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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 Ω I Dsat ( R S +R D ) + V DS = 5V – (-5V) R D = (10V – 3V)/ 100 μ A – 36.87k Ω = 33.13 k Ω 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 ....
View Full Document

This note was uploaded on 10/24/2011 for the course ELECTRICAL ee2021 taught by Professor Tan during the Spring '11 term at National University of Singapore.

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online