Internal behaivor of BJT in saturation

I'am trying to understand to internal behaivor of BJT when it's in saturation. why the current Ic is highets in saturation?

I'm mean when BJT is in saturation Vcb and Vbe are in direct. So if we see the model erbers moll for a emiter comun of npn



We can see that the emitter to the collector loads injected, and the collector to the emitter loads injected. We know the emitter has more impurities than collector so the current that the emitter injected to collector are bigger than the current that collector injected to emitter. As the currents are different sense the difference of the currents is not zero but is smaller.

So, Why the current in saturation is maxim and depends of the source and the resistors? I mean if looking the erber moll model the current flowing through transistor is the difference of the current injected por the collector and emitter and this is smaller.

Greetings.

As the collector current increases the collector voltage falls due to the external I X R drop. When the collector voltage reaches about .2V its in saturation. The currents through the transistor still obey Ohms/Kirchoffs law.
Frank

When saturated, the collector/base becomes forward biased and begins to conduct.

With a resistive load, you have to pull more current to
get to saturation than to stay linear - simple Ohm's law.

As you enter saturation your hFE drops. Some of this,
because base terminal current begins to be "stolen" by
the B-C junction and this does not contribute to base-
emitter current anymore, and goes out the collector in
the wrong direction.

The benefit of saturation is the conductivity modulation
of the collector, when you forward bias the B-C junction
you inject minority carriers to the collector and drive the
resistance down in the lightly doped region that dominates
Rc. Without this the device would make a dismal power
switch. But what you do to make a good power switch
(ohmically) often fights against good switching (storage)
times - where do you want your lifetime@current, really?

As you enter saturation your hFE drops. Some of this,
because base terminal current begins to be "stolen" by
the B-C junction and this does not contribute to base-
emitter current anymore, and goes out the collector in
the wrong direction.

Click to expand...

when you says that the currents goes out in the weong direction. This direction is oposite of the right direction of the current (which it's the oposite direction of mayority carriers of emiter) so the current goes small. Such Ic can't be biggest.

So the behaivor of BJT in saturation is like closed circuit, and for that the Ic obey Ohm's law. I can't understand that BJT in saturation is like closed circuit.

Let me see if i can explain this. Let see the next graphics



The blue arrow is the current (Icb0) by the Vcb and the red arrow is the current (Ibe0) by Vbe. The currents have opossite direcction but Ibeo>Icbo because emitter is a little more doped. When BTJ is in active zone there isn't Icbo and the collector corrent is by emitter current.

Icb0=Icss ( exp( Vcb/Vt)-1 ) , Ibe0=Iess ( exp( Vbe/Vt)-1 )

When BTJ is in saturation : Ic = Ibe0-Icb0 and Ibeo>Icbo

When BJT is in active zone: Icbo=Icss ≈0 but there is collector current that it's Ic=βIb

βIb< Ibe0-Icb0 ??? I do not think so.