I'm trying to understand how the emitter-follower follows the base voltage at the emitter (minus the base to emitter voltage drop of about 0.6 V.) I've done some googling and also consulted Horowitz & Hill without any explaination. Plenty of quantification and impedence matching advice though.
Consider the circuit. R1 (100 ohm) provides current to the Q1 (NPN) base from a 5 volt source. Q1 collector is at a 10 V source. Q1 emitter is connected to ground through a series resistor R2 (100 ohm.) Stipulate an insignificant voltage drop across base resistor R1.
The voltage drop across the load resistor R2 is the base voltage minus the base to emitter voltage drop through Q1. The emitter follows the base voltage.
My own understanding of how this works has to deal with the forward biasing of the base to emitter PN junction. The transistor will only conduct when the base is about 0.6 volts higher than the emitter. If the emitter starts becoming more positive, considering the base to emitter drop, than the base, then the transistor will begin to turn off. If the emitter starts becoming more negative, again considering Vbe, then the transistor starts turing on.
So, in a way, the transistor is turning on and off, which is why the emitter voltage follows the base. Is this correct? Obviously, the transistor will not be behaving like a perfect switch.