BJT(npn) physical operation help!

1. May 26, 2006

jinyong

I just have a few questions on the physical operations of the BJT.

1) Van Zeghbroeck's online book says, The carrier densities vary linearly between the boundary values as expected when using the assumption that no significant recombination takes place in the quasi-neutral regions. Why would the carrier density vary if there is no recombination. What causes the minority carriers to drop down in density as you move across the junction.

2) If the excess minority carrier concentration is the highest at the EBJ and lowest at the CBJ then how does the majority of the electrons get collected at the collector? At the CBJ the electron density is only np0 not np0*e^VBE/VT? How does that work? I thought most of the injected electrons from the emitter is suppose to be collected by the collector?

Thanks.

2. May 26, 2006

antonantal

The minority carriers density drops down because the excess minority carriers density drops down. In a diffusion process (from which the excess minority carriers originate) the probability to find a diffused particle at a specific distance from the source of the diffusion decreases with the growth of the distance.
If you don't apply any voltage to the junctions, there will be no diffusion process and so the density of the carriers will be flat all across the base. When you polarize the junctions in forward active regime the electrons from the emitter will diffuse (causing the electron diffusion current) and the concentration of the carriers in the base will increase (more at places near the BE junction and less at places far from the BE junction) until the state of equilibrum described in that paragraf of the book is reached. At this point the diffusion is still going on but the amount of electrons that arrive in the base equals the amount of electrons that get collected at the BC junction (or dissapear from the base by other means that we ignored). Any electron that gets collected near the BC junction will be replaced by another one that is near it, the place of the second one will be taken by another one near it, and this chain process continues up to the BE junction where it gets replaced by an electron that comes from the emitter - this is why the graph of the carrier concentration doesn't change.

There isn't the concentration at the BC junction or the concentration at the BE junction that determines the diffusion current but the difference between this concentrations. Diffusion is the natural tendency of gas particles (or other particles - electrons and holes in a semiconductor) to level their concentration. If the concentration is flat there is no diffusion.

3. May 26, 2006

Gokul43201

Staff Emeritus
This may be what anto said above, but simplistically speaking, the concentration gradient is what maintains equilibrium against a potential gradient. In "neutral" regions, the field is uniform and hence the potential gradient is linear. So, a linear concentration gradient is needed to reach equilibrium.

Alternatively, you can think in terms of the free energy,
$$dF = \mu (dN) - e(dV) = 0 \implies \mu (dN) = e(dV) \implies \mu \frac{dN}{dx} = e\frac{dV}{dx}$$

Last edited: May 26, 2006
4. May 30, 2006

jinyong

Thanks for all the help.

One more question. You say that: The minority carriers density drops down because the excess minority carriers density drops down. In a diffusion process (from which the excess minority carriers originate) the probability to find a diffused particle at a specific distance from the source of the diffusion decreases with the growth of the distance.

I don't really get this. I thought when things diffuse they will try to even out their concentrations throughout the material. Any readings that describes this diffused particle probability? Do you derive it from the continuity equation?

5. Jun 1, 2006

antonantal

What you say is true if the system is isolated. But since electrons are constantly injected from the emitter and extracted at the collector the concentration at the steady state will have a gradient. If you would someway isolate the base from the emitter and the collector at a certain moment, the concentration would then begin to level out because of the diffusion.

Here is a document which shows how the probability to find a diffused particle decreases with the growth of the distance. It refers to molecule diffusion but in principle it is the same with any particles. Only the constants differ:
http://unicorn.ps.uci.edu/243/handouts/diffusion1d.pdf

6. Jan 27, 2012

aperception

I realize that this thread is 5 years old, but I spent a bit of time thinking about it, and feel that I can add a little more physical clarity for anyone who searches and finds this thread. I will discuss a pnp transistor with geometry as per figure 5.6 in Sze (attached).

In the forward active mode, the base collector junction is reverse-biased, meaning minority carriers (holes in this PNP case) are very rapidly swept from base to collector, whereas essentially no minority carriers are coming into the base from from the collector. This is what gives rise to the boundary condition that the minority carrier density at x=W is zero (again see attached diagram). I.E there can't be any minority carriers here because they are all swept into the collector.

So the boundary conditions are that the minority carrier density at x=0 is very high, and at x=W it is zero. Then a strong diffusion current results (as antonantal already discussed), and the solution to the continuity equation with the above boundary conditions 'just happens' to give an approximately linear solution for what happens in between, assuming W is small. See section 5.2.1 in Sze for more mathematical details.

File size:
10.1 KB
Views:
144