IB (KCL)

Formula:
  IC = β × IB
  IE = IC + IB
  α = β / (β + 1)

Substitution:
  IC = 80 × 40 µA
  IE = IC + 40 µA
  α = 80 / (80 + 1)

Calculation:
  IC = 80 × 40 × 10^-6 = 3200 µA = 3.2 mA
  IE = 3.2 mA + 0.04 mA = 3.24 mA
  α = 80 / 81 = 0.9877

  Verification: IC = α × IE = 0.9877 × 3.24 mA = 3.20 mA ✓

Final Answer:
  IC = 3.2 mA
  IE = 3.24 mA
  α = 0.9877

Exam Tip: GATE frequently gives α and asks for β, or vice versa. The formula β = α / (1 - α) is critical. If α = 0.98, then β = 0.98 / 0.02 = 49. Practice this conversion until it is instant. A second common trap: when β is given as hFE in a problem, it is the same as β. Do not treat them as different quantities.

Key Properties

• IE is always the largest of the three currents. It is the sum of the other two.
• IB is always the smallest current. For β = 100 and IC = 1 mA, IB = 10 µA.
• α is always between 0 and 1. Typical values are 0.95 to 0.999. It never equals 1.
• β has no units. It is dimensionless. Its value ranges from about 20 (power transistors like TIP31) to over 500 (small-signal types like BC547).
• Both α and β decrease at very high collector currents due to high-level injection effects.
• The relationship IE = IC + IB holds regardless of the BJT region: active, saturation, or cutoff.

Quick Revision

• IE = IC + IB (KCL, always exact).
• β = IC / IB. Typical: 100 to 600 for BC547.
• α = IC / IE. Always less than 1.
• α = β / (β + 1). β = α / (1 - α).
• For large β: α ≈ 1 and IE ≈ IC. This is the common approximation.
• hFE = β = DC current gain. Small-signal hfe = β at that operating point.
• Exam trap: Writing IC = α × IB is completely wrong. α relates IC and IE, not IC and IB. This mistake appears often in student work and leads to wrong answers.