I am new to this area so please can you keep your answers simple, thanks.

From what I know for an npn transistor in the common-emitter connection the base-emitter junction is in the forward bias and the base-collector junction in the reverse. Electrons flow from the emitter to the base where some leave the base as they recombine with holes, forming the collector current and some pass into the collector forming the collector current. If we increase the base current we get an increase in collector current, why is this?

My first thought would be that the collector current would decrease since more electrons are are flowing into the base electrode rather then the collector electrode. I really am at a loss of how to explain this.


Dave is correct... I'll try to clarify some more.

In an NPN:

  • The base-emitter voltage and the doping of the base determine the rate of emitter electron current injection into the base, which is swept into the collector due to the potential drop from base to collector and the narrowness of the base region.
  • The base-emitter voltage and the doping level of the emitter also determine the rate of base hole injection into the emitter, which does reduce the collector current.
  • The ratio of the dopant densities sets the current ratio between the base and collector (Beta).

BJTs are designed with light doping in the base and a very narrow base width to maximize the diffusion of the emitter current to the collector. As a result, the base current needed to develop the Vbe for a given rate of emitter current injection is very small compared to the emitter and collector current, and so BJTs have high current gain.

Here's an online reference that goes into some detail: Modern Semiconductor Devices for Integrated Circuits, Ch. 8

  • \$\begingroup\$ Hi, thanks for your answer, I think I am getting my head around it. Does this seem about right:1.Increasing the base-emitter voltage increase the hole injection into the base i.e. the base current. 2. The increase in the potential difference between the base and emitter also results in more electrons flowing from the base to the emitter. 3. And thus more electrons are sweeped into the collector. 4. This hence increases the collector current. \$\endgroup\$ – user53915 Sep 25 '14 at 4:03
  • \$\begingroup\$ Sorry in point 2. it is ment to say from the emitter to the base. \$\endgroup\$ – user53915 Sep 25 '14 at 11:20
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    \$\begingroup\$ @Joseph Yes, that's about it... In an NPN bipolar transistor, the emitter injects electrons into the base and they are mostly swept over into the collector. The rate of injection scales up with increasing Vbe. To support the increase in base potential with increasing Vbe, the base must inject holes, or take electrons, but at a much lower rate than the emitter current: to sustain the Vbe with a very small base region does not require as many carriers. So the base current is relatively small compared to the emitter current, so most of the emitter current goes to the collector. \$\endgroup\$ – mixed_signal Sep 26 '14 at 2:00

Fundamentally, it is the voltage across the B-E junction that determines the amount of current flowing through it. It is an exponential relationship that is described by the Ebers-Moll equations.

An increase in this voltage results in increased currents in both the base and the collector, and indeed, an increase in base current causes an increase in B-E voltage. But the base and collector currents generally have a fixed ratio to each other for any particular transistor under a particular set of conditions, so the increase in collector current is β times the increase in base current.

I don't know if this description helps at all — the relationship between cause and effect can sometimes be confusing in solid-state physics. Let me know if I can improve this answer.

  • \$\begingroup\$ Sorry is it possible for you to explain it in terms of holes in the base and electrons in the emitter and changes in their concentration? thanks \$\endgroup\$ – user53915 Sep 24 '14 at 16:52

Imagine, a water tank. Water is not going to jump out - that's the same a a standard base-emitter junction.
Now imagine... tilting the tank on its side, to fill a water cup (or a hose with a funnel on top). Sure you get the water you want, BUT there will also be a LOT of other water spilling over the edge.
That "extra" water slipped over a slightly forward-biased base-emitter junction (the tilted edge), then found itself in a highly foward-biased zone (hanging in the air).
The more you tip the tank,the more water will splash over the edge.

Another image some find useful is: a large freeway, 100mph limit, and a sudden bend (most cars wouldn't make the bend). A larger current is more cars trying to take the corner at the same time.


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