2008_test2_solution

2008_test2_solution - ECE-342 Test 2 Solutions , Nov 4,...

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Unformatted text preview: ECE-342 Test 2 Solutions , Nov 4, 2008 6:00-8:00pm, Closed Book (one page of notes allowed) Please use the following physical constants in your calculations: Boltzmanns Constant: k = 8 . 62 10- 5 eV/K = 1 . 38 10- 23 J/K Electron Charge: q = 1 . 60 10- 19 coulomb Free space Permittivity: = 8 . 854 10- 14 F/cm Thermal Voltage: V T = kT/q = 25 . 8 mV at 300K Silicon Properties at 300K: Bandgap: E G = 1 . 12 eV Relative Permittivity: r = 11 . 7 Intrinsic carrier density: n i = p i = 1 . 5 10 10 /cm 3 Intrinsic Si Electron Mobility: n = 1350 cm 2 /V s Intrinsic Si Hole Mobility: p = 480 cm 2 /V s 10 15 10 16 10 17 10 18 10 19 10 20 200 400 600 800 1000 1200 1400 1600 1800 2000 Total Impurity Concentration (cm-3 ) Mobility ( cm 2 /V-sec ) Electron Mobility Hole Mobility Carrier Mobility Trends for Silicon Hole mobility in n-type material with N D =10 17 (Problem 1h) 320 cm 2 /(Vsec) 280 300 320 340 360 380 400 10 9 10 10 10 11 10 12 10 13 Absolute Temperature (K) Intrinsic Carrier Density (per cm 3 ) Intrinsic Silicon Carrier Density 1. A silicon PN junction diode is formed using an acceptor concentration of 5 10 18 / cm 3 and a donor population of 10 17 / cm 3 . The junction area is 400 m 2 . (a) Complete the diagram above. Show the depletion region, and indicate the polarity of any bound charges which remain in the depletion region. Clearly indicate which side of the depletion region is wider. Show the direction of the electric field in the depletion region. (b) Find the built-in potential for the device. Solution: V = V T ln N A N D n 2 i = (25 . 8 mV ) ln 5 10 35 (1 . 5 10 10 ) 2 = 0 . 9117 V (c) Find the depletion region width for 5 V reverse bias applied. Solution: w D 5 = s 2(11 . 7)(8 . 854 10- 14 F / cm ) 1 . 6 10- 19 C (5 V + 0 . 9117 V ) 1 5 10 18 cm- 3 + 1 10 17 cm- 3 = 2 . 79 10- 5 cm (= 0 . 279 m ) (d) Find the small-signal capacitance associated with a 5 V reverse bias. Solution: C J 5 = A w D 5 = (11 . 7)(8 . 854 10- 14 F / cm )(400 10- 8 cm 2 ) 2 . 79 10- 5 cm = 1 . 48 10- 13 F (= 0 . 148 pF ) (e) Show on the diagram where the electric field is a maximum. Find the magnitude of the electric field for a 5 V reverse bias at this location. Solution: The electric field as a function of position is as shown on the graph below. The maximum occurs at the metallurgic junction. The area under the curve gives the (known) junction voltage: Area = 1 2 E max w D 5 = 5 V + V E max = 2(5 . 9117 V ) 2 . 79 10- 5 cm = 4 . 23 10 5 V / cm ....
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This note was uploaded on 11/22/2009 for the course ENEE 313 taught by Professor Staff during the Spring '08 term at Maryland.

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2008_test2_solution - ECE-342 Test 2 Solutions , Nov 4,...

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