The ratio of the current to base vs the current to the collector

I believe that you're reading books on analog electronics - these books are not aiming to provide you with an understanding of semiconductor devices. Take any book on semiconductor physics and you'll find the complete derivation there.

In short (the below applies to NPN BJT):

It is because Collector-to-Base junction is reverse biased in the active mode of operation.

This reverse biased junction just "passes through" the carriers injected from Emitter into Base and then diffused through Base to Collector. The amount of these carriers is dependent upon BE voltage, and this whole amount is swept across reverse biased CB junction, regardless of CB voltage (as long as it is reverse biased). Therefore the current depends strongly on \$V_{BE}\$, but is independent of \$V_{CB}\$.

It is the depletion region charge which creates the electric field which "sweeps" the carriers from Base to Collector side of the junction.

Please note that there are second order effects ("weaker" effects) which contain \$V_{CB}\$ dependency (see for example Early effect).

I think you need to jump (maybe not in at the deep end but) in a bit deeper and I would recommend studying the following graphs for a BC547 transistor. The data sheet is here: -

The top-left graph shows quite a few things: -

• For a given base current, altering the collector-emitter voltage (above saturation) you can see that the transistor tends to behave like a constant current source.
• In saturation the transistor tends to behave like a variable resistor i.e. a given bas current means that the collector current is roughly proportional to collector-emitter voltage (nice for variable gain control)

The top-right graph indicates what voltage can be expected to be seen on the base for driving a certain collector current with collector-emitter voltage held at 5V.

Bottom-left indicates how much current gain you can expect from the device.

Bottom-right is another representaion of what happens in the saturation region.

So can anyone explain me how the collector voltage(V (output)) will vary with variation in base current with respect to the following figure??

The trick here is asking the question how the collector current will vary with base current and once you have that you can decide what volt-drop will be across the resistor. From this you can check that the transistor is not entering saturation (because that puts you in the linear parts of the curves where collector current is more defined by collector voltage) and you'll need to re-appraise what the collector current is to recalculate the resistor volt-drop (iterations needed).

The bottom line is - try to fully understand the top-left graph - imagine a base current of (say) 200uA - ask yourself what collector current flows when the CE voltage is between 4V and 14V - the answer is about 50mA - hence the BJT is generally regarded as producing a constant current for a given base current but, the reality/detail is that collector current will vary between about 47mA (4V on collector) and 53mA (14 volt on collector). You might also start to notice that all the "constant-current" sections have an ever increasing slope and they all "point" to a position on the X-axis that is the same - it's called the Early voltage but a little too much to go into at the moment!!

Try to get a feel for this and recognize the onset of saturation and how it reshapes what I've just said - you can see, that for lower base drives, saturation is entered when collector voltage is less than 1V but, at higher base drives, saturation is entered at 2V or more - there is no exact definition of saturation that can be derived from the graphs so you just have to develop a gut-feeling for what it means.