1)
In the first configuration (the collector is not connected to the power supply) your NPN essentially behaves just like a regular PN diode, and the circuit is equivalent to this one:
simulate this circuit – Schematic created using CircuitLab
The reason for this behavior is that NPN transistor may be seen as two PN diodes connected back-to-back:
Leaving the Collector open and connecting the Base to the highest potential leaves you with simple forward-biased PN diode, having forward voltage drop of \$\sim 0.72V\$.
BJTs are not employed in this configuration because it is cleaner and cheaper to use a proper PN diode in such cases.
BTW, the usual way to use NPN as a switch would be to put the load (in this case the LED) between the power supply and the Collector:
simulate this circuit
In the above schematic you can also see one of the approaches for utilizing the same power supply for both the Base and the Collector driving (which answers your second question). The values of the resistors should be chosen based on the parameters of the LED and the transistor. The voltage divider formed by R1 and R2 is constantly consuming power (even when the switch is open) - this is the main disadvantage of this simple scheme.
The purpose of R_B is to pull Base's voltage to ground when the switch is open - usually you don't want to leave the Base floating. The value of this resistor may be taken very big, such that its presence does not affect the voltage divider.
NOTE: as mentioned by Jippie, R2 may be completely omitted from the circuit. In this case the voltage divider will be formed by R1 and R_B when the switch is closed. When the switch is open, the absence of R2 will prevent from current to flow, therefore the OFF power consumption will be reduced. There are cases when you do need R2 though: its presence reduces the maximal voltage on the switch, and this may be desired in some cases (depends on the switch you're using).
Hope this helps.
8050 transistors are made with the Japanese and the American style pinouts- it is not a JEDEC or JIS registered part.
Looks like the one you happen to have has the Japanese pinout.
E C B
Best Answer
To make current flow through the base of the transistor, you have to make the voltage on the base about 0.7Volts higher than the emitter.
Connected as you have in the schematic marked "Figure 2," that means you only need 0.7Volts to turn the load on completely.
If you were to put the load between the emitter and ground, though, you would have to drive the base to a much higher voltage in order for the transistor to conduct and power the load.
Since the voltage across the load can vary, you can never be sure just how much voltage to put on the base to make the transistor conduct. You can also never be sure just how far you've "turned the transistor on." If you were trying to make an amplifier or motor speed control, it would be difficult to control the output since would depend in part on the load.
Another thing to consider is that if you are using a 5Volt signal from a microprocessor to control the transistor, then you might not be able to drive the base of the transistor high enough if you are using the transistor to switch a 12Volt load in the emitter circuit.