Question

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolithography and the resulting ability to print ever-smaller features. Modern circuits are made using a variety of complicated lithography techniques, with the goal to make electronic traces as small and as close to each other as possible (to reduce the overall size, and thus increase the speed). In the end, though, all-optical techniques are limited by diffraction.

Assume we have a scannable laser that draws a line on a circuit board (the light exposes a line of photoresist, which then becomes impervious to a subsequent chemical etch, leaving only the narrow metal line under the exposed photoresist). Assume the laser wavelength is 248.0 nm (Krypton Fluoride excimer laser), the initial beam diameter is 1.0 cm, and the focusing lens (diameter = 1.3 cm) is extremely 'fast', with a focal length of only 0.625 cm.
a. What is the approximate width w of the line?
b. What is the minimum resolvable line separation between adjacent lines?
c. If the laser wavelength is instead reduced to 157 nm, what is the new minimum resolvable line separation?

Answer 1

0.000003782 m

0.000001891 m

0.000001197125 m

Step-by-step explanation:

`\lambda` = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

`\theta=(1.22\lambda)/(D)`

The width is given by

`d=2\theta f\\\Rightarrow d=2(1.22\lambda f)/(D)\\\Rightarrow d=2(1.22* 248* 10^(-9)* 6.25* 10^(-2))/(1* 10^(-2))\\\Rightarrow d=0.000003782\ m`

The required width is 0.000003782 m

Minimum resolvable line separation is given by

`(0.000003782)/(2)=0.000001891\ m`

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when
`\lambda=157\ nm`

`d=2(1.22* 157* 10^(-9)* 6.25* 10^(-2))/(1* 10^(-2))\\\Rightarrow d=0.00000239425\ m`

The new minimum resolvable line separation between adjacent lines is

`(0.00000239425)/(2)=0.000001197125\ m`