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This circuit uses a PNP (top, right) to switch a load (bottom, right resistor) by tying the load to the "high" (voltage) side. This is an alternative to low-side switching using an NPN to tie the load to ground.
To turn a PNP "off", current must not flow from emitter (leg with the arrow) to base (middle leg). Since voltage flows through a diode when the voltage from cathode (positive) to anode exceeds the forward threshold voltage (~0.6V), the base must bias the junction between itself and the emitter to prevent current from flowing. This means the base must be close to supply voltage, so switching a large load with logic levels is not practical - a 3.3 or 5V signal could not turn the PNP "off".
Fortunately, a pair of resistors and an NPN transistor can be used interface the low voltage level with the high level. In this circuit, the top resistor is used to provide the 15V required at the base to keep the PNP transistor off. To turn the PNP "on", the NPN simply ties the PNP's base to ground.
The hardest part about this circuit is realizing where current flows when the load is switched on (NPN's base is logic-level high, PNP's is tied to ground via the NPN's collector-emitter path). In this scenario, the top two resistors may appear to be acting as a voltage divider, but really the PNP's emitter-base junction is supplying the junction between these resistors with a voltage close to supply (minus the voltage drop across the diode).
This means the voltage across the top resistor is only ~0.6V (supply on one leg, supply minus diode drop on the other), so according to Mr. Ohm there is only 0.6mA flowing through the resistor. Note there is a significant current flowing through the second (middle) resistor and the NPN to ground - supplied by the PNP's emitter-base path. This resistor is present to prevent destructive currents.
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