Question

Ultra-thin semiconductor materials are of interest for future nanometer-scale transistors, but can present undesirably high resistance to current flow. How low must the resistivity of a semiconductor material be, to ensure that the resistance of a 2nm-thick, 10nm-long, 100nm-wide region does not exceed 100 ohms?

Answer 1

Answer:

p = 2*10^(-7) ohm m

Explanation:

The resistivity and Resistance relationship is:

$p = \frac{R*A}{L}$

For lowest resistivity with R < 100 ohms.

We need to consider the possibility of current flowing across minimum Area and maximum Length.

So,

Amin = 2nm x 10 nm = 2 * 10^(-16) m^2

Lmax = 100nm

Using above relationship compute resistivity p:

 $p = \frac{100*2*10^(-16)}{100*10^(-9)} \\\\p = 2 * 10^(-7)$

Answer: p = 2*10^(-7) ohm m