In order to stay in the saturation, the voltage between collector and emitter has to be exactly equal to Vce in the char above, is that right?
But when the transistor is working at Vce and 0 < Ib < (Ic/beta), meaning Ic is not at the maximum capability for the transistor. The voltage drop between collector and emitter is a fixed value for the transistor made of certain material (0.7V for silicon and 0.2V for germanium). Is that right?
And, Rc, the resistor in serial with collector takes the remaining of the voltage from power supply. E.g., 5V power supply Vce is 0.7V, as long as the transistor is open. And, the voltage left for resistors in serial with collector is 5V – 0.7V = 4.3V. Is that right?
So, the value for Rc to limit the current Ic from its maximum, say 0.6A, would be 4.3V / 0.6A = 7.17 Ohm?
When using voltage divider(two resistors in series) to control the Ub, voltage at base, making Ub higher than the 0.7V voltage drop required for Ube to open the transistor, why could the Ub be higher than 0.7v? I thought Ub would always stay at 0.7V for silicon knots and 0.2V for germanium knots.
As I'm still waiting for my NPN transistors to arrive to do experiment, my breadboard experiment with PNP transistor BC557 B shows, when Ub is lower than Ue, which is required to open the transistor, it seems Ub is decided by Uc minus the voltage drop between c and b, instead of the voltage divider for base pin.