Super quick PNP tutorial please!

[TheBaconWizard]

Ok, so the short to GND from the base allows current to flow and there's R1 to limit the current..  but why the track from +12v to the base with R2? Why would you need that?

I've seen it pretty much everywhere, so clearly it's necessary... I would have thought it would work with just the short to GND (and R1) alone.

[Andy Watson]

... I would have thought it would work with just the short to GND (and R1) alone.

It would, but ...
It is a bit "belt and braces" but without R2, any leakage currents would be amplified by the transistors gain - so you would have the load being partially turned on - this can easily happen if the terminal of R1 is floating (try it - remove R2 and attach an LED as the load - just touching R1, in the presence of a presence of mains e-field will cause the LED to light).
Another advantage of using R2 is that it lowers the drive-impedance as seen by the base, allowing faster turn-on and especially faster turn-off when the transistor is switched. You don't specify the load so this may or may not be a consideration in your application.

[TheBaconWizard]

AH right, I understood most of that..

As it happens, the load is an LED string (and really quite powerful: 5 amps @ 6.8v)

I'm not relying on the PNP shown especially, just needed a run-down of the principle. It's obvious to me that I'll need the 2nd path with R2.

Thanks for your help

[orolo]

Another reason: stability, predictability and preventing runaway. If you don't use R2, and your transistor starts to heat up driving those 5 amps, Veb wil decrease, so the base current will have to increase to compensate the drop, which means that the collector current will be amplified, which means more temperature, etc. The voltage divider R1/R2 adds some stiffness to Vbe. Using a diode thermally coupled to the transistor, instead of R2, could be a crude temperature compensation trick.  Using a couple of diodes instead of R2, and a small emitter resistor would be a classical temperature compensated constant current driver. Since the emitter resistor would disipate a lot of power at 5 amps, subsititing R2 by a whole temperature compensated active load is the best option: I've seen that done in some power amplifiers.

Strictly speaking: if you don't use R2, base current is given by: Vcc = Veb + Ib * R1. This should be solved via load line analysis (or using the Lambert function), since Veb is a logarithmic function of Ib, and a rather complex function of temperature. If you use R2, this resistor will be in parallel with the dynamic resistance of the eb diode: if R2 is significantly smaller than the dynamic resistance, Veb will be more dependent on R2 than Ib. So the circuit will be more stable and predictable.

[mikeselectricstuff]

As it happens, the load is an LED string (and really quite powerful: 5 amps @ 6.8v)

In which case you should be using a P-Channel MOSFET to reduce losses in the switch

[alsetalokin4017]

As it happens, the load is an LED string (and really quite powerful: 5 amps @ 6.8v)

In which case you should be using a P-Channel MOSFET to reduce losses in the switch

D'ya think? Especially since the max continuous collector current on the 2n3906 is only -200 mA.........

[orolo]

In which case you should be using a P-Channel MOSFET to reduce losses in the switch

I don't exactly disagree, but if constant current is the objective at Vcc=12V, the transistor should be active, and the losses, more than 5 Amp * Vsat, would be about 5 Amp * ( 12 - 6.8 ) = 26W, horrendous, regardless of transistor type.