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Why divider bias is beta independent; design (24 V)

Short answer

Voltage divider bias is β independent because I2 ≫ IB fixes VB, so IE = (VB−0.7)/RE. Design with VE=2.4 V, VCE=12 V: RE=1.59 kΩ, RC=6.4 kΩ; R1=190 kΩ, R2=31 kΩ. Gain Av = −384, Rin = 2.27 kΩ.

Why voltage divider biasing is called β independent? Design a voltage divider type DC biased CE amplifier to obtain β independent biasing. Use appropriate guidelines to support your design. Given parameters are: Vcc= 24V, Ic= 1.5mA and 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 of the CE amplifier (VCC = 24 V, IC = 1.5 mA, β = 150)

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  = 1.5 mA        IB = IC/β = 1.5 mA/150 = 10 µA
IE  = IC + IB = 1.51 mA
VE  = VCC/10 = 2.4 V     RE = VE/IE = 2.4/1.51 mA = 1.59 kΩ
VCE = VCC/2  = 12 V
RC  = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/1.5 mA = 6.4 kΩ
VB  = VE + 0.7 = 3.1 V
I2  = 10 IB = 100 µA
R2  = VB/I2 = 3.1/100 µA = 31 kΩ
R1  = (VCC - VB)/(I2 + IB) = 20.9/110 µA = 190 kΩ

Calculated values: R1 ≈ 190 kΩ, R2 ≈ 31 kΩ, RE ≈ 1.59 kΩ, RC ≈ 6.4 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/1.51 mA = 16.6 Ω
rπ  = β re = 2483 Ω         gm = IC/VT = 60 mA/V
R1 || R2 = 26.7 kΩ
Rin = R1 || R2 || rπ = 2.27 kΩ
Av  = -gm RC = -60 mA/V × 6.4 kΩ = -384

Ro ≈ RC = 6.4 kΩ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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