I have this classical Schmitt trigger circuit from an old Elektor magazine.
It describes the following, when U1 is lowered (from a high state):
If the right transistor cuts off entirely and the left transistor is completely turned on, then there is on the emitter resistor about 9V / 11 = 0.8 V. Then one should add the BE-voltage, that the left transistor needs to conduct. The low threshold is thus 1.4V.
Where does the number 11 come from? Why does one need to add the emitter voltage to the BE-voltage?