emitter-follower method of operation

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.

Can I point you to an article that I found very helpful in really understanding how transistors work? It is worth reading, especially when you get to the bottom of it and get to part two.

http://amasci.com/amateur/transis.html

This link is thanks to member LvW who posted it in another thread.

Your explanation isn't bad, except for the "switching" idea. A transistor isn't like a bump car, that can be only on or off. Why not simply thinking of a balanced state?

FoxyRick, thanks for the link, folloing it I find a good article and:

Here is Hill making a post claming BJT's are voltage controlled. Probably knows what he is talking about =).

https://cr4.globalspec.com/comment/720033/Re-Voltage-vs-Current

Just because a circuit (like the one you are showing) is in the common-collector (same as emitter-follower) configuration does not necessarily mean that it will work in the manner it was intended. For this circuit to work properly, one has to make sure that the transistor itself is in the forward-active mode. Forward active mode requires that the base-emitter junction is forward biased (VBE=0.6V) AND the base-collector junction is reverse-biased (VBC<0.4V). You might see slightly different numbers for the junction potentials but the idea is the same. Ultimately, the supply voltages, the resistors and the Beta of the transistor can help us show this.

To analyze this circuit, we first assume active mode operation and calculate all necessary voltages and currents. Then we check to make sure that VBC is indeed less than 0.4V. Assuming a beta of 100.
base current IB=(Vb-Vbe,on)/(R1 + (beta+1)*R2) = 4.3/10200=422uA
base voltage VB=Vb-IB*RB=5-(422uA)*100=4.96V
base-collector voltage VBC=VB-VC=4.96-10=-5.04 < 0.4V ; therefore the circuit is in the active region.

I want to point out that VB refers to the node voltage at the base while Vb refers to the dc voltage source connected to the base circuit through R1.
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