Transistor amplifier circuits

This is an AC analysis of signals and all the clutter of DC components is removed - input and output capacitors are replaced with short circuits and power supplies are also replaced with short circuits. It's not a DC analysis or any other analysis - it's an analysis of signal amplification and AC impedances with a minimalist approach diagramatically. This method is meant to help beginners but I've never thought it was helpful - show me the full circuit any day and I'll figure it out from that has always been my approach!

And for your point 3 it is assumed that base current is zero in a lot of AC analyses.

Point 4 - this circuit can be used with the collector effectively seeing a load resistance of under 100 ohms. Try researching an emitter fed cascode amplifier - two transistors are used in common base and the collector of the 1st transistor feeds the emitter of the 2nd transistor. Most cascode amplifier circuits feed the base of the first transistor from the external input but it can work with both transistors in common-base configuration.

The input resistance is determined by the forward conduction dynamic resistance of the emitter-base junction - it's forward biased and has a dynamic (AC) impedance of 20 ohms in your question - this is presumed to be the slope of the emitter-base diode. Load resistance will not affect this.

"Common Collector" means that the Collector is a common point for both input and output signals (ie. Ground, or a fixed voltage relative to it). In a true Common Collector circuit the value of RC is zero Ohms. Larger values will still work, but with reduced output voltage swing.

In your circuit the Emitter can only go up to slightly less than 9V, because RC is then dropping 1V so the transistor is saturated (approaching zero volts between Collector and Emitter). However since you only need 2Vpp output and the quiescent Emitter voltage is ~5V, this should not be a problem.