Hello everyone
I was hoping you could be of help to me

If you see the picture attached, in the books explanation both inputs are tied so they are equal, and when Vin = 0V or low..... Qi is in saturation I get that, Furthermore Qs and Qo are in cutoff I get that too..
I just can't seem to grasp why Qp is in forward active and not saturation, as forward active would require that the current the collector of Qp be β*Ib (that is the base current of transistor Qp) however, wouldn't such a current be quite to difficult to obtain since the current has no path to follow... and if we assume that it is possible (if we were to assume that the base current is extremely small), what implies that the base current will be of such a small nature?

Thank you

 

Qp is configured as an emitter-follower.
An emitter follower can't go into saturation unless the base voltage supply is greater than the collector voltage, and that's not possible for that circuit.
Thus the minimum collector-emitter voltage when on, will be equal to the base-emitter voltage of about 0.7V, which is not in saturation.

 

Qp is configured as an emitter-follower.
An emitter follower can't go into saturation unless the base voltage supply is greater than the collector voltage, and that's not possible for that circuit.
Thus the minimum collector-emitter voltage when on, will be equal to the base-emitter voltage of about 0.7V, which is not in saturation.

Thank you, note taken .