A voltage GREATER THAN THE POWER RAIL on the base
QuoteA voltage GREATER THAN THE POWER RAIL on the base I find that explanation quite confusing. Bipolar transistor is current amplifier, you apply small current trough base-emitter junction and you getting much bigger trough emitter-collector circuit.
I think that means you need the Base to be at a voltage greater than +5V. That's just my limited understanding.
One additional thing you need to be careful with PNP high side switches is the voltage used to drive the load. Normally it is best to use the same voltage to drive the load that is used to power the microcontroller. Consider the following.: Suppose the load voltage is +12V and the microcontroller is running at 5 volts. Assume the output pin is high, at 5V, and the base resistor is 1000 ohms. Base current would be calculated byIb = (Vcc - Vp0 - Vbesat)/R1 =( 12 - 5 - .7)/1000 = .0063A = 6.3maUnless the transistor has exceptionally low gain, it will be turned on even though with the output pin high it should be turned off. Worse yet, the microcontroller output pin will be seeing more than 5V, which greatly exceeds the usual limitation of the microcontroller supply voltage plus 0.3V. The microcontroller would likely be damaged in this situation. Part 12 discusses ways of driving a PNP transistor when the load must be driven with a higher voltage than the microcontroller.