Hello,
I'm not sure if I should ask a new question or resurrect an old thread/video, so I've opted for a new question.
I watched #1116 the other day, about the "capacitance multiplier", and it is very cool. I get that it works, but I don't know how it works. To me, it seems that once the base is turned on by the resistor, the transistor is conducting. If the transistor is conducting then the voltage on the emitter is just the voltage on the collector minus 0.6 for the base-emitter junction.
Also, Dave says that a MOSFET would incur a larger voltage drop... but I would expect it to be smaller, since (assuming a use a hexfet or something similar), there's no junction voltage to overcome. Rather, it would have a portion of an ohm of resistance, and therefore less of a voltage drop.
What am I missing? Thanks. (I've read about emitter followers a bit, but they generally talk about them as amplifiers... I still don't intuitively understand this circuit).