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Design divider-bias CE amplifier with gain -90

Short answer

With IC = 1 mA, VE = 1.2 V, VCE = 6 V: RC = 4.8 kΩ, RE1 ≈ 28 Ω (unbypassed), RE2 ≈ 1.16 kΩ (bypassed). Bias resistors: R1 ≈ 92 kΩ, R2 ≈ 19 kΩ. Gain is -90, stable due to RE1.

Design voltage divider biased common emitter BJT amplifier to get voltage gain of -90. Assume β=100 and Vcc=+12V.

A bypassed CE stage has Av = -RC/re. To obtain a specific gain of -90 and a stable value, a small unbypassed emitter resistor RE1 is kept in the ac path, with RE2 bypassed: Av = -RC/(re + RE1).

Given: VCC = 12 V, β = 100, Av = -90.

Guidelines

VE = VCC/10 = 1.2 V; VCE = VCC/2 = 6 V; I2 = 10 IB; VB = VE + 0.7 V. Choose IC = 1 mA (a convenient value).

Design

IC = 1 mA ,   IB = 10 µA ,   IE = 1.01 mA
RE (total) = VE/IE = 1.2/1.01 mA = 1.19 kΩ
RC = (VCC - VE - VCE)/IC = (12 - 1.2 - 6)/1 mA = 4.8 kΩ
re = 25 mV/1.01 mA = 24.75 Ω

Gain: |Av| = RC/(re + RE1) = 90
   re + RE1 = 4800/90 = 53.3 Ω   →   RE1 = 53.3 - 24.75 = 28.6 Ω (use 27 Ω to 30 Ω)
RE2 = RE - RE1 = 1.19 k - 0.029 k = 1.16 kΩ (use 1.2 kΩ), bypassed by CE

VB = 1.2 + 0.7 = 1.9 V ,   I2 = 10 IB = 100 µA
R2 = 1.9/100 µA = 19 kΩ
R1 = (12 - 1.9)/110 µA = 91.8 kΩ

Final: R1 ≈ 92 kΩ (91 kΩ), R2 = 19 kΩ (18 kΩ), RC = 4.8 kΩ, RE1 ≈ 28 Ω, RE2 ≈ 1.16 kΩ with bypass capacitor CE.

Check

Av = -4800/(24.75 + 28.6) = -90.0
Rin = R1 || R2 || (β+1)(re + RE1) = 15.7 k || 5.39 k = 4.0 kΩ

(Here the gain is for no load. A load RL reduces it to -(RC||RL)/(re + RE1); RE1 also makes the gain stable against changes in β and temperature.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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