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?


2 Answers 2


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.

  • $\begingroup$ Except for ambiguous “gain of a transistor” wording (in fact, it ought to be called “current gain”), nice explanation. Why did Ī so underestimate this author? $\endgroup$ Oct 24, 2014 at 8:51
  • $\begingroup$ @IncnisMrsi, updated per your recommendation. Thanks! $\endgroup$
    – Art Brown
    Oct 24, 2014 at 15:27

Very roughly there are two different processes at work: diffusion and drift. Drift is the motion of the charged particles induced by a field (either the applied voltage or the field in the depletion region.) Diffusion is a function of the width of the depletion region. Drift is not. Drift is a function of the number of minority charge carriers near the edges of the depletion region. Even without an applied voltage there is a field in the depletion region.

Here is a picture from Section 5.2 of a very nice set of notes by Bart Van Zeghbroeck at University of Colorado:

charge flow in an npn transistor
(source: bart at ecee.colorado.edu)

When you forward bias the base-emitter junction you get a large diffusion current flowing from base to emitter. (Red arrow labeled $I_{E,n}$ and blue arrow labeled $I_{E,p}$). You are reverse biasing the base-collector junction, which almost completely shuts off the diffusion current from base to collector, but the drift current from collector to base is proportional to the number of electrons near the edge of the depletion region on the base side.

The emitter is injecting electrons into the base through diffusion. If the base is narrow enough the electrons won't have time to recombine with the holes in the base. (The recombination is labeled $I_{r,B}$ in the picture.) So there will be an excess of electrons at the edge of the depletion region between the base and collector. The drift current sweeps those electrons into the collector. So you get a large drift current from collector to base.

The base-emitter junction is "just a diode", so the diffusion current through the base-emitter junction is exponentially related to the voltage between emitter and base by the diode equation: $I \approx I_s (e^{V_{BE}/V_T}-1)$, where $I_s$ is a constant based on the materials and doping and $V_T$ is the thermal voltage (about 26mV at room temperature). The amount of those electrons that reach the base-collector depletion region is a function of the width of the base and the rate at which holes and electrons recombine in the base. The drift current is, (I think) linearly proportional to the number of electrons that manage to reach the edge of the depletion region. So a reasonable first approximation is that the current from emitter to collector is exponentially proportional to the voltage between emitter and base.

Practical bipolar transistors are actually N+/P/N transistors so that the diffusion current from base to emitter is almost completely electrons from emitter to base rather than holes from base to emitter.

You might also find this physics.stackexchange question relevant.


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