bjtdigital-logicsaturation
I simulated this circuit using Qucs:
With Vin = 5V, VB = 0.746V and VC = 0.024V which means that the BJT is operating in the saturation region.
But I don't understand why. Vcc = 5V and Vin = 5V. RB = RC = 1k ohm. So I expect that VB = VC and the base-collector junction is reverse biased which means that the BJT is in the forward-active region.
What is the wrong with my analysis? and is there a "trick" that I can use to determine the mode of operation of a BJT?
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.
The saturation current of a PN junction, as you correctly said, depends on the cross sectional area of the junction itself.
In fact, if you look at a datasheet \$ I_{CBO} \gg I_{EBO} \$, confirming your idea.
Moreover, Sedra/Smith (I'm looking at the 6th edition, page 361) says:
The structure in Fig. 6.7 indicates also that the CBJ has a much larger area than the EBJ.
As you said, the collector-base junction (CBJ) has a larger cross sectional area than the emitter-base junction (EBJ). They then continue:
Thus the CB diode \$ D_C \$ has a saturation current \$ I_{SC} \$ that is much larger than the saturation current of the EB diode \$ D_E \$. Tipically, \$ I_{SC} \$ is 10 to 100 times larger than \$ I_{SE} \$.
Best Answer
In saturation you'll have \$I_c < \beta I_b \$. So calculate \$I_b\$ as (5-0.7)/1k=4.3mA. Assuming beta of 100 (or even 10), you'll have an \$\beta I_b\$ in the 43-430mA range. But through the collector you can only have at most 5mA due to \$R_c\$ limiting the current. So the transistor is in saturation.