Many websites and courses list the doping profile of a BJT as "emitter heavily doped", "base lightly doped" and "collector moderately doped". However, the more reputable sources and textbooks like "Semiconductor Physics and Devices by Donald A. Neamen" state that the collector is the doped the lightest of all, introducing a lot of confusion. However, I was not able to find a good reason for the collector being lightly doped. How does collector doping impact transistor performance?

Why, is the collector the lightest doped of all 3 regions of the BJT? Please provide any references too.

The mentioned part in "Semiconductor Physics and Devices by Donald A. Neamen" is:

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  • 2
    \$\begingroup\$ I think you should take a picture of your book where it says what you have told us and embed that picture into your question. As far as I'm concerned it's the base that is doped the lightest (as per your first sentence). \$\endgroup\$
    – Andy aka
    Feb 25, 2022 at 16:36
  • \$\begingroup\$ Consider the area (volume) of the individual junctions. \$\endgroup\$ Feb 25, 2022 at 16:40
  • \$\begingroup\$ Thanks @Andyaka. I have added it \$\endgroup\$ Feb 25, 2022 at 17:22

1 Answer 1


It is to permit a high voltage to be applied to the collector without breaking down the collector-base PN junction. PN junction breakdown voltage is affected by the width of the depletion region, which in turn is affected by doping levels on either side of the junction. The base is lightly doped to minimize recombination. The emitter is heavily doped because:

A) You don't need a high Vbe and

B) It increases current gain

  • \$\begingroup\$ Thanks! This makes perfect sense, but could you provide some references? There seems to be little information on this online. \$\endgroup\$ Feb 25, 2022 at 17:54
  • \$\begingroup\$ Does early effect also have anything to do with this? Base being more doped than collector means there is less base width modulation, right? \$\endgroup\$ Feb 26, 2022 at 5:41

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