I have seen this thread and I have a question about the picture describing the movement of electrons and holes.
Basic operation of a bipolar junction transistor

Consider this picture bjt transistor Microelectronics, Naeman

Here is my step by step way of thinking. B-E is forward biased and B-C is reverse-biased. Because B-E is forward biased, electrons move to the right from the emitter region to the collector region (because p is so narrow). Holes try to "meet" with the electrons at the left pn junction, \$i_{B1}\$.

Isn't \$i_{B2}\$ exactly doing what \$i_{B1}\$ is doing: combining its holes with the electrons at the pn junction?

  • \$\begingroup\$ I personally think it is confusing to draw holes and electrons on the same image. \$\endgroup\$
    – jippie
    Feb 23, 2016 at 17:35
  • 1
    \$\begingroup\$ Base region is indeed very thin, also consider the fact the collector voltage. The electrons in the emitter an base region "see" a high collector voltage "in the distance", making them all happy and wanting to travel over there. Some of these electrons can gain enough energy from that distant voltage to be pulled through the base-collector depletion region. \$\endgroup\$
    – jippie
    Feb 23, 2016 at 17:43

1 Answer 1


Two hole current components:

  1. Since B-E junction is forward biased, electrons move from emitter to base and holes move from base to emitter. (The majority carriers are injected to the other side). The current due to injection of holes is \$i_{B1}\$.

  2. The electrons injected from emitter move to the collector region but a small recombination happens in the base while this travel. This base-recombination current is \$i_{B2}\$.

It is true that these excess holes will get recombined. But the holes in \$i_{B1}\$ recombines in emitter region where as the \$i_{B2}\$ holes recombination happens in the base. Hence they are shown as two different components of currents.


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