Help with BJT regions of operations

[iScience]

i understand this graph

the ordering of the regions make sense to me;

cutoff: little/no base current ----> no current anywhere
active: base current moderately past V_be -----> linear current amplification
saturation: etc

but here..

http://[URL='http://i.imgur.com/JpGqXUQ.png]']http://i.imgur.com/JpGqXUQ.png[/URL]

what is going on?? the base current is a third dimension where if i increase it further, eventually it should lead to saturation. but this graph says saturation is inversely proportional to (basically) V_cc .

i thought active region meant that the number of electrons being combined with holes at the base is linearly proportional to the electrons passing to the collector; and saturation was where the number electrons combining at the base was non-linearly proportional to that passing onto the collector.

so why is saturation at the region where V_ce is small and active where V_ce is large, shouldn't it be backwards??

clarification please!

thanks

 

[cabraham]

Increasing Ib results in an increase in Ic only if the power supply and collector resistor allow it. Suppose the supply is 12 volts, and the collector resistor is 1.0 kohm. The Ic value is limited to 12 mA, cannot exceed this value. If the beta value (hfe) is 100 and the base current Ib is 50 uA, then Ic = 5.0 mA. Increasing Ib to 100 uA results in Ic increasing to 10 mA. With 10 mA value of Ic, the 1.0 kohm resistor drops 10 volts, leaving 2.0 volts for Vce.

Now if we increase Ib to a value of 200 uA (0.20 mA), the value of Ic cannot be 100*0.20 mA = 20 mA, since the collector resistor Rc limits Ic to 12 mA. The bjt saturates, Vce attains a value of around 0.20 volts, and the Ic value is 11.8 mA.

If, however, the Rc value was reduced to 470 ohms, then with 0.20 mA of Ib, the Ic value would be 100 times that or 20 mA. The Rc voltage drop is 9.4 volts, and Vce is 2.6 volts. Did I help?

Claude