Emitter Bias

When a single supply voltage is unavailable, the emitter bias circuit uses two power supplies to fix the transistor operating point without a voltage divider. This configuration is found in dual-supply op-amp driver stages and precision analog front ends.

Core Concept

In emitter bias, a negative supply rail (VEE) replaces the lower half of a voltage divider. The emitter resistor RE connects directly to the negative supply instead of ground. This makes the base voltage essentially zero (grounded through RB), and the emitter sits slightly above VEE.

The current through RE sets the emitter current IE. Since the base is near ground, VE = VEE + IE * RE approximately equals -0.7V in silicon transistors like the BC547. The collector current IC follows directly from IE and the transistor beta.

This bias scheme is highly stable. If temperature rises and IC tries to increase, the voltage across RE increases, pulling VBE down and reducing IC. This is the same feedback mechanism as emitter degeneration. The stability factor S is lower than single-supply fixed bias, making emitter bias preferred in lab-grade instrumentation.

Key Equations

Emitter current: IE = (VEE - VBE) / RE where VBE = 0.7V for silicon, VEE is the magnitude of the negative supply, RE is in ohms.

Collector current: IC = alpha * IE = (beta / (beta + 1)) * IE where alpha is typically 0.98 for beta = 50.

Collector-emitter voltage: VCE = VCC + VEE - IC*(RC + RE) with VCC positive and VEE the magnitude of negative rail.

Stability factor: S = (1 + beta) / (1 + beta * RE / (RE + RB)) which approaches 1 when RE >> RB/beta, meaning near-perfect stability.

Given:
  VCC = +12V, VEE = 12V (magnitude of -12V rail)
  RE = 2.2 kΩ, RC = 4.7 kΩ, RB = 100 kΩ
  VBE = 0.7V, beta = 100

Why this formula:
  Base is connected to ground through RB with negligible base current drop,
  so VB ≈ 0V. Hence VE = -0.7V, and IE = (VEE - VBE) / RE.

Formula:
  IE = (VEE - VBE) / RE

Substitution:
  IE = (12 - 0.7) / 2200
  IE = 11.3 / 2200

Calculation:
  IE = 0.00513 A = 5.13 mA
  IC ≈ IE (since beta >> 1)
  IC ≈ 5.13 mA

  VCE = VCC + VEE - IC*(RC + RE)
  VCE = 12 + 12 - 0.00513 * (4700 + 2200)
  VCE = 24 - 0.00513 * 6900
  VCE = 24 - 35.4 → Check: reduce RC or RE
  [With RC = 1kΩ, RE = 1kΩ for demonstration]
  VCE = 24 - 0.00513 * 2000 = 24 - 10.26 = 13.74V

Final Answer:
  IE ≈ 5.13 mA, IC ≈ 5.13 mA, VCE ≈ 13.74V (with RC=RE=1kΩ)

Exam Tip: GATE often gives a dual-supply emitter bias circuit and asks for IC or VCE. The key step students miss is that VB = 0V (base grounded through RB), so VE = -0.7V and IE = (VEE - 0.7)/RE. Do not apply the voltage divider formula here — there is no divider. Also, VCE spans from VCC to VEE, so it equals VCC + VEE - IC*(RC+RE), not just VCC - IC*RC.

Key Properties

• The Q-point is set almost entirely by VEE and RE, not by beta. This makes the circuit insensitive to transistor replacement.
• Stability factor S approaches 1 when RE is large relative to RB/beta. For RB=100kΩ and beta=100, S is around 1.9 with RE=2.2kΩ.
• The base resistor RB in emitter bias carries only the base current (IB = IC/beta). It does not need to provide a voltage divider ratio.
• A BC547 in emitter bias with dual 12V supplies can deliver 5-10 mA of collector current with good linearity for audio preamplifier stages.
• Because VB ≈ 0V, this circuit is easy to interface with signal sources that have a ground reference, avoiding coupling capacitor mismatch issues.
• The negative supply rail must be stable. A noisy VEE rail directly modulates IE and adds hum to the output.

Quick Revision

• Emitter bias uses dual supplies: +VCC and -VEE.
• VB ≈ 0V because the base connects to ground through RB with tiny IB voltage drop.
• IE = (VEE - VBE) / RE, using the magnitude of the negative supply.
• VCE = VCC + VEE - IC*(RC + RE): both supply rails contribute.
• Stability factor S is much lower than fixed bias, closer to voltage divider bias performance.
• Preferred in dual-supply op-amp systems and instrumentation where a negative rail already exists.
• Exam trap: Students subtract only VCC in the VCE formula and forget the VEE contribution, giving an answer that is 10-12V too low.