bjtsaturation
In below circuit I sweep Vb from zero to 5V Vcc.
And I obtain the following plots for the currents:
I have a bit problem with understanding two things in saturation region:
1-) Why/how does the Ic decreases with Vb after the saturation point?
2-) How can I quantify/formulate Ib at saturation region?
It can be confusing because in a MOSFET the saturation region is something else and they call the "linear" region what would be the "saturation" region in a BJT. Why oh why?
Here's my simplified picture of things for a BJT: -
Note that all the curves for different base currents do not overlap as is commonly shown. If they did overlap there would be no BJT based 4-quadrant multipliers (Gilbert cell). They rely on the saturation region being able to modulate the current for a given CE voltage. Anyway, that's a bit off the mark for your question.
The saturation region does include the scenario when CB is forward biased but I don't think this is particularly helpful - the saturation region (or close to it) must still encompass normal transistor amplification and, as far as I know, this cannot happen when collector and base are forward biased.
Why doesn't further increase in base current cause changes in collector current?
It does up till the point when the collector-base junction is forward biased. The curves look bunched in your diagram (and this is an error basically) but they are still different and for a given low voltage across C-E, the current is proportional to that voltage AND the base current.
Hope this helps.
why the saturation region of the MOSFETs has this name?
The post you link to does explain this, but in case it needs repeating and perhaps backing up with a textbook reference, the saturation region for a MOSFET is called so because the drain current saturates, i.e. basically stops increasing as Vds increases further.
You are correct that the active region of a BJT corresponds to the saturation region of a MOSFET when these devices are used as amplifiers.
The saturation region of a BJT (e.g. when turned on as a switch) corresponds to the triode/ohmic region of a MOSFET.
Some authors also call the saturation region of a MOSFET the "active mode", which does match the terminology used for BJTs. But they also call the triode/ohmic region the "linear mode" which perhaps doesn't help that much because "linear" suggests an amplifier rather than a switch. Linear here again refers to how the MOSFET characteristic looks like in that region rather than any external/use considerations. (Luckily, it seems nobody calls the BJT saturation region "linear mode".)
The only thing that's not confusing about this terminology is the cut-off region, which is the same for both. Here's a summary table for the correspondence (from an external/use viewpoint):
This summary also includes the reverse active region for BJTs, which is seldom used, but it doesn't include synonyms for the triode region; as I said "linear mode" or "ohmic region" are also used to denote the MOSFET triode region.
Best Answer
The collector current in your circuit decreases because, when the BJT is in saturation, by definition, the junction B-C is forward biased.
Therefore you'll have that the collector voltage will be just \$V_C=V_B-V_{BC}\$.
As a result, the current in the resistor Rc will be:
$$I_C=\frac {V_{CC}-V_C}{R_C} = \frac{V_{CC}-(V_B-V_{BC})}{R_C}$$
Therefore the higher \$V_B\$, the smaller \$I_C\$.
To answer the second question, simply use KCL at the BJT. You know that \$I_C+I_B-I_E=0\$ (assuming base and collector currents entering the BJT. Emitter current exits the BJT). Hence \$I_B=I_E-I_C\$. In fact in your plot, \$I_B\$ is the difference of the other two currents.
As a first order approximation, you can assume constant \$V_{BC}\$ and \$V_{BE}\$. This allows you to calculate \$I_C\$ and \$I_E\$ easily.