EE331HW4solutions

# EE331HW4solutions - EE-331 Devices and Circuits 1 Prof R B...

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Problem 2: Temperature dependence of a diode's reverse saturation current. A1 0 2 cm 2 := N A 10 18 cm 3 := N D 10 16 cm 3 := L n 5.0 µ m := L p 5.0 µ m := From Fig. 2.8 of the text: µ n 250 cm 2 volt 1 sec 1 := µ p 400 cm 2 volt 1 sec 1 := n i T ( ) 1.08 10 31 cm 6 K 3 T 3 exp 1.12 eV k B T 0.5 := V Th T () k B T q := I S T () qA V Th T µ n L n N A µ p L p N D + n i T 2 := (a) V Th 273.15 K ( ) 23.5 mV = n i 273.15 K ( ) 6.953 10 8 × cm 3 = I S 273.15 K ( ) 1.468 10 15 × amp = (b) V Th 298.15 K ( ) 25.7 mV = n i 298.15 K ( ) 5.825 10 9 × cm 3 = I S 298.15 K ( ) 1.125 10 13 × amp = (c) V Th 323.15 K ( ) 27.9 mV = n i 323.15 K ( ) 3.547 10 10 × cm 3 = I S 323.15 K ( ) 4.520 10 12 × amp = The above temperatures are for 0 C, 25 C, and 50 C, after converting each to the Kelvin scale. EE-331 Devices and Circuits 1 Prof. R. B. Darling Homework # 4 Solutions First, define some useful units and constants: µ m1 0 6 m := eV 1.602 10 19 joule := k B 8.62 10 5 eV K 1 := q 1.602 10 19 coul := ε o 8.854 10 14 farad cm 1 := n i 10 10 cm 3 := V T 0.025 volt := ε r 11.7 := Problem 1: Problem 3.32 (a,b) of Jaeger and Blalock. I S 1.0 10 15 amp := i D 100 µ A := T 273.15 K 25 K + := T 298.15 K = V T k B T q := V T 25.7 mV = (0 C = 273.15 K) (a) v D V T ln 1 i D I S + := v D 0.651 volt = (Note that the voltage has decreased as the temperature has increased.) (b) v Dnew v D 1.8 mV K 1 50 K 25 K := v Dnew 0.606 volt = EE-331 HW#4 Solutions Page 1 R. B. Darling

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C d 41 0 7 × farad = (c) i D 100 mA := Q d i D τ := Q d 11 0 9 × coul = C d Q d V T := C d 0 8 × farad = Problem 5: Problem 3.49 of Jaeger and Blalock.
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## This note was uploaded on 09/10/2011 for the course EE 331 taught by Professor Taicheng during the Winter '08 term at University of Washington.

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EE331HW4solutions - EE-331 Devices and Circuits 1 Prof R B...

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