Calculate the built-in voltage of a junction in which the P and n regions are do

Question

Calculate the built-in voltage of a junction in which the P and n regions are doped equally with 10 16 atoms/cm3. Assume n1=1.5 times 10 10/cm3. With the terminal left open, what is the width of the depletion region, and how far does it extend into the P and n regions?. If the cross-sectional area of the junction is 100 mum2, find the magnitude of the charge stored on either side of the junction. If, for a particular junction, the acceptor concentration is 10 16/cm3 and the donor concentration is 10 15/cm3, find the junction built-in voltage. Assume n1=1.5 times 10 10/cm3. Also, find the width of the p and n regions when (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 400 mum2. If a 5-V reverse-bias voltage is applied across the junction specified in Problem 3.13, find W and QJ.

Explanation / Answer

Acceptor conc. is 1016/cm3==>Donor conc. is 1015/cm3==>ni=1.5E10/cm3

junction area is 400 m²==>VR=5 V=reverse bias voltage==>s=1.04E-12

VO=0.0259*Ln((E16*E15)/(1.5E10)²)0.635003 V0.635 V==>q=1.6E-19

W=xn+xp=((2*s/q)*(1/NA+1/ND)*(VO+VR))0.00028387 cm0.0000028387 m2.84 m

junction area is 400 E-6 m²=4 E-6 cm²==>

QJ=A*(2*s*q*(NA*ND)/(NA+ND)*(VO+VR))1.652 E-13 C

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