I have read some answers about how BJTs work, but I'm still not convinced.

To show my doubts better, suppose we extract base and collector from BJT NPN transistor with all its construction details (thin base, doping things, etc.). Let's treat it like odd diode and connect it reverse biasaed way to a battery.

Would electrons from battery minus clip diffuse through P material (is there chance this strange diode would conduct)? I mean why in diode reverse biasing stops current, but in BJT doesn't do the same? What helps transferring electrons through depletion region?

  • \$\begingroup\$ The fact that the depletion region has been crushed to nothing against the emitter. \$\endgroup\$ Jan 9, 2015 at 20:30
  • \$\begingroup\$ youtube.com/watch?v=f5MgK21bON8 Imagine the race cars are the electrons, the track is your emitter. Then collector is the gravel. Some cars just drive too fast to make the corner ;o) \$\endgroup\$
    – jippie
    Jan 9, 2015 at 20:37
  • \$\begingroup\$ So would this extracted base-collector diode conduct when reversed biased? \$\endgroup\$
    – Lord_Volt
    Jan 9, 2015 at 20:42
  • \$\begingroup\$ Only if the voltage is high enough to overcome the depletion region. \$\endgroup\$ Jan 9, 2015 at 21:25

1 Answer 1


The key to it all is the minority carriers in the base.

Your suspicion is correct that if all you had was the CB junction it would just become a diode. Reverse biasing this diode does not give you any current. The p-base of an npn is full of holes and the n-collector is has lots of electrons. In reverse bias the majority carriers move away from the junction on both sides and you do not get any current, just like an normal diode.

The tricky part happens when you forward bias the base-emitter junction. The holes in the p-base move towards the BE junction and the electrons in the emitter also move towards the junction. Some of them annihilate each other but because of the doping inequity a lot of the electrons from the emitter pop through into the base!!! As a result they can keep propagating through the base to the collector and you get the collector-emitter current that you were hoping for.

You should take long look at the diagram labelled Lecture 7 - Slide 12 http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-012-microelectronic-devices-and-circuits-fall-2009/lecture-notes/MIT6_012F09_lec07.pdf Holes are green and electrons are blue.


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