Homework2_Solution

# Homework2_Solution - ECE 3040 Dr Alan Doolittle...

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ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Homework #2 Solutions NOTE: Some reference materials including tables, figures etc. .. handed out in class or in your text may be needed to solve these problems. Unless otherwise instructed, assume room temperature (300 K). 1.) A GaN semiconductor (E G =3.45 eV, intrinsic concentration is n i =1e-14 cm -3 ) is initially barely doped n-type so that the electron concentration is 1 electron/cm 3 . Acceptors are added to the material. How far (in energy) and in what direction (in energy) must the fermi energy move to result in 1 hole per cm 3 . n=ni e {(Ef1-Ei)/kT} k=8.6e-5 eV/deg , kT=0.0259 eV 1=1e-14 e {(Ef1-Ei)/0.0259} E f1 -E i =0.0259 ln(1/1e-14) E f1 -E i =0.83 eV in the second case, p=ni e {(Ei-Ef2)/kT} 1=1e-14 e {(Ei-Ef2)/0.0259} E i -E f2 =0.0259 ln(1/1e-14) E i -E f2 =0.83 eV so E f1 -E f2 = (E f1 -E i ) +(E i -E f2 ) = 1.63 eV downward toward the valance band.

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ECE 3040 Dr. Alan Doolittle Microelectronic Circuits 2.) Pierret 2.4
ECE 3040 Dr. Alan Doolittle Microelectronic Circuits 3.) Find the electron and hole concentrations as well as the fermi level and intrinsic energy positions for a silicon sample with the following conditions: (You may need data from tables 2.1 and 2.3 and assume Total ionization and a 27 C temperature) a) 10 15 cm -3 B Nd=0 and Na=10 15 cm -3 since boron, B, is a group III element in silicon (you could have also determined this from the listing in Table 2.3). Thus, since N a >>n i and N a >>N d , p~N a => p=10 15 cm -3 n=(n i ) 2 /p = (10 10 ) 2 /10 15 = 10 5 cm -3 E i =0.5(Eg)+0.75 kTln( m p * / m n * ) = 0.5(1.12)+0.75 (0.0259)ln( 0.81 /1.18 )=0.553 eV E f =E i -kT ln( N a / n i )=0.553-0.0259 ln(10 15 / 10 10 ) = 0.2548 eV b) 5x10 17 cm -3 B and 4.95x10 17 cm -3 P Nd=4.95x10 17 cm -3 and Na=5x10 17 cm -3 since boron, B, is a group III element in silicon and phosphorous, P, is a group V element in silicon (you could have also determined this from the listing in Table 2.3). Thus, since assumptions (N a >>n i and N a >>N d ) are NOT valid, p=0.5 (5.0x10 17 - 4.95x10 17 )+[(0.5 (5.0x10 17 - 4.95x10 17 )) 2 + (10 10 ) 2 ] = 5x10 15 cm -3 n=(n i ) 2 /p = (10 10 ) 2 /(5x10 15 ) = 20,000 cm -3 E i = 0.553 eV E f =E i - kT ln( p / n i )=0.553-0.0259 ln(5x10 15 / 10 10 ) = 0.213 eV

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• Spring '07
• HAMBLEN
• Condensed matter physics, Dr. Alan Doolittle, Doolittle Microelectronic Circuits, Alan Doolittle Microelectronic

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Homework2_Solution - ECE 3040 Dr Alan Doolittle...

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