So, I understand that for a NPN transistor to work the emitter-base junction needs to be forward biased and the collector-base junction needs to be reverse biased. I understand how current flows from the emitter to the base when forward biased, but I can't seem to wrap my head around how it flows from the emitter, through the base, and into the collector.
From what I understand about the depletion region of a PN junction diode, when forward biased, the electrons in the N-type material are repulsed by the negative charge of the supply (say a battery) and the holes in the P-type material are repulsed by the positive charge from the supply, thus narrowing the depletion region and allowing electrons to flow from the N region to the P region. I also understand that the holes and electrons are attracted to the source charges when reverse biased, widening the depletion region and blocking any electrons from flowing.
What I don't understand is what happens to the depletion region of the collector-base junction that allows electrons to flow from the emitter to the collector even though the collector-base junction is reverse biased.
Take a simple common emitter circuit for example. I apparently am not cool enough to post pictures yet so you'll have to use your imagination. So pretend we have the base connected to the positive terminal on a 1v battery (Vbb). The collector is connected to the positive terminal on a 10v battery (Vcc). The emitter and both negative terminals are connected to ground. Also pretend that there is some kind of current limiting. How does current get from the emitter through to the collector? I can see how they are at different potentials, but I don't understand how this overcomes the reverse biasing of the collector-base junction.
Can anyone tell me what actually happens at the junction to allow current flow and how changing Vbb will affect that current?
Best Answer
As explained in Wandering Logic's answer, electrons are injected from the emitter to the base when that junction is forward-biased. Each such electron will suffer one of two fates: either it recombines with a hole in the base region (in a recombination time $\tau_r$, more or less), contributing to the base current $I_b$, or it's swept into the collector by the electric field at the reverse-biased collector-base junction (in a drift time $\tau_d$), contributing to the collector current $I_c$.
In fact, the current gain of a transistor is just $I_c/I_b = \beta = \tau_r/\tau_d$, since that many electrons get swept into the collector for each that recombines in the base. The trick to making a good transistor is to maximize that ratio; hence the thin-ness of the base region.