An abrupt Si p-n junction has doping conncentration on p-side and n-side as NA-1018-cm3 and ND-5-1015-cm3 respectively- Then the contact potential at the junction isAssume intrinsic carrier concentration- ni-1-0-1010-cm3-and kT-0-026V

Given, Doping concentrations, on p side and n side N_A=10^18/cm^3 N_D=5×10^15/cm^3 Contact potential,V_bi=KTIn(N_A N_Dn_i^2) ∴ V_bi=0.026In(10^18×5×10^15(10.× ...

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PN Junction in Equillibrium

An abrupt Si ...

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$An abrupt Si p-n junction has doping conncentration on p-side and n-side as N_{A} = 10^{18} / c m^{3} and N_{D} = 5 \times 10^{15} / c m^{3} respectively. Then the contact potential at the junction is (Assume intrinsic carrier concentration, n_{i} = 1.0 \times 10^{10} / c m^{3}, and k T = 0.026 V)$

0.65 V

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0.78 V

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0.82 V

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1.20 V

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Solution

The correct option is C $0.82 V$
$Given, Doping concentrations, on p-side and n-side N_{A} = 10^{18} / c m^{3} N_{D} = 5 \times 10^{15} / c m^{3} Contact potential, V_{b i} = K T I n (\frac{N_{A} N_{D}}{n_{i}^{2}}) ∴ V_{b i} = 0.026 I n (\frac{10^{18} \times 5 \times 10^{15}}{(10. \times 10^{10})^{2}}) ∴ V_{b i} = 0.82 V$

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An abrupt Si p-n junction has doping conncentration on p-side and n-side as NA=1018/cm3 and ND=5×1015/cm3 respectively. Then the contact potential at the junction is(Assume intrinsic carrier concentration,ni=1.0×1010/cm3,andkT=0.026V)

The correct option is C 0.82VGiven,Doping concentrations, on p-side and n-sideNA=1018/cm3ND=5×1015/cm3Contact potential,Vbi=KTIn(NANDn2i)∴Vbi=0.026In(1018×5×1015(10.×1010)2)∴Vbi=0.82V