In what way do I have to look at this to be able to understand?
The base current increase (decrease) is due to an increase (decrease) in \$v_{BE}\$.
The increase (decrease) in \$v_{BE}\$ increases (decreases) the injection of carriers from the heavily doped emitter.
Most of these carriers cross the thin base region without recombining and are then swept across the base-collector junction into the collector region. A small percentage don't and these form the base current.
Update to answer a comment:
Once it has been biased isnt Vbe=0.7V. Now when we apply a small ac
signal of the order of milli volt isnt the change negligible?
No, the collector current is exponential in the base-emitter voltage:
\$i_C = I_Se^{(v_{BE}/V_T)}\$
To get a feel for this, consider this question: to double the collector current, how much would \$v_{BE}\$ need to increase?
For example, assuming the bias value is \$V_{BE} = 0.7V \$, increasing this voltage by a mere \$17 mV \$ (an increase of just under 2.5%) will double the collector current.
Another approach:
According to the collector current equation, if we change the base-emitter voltage from its quiescent value by some small amount, the change in collector current is approximately:
\$\Delta i_C = \dfrac{I_C}{V_T} \Delta v_{BE}\$
As a typical example, let the quiescent collector current \$I_C = 1mA \$. At room temperature, \$V_T = 25mV\$.
Then, for these numbers, the collector current changes by 4% when the base-emitter voltage changes by 1mV.
Answering questions:
@LvW Just out of curiosity: What if VCE = VBE? C-B pn junction won't
be reverse-biased then, so it won't attract electrons in base region.
Thus, IC will be zero, and IE will be equal to IB?
But the C-B diode is not forward biased. This is an application where the BJT is used as a diode and no "classical" amplification is possible (transition region between saturation and amplifying region).
IC and IE are controlled and only controlled by VBE; IB is just a side
product; Once VCE is greater than VBE, its specific value does not
matter, because E-B junction is reverse-biased. Am I right?
It does not matter too much - on the other hand: Look at the Ic=f(VCE) curves. Ic slowly rises with VCE because of the Early-effect.
Given VBE, IE is fixed, and as a result, the sum of IB and IC is
fixed. When VCE < VBE, what IB and IC are depend on VCE. The greater
VCE is, the greater IC/IB is. However, the value of IC/IB is capped by
"beta", which is reached when VCE = VBE. " Is this right?
In this case (VCE < VBE) the C-B diode is open and there is a small current Ic which has a direction opposite to the "normal" Ic direction. Example: For VCE=0 we have a current Ic which is negative (The Ic=f(VCE) curves do NOT cross the origin!).
Best Answer
It's much better to think of a bipolar junction transistor (BJT) as being current source, rather than having resistance.
Why you may think? Between the collector and emitter, you can have a voltage across them, and a current flowing between them, and V/I = ohms, right?
The difference between describing a transistor as having resistance, and having a current source (we tend not to make a difference between the terms current source and current sink when using these models), is what stays constant when you change other things. In this case, when you change the collector voltage, the collector current stays more or less constant.
It's not very useful describing the V/I relationship as a resistance, if the resistance varies with the collector voltage. It's always most useful to find something that's constant.
Let's take two good DMMs which, when on the resistance range, one applies 1v to components and the other applies 2v. If we set them to measuring a 1k resistor, they would both read 1k. If we set them to measuring a transistor whose base current was such that the collector current was 10mA, one would read 100 ohms, the other would read 200 ohms.
This is why although a transistor has a defined V and I, and their ratio has units of resistance, we do not say that it has a resistance, because it will be measured differently by different resistance-measuring devices. Two current measuring devices will give substantially the same reading as each other.
However it's just occured to me that DMM tend not to work like that when measuring resistance. They don't apply a voltage and measure the current, they apply a current and measure the voltage. With an excitation current, things get even worse for our putative 'transistor has a resistance' model. Consider two DMMs, one that measures voltage with a 1mA current, and one that uses 2mA. Let's say the transistor IB has been adjusted to pull 1.5mA when the collector has a sufficient bias. The 1mA DMM will be pulled down to VCEsat, somewhere in the 0.1v to 0.3v range depending on the transistor detail, so will show a 'resistance' of perhaps 0.2v/1mA = 200 ohms. The other DMM's 2mA will not be pulled low by the transistor, and most DMMs read OL (overload) when insufficient current is drawn from the test leads to reduce their output voltage.