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EE2005-Tutorial4-Solution

# EE2005-Tutorial4-Solution - NATIONAL UNIVERSITY of...

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NATIONAL UNIVERSITY of SINGAPORE Department of Electrical and Computer Engineering EE2005 – Electronics Tutorial 4 Solution Q1. From the saturation drain current model, 2 ) ( ' 2 1 th GS ox n Dsat V V L W C I = μ For 2 / 120 ' 2 1 V A L W C K ox n n μ μ = = and A I Dsat μ 100 = , we have V GS -V th = [100 μ A/(120 μ A/V 2 )] 1/2 = 0.913V See that V GS > V th and V DS > ( V GS -V th ). The device operates in saturation region. V GS - 0.4V= 0.913V V GS = V G V S =1.313V Since V G = 0V, V S = -1.313V = -5V + I Dsat R S Hence, R S = (-1.313V + 5V)/100 μ A = 36.87k I Dsat ( R S +R D ) + V DS = 5V – (-5V) R D = (10V – 3V)/ 100 μ A – 36.87k = 33.13 k Q2. Since gate and drain regions are connected, V G =V D Æ V GS =V DS Therefore, V DS > ( V GS -V th ). The device operates in saturation region. (1 mark) 2 ) ( ' 2 1 th GS ox n Dsat V V L W C I = μ L W C I V V ox n Dsat th GS ' 2 1 μ = = 5 . 0 3 ) / 250 ( 2 1 100 2 V A A μ μ = 0.365 V (2 marks) V GS = 0.365 V + V th = 0.865 V Since V S = 0V, V D = V G = 0.865 V

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