This is a lecture on Class A Output Amps by Upenn. Link here: http://www.seas.upenn.edu/~ese319/Lecture_Notes/Lec_20_PowerAmps1_10.pdf
I can't understand how when Vi is negative, it causes VCE2 to turn on. Vi is only connected to Q1 right, so if it's in negative mode, Q1 should be off, and hence behave like an open circuit. How is it, that Vi is still able to have an effect on Vo while negative?
I can get the equation Vo > -Vcc + Vce2-sat, but how does the second equation where Vi > -Vcc + Vce2-sat + 0.7 come about. Vi is not even connected to Q2, and that part of the circuit should be open, how can we deduce this?