I'm confused as to where he gets the equation for Vo from? (Vo = VCE + ΔVo) Vbe is out input signal being applied to the base and has a peak voltage of 5mV. Since this is an npn transistor, to remain in active mode Vcb > -0.4. Vcb = Vo-Vbe > -0.4 So Vo > -3.95V ? I don't see how he gets that Vo >= 0.3. Can someone clarify?