I dont understand the theory behind the two transistor switching circuit . This is because I have used the parallel port to switch on a BC338 witch lit up a lamp from a secondary power source.
In the picture, if it helps could you describe the operation of the circuit please.
The two transistors are connected as a darlington transistor that has a high current gain.
When the input voltage to the 10k resistor goes positive more than about 2.0V then the two transistors turn on and the output device gets the supply voltage minus 0.7V.
When the input voltage to the 10k resistor goes down to 1.0V or less then the two transistors turn off.
The diode arrests the inductive voltage spike when the output device turns off.
If the load (in your case a light or LED) can be driven by a single transistor (usually Ic is less then 100mA) you don’t need to add another one, but if you want (or need) to increase the output current and in the same time maintain the input current at very low level you just add another transistor (darlington configuration) or use a relay driven by a single transistor ..
Since the schematic includes an inductive kickback diode I will assume that it is intended to sink high current. The darlington configuration has a current gain of the first transistor beta multiplied by the second transistor beta, i.e. it's high.