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Question

Physics Question on kinetic theory

A pnp-n junction has acceptor impurity concentration of 1017cm310^{17}\, cm ^{-3} in the PP side and donor impurity concentration of 1016cm310^{16} cm ^{-3} in the NN side. What is the contact potential at the junction? (kT=(k T= thermal energy, intrinsic carrier concentration ni=1.4×1010cm3)\left.n_{i}=1.4 \times 10^{10}\, cm ^{-3}\right)

A

(kT/e)ln(4×1012)( kT / e ) \ln \left(4 \times 10^{12}\right)

B

(kT/e)ln(2.5×1023)( kT / e ) \ln \left(2.5 \times 10^{23}\right)

C

(kT/e)ln(1023)(k T / e) \ln \left(10^{23}\right)

D

(kT/e)ln(109)( kT / e ) \ln \left(10^{9}\right)

Answer

(kT/e)ln(4×1012)( kT / e ) \ln \left(4 \times 10^{12}\right)

Explanation

Solution

Constant potential at the junction Vconstant kTeln(nandni2)V_{\text {constant }} \frac{k T}{e} \ln \left(\frac{n_{a} n_{d}}{n_{i}^{2}}\right) Vconstant =kTeln(1017×1016(1.4×1010)2)\therefore V_{\text {constant }} =\frac{k T}{e} \ln \left(\frac{10^{17} \times 10^{16}}{\left(1.4 \times 10^{10}\right)^{2}}\right) =kTeln(4×1012)=\frac{k T}{e} \ln \left(4 \times 10^{12}\right)