Transistors why does increasing base current increase collector current?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

To put a slightly different spin on Dave Tweed's answer (sometimes looking at it the other way up helps).

It's the base emitter voltage that causes a current to flow out of the emitter, in an exponential relationship described by the Ebers-Moll equations.

As the base region is so thin, most of this current flows through to the collector, but a small fraction of it, in the order of 1%, flows into the base terminal.

If the base were open circuit (the way that you can leave the gate of a FET open circuit), then this 1% of emitter current arriving at it would rapidly discharge its self capacitance, dropping its voltage, stopping the flow of current.

In order for the emitter current to be maintained, a current must flow into the base to recharge the BE junction as fast as the emitter-base current is discharging it.

This balancing act of net charge flowing into or out of the BE junction changing its voltage which changes the emitter current happens so quickly that we might as well say it's the base current that causes the emitter current.