On my Transistor issue:
Here it is cause it just happened again with an NPN. Does the transistor "START" to conduct with the .7v on its base or is it fully on with .7? I think my brain lock is the voltage range at which it amplifies and isn't just a switch?
BJTs are current devices, not voltage devices. There is no simple easy relationship between a BJT's "input" and "output" voltage, so if you start there you are bound to be perplexed.
There is a single simple relationship between "input" current and "output" current; IC=hFEIB
So, start by working out the input (base) current, that will give you the output (collector) current. That current flowing through the load will determine the output voltage.
Let's not start that.
In applications where the BJT is being used as a linear amplifier, it's always right to consider it to be a voltage controlled device, with IC being dependant on VBE.
Current control only makes sense when dealing with a saturated switch, but even then, in order to ensure a fast turn-off time, it's necessary to take the base voltage below the normal 0.6V diode drop, preferably even negative.
Ah, let's get back to basics: does the current through a resistor causes the voltage across it, or the voltage across a resistor causes the current through it. Answer: mu.
Realise that the OP is not considering BJTs on their own, he is trying to understand how they work (i.e. design equations) for typical simple amplifiers.
If you look at most simple amplifiers, linear or saturated, the input voltages is much more than V
BE. That enables design patterns which more-or-less remove the variation in V
BE from that equations defining the amplifier's operation. The current relationships, however, remain.
For common collector amplifier, consult TAoE section 2.2.2-2.2.5; for common emitter, 2.2.8-2.2.9. Any other introductory textbook ought to be similar.
For more comprehensive analysis, consult TAoE chapter 2
x, but the OP won't be interested in that.