Q44 / 88 · NTA Abhyas (JEE Main) 2020 · Physics · Semiconductors

A p-n junction has acceptor impurity concentration of 1017 cm-3 in the p side and donor impurity concentration of 1016 cm-3 in the n side. The contact potential at the junction is (kT= thermal energy, intrinsic carrier concentration ni=1.4× 1010 cm-3 )

  • Correct answer
    A
    kTe ln (4×1012)
  • B
    kTe ln (2.5×1023)
  • C
    kTe ln (1023)
  • D
    kTe ln (109)

Solution

Constant potential at the junction

Vconstant=kTe ln nandni2

Vconstant=kTe ln 1017×10161.4×10102=kTe ln (4×1012)

Hence, the correct answer is option: 'A'.

More Semiconductors Questions