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Design CE amplifier with divider bias (20 V, 10 mA)

Short answer

Using VE=VCC/10=2V, VCE=VCC/2=10V, RE=198Ω, RC=800Ω, R1=23.1kΩ, R2=4.32kΩ. RE is unbypassed for high input impedance, giving Rin≈2.97kΩ and Av≈-4.0.

Design a Common Emitter Amplifier using voltage divider biasing circuit having high input impedance using appropriate design guidelines. Given Parameters: VCC=20V, IC=10 mA, β=80.

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 = 5 IB, greater than IB. This is the lower end of the 5 to 10 IB range, which makes R1 || R2 as large as possible for a high input impedance while still keeping the base voltage fixed and the bias independent of β.
  4. VB = VE + 0.7 V.
  5. For a high input impedance, RE is left unbypassed, so that (β+1)RE appears in the input resistance instead of only rπ.

Design

IC  = 10 mA        IB = IC/β = 10 mA/80 = 125 µA
IE  = IC + IB = 10.12 mA
VE  = VCC/10 = 2 V     RE = VE/IE = 2/10.12 mA = 198 Ω
VCE = VCC/2  = 10 V
RC  = (VCC - VE - VCE)/IC = (20 - 2 - 10)/10 mA = 800 Ω
VB  = VE + 0.7 = 2.7 V
I2  = 5 IB = 625 µA
R2  = VB/I2 = 2.7/625 µA = 4.32 kΩ
R1  = (VCC - VB)/(I2 + IB) = 17.3/750 µA = 23.1 kΩ

Calculated values: R1 ≈ 23.1 kΩ, R2 ≈ 4.32 kΩ, RE ≈ 198 Ω, RC ≈ 800 Ω (in practice, use the nearest standard resistor values). C1 and C2 are the coupling capacitors.

Check and small-signal results (RE unbypassed)

re  = VT/IE = 25 mV/10.12 mA = 2.47 Ω
R1 || R2 = 3.64 kΩ
Rib = (β+1)(re + RE) = 81 × (2.47 + 198) = 16.2 kΩ
Rin = R1 || R2 || Rib = 3.64 k || 16.2 k = 2.97 kΩ
Av  = -RC/(re + RE) = -800/200.5 = -4.0

Ro ≈ RC = 800 Ω.

Why RE is not bypassed: if RE were bypassed by a capacitor, the input resistance would fall to R1 || R2 || rπ = 3.64 k || 197.5 Ω ≈ 187 Ω (rπ = β re = 197.5 Ω), which is low, although the gain would rise to -gm RC = -400 mA/V × 800 Ω = -320. With a large collector current of 10 mA, rπ is small, so a high input impedance is obtained only by keeping RE in the ac path. Rin is then limited mainly by the divider (about 3 kΩ); a still higher value would need a smaller IC or a larger β.

✓ Answer checked by Claude Opus 5.5 (high effort)

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