ECE 4420 – Spring 2004 Page 1
Homework No. 12 – Solutions
Problem 1 – P10.1
a) 0.18µm, Al, Metal 5
L=20mm, W=0.4µm, Rsq=27 mΩ/□
Rwire=Rsq×L/W
Rwire=27 mΩ/□×20mm /0.4µm=1.35 KΩ
b) 0.13µm, Cu, Metal 8
L=20mm, W=0.4µm, Rsq=21 mΩ/□
Rwire=Rsq×L/W
Rwire=21 mΩ/□×20mm /0.4µm=1.05 KΩ
Cint=0.1fF/µm
Cwire=Cint×L
Cwire=0.1fF/µm×20mm=2pF
τ = 0.38×Rwire×Cwire
τAl = 0.38×Rwire×Cwire= 0.38×1.35 KΩ×2pF = 1.026 ns
τCu = 0.38×Rwire×Cwire= 0.38×1.05 KΩ×2pF = 0.798 ns
Problem 2 – P10.3
The wire is shielded above and below so it has all the Lateral, Area and Fringing
capacitance.
T=1µm, H=0.5µm, W=0.4µm, λ=0.1µm
εox= 4εo= 4( 88.5×10-4 fF/µm) = 354×10-4 fF/µm
CLateral= εox×T/S = (354/S)×10-4 fF/µm
CArea= εox×W/H = 142×10-4 fF/µm
CFring= εox×Ln(1+T/H) = 389×10-4 fF/µm
CTotal= CLateral+CArea+CFring
S: 4λ … 40λ, Step = 4λ (0.4µm … 4µm)
