Calculate Electron Concentration in IC Chip Conducting Line

The problem asks for the electron concentration (n) in a conducting line on an IC chip, given its physical dimensions, the current flowing through it, the voltage drop across it, and the electron mobility.

Given Parameters

• Length, L = 2.8 mm = 2.8 $\times$ 10^-3 m
• Width, W = 1 $\mu$m = 1 $\times$ 10^-6 m
• Thickness, T = 4 $\mu$m = 4 $\times$ 10^-6 m
• Current, I = 5 mA = 5 $\times$ 10^-3 A
• Voltage Drop, V = 100 mV = 0.1 V
• Electron Mobility, $\mu_n$ = 500 cm²/V.s = 5 $\times$ 10^-2 m²/V.s
• Elementary Charge, q = 1.602 $\times$ 10^-19 C

Determine Cross-Sectional Area

The cross-sectional area (A) is the product of width and thickness:

A = W $\times$ T = (1 $\times$ 10^-6 m) $\times$ (4 $\times$ 10^-6 m) = 4 $\times$ 10^-12 m²

Apply Current Density Formula

The current density (J) is related to conductivity ($\sigma$), electron concentration (n), elementary charge (q), electron mobility ($\mu_n$), and electric field (E) by:

J = $\sigma E$ = (n q $\mu_n$) ($V/L$)

Current density is also defined as J = I / A. Equating the two expressions for J:

$\frac{I}{A} = n q \mu_n \frac{V}{L}$

Solve for Electron Concentration

Rearrange the formula to solve for n:

n = $\frac{I L}{A q \mu_n V}$

Calculate the Value of n

Substitute the given and calculated values into the formula:

n = $\frac{(5 \times 10^{-3} \text{ A}) \times (2.8 \times 10^{-3} \text{ m})}{(4 \times 10^{-12} \text{ m}^2) \times (1.602 \times 10^{-19} \text{ C}) \times (5 \times 10^{-2} \text{ m}^2/\text{V.s}) \times (0.1 \text{ V})}$

n = $\frac{14 \times 10^{-6}}{3.204 \times 10^{-33}}$ m^-3

n $\approx$ 4.3695 $\times$ 10²⁷ m^-3

Convert Units to cm^-3

Since the options are in cm^-3, convert the result:

1 m^-3 = 10^-6 cm^-3

n $\approx$ 4.3695 $\times$ 10²⁷ $\times$ 10^-6 cm^-3

n $\approx$ 4.3695 $\times$ 10²¹ cm^-3

This value is approximately 4.38 $\times$ 10²¹ cm^-3.