HW SOLN 5 - ECE331 Homework #5 Solution 1. Streetman, 3.2...

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1 ECE331 Homework #5 Solution 1. Streetman, 3.2 The probability that a state at E above E f is occupied is ) / exp( 1 1 ) / ) exp(( 1 1 ) ( kT E kT E E E E E f f f f Δ + = Δ + + = Δ + The probability that a state at E below E f is empty is ) / exp( 1 1 ) / ) exp(( 1 1 1 ) ( 1 kT E kT E E E E E f f f f Δ + = Δ + = Δ
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2 2. Since Ec-Ed >> kT, not all donors are ionized. The ionization probability is 1-f(Ed). Then the ionized donor ion concentration is 3 14 15 10 733 . 8 ) 0259 . 0 / 05 . 0 exp( 1 1 1 10 ) / ) exp(( 1 1 1 ) ( 1 ( + × = + = + = = cm kT E E N E f N N f d d d d d For carrier concentrations, we have kT E E N n F C C ) ( exp 0 = kT E E N p V F V ) ( exp 0 = So, 10 0 0 10 15 . 1 ) 0259 . 0 / ) 25 . 0 1 . 1 ( exp( ) 0259 . 0 / 25 . 0 exp( ) / ) ( exp( ) / ) ( exp( × = = = kT E E kT E E p n V F F C Based on the space charge neutrality, we have n 0 = p 0 + N d + So, we can solve the carrier concentrations based on the above two equations, n 0 = 8.73×10 14 cm -3 p0 = 7.59×10 4 cm -3 So the intrinsic carrier concentration, 3 9 4 14 0 0 10 14 . 8 10 59 . 7 10 73 . 8 × = × × × = = cm p n n i 3. Since ) exp( 0 kT E E n n i F i = , so Ef-Ei = kTln(n0/ni) = 0.0259×ln(10 16 /1.5×10 10 ) = 0.347 eV 4.
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3 5. This material has N c =10 19 cm -3 and N v =5X10 18 cm -3 , and E g =2eV at 300 K. At T = 900 K (627°C), From 2 / 3 2 * 2 2 = h kT m N n c π , we have 3 19 2 / 3 19 2 / 3 1 2 1 2 10 20 . 5 ) 300
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This note was uploaded on 05/05/2010 for the course ECE 331 taught by Professor Rajan during the Spring '09 term at Ohio State.

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HW SOLN 5 - ECE331 Homework #5 Solution 1. Streetman, 3.2...

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