bjtcommon-emittersaturation
In BJT for working in active region for a common emitter configuration emitter and base should be in forward bias and base and collector should be in reverse bias. But in case of voltage we say that base-collector voltage should be less than 0.4V.
But if V(base)-V(collector)=0.1 (suppose) then base-collector junction will be in forward bias that means it will enter into saturation mode. But book says it will remain in active mode. How is it possible?
In general, it is the states of the PN junctions inside the transistor which will determine what operation region it is in. However, after gathering some experience, one can deduce the states of the above junctions by inspecting the circuit itself without actually measuring the voltages at the terminals.
An example:
Lets analyze the circuit you've referenced.
Once the switch is closed a current of approximately \$1mA\$ will flow into the base, which will cause:
$$V_{BE} \approx 2V$$
Since this is higher than the minimum of \$0.6V-0.7V\$ for being out of cut-off - the transistor is in one of its operational modes. In reality, the Base-to-Emitter voltage will not rise much beyond \$0.6V-0.7V\$ (due to presence of protection resistor R1), which means that the Base current will be a bit higher than \$1mA\$.
Knowing that the motor is \$12V, 100mA\$, and that the transistor is capable of handling \$100mA\$ Collector-to-Emitter current, we can deduce that:
$$I_C = I_{Motor} \approx 100mA$$
Given that we know (from motor's specs) that the motor will consume \$100 mA\$ at \$12V\$, the voltage on the motor:
$$V_{Motor} \approx 12V$$
Which leads to:
$$V_C \approx 0V$$
But this means that Collector-to-Base junction is forward biased which implies that the transistor in saturation.
The above analysis is quite general for this configuration (full voltage rated motor switched by matching BJT), therefore, in majority of circuits like this one, the transistor will be in saturation.
Experienced engineers perform the analysis above at a glance, knowing that the transistor in saturation a second after they see the schematics.
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.
Best Answer
Here is a bipolar I_V plot; saturation is the far left region; notice all the lines merge in "saturation", not good for linear amplification but good for switches.
Here is the circuit used to produce that plot
simulate this circuit – Schematic created using CircuitLab