Question

of aluminum and is 1000µm long, 20nm wide, and 40nm thick, what would be the resistance of the interconnect?

Answer 1

Answer : The interconnect resistance will be, 35250 Ω

Explanation :

The formula used to calculate the interconnect resistance is:

`R=\rho (L)/(A)\\\\R=\rho (L)/(TW)`

where,

R = resistance = ?

L = length of wire = 1000 μm = 1000 × 10⁻⁶ m

T = thickness of wire = 40 nm = 40 × 10⁻⁹ m

W = width of wire = 20 nm = 20 × 10⁻⁹ m

`\rho` = resistivity of aluminum = 2.82 × 10⁻⁸ Ωm

A = area

Now put all the given values in the above formula, we get:

`R=(2.82* 10^(-8)\Omega m)* (1000* 10^(-6)m)/((40* 10^(-9)m)* (20* 10^(-9)m))`

`R=35250\Omega`

Thus, the interconnect resistance will be, 35250 Ω