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Design CE divider bias (12 V, 2 mA, beta 150)

Short answer

Voltage divider bias: VE = VCC/10 = 1.2 V, VCE = 6 V, I2 = 10 IB. RE = VE/IE ≈ 596 Ω, RC = (12 – 1.2 – 6)/2 mA = 2.4 kΩ. VB = 1.9 V gives R2 ≈ 14.2 kΩ, R1 ≈ 68.9 kΩ. With RE bypassed, Av ≈ –192.

Design a voltage divider type dc biased CE amplifier to obtain β independent biasing. Use appropriate guidelines to support your design. Given VCC = 12 V DC, IC = 2mA and β =150.

Design of a voltage divider biased CE amplifier (VCC = 12 V, IC = 2 mA, β = 150).

Why divider bias is β independent

Voltage divider bias is called β independent because the divider holds the base voltage almost fixed, so the collector current is set only by the resistors and VBE, not by β.

With the divider current I2 ≫ IB (I2 = 10 IB), the base voltage is VB ≈ VCC R2/(R1 + R2), which does not depend on β. Then

IE = (VB - VBE)/RE ,    IC = α IE ≈ IE

Neither equation contains β. If β changes (a different transistor or a change in temperature), IB changes but VB stays constant, so IC and the Q-point remain stable. (Exact Thevenin form: IC = β(VTH - 0.7)/(RTH + (β+1)RE); with RTH ≪ (β+1)RE, β cancels.)

Design guidelines used

  1. Emitter voltage VE ≈ VCC/10 (large enough for thermal and bias stability, small enough to leave room for the output swing).
  2. Collector-emitter voltage VCE ≈ VCC/2, so the output can swing equally up and down (Q-point at the centre of the load line).
  3. Divider current I2 = 10 IB, much greater than IB, so the base voltage is fixed and the bias is independent of β.
  4. VB = VE + 0.7 V.

Design

IC  = 2 mA        IB = IC/β = 2 mA/150 = 13.3 µA
IE  = IC + IB = 2.013 mA
VE  = VCC/10 = 1.2 V     RE = VE/IE = 1.2/2.013 mA = 596 Ω
VCE = VCC/2  = 6 V
RC  = (VCC - VE - VCE)/IC = (12 - 1.2 - 6)/2 mA = 2.4 kΩ
VB  = VE + 0.7 = 1.9 V
I2  = 10 IB = 133.3 µA
R2  = VB/I2 = 1.9/133.3 µA = 14.2 kΩ
R1  = (VCC - VB)/(I2 + IB) = 10.1/146.7 µA = 68.9 kΩ

Calculated values: R1 ≈ 68.9 kΩ, R2 ≈ 14.2 kΩ, RE ≈ 596 Ω, RC ≈ 2.4 kΩ (in practice, use the nearest standard resistor values). The emitter resistor RE is bypassed by a large capacitor CE; C1 and C2 are the coupling capacitors.

Check and small-signal results (RE bypassed)

re  = VT/IE = 25 mV/2.013 mA = 12.4 Ω
rπ  = β re = 1863 Ω         gm = IC/VT = 80 mA/V
R1 || R2 = 11.8 kΩ
Rin = R1 || R2 || rπ = 1.61 kΩ
Av  = -gm RC = -80 mA/V × 2.4 kΩ = -192

Ro ≈ RC = 2.4 kΩ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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