A silicon p+n junction has doping concentrations of Na 10^18 and Nd=5x10^15. The cross sectional area...

 

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A silicon p+n junction has doping concentrations of Na 10^18 and Nd=5x10^15. The cross sectional area of the junction is A=5x10^-5 cm^2. a. Calculate the junction capacitance for Vr=0 V b. Vr=3 V c. Vr=6 v d. Plot 1/C^2 vs. Vr. Show that the curve can be used to find Nd and that the intersection with the voltage axis yields Vbi The peak electric field in a reverse biased pn junction is Emax=3x10^5 V/cm. the doping concentrations are Nd 4x10^15 and Na=4x10^17. a. Find the magnitude of the reverse bias voltage b. Repeat part a for Nd=4x10^16 and Na for 4x10^17 c. Repeat part f for Nd-Na-4x10^17 A silicon p+n junction has doping concentrations of Na 10^18 and Nd=5x10^15. The cross sectional area of the junction is A=5x10^-5 cm^2. a. Calculate the junction capacitance for Vr=0 V b. Vr=3 V c. Vr=6 v d. Plot 1/C^2 vs. Vr. Show that the curve can be used to find Nd and that the intersection with the voltage axis yields Vbi The peak electric field in a reverse biased pn junction is Emax=3x10^5 V/cm. the doping concentrations are Nd 4x10^15 and Na=4x10^17. a. Find the magnitude of the reverse bias voltage b. Repeat part a for Nd=4x10^16 and Na for 4x10^17 c. Repeat part f for Nd-Na-4x10^17

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Given values Na 1018 cm3 Nd 5x1015 cm3 A 5x105 cm2 118 0 permittivity of silicon 0 8854 x 1014 Fcm v... View the full answer