3 transistors connected in series principle?
simulate this circuit &ndash; Schematic created using CircuitLab

Let's make some assumptions first for making the analysis quite simple.


The LED requires current of 100mA for glowing.
All three transistors are identical with gain = 100 for each transistor.


3.The base-emitter (Vbe) is 0.7V for each of the transistors.


I am going to give a theoretical insight of how this circuit works. 


Lets start our analysis from Q3. The current needed to glow the LED up is 100mA. This is the collector current for Q3. Since the transistor has gain of 100, the base current required is 100mA/100 (=1mA). This current should be supplied by the transistor Q2.

Into Q2 now. The emitter current should be 1mA for this transistor. The emitter current is a sum of collector current and base current. We can neglect the base current. So, the emitter current is equal to collector current now. And you can see the colletor current for Q2 is 1mA. Since this transistor also has gain of 100, the base current is now 1mA/100 (=0.01mA). This base current is supplied by Q1.

Repeating the same analysis for Q1, you can see that the collector current for Q1 is 0.01mA and the base current is 0.01mA/100 (= 0.0001=10uA). The current through the R4 resistor is 100nA. Note this current.

Now, lets see what is the voltage at B1. 

The transistor turns on only when the base emitter voltage is just enough to allow required current to flow into base of the transistor. I have mentioned above this voltage to be 0.7V. So, for Q3 to turn on, the node B3 must be 0.7V higher than its emitter. Since the emitter is connected to 0V, the value for B3=0.7V.

Now, looking at Q2, it has its emitter tied to B3 which is at 0.7V. So the transistor can turn on only when the base voltage is 0.7V greater than the emitter voltage. i.e B2=1.4V.

For Q3, the emitter is tied to B2 so, the emitter is sitting at 1.4V. Repeating the same procedure as above, the base has to be at 2.1V. So, B1=2.1V.

The resistor R4 comes to play now. It needs to supply 10uA current. So, the voltage across it has to be 10uA * 1k = 10mV. Since one end of resistor is tied to 2.1V (=2100mV), another end should be connected to 2.11v(=2110mV). You can argue that it is almost impossible to have such precise voltage supply and it's valid if we look at the real-world scenario. But, I am just considering a theoretical aspect of the problem.