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.
In such a circuit the emitter current equals the collector current PLUS the base current:
Ie = Ic + Ib
In normal situations the collector current is much larger than the base current, hence by good approximation
Ie = Ic
But in your case the collector current is zero, so the first equation degrades to
Ie = Ib
Which is exactly what you see: a very small current through the LED.
Another way to look at your circuit is to replace the b-e of the transistor with a diode. This gives the same results.
Best Answer
With 0 on Q1's base, Q1 will saturate to probably within 100mV of the 5V rail i.e about 4.9V on the collector. If it were an NPN emitter follower and you applied 5V to the base, the emitter would be at 4.3V - maybe that's where you are having a minor bit of confusion?
With Q2's base sitting at 4.9V and its collector being at (say) 3.3V you'll have a heavily forward biased base-collector diode (un-natural operations) and may indeed damage Q2 due to excessive current flow.
Iif you had (say) a 10k resistor in series with Q2's base this would protect the device and you would see a voltage of nearly 3.3V on the emitter because Q2 would be saturated-on.