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Bypass capacitor significance; CE small signal analysis

Short answer

Bypass capacitor CE shorts RE at signal frequencies, removing AC negative feedback to raise gain. Small-signal model: R1||R2||rπ at input, current source gm vπ at output. Zin = R1||R2||rπ, Zo ≈ RC, Av = –gm(RC||RL).

What is the significance of bypass capacitor in CE amplifier? Draw the small signal model of voltage divider bias for emitter bypassed capacitor CE amplifier circuit and find its input impedance, output impedance and voltage gain.

Significance of the bypass capacitor

The bypass capacitor CE is connected across the emitter resistor RE. At dc it is open, so RE gives bias stability (negative feedback). At signal frequencies its reactance is very small, so it shorts RE to ground for ac. This removes the ac negative feedback, so the voltage gain is large (Av = -gm RC) instead of being reduced to -RC/RE. It also needs to be large: XC ≪ re at the lowest signal frequency.

Small-signal model (hybrid-π) of divider-bias CE amplifier with CE

 vs--Rs--+--B--rπ--E(gnd)      C
         |      |   \         |
       R1||R2   vπ   (gm vπ)--+--RC||RL--GND
         |
        GND
  • Input side: R1 || R2 in parallel with rπ between base and ground (emitter grounded by CE).
  • Output side: current source gm vπ in parallel with ro, RC and RL.

With rπ = β/gm = β VT/IC and gm = IC/VT.

Input impedance

Zin = R1 || R2 || rπ

Output impedance

Set vs = 0, so vπ = 0 and the source gm vπ = 0:

Zo = RC || ro  ≈ RC     (looking into the collector, load excluded)

Voltage gain

vo = -gm vπ (RC || RL || ro) ,   vπ = vin

Av = vo/vin = -gm (RC || RL)

For example, IC = 1.5 mA gives gm = 60 mA/V; with RC||RL = 4 kΩ, Av = -240. The negative sign shows a 180° phase shift.

✓ Answer checked by Claude Opus 5.5 (high effort)

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