Electronic – What’s the difference between NPN and PNP transistors

I am very sure that the noise generated by holes and electrons is quite the same. If there are differences in the NPN and PNP transistor, they must be an indirect cause and not caused by the type of carriers.

Essentially there are only two dominant types of (white) noise sources

1) shot noise (this is due to the discreteness of carrier flow when passing a potential barrier such as a PN junction). This is independent on the device structure, doping levels, material compostition, carrier type and so on. It is fundamental and the same for holes and electrons (independent on temperature, only depends on current!).

2) thermal or resisitive noise (due to brownian motion of carriers in a resisitive layers) This is also fundamental and only depends on temperature and overall resistance.

Hence transisitors, having e.g. different layer resistances will also generate different amount of noise. This may be the cause for different behavior of PNP and NPN transisors (or more precisely, between any pair of transistors, not just different polarity). Shot noise is quite the same in all bipolar transistors.

BTW: The noise generated by surface states (as mentioned by rawbrawb) is always some kind of 1/f noise and vanishes at a certain frequency (in contrast to the above two white noise sources). Contrary to white noise (which is quite easily explained) 1/f noise is a vrey complicated phenomenon with many different effects that can cause it. It strongly depends on material quality, device strucure and so on. Strong differences between devices are expected for 1/f noise. So always look for the frequency the noise is specified at!

Remove R5 and you will have what you describe. The configuration of Q5 is called common collector or emitter follower. Essentially, the voltage at the emitter is the voltage at the base minus 0.6V, but the emitter current can be much more than the base current, because the gain of the transistor will draw more current from the collector. Thus, it's a current amplifier.

Remember, the base-emitter junction is a diode. So, the emitter will be about 0.6V below the base if you forward bias it. With R5 removed, you can pull the emitter up to \$V_{cc} - 0.6V\$. With R5 present, you won't get it as high, since some voltage will be dropped when current flows in R5.

Since there are things that will limit the current in the emitter leg of Q5, you don't need R5 to limit the base current, which isn't true of Q2 or Q4, which have their emitters shorted to ground, or Q1, with its emitter shorted to \$V_{cc}\$.

See Why would one drive LEDs with a common emitter?

There isn't much difference in performance. In circuit 1, the anode of D1 will be at \$V_{cc} - 0.2V\$, whereas in circuit 2, it will be at \$V_{cc} - 0.6V\$, so the LED current is a bit higher in circuit 1, assuming R1 and R4 are the same value.

Circuit 2 has the advantage that the base current goes towards powering the LED, but since the base current is small, this isn't a big effect.

The last subtle difference is that in circuit 1, Q1 enters saturation, which will charge the base-emitter capacitance. When you then turn it off, this capacitance has to discharge before Q1 really goes off, adding a bit of delay from when your MCU output goes low to when the diode gets switched off by Q1. Q5 never enters saturation, because the emitter voltage is brought up to just the point where the transistor enters saturation, but not more. So, no turn-off delay. The delay is very short, and probably not significant until you are switching at least 50kHz.