I have been recently asked the following question:
Why is the base of a transistor kept thin?
MY ASSUMPTION
Base is kept thin so that not many charge carriers could be neutralized in the process of biasing ...
AM I CORRECT?
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$\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$– Qmechanic ♦Commented Feb 27, 2018 at 10:06
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3$\begingroup$ @Qmechanic - this is within scope as semiconductor physics. Many EEs never really learn the details of how junction transistors (as opposed to FETs) really work. $\endgroup$– Jon CusterCommented Feb 27, 2018 at 14:29
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$\begingroup$ @JonCuster - I completely agree with you! $\endgroup$– freecharlyCommented Feb 27, 2018 at 19:51
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2 Answers
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Your assumption is superficially OK, but some key specifics are not there. For example, which charge carriers are you thinking about - those injected from the emitter, or those injected through the base contact? Next, what do you mean by 'neutralized' - are you thinking of recombination and detailed balance of the majority and minority carriers or not?
So: with no base current, forward biasing the emitter-base junction will inject majority carriers from the emitter into the base, where they are minority carriers. The base width is chosen such that those excess minority carriers will recombine before reaching the collector. Here, the wider the base the better to some extent.
Now add in base current - what happens now? There are additional minority carriers and these can dramatically shift, through detailed balance describing carrier recombination, the steady state minority carrier concentration. Thus, some of the minority carriers will survive the base to get to the collector. Small changes in base injection will cause large shifts in the steady state minority carrier population, accounting for the gain of the device. Under these conditions, a narrower base is advantageous - a greater fraction of the emitter-base current will make it to the collector at a given steady-state minority carrier population.
So, the base has to be wide enough to not have complete recombination of minority carriers injected from the emitter when there is no base current. And, it has to be thin enough so that minority carriers will transit the base under appropriate base currents. But, it is all about carrier recombination and the difference between equilibrium (no injection) and steady-state (with injection) minority carrier concentrations.
answered Feb 27, 2018 at 14:41
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$\begingroup$ Your last paragraph probably needs some modifications. It is difficult to understand what you mean by the first sentence as to why the base has to be wide enough. You cannot make the base of a transistor arbitrarily thin because you have to avoid the the so-called punch-through, where the depletion zones of emitter-base and base-collector junctions touch, which ends the transistor operation. $\endgroup$ Commented Feb 27, 2018 at 20:04
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$\begingroup$ @freecharly - indeed, that is yet another limitation on how narrow the base width can be. I tried to keep it to first order in carrier recombination effects. $\endgroup$ Commented Feb 27, 2018 at 21:39
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$\begingroup$ Jon, I still cannot understand what you mean by "the base has to be wide enough to not have complete recombination of minority carriers injected from the emitter when there is no base current". The recombination of minority carriers in the base produces part of the base current. $\endgroup$ Commented Feb 27, 2018 at 22:12
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$\begingroup$ @freecharly - assume one did not have a base contact at all. If you magically injected carriers at the emitter-base junction, the base has to be thick enough that essentially no carriers will make it to the base-collector junction and into the collector. If the 'base' is one atomic layer thick, well, there isn't enough time for recombination (one could argue that this is sort of what goes on in punch-through, but its an argument that sweeps a lot under the rug). $\endgroup$ Commented Feb 28, 2018 at 0:07
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$\begingroup$ If you have no base contact at all, you have a zero base current, and the emitter current is equal to the collector current. Almost all minority carriers injected from the emitter into the base will reach the collector. They don't have to recombine in the base. Actually, in this situation a net generation of majority carriers occurs in the base which produces most of the carrier flow from the base into the emitter. This corresponds to the $I_C vs. V_{CE}$ output characteristic for $I_B=0$ of the bipolar transistor in CE configuration. There is no lower limit on base width due to this. $\endgroup$ Commented Feb 28, 2018 at 1:23
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The base of bipolar transistors is kept thin mainly to improve their speed of operation, which is related to the transit time $t_T$ of the electrons from the emitter to the collector which decreases with decreasing thickness of the base. In modern transistors, recombination in the base is practically negligible. The so-called transit frequency $$f_T=\frac {1}{2\pi t_T}$$ is an important figure of merit of a bipolar transistor which is related to the upper frequency of operation of a bipolar transistor.
answered Feb 27, 2018 at 19:48