The first configuration is useful if you want to easily saturate (fully turn on) the transistor so it basically "gets out of the way" (becomes almost a short) so that you end up with a lit LED, current-limited by a 1K resistor.
In the second configuration, you have an emitter-follower. Approximately speaking, the voltage at the top of the 200 ohm resistor above the LED is about 0.7 volts below the voltage at the base of the transistor. The current through the LED then follows from this in a straightforward way.
In this configuration, it behooves you not to have the collector resistor. So hereafter let us assume that it has been removed from the circuit.
The transistor just acts as a current source: that is to say, roughly speaking, the voltage at the base "programs" the voltage on the resistor-LED stack, and the transistor supplies the current via its collector.
The circuit basically acts as a buffer, making it look like the resistor-LED stack has a much higher impedance than it really does. The circuit driving the base "thinks" it is driving the LED with only a fraction of the current that it actually requires.
The circuit is useful if you want to be able to vary the intensity of the LED by varying the input voltage. It's also useful if you want a high turn-on voltage. (Remember, to get X volts on the resistor-LED stack, you need to input X + 0.7V).
For instance, we could use a circuit based on an emitter-follower if we had an DAC (digital to analog converter) which we wanted to use to drive the LED, but discovering that it doesn't have enough current-driving ability to do the job directly. (Of course, a better way to control LED intensity from the digital realm is to use pulse width modulation (PWM): and that basically calls for the first circuit, rapidly turned on and off with varying duty cycles.)
That quote is just wrong.
That circuit is correct, but not for your application as it will require about 8V minimum input to give 6V at the output.
It works by supplementing the output current of the 78xx (7806 for 6V) regulator if the regulator takes significant current (about 600mA) with additional current through the PNP transistor.
Your Lipo batteries will not have a constant output voltage, they will probably vary between 8.4V when fully charged down to about 6V or even less when discharged.
What do you need the 6V for? How much current do you need? What is the acceptable voltage range for the load?
Best Answer
No. The amount of conventional current (not electron current) entering the base and flowing out through the emitter determines whether an NPN turns on (i.e. how much current passes from the collector to the emitter). For a PNP, it is the amount of conventional current flowing into the emitter and out of the base that determines how much current passes from the emitter to the collector). The base-emitter junction behaves like a diode which is where the typical 0.7V value across the base-emitter comes from.
Note that things like "the voltage on the base" have no meaning for a transistor. The voltage relative to what? Relative to ground? What if neither of the other two pins are at ground? Voltage is relative and the transistor does not know what the voltage is anywhere except between two of its pins.
Saying "the voltage at X pin" is like saying "the distance at your house". It doesn't mean anything.