As long as within the active region the collector-emitter junction is reverse biased, is there a purpose for which the collector must be doped? Moreover, not only that it is doped, but it is moderately doped, thus more than slightly doped like the base region.
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3\$\begingroup\$ For low resistance for efficiency, as much for creating the junction with the base. \$\endgroup\$– Neil_UKMar 28, 2021 at 9:46
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\$\begingroup\$ If the collector is not doped the resistance is lower. \$\endgroup\$– paukMar 28, 2021 at 9:52
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7\$\begingroup\$ @pauk no, undoped silicon is (effectively) an insulator. \$\endgroup\$– Marcus MüllerMar 28, 2021 at 10:33
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\$\begingroup\$ If the collector wasn't doped the transistor wouldn't work. \$\endgroup\$– Miss MulanMar 28, 2021 at 12:55
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\$\begingroup\$ Typically the base is more heavliy doped than the collector, and I wouldnt consider the base "slightly" doped, its more into "moderate" territory. \$\endgroup\$– MattMar 28, 2021 at 14:10
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1 Answer
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The essence of transistor behavior in the active region is as follows.
- The base-collector pn junction is reverse biased
- majority carriers have a difficult time crossing a reverse biased pn junction
- minority carriers easily cross a reverse biased pn junction
- unless the emitter injects minority carriers into the base, there are few minority carriers in the base
- when the base-emitter pn junction is forward biased, majority carriers from the emitter cross the base-emitter junction into the base, where they become minority carriers, (where they easily cross over the reverse biased base-collector junction).
Now:
As long as within the active region the collector-emitter junction is reverse biased, is there a purpose for which the collector must be doped?
The conductivity of the collector depends upon doping. Undoped silicon is a poor conductor.
Moreover, not only that it is doped, but it is moderately doped, thus more than slightly doped like the base region.
The base is lightly doped to reduce the number of majority carriers in the base. The concentration of majority carriers is directly related to the doping concentration. The higher the concentration of majority carriers, the greater the chance of the minority carriers recombine in the base with majority carriers. This increases base current, and decreases collector current.
The collector is doped less than the emitter to make the transistor asymmetric. The more symmetrical a transistor, the poorer it performs in the forward active region (but the better it performs in the reverse active region).
Those are the reasons why the base and collector are both doped less strongly than the emitter. Why the base might be doped even less than the collector I do not currently know. Perhaps someone with more expertise in transistor design can answer.
answered Mar 28, 2021 at 15:11
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\$\begingroup\$ Thus, the minority carriers could pass through a reverse-biased junction faster than through an undoped semiconductor? \$\endgroup\$– paukMar 28, 2021 at 18:28
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\$\begingroup\$ @pauk If by "faster" you mean higher current, I think the answer is yes. \$\endgroup\$ Mar 28, 2021 at 20:20
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\$\begingroup\$ Thanks. Also, could you explain to me why is the collector doped "heavier" than the base? \$\endgroup\$– paukMar 28, 2021 at 21:10
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\$\begingroup\$ I'm not an expert on transistor design, but I believe the reason is as follows. Minority carriers in the base will tend to recombine with the majority carriers in the base. One doesn't want that. One wants the minority carriers to cross the base and enter the collector. So, to keep the number of majority carriers down, the base is only lightly doped. Also, it is physically thin, so that minority carriers can cross it quickly before they recombine. The number of majority carriers in the base is directly related to the doping concentration. Maybe someone more expert can explain more fully? \$\endgroup\$ Mar 28, 2021 at 21:30