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Does the base-emitter connected BJT have the same built in voltage as the base collector connected BJT? Why or why not?

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  • \$\begingroup\$ There is no built-in voltage in a transistor. \$\endgroup\$
    – Andy aka
    Commented Nov 30, 2015 at 8:43
  • \$\begingroup\$ As I recall, the IV curve for base-emitter junction is typically not the same as the IV curve for the base-collector junction because the doping is different in the emitter and collector. And maybe some other reasons I can't remember anymore. \$\endgroup\$
    – user57037
    Commented Nov 30, 2015 at 8:57
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    \$\begingroup\$ A P-N junction actually does have a built-in voltage, and this exact term is used in academic courses teaching semi-conductor physics. This does not violate the laws of physics. The voltage will disappear if the current is not zero. I believe the down-voting of this question is not justified. If a few people could help vote it back up to zero at least it might receive a decent answer. \$\endgroup\$
    – user57037
    Commented Nov 30, 2015 at 16:41

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No, the built-in voltage of the base-collector (BC) and base-emitter (BE) junctions will normally not be the same. The built-in potential of a P-N junction is determined by the equation:

Vbi = kT/q * ln(Nd * Na / (ni * ni))

Source: http://www.eecs.berkeley.edu/~hu/Chenming-Hu_ch4.pdf

Vbi is the built-in voltage. Nd is the donor concentration, Na is the acceptor concentration ni is the intrinsic charge carrier concentration of silicon, k is the Boltzman constant, T is the temperature in Kelvin, and q is the charge of an electron.

So, assuming constant temperature, the built-in voltage is determined by the dopant levels in the two silicon regions (P and N). Normally, I believe the emitter will be more heavily doped. Thus, the built-in voltage of the BE and BC junctions will normally not be the same.

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