Question

Consider an InSb NW with ballistic mean free path of 150nm. Calculate the current through a 250nm long InSb NW when a 100mV bias is applied. Assume that valley degeneracy 1.

Answer 1

Answer:

$I =38.46KA$

Explanation:

Given parameters include:

V = 100mV = 100 × 10⁻³ V

mean free path(W) = 150nm = 150 × 10⁻⁹ m

Length = 250 nm

From Ohm's Law ;

V = IR

$I = \frac{V}{R}$    ---------- equation (1)

Also;

$R = \rho__0\frac{L}{A}$

where:

$\rho __o$ = resistivity  of InSb = 4 × 10⁻¹³ Ω-m

L = length

A = area of cross section

Replacing our values in the above equation; we have:

$R = 4*10^{-13}*\frac{250nm}{150nm*250nm}$

$R = 4*10^{-13}*\frac{1}{150*10^{-9}m}$

$R = 2.6*10^{-6}$ Ω

$R = 2.6 \mu$Ω

From equation (1):

$I = \frac{V}{R}$  

Therefore;

$I = \frac{100*10^{-3}}{2.6*10^{-6}}$

$I =38461.54 V/$Ω

Since 1 V/Ω = 0.001 kilo ampere (KA)

Then;

$I =(38461.54 *0.001)KA$

$I =38.46KA$