Hello.I am having a problem in understanding the Collector Characteristic Curves of a transistor.

In the figure above,why is Ib=0 till the cut-off Region?I have read my text book but I do not understand the relationship of Ib with this graph.Please help me out.Thank you

 

The graph says that, with no base current, the collector leakage current is about 5 or 6 mA.

 

You can imagine this as a 3-dimensional graph. A relation among base current, collector current and Vce. The blue lines are points that satisfy that location.

The red line is a load line and is part of another graph, the one that shows voltage and current of the load.

The equilibrium (Q) point is where all those quantities can coexist.

 

Why does saturation region increases as Ib increases?

 

I don't quite understand the phrase "the saturation region increases". Please elaborate a bit.

 

Saturation as far as voltage is concerned, is when the collector voltage is less than the base voltage, and so now the VCB is forward biased rather than reversed biased.

So looking at the graph if you apply 120uA. of base current your going to have a lot of collector current as the graph shows, however if the voltage at the collector is small VCE around 1v. as shown then the base is more positive than the collector meaning the transistor is saturated, but if you bring the collector voltage VCE higher with that amount of base current, now the collector is getting into the region of making the VCB to be reversed biased as it should be to work as an amplifier.

So the larger the base current the more VCE needs to be to keep the transistor in its linear region of amplification, characteristics.

 

I don't quite understand the phrase "the saturation region increases". Please elaborate a bit.

He's talking about where all the Ib curves converge to a single line. Hobbyist explained it pretty well.

 

What is the chief physical explanation for this effect? Voltage drops in extrinsic / ohmic resistances?