Output Charecteristic curve of BJT's

[Badar]

These curves are drawn between Ic and Vce at different constant Values of Ib.
Suppose at 10 microampere Ib if we increase Vce Ic increase rapidly called saturation region after that Ic becomes almost constant till breakdown region called active region.
Load line is THE SUMMERY OF ALL OPERATING POINTS OF TRANSISTOR.
Q point is chosen so that it may have maximum swing up and down for different values of Vce in A.C.
That is what i know about transistors.

My question is that we have a big active region for a constant value of Ib.Shouldn't these whole active regions for different Ib's be the operating points.
But we call only the load line the summery of operating points while it just takes one point per active region (at different values of Ib).While transistor operates at all other points.
Is this definition of Load Line wrong or i misunderstood it.
I think i've conveyed my idea.

NOTE:- b, ce, c WITH V AND I ARE IN CAPITAL. I TYPED THEM IN SMALL LETTERS SO THAT YOU TAKE IT IN SUBSCRIPT

 

[Roff]

The load line plots the resistance in the collector-emitter circuit. The intersections of the load line and the transistor curves constrains the operating points for that particular load. The transistor can indeed operate at any point on the graph, but only if you change the load line so that it passes through that point. This will require changing the load resistance and/or the supply voltage.
A google search for "load line concept" brings up several sites that might help in your understanding.

 

[Papabravo]

The load line, like any linear function can be specified either by two points, or by a slope and an intercept. Let's take the first case in a common emitter configuration.

1. If no current is flowing through the transistor, or the load, then the voltage at the collector must equal the supply voltage. Why? because Ohm's law requires that if there is no current in the load resistor there must be no voltage drop across it. The transistor is in cutoff. This is the x-intercept of the load line.
2. If the transistor was a dead short, what is the maximum current that could flow? Ohms law again -- it's Vcc/R, where R is the load resistor. This is the y-intercept of the load line.

Now ask yourself what is the slope of the load line? According to logic, reason, and the phase of the moon, it must be: -(1/R). As you already know a transistor has some voltage Vce across it represented by the separation of the characteristic curves from the y-axis.