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CE with unbypassed RE; application of CB amplifier

Short answer

With unbypassed RE, output impedance is approximately RC. Voltage gain is Av = -(RC||RL)/(re+RE). Common base amplifiers are used in high-frequency/RF circuits, current buffers, and cascode stages due to low input impedance and no Miller effect.

Draw common emitter transistor amplifier circuit (emitter bias with unbypassed emitter capacitor) and find its output impedance and voltage gain. Write application of common base amplifier.

Circuit (voltage divider bias, RE unbypassed)

              +VCC
               |
         +-----+-----+
         |           |
         R1          RC
         |           |         C2
 vs--||--+------B    C----+----||----+---- vo
     C1  |       \  /     |          |
         R2       NPN     (vo node)  RL
         |         |                 |
        GND        E                GND
                   |
                   RE  (no bypass capacitor)
                   |
                  GND

C1 and C2 are coupling capacitors; RE is not bypassed, so it appears in the ac path.

Output impedance

Set vs = 0 and look into the collector. The unbypassed RE (series feedback) raises the resistance looking into the collector:

Zout(collector) = ro [1 + gm (RE || rπ)]  →  ∞  (ro = ∞)
Zo = RC || Zout(collector) ≈ RC

So the output impedance of the stage is approximately RC (RE only raises the collector impedance, which is already very large).

Voltage gain

With the T-model: vb = ie (re + RE) and vo = -ie (RC || RL), so

Av = vo/vb = -(RC || RL)/(re + RE)

(If RE ≫ re: Av ≈ -(RC||RL)/RE, which is set by the resistors alone.) The input impedance is Zin = R1 || R2 || (β+1)(re + RE).

Applications of the common base amplifier

  • High-frequency and RF amplifiers (VHF/UHF), because there is no Miller effect and the bandwidth is wide.
  • Current buffers, and stages that need a low input impedance (e.g. matching to low-impedance sources such as microphones or antennas).
  • The upper stage of a cascode (CE-CB) amplifier.
  • Constant current sources and current mirrors.

✓ Answer checked by Claude Opus 5.5 (high effort)

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