Transistors

When trying to work out collector and emitter voltages start with the base current Ib.

The transistor will try to make Ic equal to Ib * beta. This means that if there is a collector resistor Rc then Vc will be pulled down to whatever level makes the voltage drop across Rc produce the desired current. That level cannot be less than 0 - if it is, then Vc ~ 0 and Ic/Ib < beta (saturation).

If there is no Rc then Vc will be Vcc, it seems. But this does not mean Ic = 0; the collector works like a current sink (a hypothetical thing that absorbs current without obeying Ohm's Law).

Similarly, Ve will be whatever level results in the desired current (Ib + Ic) through the emitter resistor. Bear in mind that BE works like a diode, so if it is forward biased then Vb - Ve = 0.6 V. (And this also means that it will break down if reverse biased by too great a difference)

Btw there's nothing magic about the -4.4/-5V cutoff in the Art of Electronics emitter follower example. The figure is -5V because that is Ve when Ic = 0; Ve is -5V in that situation because the emitter is between two resistors of equal value. Beyond the load resistor is 0V and beyond the emitter resistor is -10V, so Ve is halfway between in accordance with the voltage divider principle. If for example Re is only half the value of RL, then the figure is -6.67V.