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This question already has an answer here:

Is doping the emitter more than the collector crucial? Would current not flow at all if both regions were equally doped? Why? Is there a way to explain this in a simple fashion?

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marked as duplicate by Voltage Spike, JRE, Dave Tweed Dec 25 '16 at 19:01

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    \$\begingroup\$ Come on, they are not as heavily doped as most athletes... \$\endgroup\$ – dim Dec 24 '16 at 13:21
  • \$\begingroup\$ should i call it highly doped? \$\endgroup\$ – user16307 Dec 24 '16 at 13:23
  • \$\begingroup\$ No, it was just a bad joke, sorry \$\endgroup\$ – dim Dec 24 '16 at 13:24
  • \$\begingroup\$ @RoyC I dont uderstand anything from that 1 sentence explanation in that question. Why having more charge carriers matter is mystery to me.. \$\endgroup\$ – user16307 Dec 24 '16 at 13:30
  • \$\begingroup\$ Ok fair enough. I am sure you will get a better explanation here. \$\endgroup\$ – RoyC Dec 24 '16 at 14:57
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The emitter, as the name suggests, emits carriers. In case of an NPN transistor those will be electrons.

A higher doping level means more of these carriers will be generated (per volume and time) compared to a region with less doping. These "shallow doped" areas are the base and the collector.

A BJT works by the carriers from the emitter "overwhelming" the amount of (opposite) carriers (in case of an NPN: holes) in the base region. Also, the base region is narrow. This makes the chance that a carrier originating from the emitter recombines in the base, small. If that chance was large (due to a highly doped base) then the base current (to fill up all those used-up holes) would be much larger !

So: the Base needs to have a lower doping level (compared to the emitter) so that the beta (current amplification) will be large, which is what we want.

The collector could be doped more and the beta could then still be high. Remember that beta is related to the doping ratios of emitter and base. In an NPN under non-saturated operation the collector is at a high voltage so all the carrier electrons will be "sucked out" and travel to the positive supply. The electrons originating from the collector therefore cannot influence what happens at the emitter-base junction.

However a low doping level on the collector does increase the size of the depletion region and this increases the maximum collector-base voltage.

Also it is much easier to manufacture a BJT with a shallow doped collector, then the base and emitter regions are doped "on top of" the doping of the collector.

Also the collector region is where the Vcb voltage drop occurs and power is dissipated. The heat resulting of this has to be taken away and having the collector directly sit on the metal case of the transistor helps with this.

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