Fixed Bias Circuit

Fixed bias is the simplest way to set the DC operating point of a BJT amplifier. A single resistor RB connects the supply VCC to the base and determines IB directly.

Core Concept

In fixed bias, a single resistor RB connects VCC directly to the base. The base current IB is set by KVL: IB = (VCC - VBE) / RB. Since VCC and RB are constants, IB is fixed. The collector current IC = β × IB.

The main weakness of fixed bias is that the Q-point (quiescent operating point) depends directly on β. If β changes due to temperature or transistor replacement, IC changes proportionally. A transistor with β = 100 gives IC = 2.4 mA; replace it with one having β = 300 and IC triples to 7.2 mA. The amplifier is no longer at the designed operating point.

Despite this instability, fixed bias is useful for studying BJT theory because the equations are simple and direct. It also works in discrete circuits where the transistor β is tightly controlled or where exact biasing is not critical, such as simple switching circuits.

Key Equations

Base current: IB = (VCC - VBE) / RB. VBE = 0.7V for silicon. VCC is the supply voltage. RB is the base resistor in ohms.

Collector current (active region): IC = β × IB. Collector-emitter voltage: VCE = VCC - IC × RC. This is from KVL around the collector-emitter loop.

Stability factor: S = dIC / dICO = (β + 1). For fixed bias, S = β + 1. A high S means the Q-point shifts significantly when temperature increases. For β = 100, S = 101, which is very poor stability.

Given:
  VCC = 12V
  RB = 470 kΩ
  RC = 2.2 kΩ
  VBE = 0.7V
  β = 100 (BC547 minimum)

Why this formula:
  KVL base loop: VCC - IB×RB - VBE = 0
  Active region assumed; verify with VCE check.

Formula:
  IB = (VCC - VBE) / RB
  IC = β × IB
  VCE = VCC - IC × RC

Substitution:
  IB = (12 - 0.7) / 470000
  IC = 100 × IB
  VCE = 12 - IC × 2200

Calculation:
  IB = 11.3 / 470000 = 24.04 µA
  IC = 100 × 24.04 × 10^-6 = 2.404 mA
  VCE = 12 - (2.404 × 10^-3 × 2200)
      = 12 - 5.289
      = 6.71V

  Verify active region: VCE = 6.71V > 0.2V. Confirmed.

Final Answer:
  IB = 24.04 µA
  IC = 2.4 mA
  VCE = 6.71V
  Q-point: (VCE, IC) = (6.71V, 2.4 mA)

Exam Tip: GATE problems on fixed bias often ask you to find the new Q-point when β changes. If β doubles from 100 to 200, IC doubles and VCE drops. If VCE goes below 0.2V, the transistor enters saturation. Always check. The stability factor S = β + 1 is the other tested formula. Know that fixed bias has the worst stability of all biasing schemes.

Key Properties

• Uses a single resistor RB from VCC to the base. Circuit is simple but Q-point is β-dependent.
• Stability factor S = β + 1 for fixed bias. For BC547 with β = 100, S = 101 (very unstable).
• The ideal Q-point for maximum symmetrical swing is at VCE = VCC/2, which is VCE = 6V for a 12V supply.
• If β changes by a factor of 2 (common between transistors of same part number), VCE changes significantly, often driving the transistor into saturation.
• No emitter resistor means no DC negative feedback, which is why the Q-point is unstable.
• The DC load line equation is VCE = VCC - IC × RC. It passes through (VCC, 0) and (0, VCC/RC) on the output characteristics.

Quick Revision

• Single RB from VCC to base. IB = (VCC - VBE) / RB.
• IC = β × IB. VCE = VCC - IC × RC.
• Stability factor S = β + 1. Worst among all bias circuits.
• Q-point shifts directly with β. No thermal stability.
• DC load line: endpoints (VCC, 0) and (0, VCC/RC).
• Ideal Q-point for amplifier: VCE = VCC/2 for maximum swing.
• Exam trap: Forgetting to subtract VBE = 0.7V when finding IB. Writing IB = VCC / RB instead of IB = (VCC - VBE) / RB. This gives an IB about 6% too high for a 12V supply, and more for lower supplies.