The input and output of an emitter follower are the base and the emitter, respectively, and the collector is at AC zero. The circuit is therefore a common-collector circuit (for AC).

The negative feedback effect due to can be shown qualitatively:

(107) |

(108) |

(109) |

(110) |

(111) |

**AC small-signal equivalent circuit**

The AC equivalent circit (based on small signal model of the transistor) of the emitter follower can be found as:

Based on this small signal model, the three system parameters: voltage gain, input resistance, and output resistance can be obtained as shown below.

**AC voltage gain:**As is significantly greater than and , it is neglected in the analysis below.

(112) (113) **Input resistance:**The input resistance is the parallel combination of and the resistance of the circuit to right of the base of the transistor, including the load , which can be found as the ratio of the voltage and the current .

(114) (115) (116) **Output resistance:**The output resistance is the parallel combination of and the resistance of the circuit to the left of the emitter of the transistor (including ), which can be found as the ratio of the open-circuit voltage (with ) and the short-circuit current (with ).

- Find open-circuit voltage :
is approximately the same as the source voltage , as the voltage gain of the emitter follower is close to unity.

- Find short-circuit current :
(117)

(118) (119) - Find open-circuit voltage :

**Conclusion:**

The emitter follower is a circuit with strong negative feedback, i.e., the full output is fed back to become part of its input . Due to this deep negative feedback, it has the following properties:

- The voltage gain is smaller than but close to unity with .
- The input resistance is large .
- The output resistance is small .

Although the emitter follower circuit does not amplify the signal voltage, it drastically improves the input and output resistances, when compared with the common-emitter circuit with and output resistances .

Due to its high input resistance , it draws little current from the source, causing little internal voltage drop across the internal resistance of the source, also due to its low output resistance , it can drive heavy load (low ) without lowering its output voltage. It is therefore widely used as a buffer between two stages of a multi-stage amplification circuit.

**Example**

Assume , , . Find so that the DC operating point is in the middle of the load line.

For to be in the middle of the load line, we need to have , i.e., :

(120) |

(121) |

(122) |