4 pn Diode Principle - Soultion

# 4 pn Diode Principle - Soultion - Tutorial (pn-Diode...

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Unformatted text preview: Tutorial (pn-Diode Principle) – Solution 15. Using Eq.(1.1), determine the diode current at 20 o C for a Si diode with I S = 50nA and an applied forward bias of 0.6V. An. T K = 20 + 273 = 293 k = 11,600/ n = 11,600/2 = 5800 ( Si - η = 2 ) I D = I s 1 D K kV T e  -       = 50 × 10- 9 (5800)(0.6) 293 1 e  -     = 50 × 10- 9 ( e 11.877- 1) = 7.197 mA 17. (a) Using Eq.(1.1), determine the diode current at 20 C for a Si diode with I S = 0.1 μ A at a reverse-bias potential of -10V. (b) Is the result expected? Why? An. (a) T K = 20 + 273 = 293 k = 11,600/ n = 11,600/2 = 5800 I D = I s 1 D K kV T e  -       = 0.1 μ A (5800)( 10 V) 293 1 e-  -     = 0.1 × 10- 6 ( e- 197.95- 1) = 0.1 × 10- 6 (1.07 × 10- 86- 1) ≅ 0.1 × 10- 6 0.1 μ A I D = I s = 0.1 μ A (b) The result is expected since the diode current under reverse-bias conditions should equal the saturation value....
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## This note was uploaded on 01/11/2011 for the course EE 2106 taught by Professor Ytchow during the Spring '07 term at City University of Hong Kong.

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4 pn Diode Principle - Soultion - Tutorial (pn-Diode...

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