assignment1solutions

# assignment1solutions - from the doping if the bar is to...

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ECED 3201.03A Introduction to Electronics Assignment #1 solution http://www.dal.ca/~jgu/3201/assignments.html 1) The nucleus of a gallium atom has 31 protons. Assuming that each shell must be filled before a succeeding one can contain electronics, determine the number of electrons in each of its shell and in the sub shell of each shell. Ans: K L M N S S P S P D S P D F 2 2 6 2 6 10 2 1 0 0 2) If n i =1.6*10 16 electrons/m 3 in a bar of intrinsic silicon in which the electron and hole velocities are 140m/s and 50m/s, respectively, find a. The component of current density in the bar due to hole flow b. The component of current density in the bar due to electron flow c. The total current density in the bar Ans: n i =1.6*10 16 electrons/m 3 , v n =140m/s, v p =50m/s a. J p =n i *q n *v p =1.6*10 16 1.6 10 -19 * 50 = 128 mA/m 2 b. J n =n i *q n *v n =1.6*10 16 1.6 10 -19 * 140 =358.4 mA/m 2 c. J=J n +J p =486.4 mA/m 2 3) A bar of silicon 0.1cm long has a cross-sectional area of 8*10 -8 m 2 and is heavily doped with phosphorus. What should be the majority carrier density resulting

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Unformatted text preview: from the doping if the bar is to have a resistance of 1.5 K ? ? Assume nominal room-temperature values for all silicon parameters. Ans: Because the material is doped with Phosphorus, it is N-type For N-type conductivity ? ? n*u n *q n , u n =0.14m 2 /vs Then R= ? *l/A=l/(A* ? )=l/(A*n*u n *q n ) n=l/(R*A*u n *q n )=0.1*10-2 /(1.5*10 3 *8*10-8 *0.14*1.6*10-19 ) =3.72*10 20 Carriers/m 3 4) A silicon PN junction is formed from N material doped with 2.5*10 21 donors/m 3 and P material doped to have the same impurity density. Find the thermal voltage and barrier voltage at 20 ? c. Ans: V T =KT/q=1.38*10-23 *(273+20)/(1.6*10-19 )=25.27mv V ? =V T ln(N A* N D /n i 2 )=25.27*ln((2.5*10 21 ) 2 /(1.5*10 16 ) 2 )=0.607v 5) A diode has a saturation current of 45PA at a temperature of 273k, what is the approximate value of I s at T=373k? Ans: I s =2 10 *45PA=46080*10-12 A=4.6*10-8 A...
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