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How does NPN transistors actually work? In the picture below, I understand that electrons flow in A as the n-p junction of the Emitter and Base is forward-biased. However, the p-n junction of Base and Collector is reverse-biased, so how does the electrons in B flow?

enter image description here


marked as duplicate by Leon Heller, PeterJ, Andy aka, Ricardo, Scott Seidman Jun 30 '15 at 23:46

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  • \$\begingroup\$ Do some youtube search, there are some people that explain it quite ok there, as it works best with some images. \$\endgroup\$ – PlasmaHH Jun 30 '15 at 12:13
  • \$\begingroup\$ A very simplistic view on this: The E-B junction is in forward, the B region is very narrow/short. So electrons that happily cross the E-B junction get too close to the collector and are sucked into it (the collector). As I said, this is a simplistic view, not scientific in any way but maybe helps you to understand. Dave Jones is the man you need to see: youtube.com/watch?v=qUeK7pHe0rI Just watch that video (and all his other videos as well) and become a 1st rate engineer :-) \$\endgroup\$ – Bimpelrekkie Jun 30 '15 at 12:16
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    \$\begingroup\$ Such questions that can easily be answered with a Google search are closed. \$\endgroup\$ – Leon Heller Jun 30 '15 at 12:45
  • \$\begingroup\$ If you really want to know how it works, I suggest starting with a good book on Solid-State physics. I used Kittel, which dates back more than 50 years. If you just want to design circuits, you should study the characteristics. \$\endgroup\$ – Spehro Pefhany Jun 30 '15 at 16:01

Very short (and incomplete) explanation:
There are many things to understand first but I think the most important thing is to know that currents can be differentiated by their cause:
Drift current is caused by a gradient of the electrical field (voltage).
Diffusion current is caused by a gradient of concentration of carriers (i.e. electrons or holes).

Minority carriers (electrons in the p-region) reach the collector by diffusion (→ diffusion current). It requires the base to be thin enough that only a small percentage of minority carriers are lost by recombination (i.e. thickness of base << diffusion length; that's why it doesn't work with two discrete diodes connected back to back).

To understand the whole process I suggest to understand drift and diffusion current, minority carriers, diffusion length, recombination, pn-junction, ...


In short:

  • B-E junction is forward biased and causes electrons (better: negative charges) to move from the emitter (it "emittes") to the base region (forward biased pn junction) which is very thin.

  • Therefore, the majority of the carriers has enough energy to cross this depletion area and will be "collected" by the collector potential which is larger than the base potential (that means: The C-B junction is reverse biased). This is the collector current.

  • A (small) rest of the charged carriers forms the base current.

  • Therefore: The B-E voltage causes the emitter current IE which is split into the collector current IC and the (small) base current IB: IE=IC+IB.

  • That means: The voltage VBE controls/determines both currents: IC and IB. And the ratio IC/IB is nearly constant (called "current gain").


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