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# hw4s - i ox s s d i i d qN 2 2 2 = 0.938 0.064 = 1.002 V...

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Question 1 For a silicon Schottky diode with B = 1 V and N d = 10 18 cm -3 , calculate: a) The built-in potential, depletion layer width and the maximum electric field in thermal equilibrium. b) The barrier lowering in thermal equilibrium. c) The change in the built-in potential,  as obtained from a Capacitance-Voltage measurement, if there is a 3 nm interfacial SiO 2 layer between the metal and the semiconductor. d) The ideality factor, n , of the diode as obtained from a Current-Voltage measurement, if there is a 3 nm interfacial SiO 2 layer between the metal and the semiconductor. The built-in potential is obtained from: q E E n F c B i , = 1 – 0.0616 = 0.938 V The depletion layer width for V a = 0 V equals: d a i s d qN V x ) ( 2 = 3.52 x 10 -6 cm The electric field at the interface equals: s d d x qN x ) 0 ( E = - 53.4 kV/cm The barrier lowering in thermal equilibrium equals: s B q  4 max E = 80.3 mV The built-in potential with the interfacial layer equals:

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Unformatted text preview: i ox s s d i i d qN 2 2 * 2 = 0.938 + 0.064 = 1.002 V The ideality factor is obtained from: i n / 1 1 = 1.34 Question 2 Derive equation 3.7.3 in section 3.7 of the on-line text. Starting from ox d d s d d a i d x qN x qN V 2 2 One finds the depletion layer width: B V A A x a i d ) ( 2 where ox s d A and d s qN B 2 And the capacitance becomes So that ) ( 2 ) ( * 2 a i t D s a i d a n d a n V V L B V A B qN dV dx qN dV dQ C with i ox s s d i i i d qN B A 2 2 2 * 2...
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hw4s - i ox s s d i i d qN 2 2 2 = 0.938 0.064 = 1.002 V...

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