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What is the input impedance of a common-emitter stage with a bypassed emitter?

  • A.Beta times the intrinsic emitter resistance, in parallel with the divider
  • B.The collector load divided by the voltage gain of the stage
  • C.The emitter resistor multiplied by one plus the current gain
  • D.The divider's Thevenin resistance taken by itself alone

Answer: A

Looking into the base gives beta times the emitter resistance, and the bias divider appears in parallel with it - often dominating, since a stiff divider is deliberately made low. The third option applies when the emitter resistor is unbypassed. Remembering that the bias network is part of the signal path is what stops input impedance being overestimated by a wide margin.

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