How do BJT transistors work in a saturated state?

This is what I know about NPN BJTs (Bipolar Junction Transistors):

• The Base-Emitter current is amplified HFE times at Collector-Emitter, so that Ice = Ibe * HFE


• Vbe is the voltage between Base-Emitter, and, like any diode, is usually around 0,65V. I don't remember about Vec, though.


• If Vbe is lower than the minimum threshold, then the transistor is open and no current passes through any of its contacts. (okay, maybe a few µA of leak current, but that's not relevant)

But I still have some questions:

• How the transistor works when it is saturated?


• Is it possible to have the transistor in open state, under some condition other than having Vbe lower than the threshold?

In addition, feel free to point (in answers) any mistakes I made in this question.

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